# 9 G P H

```2010 First Edition Rev 2.0 9TH GRADE PHYSICS AT HACKLEY SCHOOL Created, written, and edited by | William S. McLay III Introduction ......................................................................................... 10 Chapter 1 – Basic Units of Mechanics and Graphing ............... 13 1.1 Basic Units of Mechanics ........................................................................................................ 13 1.2 History of Two Systems of Units ............................................................................................. 13 1.3 English System vs. Metric System ........................................................................................... 14 Foot vs. Meter ............................................................................................................................... 15 Pound vs. Kilogram ....................................................................................................................... 15 Second vs. Second ......................................................................................................................... 16 1.4 Metric Conversions ................................................................................................................. 17 1.5 Dimensional Analysis .............................................................................................................. 21 1.6 Graphing in Physics ................................................................................................................. 23 1.7 Slope ........................................................................................................................................ 30 A few important concepts to remember ................................................................................................ 34 Questions and Problems ......................................................................................................................... 34 Chapter 2 – Vectors ............................................................................ 40 2.1 Introduction to Vectors ........................................................................................................... 40 2.2 Adding Vectors Geometrically ................................................................................................ 40 2.3 Adding Vectors Mathematically .............................................................................................. 42 A few ideas to remember when adding vectors ..................................................................................... 49 Questions and Problems ......................................................................................................................... 49 Chapter 3 – Linear Motion ............................................................... 51 3.1 Distance vs. Displacement ...................................................................................................... 51 3.2 Speed vs. Velocity ................................................................................................................... 52 3.3 Constant Velocity .................................................................................................................... 54 3.4 Acceleration ............................................................................................................................ 55 3.5 Kinematic Equations (aka Linear Motion Formulas) ............................................................... 63 3.6 Using Formulas and Equations to Solve Problems .................................................................. 64 A few important concepts to remember ................................................................................................ 67 Questions and Problems ......................................................................................................................... 68 Chapter 4 – Acceleration – Velocity – Displacement Graphs 70 4.1 Introduction ............................................................................................................................ 70 4.2 Determining Velocities from a Displacement – Time Graph .................................................. 71 4.3 Creating a Velocity–Time Graph from an Acceleration–Time Graph ..................................... 74 2 4.4 Creating a Displacement–Time Graph from a Velocity–Time Graph ..................................... 75 4.5 Getting Everything from the Versatile Velocity – Time Graph ............................................... 77 A few important concepts to remember ................................................................................................ 79 Questions and Problems ......................................................................................................................... 80 Chapter 5 – Vertical Linear Motion ............................................... 86 5.1 Introduction to Free‐Fall ......................................................................................................... 86 5.2 Vertical Linear Motion ............................................................................................................ 86 5.3 Using Motion Equations in Vertical Motion ............................................................................ 89 5.4 Terminal Velocity .................................................................................................................... 90 A few important concepts to remember ................................................................................................ 93 Questions and Problems ......................................................................................................................... 93 Chapter 6 – Projectile Motion ......................................................... 95 6.1 Introduction to Projectile Motion ........................................................................................... 95 6.2 Shooting the Monkey .............................................................................................................. 97 6.3 Horizontal Launch ................................................................................................................... 98 6.4 Angled Launch ....................................................................................................................... 101 A few important concepts to remember .............................................................................................. 104 Questions and Problems ....................................................................................................................... 104 Chapter 7 – Forces ............................................................................ 107 7.1 Introduction to Forces .......................................................................................................... 107 7.2 Newton’s Laws of Motion ..................................................................................................... 107 7.2.1 Newton’s 1st Law of Motion – Law of Inertia ............................................................... 107 7.2.2 Newton’s 2nd Law of Motion ........................................................................................ 109 7.2.3 Weight ........................................................................................................................... 110 7.2.4 Net Force ....................................................................................................................... 111 7.2.5 Newton’s 3rd Law of Motion –Law of Action‐Reaction ................................................ 112 7.3 Free Body Diagrams .............................................................................................................. 113 7.4 Equilibrium ............................................................................................................................ 118 7.5 Pressure ................................................................................................................................ 123 A few important concepts to remember .............................................................................................. 126 Questions and Problems ....................................................................................................................... 127 Chapter 8 – Friction ......................................................................... 130 8.1 Introduction to Friction ......................................................................................................... 130 3 8.2 Forces of Friction ................................................................................................................... 130 A few important concepts to remember .............................................................................................. 139 Questions and Problems ....................................................................................................................... 139 Chapter 9 – Work and Energy ....................................................... 142 9.1 Work ...................................................................................................................................... 142 9.2 Work in the Vertical .............................................................................................................. 144 9.3 Energy ................................................................................................................................... 145 9.3.1 Kinetic Energy ................................................................................................................ 145 9.3.2 Potential Energy ............................................................................................................ 146 9.4 Conservation of Energy ......................................................................................................... 147 9.5 Pendulum .............................................................................................................................. 151 9.6 Roller Coasters ...................................................................................................................... 152 9.7 Elastic Potential Energy ......................................................................................................... 154 9.8 Non‐conservative Forces ...................................................................................................... 157 9.8.1 Air Resistance ................................................................................................................ 158 9.9 Power .................................................................................................................................... 159 A few important concepts to remember .............................................................................................. 161 Questions and Problems ....................................................................................................................... 161 Chapter 10 – Momentum and Collisions ................................... 166 10.1 Introduction to Momentum .............................................................................................. 166 10.2 Linear Momentum ............................................................................................................ 166 10.3 Impulse .............................................................................................................................. 167 10.4 Conservation of Momentum ............................................................................................. 168 10.5 Introduction to Collisions .................................................................................................. 170 10.5.1 Elastic Collision .............................................................................................................. 170 10.5.2 Inelastic Collision ........................................................................................................... 172 10.5.3 Perfectly Inelastic Collision ............................................................................................ 172 10.6 Energy Lost in a Collision ................................................................................................... 173 A few important concepts to remember .............................................................................................. 174 Questions and Problems ....................................................................................................................... 175 Chapter 11 – Circular Motion ....................................................... 178 11.1 Introduction to Angular Motion ........................................................................................ 178 11.2 Angular Displacement ....................................................................................................... 179 4 11.3 Angular Velocity ................................................................................................................ 180 11.4 Angular Acceleration ......................................................................................................... 181 11.5 Centripetal Force ............................................................................................................... 181 11.6 Centripetal Acceleration ................................................................................................... 184 11.7 Vertical Circular Motion .................................................................................................... 184 11.8 Critical Speed .................................................................................................................... 188 A few important concepts to remember .............................................................................................. 190 Questions and Problems ....................................................................................................................... 190 Chapter 12 – Gravitation and Planetary Motion .................... 193 12.1 Introduction ...................................................................................................................... 193 12.2 Newton’s Law of Gravitation ............................................................................................ 194 12.3 Gravitational Field Strength (aka What You Earthlings Call Gravity) ................................ 196 12.4 Escape Velocity ................................................................................................................. 197 12.5 Orbital Velocity ................................................................................................................. 199 12.6 Introduction to Kepler’s Laws ........................................................................................... 200 12.6.1 Kepler’s 1st Law of Planetary Motion – Law of Elliptical Orbits ................................... 201 12.6.2 Kepler’s 2nd Law of Planetary Motion – Law of Equal Areas........................................ 201 12.6.3 Kepler’s 3rd Law of Planetary Motion – Law of Harmony ............................................ 202 A few important concepts to remember .............................................................................................. 203 Questions and Problems ....................................................................................................................... 204 Chapter 13 – Rotational Mechanics ............................................ 206 13.1 Introduction to Rotational Mechanics .............................................................................. 206 13.2 Torque ............................................................................................................................... 206 13.3 Torque and Equilibrium .................................................................................................... 208 13.4 Torque and Rotational Acceleration ................................................................................. 213 13.5 Moment of Inertia ............................................................................................................. 215 13.6 Rotational Kinetic Energy .................................................................................................. 217 13.7 Angular Momentum .......................................................................................................... 219 13.8 Conservation of Angular Momentum ............................................................................... 220 A few important concepts to remember .............................................................................................. 221 Questions and Problems ....................................................................................................................... 221 Chapter 14 – Electrostatics (aka Static Electricity) ............... 225 14.1 Introduction to Electrostatics ........................................................................................... 225 5 14.2 Matter and Electric Charge ............................................................................................... 225 14.3 Conservation of Charge ..................................................................................................... 226 14.4 Electric Force ..................................................................................................................... 228 14.5 Electric Field ...................................................................................................................... 230 14.6 Rules of Electric Field Lines ............................................................................................... 231 14.7 Electric Potential Energy ................................................................................................... 234 A few important concepts to remember .............................................................................................. 235 Questions and Problems ....................................................................................................................... 236 Chapter 15 – Electric Circuits ....................................................... 240 15.1 Introduction to Electric Circuits ........................................................................................ 240 15.2 Current (I) .......................................................................................................................... 240 15.3 Voltage (V) ......................................................................................................................... 241 15.4 Resistance (R) .................................................................................................................... 241 15.5 Ohm’s Law ......................................................................................................................... 242 15.6 Joule’s Law ........................................................................................................................ 243 15.7 Cost of Electricity .............................................................................................................. 244 A few important concepts to remember .............................................................................................. 245 Questions and Problems ....................................................................................................................... 245 Chapter 16 – DC Circuit Analysis ................................................. 248 16.1 Introduction to DC Circuit Analysis ................................................................................... 248 16.2 Conservation of Charge in a Circuit ................................................................................... 249 16.3 Conservation of Energy in a Circuit ................................................................................... 250 16.4 Series Circuit ..................................................................................................................... 251 16.5 Parallel Circuit ................................................................................................................... 254 16.6 Complex Circuits ............................................................................................................... 257 A few important concepts to remember: ............................................................................................. 269 Questions and Problems ....................................................................................................................... 269 Chapter 17 – Magnetism ................................................................. 274 17.1 Introduction to Magnetism ............................................................................................... 274 17.2 Magnetic Fields ................................................................................................................. 275 17.3 Magnetic Field Lines .......................................................................................................... 275 17.4 Magnetic Force ................................................................................................................. 276 17.5 Magnetic Fields and the Right Hand Rule ......................................................................... 277 6 17.6 Magnetism and Circular Motion ....................................................................................... 281 17.7 Magnetic Force on a Current Carrying Wire ..................................................................... 282 A few important concepts to remember .............................................................................................. 283 Questions and Problems ....................................................................................................................... 284 Chapter 18 – Waves and Vibrations ........................................... 287 18.1 Simple Harmonic Motion, Periodic Motion, and Springs ................................................. 287 18.2 Pendulums ........................................................................................................................ 290 18.3 Wave Motion in Mechanical Waves ................................................................................. 291 18.3.1 Traveling Waves ............................................................................................................ 291 18.3.2 Reflection ...................................................................................................................... 293 18.3.3 Diffraction ..................................................................................................................... 295 18.3.4 Refraction ...................................................................................................................... 296 18.3.5 Interference ................................................................................................................... 297 18.4 Sound ................................................................................................................................ 298 18.5 Intensity and the Decibel Scale ......................................................................................... 299 18.6 Doppler Effect ................................................................................................................... 300 18.7 Speeds Relative to Sound .................................................................................................. 301 18.8 Standing Waves ................................................................................................................. 303 18.8.1 Standing Waves in Strings ............................................................................................. 303 18.8.2 Standing Waves in Open Tubes ..................................................................................... 304 18.8.3 Standing Waves in Closed Tubes ................................................................................... 305 18.9 Resonance ......................................................................................................................... 306 18.10 Electromagnetic Waves .................................................................................................... 307 18.11 EM Spectrum ..................................................................................................................... 309 18.11.1 Low Energy Waves – Radio Waves ............................................................................... 310 18.11.2 Medium Energy Waves – Light Waves .......................................................................... 310 18.11.3 High Energy Waves – Radiation .................................................................................... 311 A few important concepts to remember .............................................................................................. 312 Questions and Problems ....................................................................................................................... 313 Chapter 19 – Light ............................................................................ 316 19.1 Introduction to Light ......................................................................................................... 316 19.2 Reflection .......................................................................................................................... 317 19.3 Refraction .......................................................................................................................... 319 19.4 Critical Angle (θc) .............................................................................................................. 324 7 19.5 Introduction to Color ........................................................................................................ 326 19.5.1 Emission ........................................................................................................................ 326 19.5.2 Reflection ...................................................................................................................... 326 19.5.3 Transmission ................................................................................................................. 327 A few important concepts to remember .............................................................................................. 327 Questions and Problems ....................................................................................................................... 328 Chapter 20 – Optics .......................................................................... 330 20.1 Introduction to Optics ....................................................................................................... 330 20.2 Introduction to Mirrors ..................................................................................................... 330 20.3 Plane Mirror ...................................................................................................................... 331 20.4 Introduction to Curved Mirrors ........................................................................................ 335 20.4.1 Concave Mirrors ............................................................................................................ 335 20.4.2 Ray Diagrams for Curved Mirrors ................................................................................. 336 20.4.3 The Math Way for Mirrors ............................................................................................ 343 20.4.4 Convex Mirrors .............................................................................................................. 345 20.5 Introduction to Lenses ...................................................................................................... 351 20.5.1 Converging Lens ............................................................................................................ 352 20.5.2 Ray Diagrams for Lenses ............................................................................................... 353 20.5.3 The Math Way for Lenses .............................................................................................. 359 20.5.4 Diverging Lenses ............................................................................................................ 361 20.6 Human Eye ........................................................................................................................ 367 20.6.1 Myopia .......................................................................................................................... 369 20.6.2 Hyperopia ...................................................................................................................... 370 A few important concepts to remember .............................................................................................. 370 Questions and Problems ....................................................................................................................... 371 Answers to Calculated Questions ................................................ 377 Chapter 1 – Basic Units and Graphing .................................................................................................. 377 Chapter 2 – Vectors............................................................................................................................... 377 Chapter 3 – Linear Motion .................................................................................................................... 378 Chapter 4 – A‐V‐X Graphs ..................................................................................................................... 379 Chapter 5 – Vertical Linear Motion ....................................................................................................... 379 Chapter 6 – Projectile Motion ............................................................................................................... 379 Chapter 7 – Force .................................................................................................................................. 380 8 Chapter 8 – Friction ............................................................................................................................... 381 Chapter 9 – Work and Energy ............................................................................................................... 381 Chapter 10 – Momentum and Collisions .............................................................................................. 382 Chapter 11 – Circular Motion................................................................................................................ 382 Chapter 12 – Gravitation and Planetary Motion .................................................................................. 383 Chapter 13 – Rotational Mechanics ...................................................................................................... 383 Chapter 14 – Electrostatics ................................................................................................................... 384 Chapter 15 – Electric Circuits ................................................................................................................ 385 Chapter 16 – DC Circuit Analysis ........................................................................................................... 385 Chapter 17 – Magnetism ....................................................................................................................... 387 Chapter 18 – Waves and Vibrations ...................................................................................................... 387 Chapter 19 – Light ................................................................................................................................. 388 Chapter 20 – Optics ............................................................................................................................... 389 9 Introduction To the students, parents, and teachers who decide to use this text, let me explain a few things about it. First, why did I bother to create this? As I often explain to my students at the beginning of every year, I was never a textbook learner. If I tried to learn anything by simply reading the textbook, I usually got distracted at best, or fell asleep at worst. I found I learned the best from my time in the classroom, listening to someone explain the topic and working with my fellow students. As I was never a textbook learner, I have never been a textbook teacher. I rarely assigned any reading from any Physics textbook I had been required to use. Again as I explain to my students, I view a textbook as a tool, much like a calculator, so it is only as good as the person who is using it. Throughout my career, many of my students asked me why I did not write my own textbook since I was never really happy with the ones I had been using. My answer was always the same, that my textbook would be no different than any other textbook. The reason I believed they were so successful in Physics was due to what we did in class, not what was in the book. However with the rise of the internet and school’s adopting online sites where teachers could post information, I started posting my lecture notes. I found many of my students really liked this practice so that if they missed a class or just wanted to verify their own notes, they had a resource. Thus began the idea of maybe turning my lecture notes into an organized text of some sort. In my career, I have taught grades 10 to 12, depending on the school, AP Physics B and AP Physics C. Some classes were called regular, advanced, or honors Physics. I always joked that I would never want to teach 9th graders and felt, unfortunately like many Physics teachers, that you could not teach any real Physics to 9th graders because they lacked the mathematical capabilities and the maturity to grasp the subject. However, upon reading a study by Fermilab, one of the premier physics research labs in the U.S., concerning something called Physics First, I became convinced that of the three major sciences, Biology, Chemistry, and Physics, more than any, Physics was perfect for the 9th grade student. Let’s face it, Biology has become more and more complex and the need for understanding Chemistry before taking Biology is now more important than ever. I believe the average 9th grade student doesn’t really have it in him or her to sit day after day in Biology lectures and grasp it all. It seems that an 11th grader can certainly do this better and more easily grasp all the material, especially after taking a Chemistry class. The kinetic nature of Physics seems to fit well with the hyperkinetic 9th grader. Now this was all really theory to me until I came to work at the Hackley School in Tarrytown, NY which had embraced the idea of Physics First, requiring all 9th graders to take Physics. In addition, there is no separation by abilities, grade point averages, or level of math, everyone is lumped in together. As I discussed textbooks for this type of class with my colleagues, we found that what was available was either too conceptual, not giving enough of the mathematical connections, or else too complex for the average 9th grader to understand. During my first year at Hackley with a group of 9th graders, I continued my practice of posting my lecture notes, 10 much of which I had changed to accommodate the level of the class I was teaching. Again my students found these notes more helpful than their textbook, which seemed at many times too complicated. So I approached my Department Chair with the idea of taking my lecture notes and forming them into a textbook geared directly towards 9th graders taking Physics. There were three people teaching Physics to the 9th grade so we piloted this text along with the usual textbook. Overall the feedback from the teachers and the students was very favorable, so it has now become the 9th grade Physics text going forward. Now a couple of things about this textbook. First, it does not cover every aspect of Physics. There are no sections on fluids, thermodynamics, magnetic induction, or Modern Physics. It contains the material that we are able to successfully cover and cover well in one school year. That does not mean we cover the same things year to year. For example, lately we have cut back on rotational mechanics to allow more time so that optics is not rushed. Even in the first year of using this text, my colleagues and I skipped a couple of things, but I wanted to allow for more flexibility in case others wished to go farther. Second, it takes shortcuts along the way. For example, I give the formula for Work as W = Fx. Now I am aware that the actual formula is W = Fxcosθ. However, is that angle really that important when it comes to getting a 9th grader to understand the basic concepts and the math around the idea of Work and how it relates to energy. I did not really think so, but to my fellow teachers, if you want to teach it the other way, please do so. That is the great thing about being the teacher, in the end you get to add and subtract what you feel fits with what you are trying to accomplish. Like I said before, a textbook is a tool, use it to your best ability. Finally, what I have tried to do a lot of is mathematical examples, showing them from beginning to end. I have often found that the most frustrating thing about example problems in textbooks is that they lack clarity on how someone gets from one step to the next. I have done my best in this area. This can make some things such as electric circuits and optics rather lengthy, but I hope it does the job. You will notice that the answer section at the back only gives the answers to calculated problems. I have always given my students the answers ahead of time to calculated problems so they would know if they were on the right track. Of course, they also knew to never hand in an assignment with no work and just answers. I have never provided the answers to what I call conceptual questions as they should be able to find those in their notes or in the text. Finally, I do discuss converting units and dimensional analysis, but only in the context of the metric system. I try to avoid units in the English system, as much as possible, from beginning to end. My own personal quest to get rid of the English system of measurement. So to everyone who uses this, good luck, I hope it serves you well. I would be remiss if I did not include a few acknowledgements. First, to Manna Ohmoto‐
Whitfield, for all her support in actually creating this text, her suggestions, and her willingness to take a chance on this for the 9th grade. Also for being the best Science Department Chair ever. To Seth Karpinski and Dr. Andrew Ying, my Physics teaching colleagues, who agreed to be part of the pilot and subject their classes to this text. Also for their suggestions on improvements. To Walter Johnson, Headmaster of the Hackley School and Andy King, Upper School Director of the Hackley School, for supporting this effort. To all the 9th graders at The Hackley School in the 2009‐2010 school year (also known as the class of 2013) who were the 11 first to experience this text and their acceptance, criticisms, and suggestions; after all, this was created for them. To my son Will, who at 3 to 5 years old had to deal with a certain lack of attention at times when I was working on this, I hope one day you can say you were proud of what Dad did. Finally to my wife, Shannon, for her love, support, listening to me complain, telling me to shut up and deal, and for being proud of the results. Bill McLay The Hackley School Tarrytown, NY June 30, 2011 If you have any comments, critiques, or questions, please feel free to email me at [email protected] or [email protected] . © William McLay, Nicodemos Enterprises, August 2010. Photograph on the cover is courtesy of Bill McLay, it was taken with his Nikon camera while he was observing stars at the Mt. Kitt Observatory. 12 Chapter 1 – Basic Units of Mechanics and Graphing Before we get started with some actual Physics, we need to take care of a few things that will be important to understanding and using physics and in solving physics related problems or scenarios. This first part will concern understanding the types of units used in physics and converting from given or measured units into standard units in physics, and the second part will be a review of graphing data in a scientific way. 1.1
Basic Units of Mechanics If I were to describe myself to you using some measured values and from these values I told you that I am about 88, about 182, and about 42, what have I told you about myself? Well, I just gave you a bunch of numbers that really don’t mean much of anything. I have often begun this unit in class with this very question. Now most of my students correct guess that I have given my height, weight, and age, and they are correct. However, which number is which? Well, most say that the 88 is my height, as in 88 inches tall; 182 is my weight, as in 182 pounds; 42 is my age as in 42 years old. However, if I were actually 88 inches tall, then I would be 7 feet 3 inches tall, and I assure you that I am not; otherwise I would have pursued a different career. In addition, I have not been 182 pounds since I was 32. Finally, I ask if I really look 42 years old, which I am, and then cry. Just kidding. The issue is that when I gave those numbers, I was missing some important pieces of information, something called units. Now if I told you that I was about 88 kg (195 pounds), you would now know something related to my mass or weight. If I told you I was 182 cm (6 feet), you would now know something about my height. As I said, I am 40 years old, but by adding years to that 40, you would now know my actual age. So units are very important to any value, particularly in physics. The main reason, of course, is communication, units communicate something about the number you have measured or have been given. So let’s look at a little history of units and why we use certain units in physics. 1.2
Metric Conversions In many physics classes, time is often spent on how to convert from the English System of measurement to the Metric System of measurement. Though many a teacher will disagree with me, I say, “Why bother?” As we have just discussed, the English System is outdated and not useful in science and we need to our future scientists better acquainted with the Metric System. So throughout this text, you will encounter little to no English units, everything will be in metric measurements. That being said, we often encounter metric measurements that are not in “standard units” (i.e. meter, kilogram, second, etc.). In Physics, it is necessary to have all 17 measurements in standard units before doing any calculations; otherwise the answers will be incorrect. There are a few exceptions, but the plan should be to always convert any measurement to standard units before doing anything else. The great thing about converting in the metric system is that it is based on the number 10. Unlike the English System that has strange conversions such as 5280 feet in a mile or 16 ounces in a pound or 12 inches in a foot, the Metric System does everything in multiples of 10. For example there are 100 cm in a meter, 1000 meters in a kilometer, 0.001 (1/1000) kilograms in 1 gram, etc. This makes converting in the metric system much easier. Now every teacher has his or her own way of teaching conversions in metric, so my way is not the only way and may not even be the best way, but I have found it easy to use. First, we need to understand what are known as prefixes in the Metric System. Then we will see how to go from a non‐standard unit to the standard unit. Here is where you will have to unlearn a few things you may have been taught about the Metric System. Some common prefixes you may have learned are: Kilo (k) = 1000 Hecto (H) = 100 Deca (D) = 10 Deci (d) = 0.1 (1/10) Centi (c) = 0.01 (1/100) Milli (m) = 0.001 (1/1000) These are all perfectly legitimate prefixes, however in Physics, we never really use Hecto, Deca, and Deci, and many don’t use Centi, though we often will. In fact, we tend to go even farther. Here is a list of the more common prefixes used in Physics and their symbols. Metric Prefixes Prefix Name giga
mega
kilo
BASE
centi
milli
micro
nano
Prefix Symbol Prefix Value Scientific Notation
G
M
k
---c
m
μ
n
1,000,000,000
1,000,000
1,000
1
0.01
0.001
0.000001
0.000000001
109
106
103
100
10-2
10-3
10-6
10-9
If we ignore centi for a moment and look to the far right under Scientific Notation, we can see that each prefix as you move from the bottom to the top is 1000 times greater than the previous prefix. Thus as you go from one prefix to the next, you either add or remove three 18 zeroes. The easy way to do this in terms of any number is to simply move the decimal point within that number left of right. So I like to look at this in terms of a ladder than runs from right to left as it goes up as shown below. G
M
k
Base
c
m
µ
n
Here is how this works, with the exception of one unit (mass), all standard units are located at the Base on the ladder, which means they have no prefix. So we are almost always trying to get to the base. If we begin with a measurement below the Base on the ladder, such as 100.0 mm (100 millimeters), to get to the Base, we have to go one big step up the ladder where each big step is equal to 1000 or three decimal places. As the Base is to the left of the milli (m) step, we must move the decimal to the left. Thus 100.0 mm = 0.1 m (100 millimeters equals 0.1 meters, the standard unit of length.) If we begin with a measurement above the Base on the ladder, such as 25.0 km (25 kilometers), to get to the Base, we have to go one big step down the ladder where again each big step is equal to 1000 or three decimal places. As the Base is to the right of the kilo (k) step, we must move the decimal to the right. Thus 25.0 km = 25,000 m (25 kilometers equals 25,000 meters, the standard unit of length.) Let’s try another one, how many seconds is 2000 μs (2000 microseconds)? The prefix micro (μ) is two big steps below the Base, so to get to the base, we need to move six (6) decimal places to the left, thus 2000 μs = 0.002 s (2000 microseconds equals 0.002 seconds, the standard unit of time.) Not too complicated right? Well, there always seems to be some exception we have to deal with, and in this case there are two you need to be aware of. The first, and most important, deals with the standard unit of mass, the kilogram. The kilogram is the only standard unit with a prefix and thus not located at the Base on the ladder. So whenever we convert to the standard unit of mass, we must be on the kilo (k) step, not the base. So if you measure something to be 350 g (350 grams) which is a Base unit (it does not have a prefix) we must 19 convert it to kilograms. As kilo (k) is one big step above the Base, we move the decimal three places to the left. Thus 350 g = 0.35 kg (350 grams equals 0.35 kilograms, the standard unit of mass.) If you measured 467 mg (467 milligrams) to convert to kilograms, that would be two big steps up the ladder or six (6) decimals to the left, thus 467 mg = 0.000467 kg Again, this is all about moving decimal places. Now the other little hitch in our giddy‐up is that quite often we measure small distances in terms of centimeters. One reason is that the centimeter is close to the English unit the inch (1 inch = 2.54 centimeters). The prefix centi is not on a big step as it is 1/100 of the base (two decimal places) not 1/1000 (three decimal places). So whenever you encounter a unit with centi, you will move two decimal places to the left to reach the Base. For example, 45 cm = 0.45 m (45 centimeters equals 0.45 meters) 560 cg = 0.0056 kg (One small step to the Base and one big step to kilo.) Here are a few to practice, answers follow: 1. 20 cm = ____________ m 2. 500 g = ____________ kg 3. 30 ms = ____________ s 4. 80 km = ____________ m 5. 200 cm = ___________ m 6. 5 g = ______________ kg Answers: 1.
2.
3.
4.
5.
6.
20 cm = ___0.2______ m 500 g = ____0.5______ kg 30 ms = ____0.03_____ s 80 km = __80,000____ m 200 cm = ___2.0_____ m 5 g = ____0.005______ kg 20 Got it? I hope so because you will have to do this quite a bit this year. Here is a list of many of the standard units we may encounter in this text. Standard Metric Units Type of Measurement Length, width, height,
distance, displacement
Mass
Time
Velocity
Acceleration
Force (All forces)
Energy (All energies)
Power (All power)
Momentum
Angular displacement
Angular velocity
Angular acceleration
Torque
Angular Momentum
Moment of Inertia
Pressure
Temperature
Charge
Electric field strength
Potential Difference
(Voltage or EMF)
Current
Resistance
Magnetic field strength
Electromotive force
(EMF)
Period
Frequency
Intensity (sound)
Unit Name
Unit meter
m
kilogram
second
meter per second
meter per second squared
Newton
Joule
Watt or Joule per second
kilogram meter per second or
Newton second
Newton meter
kilogram meter squared per second
kilogram meter squared
Pascal
Degree Celsius*
Coulomb
Newton per Coulomb
Volt
kg
s
m/s
m/s2
N
J
W or J/s
kgm/s or Ns
Ampere (Amp)
Ohm
Tesla
Volt
A
Ω
T
V
second
Hertz
Watt per square meter or Decibel
S
Hz
W/m2 or dB
Nm
kgm2/s
kgm2
Pa
o
C
C
N/C
V
*Degree Kelvin (K) is the true scientific unit of Temperature, but Celsius is commonly used. 1.5
Dimensional Analysis Using the metric conversion method described in the previous section works well for basic conversions. However, we often have more complex conversions. For example, the standard unit for the measurement of area is in m2 (square meters). If we were given the area of a rectangle 21 drawn on a piece of paper, it would more likely be given in cm2 (square centimeters). Therefore we would need to convert cm2 to m2. This would not be so easy to do with just moving decimals. In addition, the speedometer of a car gives speeds in km/hr rather than the physics standard of m/s. As time conversion is not a simple decimal movement, we need to employ a different method of converting called dimensional analysis. This is a handy thing to know how to do as you can also use this when calculating problems to see if your answer has the correct units. (It will also be necessary to understand when you hit Chemistry). The basic idea behind dimensional analysis is choosing one or more conversion factors. These are multipliers that are equal to 1. For example, 1 kg = 1000 g, or 1 m = 100 cm, or 1 minute = 60 s. As you can see, each one of these factors has a 1 value. To begin the process, you will need the initial value and it’s current units and you will need to know the necessary units of your final answer. Example: Convert a speed of 80 km/hr into the standard unit of m/s. Start by setting up 80 km over 1 hour. Since km is on top, we need to make sure it is on the bottom of the conversion. Now we can cancel km and that leaves us with m. Now we can work on hours. This requires two conversions, 1 hr = 60 min, and 1 min = 60 s. ,
We know that 1 km = 1000 m. So we need to cancel km and leave meters. We need to convert so that km becomes m on top and hr becomes s on bottom. . As hr was on the bottom, we had to ensure it was on top in the coversion factor. The same for min to sec. Once we cancel out the units, we multiply along the top and bottom and reduce. This leaves us with our converted answer and in the correct units. Example: Convert 60 cm2 to m2. .
,
This looks different. There is no easy direct conversion from cm2 to m2. So we need to get rid of cm twice and have m multiplied twice to give m2. This involves using the 1 m = 100 cm conversion factor twice. You can see that all cm are gone and we are left with m2. Here are a few to practice; answers to follow: 1. 80 cm/min = _______________ m/s 2. 20 km/hr = _______________ m/s 22 3. 5,000 mm2 = _______________m2 4. 70 g/cm = ________________ kg/m 5. 2,000 cm3 = ______________m3 Answers: 1.
,
.
2.
,
,
.
,
3.
,
,
,
,
,
.
4.
,
,
. 5.
,
,
,
,
.
Notice that on example number 5, we had to use the 1 m = 100 cm conversion factor three times to get cm3 to m3. 1.6
Graphing in Physics It is often very useful, particularly in labs, to represent a series of measurements, information, or data points on a graph. This gives you a picture of your information and enables you to spot trends or determine other information or results from your graph. It is therefore very important that you understand how to construct a proper graph. These days, computers can easily draw graphs for you, but often you may find that you need to do one on your own. So there are a few basic rules you need to follow. 1. Draw a set of axes. 2. Title your graph in terms of y vs x (e.g. Displacement vs Time). 3. Label each axis in terms of the symbol of what is measured and the units in which it is measured. 4. Draw and label equal, reasonable intervals on each axis. 5. Plot your data points. 6. Draw the best fit line or curve through your data points. 23 Let’s take an example of measuring the distance (x) a ball rolls on the floor from a certain starting point every second (t). We start the ball rolling and start a stopwatch marking points on the floor where the ball passes each second (s) and then measuring these points from the starting point in meters (m). In our experiment we end up with the following table of values: x (m) t (s) 0 0 2 1 4 2 6 3 8 4 Now let’s graph this data. First we draw a set of axes, they can be crossed or look like an “L”. or I like the “L” shape for this one as there are no negative values. Next we label each axis in terms of the symbol and unit of measurement. In this case, one axis will have distance in terms of meters and the other time in terms of seconds. But which axis do I label with which value? To determine this we have to know which is the independent measurement and which is the dependent measurement. The dependent measurement is always on the y‐axis (the vertical axis) and the independent measurement is always on the x‐axis (the horizontal axis). What does that really mean, dependent measurement and independent measurement? The independent measurement is the one that is unaffected by any other measurement. The most common independent measurement is time. The time on the stopwatch has nothing to do with where the ball is located. Once the stopwatch begins, each second will tick off regardless of where the ball actually is located, so it is independent of the ball’s location. However, as we are measuring the location of the ball at specific times, every second, its location depends on the specific time that we want. This means that the distance is the dependent measurement as it is based on another measurement. So our distance in meters goes on the y‐axis and our time in seconds goes on the x‐axis. We also title the graph in terms of the measurement on the y‐axis vs the measurement on the x‐axis. 24 x(m) Distance (x) vs Time (t) t (s) Our next step is to divide up each axis into reasonable equal intervals. This is the one part that people mess up the most. Not so much the dividing up the axis, but in picking reasonable and equal intervals. I have seen graphs that look like this: x (m) Distance (x) vs Time (t)
65 42 27 14 8 3 10 30 60
150 t (s)
Basically this graph was made using all the collected numbers. Equal intervals means that the difference between each line on the graph must be the same value. You also want to pick reasonable values for each axis, though they do not have to be the same intervals on each axis. For example, for the data we collected, the reasonable intervals for the x‐axis would be 1.0 second intervals as the time was measured each second. For the y‐axis, you could choose 1.0 meter intervals or 2.0 meter intervals, either one are reasonable. I will choose 2.0 meter intervals for my graph. x (m) Distance (x) vs Time (t)
10 x (m) 0 2 4 6 8 8 6 4 2 1 2 3 4
t (s)
t (s) 0 1 2 3 4 Next we need to plot our data points. (Note: it is always best to do graphs on graph paper.) 25 x (m) Distance (x) vs Time (t)
1
8 x (m) 0 2 4 6 8 6 4 2 t (s) 0 1 2 3 4 1 2 3 4
t (s) Finally, draw the best fit line or curve through the points. The best fit line or curve is a straight line or continuously curved line that best encompasses all the points on the graph. It is not a game of connect the dots and in some cases the line or curve may not touch a single point, but it is a reasonable representation of the data. In our example, however, the best fit line does pass through all the points on the graph. x (m) Distance (x) vs Time (t)
10 8 x (m) 0 2 4 6 8 6 4 2 t (s) 0 1 2 3 4 1 2 3 4
t (s) This graph reveals certain trends, in this case the ball rolled at a constant rate of gaining 2.0 meters every second. That indicates that it traveled at a very constant speed across the floor. Constant relationships tend to be graphed as linear relationships. Example: Let’s take a different data set and graph it, given the following data: x (m) t (s) 0 0 1 1 4 2 9 3 16 4 26 Our initial setup should look like this: x (m) Distance (x) vs Time (t)
16 14 12 x (m) 0 1 4 9 16 10 8 6 4 t (s) 0 1 2 3 4 2 1 2 3 t (s)
4
We draw and label the axes and chose reasonable equal intervals for each axis. Now we can plot our points. x (m) Distance (x) vs Time (t)
16 14 12 x (m) 0 1 4 9 16 10 8 6 t (s) 0 1 2 3 4 4 2 1 2 3 4
t (s)
27 We draw our best fit line or curve. In this case, we see an upward curving trend. x (m) Distance (x) vs Time (t)
16 14 12 x (m) 0 1 4 9 16 10 8 6 t (s) 0 1 2 3 4 4 2 1 2 3 4
t (s)
A simple straight line would not have really given us an accurate picture of the trend, so a curve was more appropriate. Each value on the y‐axis is the square of the value on the x‐axis, giving us what is known as a quadratic relationship instead of a linear relationship. All quadratic relationships are graphed as curves. Example: Let’s try one last one. Create a graph of the following set of data: x (m) t (s) 0 0 3 2 4 4 7 6 10 8 28 If we setup our axes, label them, create our equal reasonable intervals and plot the points, we should have something like this: Distance (x) vs Time (t)
x (m) 10 8 x (m) 0 3 4 7 10 6 4 2 2 4 6 8
t (s) 0 2 4 6 8 t (s) Look carefully, our trend appears to be linear, yet not a perfect line. As we cannot just play connect the dots, we need to draw a line that best fits the data and gives us a reasonable linear trend. I like to take a ruler and holding it vertically and long ways against the graph, I can move it around until I have a line that seems to have an equal number of points above and below the line. This would give me something like this: Distance (x) vs Time (t)
x (m) 10 8 x (m) 0 3 4 7 10 6 4 2 t (s) 0 2 4 6 8 2 4 6 8 t (s)
Quite often in a situation like this, three different people may have three slightly different lines as we are really just taking our best estimate of the line that appears to best fit the data. In the end, though we will all be reasonably close enough. This best fit line indicated that the motion of the object was close to being constant. 29 1.7
Slope If we go back to our first graph we created, it looked like this: x (m) Distance (x) vs Time (t)
10 8 x (m) 0 2 4 6 8 6 4 2 t (s) 0 1 2 3 4 1 2 3 4 t (s) We discussed the fact that this linear relationship indicated that the ball was rolling at a constant rate which was indicated by the graph. How do we really know, and how can we determine that constant rate. The best way is to look at the slope of the line that is created. I know everyone remembers how to get the slope of a line from math class, but let’s do a quick review. The slope of a line is often defined as the rise over the run. The slope of a line is represented by the letter “m” for some reason. It is calculated as: ∆
∆
m – slope Δy – change in y values Δx – change in x values In case you do not remember, Δ or change in, indicates the difference between the final value and the initial value. So slope can look like this: ∆
∆
Basically you pick two convenient points on your graph and plug their y and x values into the equation. In the graph above, we could choose any two points we want, for example (2, 1) and (6, 3). As the line is moving upward, the (6, 3) point would be considered the final point and the (2, 1) point would be considered the initial point. 30 ∆
∆
6 2
3 1
4
2
m = 2 The 2 that was calculated is not the entire answer because it lacks units. As we discussed earlier in this chapter, units are an important part of an answer. We can get our units right off the graph itself. Distance (x) vs Time (t)
x (m) 10 8 x (m) 0 2 4 6 8 6 4 2 t (s) 0 1 2 3 4 1 2 3 4
t (s) As the y‐axis is measured in meters (m) and the x‐axis is measured in seconds (s) and slope is Δy over Δx, then our units are the y‐axis units over the x‐axis units, in this case m/s. So the slope of our line is: m = 2 m/s As we will learn later, m/s is the unit of speed or velocity. This means that our ball is traveling at a constant speed of 2.0 m/s. Slopes can be either positive or negative values. An upward moving line will have a positive slope; a downward moving line will have a negative slope. In addition, the steeper the line, the higher the slope value will be. 31 Example: Given the following graph with the three lines (1. Red line, 2. Blue line, 3. Green line), determine the slope of each line. x (m) Distance (x) vs Time (t)
16 1. 14 12 2. 10 8 6 4 3.
2 1 2 3 4
t (s)
In each case, I have marked off reasonable points to use in determining the slope as each of these points are easy to find. Note that whenever a line begins or ends at the origin (0, 0) you can usually use that point in determining your slope. 1. Red line – (16, 2) and (8, 1) ∆
∆
m = 8 m/s 2. Blue line – (12, 4) and (6, 2) ∆
∆
m = 3 m/s 3. Green line – (2, 3) and (0, 0) ∆
∆
m = 0.67 m/s As you can see the red line which was the steepest line also had the highest slope. The green line which was the shallowest line had the lowest slope. Let’s do one more. 32 Example: Given the following graph with the two lines (1. Red line, 2. Blue line), determine the slope of each line. F (N) Force (F) vs Distance (x)
16 1. 14 12 10 8 6 2.
4 2 1 2 3 4 x (m)
Now in this case, the red line is moving upward and I have chosen (16, 4) and (6, 0) as my two points. The blue line, however, is moving downward and I have chosen (6, 3) and (10, 1) as my two points. Now determine the slope of each line. 1. Red line – (16, 4) and (6, 0) ∆
∆
m = 2.5 N/m 2. Blue line – (6, 3) and (10, 1) ∆
∆
m = ‐2.0 N/m These two slopes tell us two things, one is that F/x, measured in N/m (did you catch the fact that the units changed?), is the red line is constant and positive, but the blue line is constant and negative. In addition, the increase is greater on the red line than the decrease on the blue line. 33 So that is the basics of graphing. You will be dealing with a lot of graphs this year and you will need to know how to create and/or interpret these graphs. Good luck! A few important concepts to remember 1.
2.
3.
4.
5.
6.
7.
8.
9.
All answers require units as units are important for communication. The metric system (SI) has units that are universal in nature. The standard meter is based on the speed of light in a vacuum. The standard kilogram is based on a specific platinum‐iridium alloy housed in Sèvres, France. The standard second is based on the period of oscillation of a Cesium‐133 atom. You can easily convert within the metric system by moving the decimal point of a number the proper number of places. When converting more than one unit, use dimensional analysis and conversion factors such as 1 m = 100 cm. Follow your graphing rules at all times: a. Draw a set of axes. b. Label each axis in terms of the symbol of what is measured and the units in which it is measured. c. Title your graph in terms of y vs x. d. Draw and label equal, reasonable intervals on each axis. e. Plot your data points. f. Draw the best fit line or curve through your data points. The slope of a graph indicates information and trends. Questions and Problems Convert the following: 1. 20 cm = ____________ m 2. 50 g = ____________ kg 3. 30 ms = ____________ s 34 4.
5.
6.
7.
8.
9.
80 km = ____________ m 5 g = ______________ kg 4 cm = _____________ m 80 µs = ____________ s 700 nm = ___________ m 900 cm = ___________ m 10. 0.6 km = ___________m 11. 30 km/min = ___________ m/s 12. 60 km/hr = _____________ m/s 13. 5,000 cm3 = ____________ m3 14. 700 g/cL =_____________ kg/L 15. 0.04 km2 = _____________ m2 Questions 16 – 21 – Graph the following data sets (should be on graph paper). 16. Distance vs. time x (m) t (s) 0 0 4 6 8 12 12 18 16 24 17. Velocity vs. time v (m/s) t (s) 0 0 12 10 16 20 25 30 28 40 35 18. Distance vs. time x (m) 0 16 64 144 256 t (s) 0 4 8 12 16 v (m/s) 5 10 15 20 25 t (s) 0 10 20 30 40 19. Velocity vs. time 20. Force vs. time F (N) 0 ‐ 2 ‐ 4 ‐ 6 ‐ 8 21. Force vs. distance F (N) 10 20 30 40 50 t (s) 0 3 6 9 12 x (m) 0 5 10 15 20 36 Questions 22 – 26 – Determine the slope of the lines in the following graphs. 22. Distance (x) vs. Time (t)
x (m)
35
30
25
20
15
10
5
0
0
2
4
6
8
10
12
t (s)
23. Distance (x) vs. Time (t)
x (m)
120
100
80
60
40
20
0
‐20
0
5
‐40
10
15
20
25
30
t (s)
37 24. Velocity (v) vs. Time (t)
30
v (m/s)
20
10
0
‐10
0
5
10
15
20
25
30
‐20
‐30
‐40
‐50
t (s)
25. Force (F) vs. Distance (x)
F (N)
30
25
20
15
10
5
0
0
2
4
6
8
10
12
x (m)
38 26. Force (F) vs. Time (t)
F (N)
35
30
25
20
15
10
5
0
0
2
4
6
8
12
10
t (s)
Adding Vectors Geometrically One of the issues we have to deal with when it comes to vectors is how to combine multiple vectors together. If someone were to tell you to travel 30 m, 20o NE and then from there travel 40 m, 30o NW; rather than doing that, you could travel the shorter distance from your beginning point to your ending point. The difficult part is finding that path. One method is to geometrically add the two vectors together (and don’t we all love to do geometry)! To do this, you take a ruler and a protractor, accurately draw the first vector and then attach the beginning of the second vector (also drawn accurately) to the end of the first vector. 40 We will call the 30 m, 20o NE vector A, and the 40 m, 30o NW vector B. Like this: B
A 30 m 40 m
30o
20o We then attach them together like this: B
A
Finally we connect the beginning of vector A with the end of vector B and then using a ruler and protractor, accurately measure the resulting arrow (we will call this vector R) for both magnitude and angle. B
R A
Thus the sum of vector A and vector B (as measured with a ruler and protractor) is equal to 29 m, 77o NW. Now the odds of that being exactly right are pretty slim considering the human error involved. (It is hard enough on paper; you can imagine what it is like doing it on a computer screen). This may seem the easy way with two vectors, but if you have four or six, every time you add another, you increase your chance of error, so this is clearly not the best way. 41 2.3
Adding Vectors Mathematically The second way is mathematical (really trigonometry) and at first it looks hard, but it is really not if you follow some basic rules. First, let’s consider vector A by itself. If we draw vector A on an axis it looks like this (note it is not necessary to draw it accurately): y A
30 m 20o
x
There is a part of vector A that could be said to point in the x‐direction. This is known as the x‐
component of A (Ax) as shown below on the left. In addition, there is a part of vector A that could be said to point in the y‐direction. This is known as the y‐component of A (Ay) as shown below on the right. (The word “component” means “part”). y y A A
Ax x
Ay x All together, A, Ax, and Ay form a right triangle. y A
20o
30 m Ay Ax x
42 The length of Ay is based on the trig function Sin where θ is the indicated angle and then sin
(sin θ is equal to the length of the side opposite the given angle divided by the hypotenuse of the triangle). In this case, sin
. Since I know the value of the angle (20o) and the hypotenuse (30 m), I can get the value of Ay by: Ay = Asin(θ) Ay = 30 sin(20) = 10.26 The length of Ax is based on the trig function Cos where θ is the indicated angle and then cos
(sin θ is equal to the length of the side adjacent the given angle divided by the hypotenuse of the triangle). In this case, cos
. Since I know the value of the angle (20o) and the hypotenuse (30 m),, I can get the value of Ax by: Ax = Acos(θ) Ax = 30 cos(20) = 28.19 I can then repeat the same process for vector B. y 40 m B 30o x
y 40 m By 30o Bx x
The length of By is based on the trig function Sin where θ is the indicated angle and then sin
(sin θ is equal to the length of the side opposite the given angle divided by the hypotenuse of the triangle). In this case, sin
. Since I know the value of the angle (30o) and the hypotenuse (40 m), I can get the value of By by: 43 By = Bsin(θ) By = 40 sin(30) = 20 The length of Bx is based on the trig function Cos where θ is the indicated angle and then cos
(sin θ is equal to the length of the side adjacent the given angle divided by the hypotenuse of the triangle). In this case, cos
. Since I know the value of the angle (30o) and the hypotenuse (40 m), I can get the value of Bx by: Bx = Bcos(θ) Bx = 40 cos(30) = 34.64 Of course, we all know that, we all took trigonometry, right? Anyone? OK, maybe we don’t remember that so well or never even heard it before. So here is a way to remember what to do, follow these two rules: 1. X goes with Cos, Y goes with Sin. 2. (Number)(Function)(Angle) In other words, rule #1 says that the x‐component will almost always go with the Cos function and the y‐component will almost always go with the Sin function. (Note: I say almost always as there are exceptions, but this will work well for our purposes and we can deal with exceptions later. Also, don’t say this to your math teacher as he/she will not like me setting up math rules that are not always true, so just use these in Physics class). Rule #2 says that to calculate either component, we write the equation in the order of (Number)(Function)(Angle). Now if we were to place the x‐component of vector A together with the x‐component of vector B (geometrically) and the y‐components of vectors A and B together (geometrically) we would have this: y By Ay Bx Ax x
44 Since the two y‐components are on the same axis and in the same direction, they can be added together. Since the two x‐components are on the same axis, but in opposite directions, they should be subtracted. Thus would lead to this: y x
But more accurately could be made to look like this: y Ry Rx x
These are now the x‐component (Rx) and y‐component (Ry) of what we will call the resultant vector R. By making a right triangle out of this, we will have the sum of the vectors and R along with its angle θ. R Ry θ Rx To get R, well we have the two sides of a right triangle so to get the hypotenuse we apply a version of the Pythagorean Theorem (usually seen as a2 + b2 = c2) shown below. 45 However that only gives us the magnitude, the direction requires an angle. To get θ we use a special trig relationship called the Inverse Tangent: So in our previous example of: Vector A of 20 m 20o NE Vector B of 30 m 30o NW The sum of Ay and By is 10.26 + 20 = 30.26 (Positive indicates a North direction) (Negative indicates a West direction) The sum of Ax and Bx is 28.19 – 34.64 = ‐6.45 6.45 (Note: We drop all negatives for this.) √30.26
R = 30.9 m tan
tan
.
.
√106.6
√11363.56 1656.49 √13020.05 R = 114.1 m tan
θ = 69.1o tan
.
.
= tan‐1(2.62) So the final value for R is 114.1 m 69.1o NE (NE since both components were positive) Try finding the resultant of these sets of vectors for practice. 1. A 50 m S B 20 m 45o NW C 90 m E (Ans. 143.8 m 27.3o SE) D 60 m 30o SE 2. A 70 m W B 50 m 60o SW C 30 m N D 80 m 30o NW (Ans. 166.4 m 9.2o NW) 48 3. A 30 m E B 60 m 60o SE C 80 m S D 40 m 45o NE (Ans. 135.6 m 49.4o SE) A few ideas to remember when adding vectors 1. Create a Y vs. X chart. 2. Separate the easy vectors (those in along an axis) into components first; one will always be zero. 3. Separate the remaining vectors using our two rules: a. X goes with Cos, Y goes with Sin. b. (Number)(Function)(Angle) 4. Check all directions for the proper signs, North and East are positive, South and West are negative. 5. Add both columns to get the components of your resultant vector (R). 6. Use the Pythagorean Theorem and the Inverse Tangent formulas to solve for the resultant vector and angle. 7. Use the signs of the components of R to determine the compass direction. Questions and Problems 1. Describe the difference between a scalar number and a vector. Find the resultant of each set of vectors, including an angle if necessary. 2. 400 m W 300 m E 3. 500 m N 800 m S 200 m N 100 m S 4. 400 m N 300 m W 49 5. 800 m E 600 m S 6. 400 m S 600 m N 200 m E 300 m W 7. 900 m E 700 m S 200 m W 500 m N 8. 200 m 300 NE 400 m 45 0 SW 9. 50 m 600 SE 80 m 300 SW 10. 40 m/s 300 NW 60 m/s W 90 m/s 600 NE 30 m/s S 11. 6.0 N 45o SE 8.0 N S 5.0 N 200 SW 7.5 N E 12. 40 Ns 25o NE 10 Ns W 50 Ns 800 NW 25 Ns E 13. 24 m 35o NE 86 m S 52 m 800 SW 78 m E 14. 0.4 m/s2 300 SW 0.7 m/s2 W 0.2 m/s2 S 0.9 m/s2 400 E 50 Chapter 3 – Linear Motion So let’s finally get started with some physics. We are going to begin with linear motion, motion along a straight line. We will start with horizontal linear motion and later look at vertical linear motion, often called free‐fall. To begin with, we need to understand some basic terms. There are terms we are used to using in our daily lives versus terms that are used in physics and their meanings. 3.1
Distance vs. Displacement Distance Displacement x [m] Scalar value Vector Defined as how far. Defined as distance relative to a specific point. Measured in meters [m] Measured in meters [m] Consider going out to the track and running one complete lap returning to where you began. You have now run 400 m. However, that is distance, how far you have run. Physics relies on the idea of displacement. As displacement is your distance relative to a specific point, your total displacement is 0 m, as relative to your starting point you have gone nowhere. One lap around a track x = 400 m in distance
400 m
0m
One lap around a track x = 0 m in displacement
The symbol we use for displacement is the letter “x”. Unlike distance, displacement can be a positive or negative number, distance is always positive. Depending on the starting point and which directions are defined as positive and negative determines the sign of a displacement. 51 In Figure 1 below, anywhere to the left of the line is defined as a negative displacement and anywhere to the right of the line is defined as a positive displacement. As the indicated point is to the right of the line, it has a positive displacement (+x) relative to the line. In Figure 2 below, anywhere below the line is defined as a negative displacement and anywhere above the line is defined as a positive displacement. As the indicated point is below the line, it has a negative displacement (‐x) relative to the line. - Displacements
+ Displacements
+ Displacements
+x
-x
Figure 1
- Displacements
Figure 2
One of the keys to physics is the ability to define positive and negative directions. The fun part is that it is really up to you which way you call positive. Typically we call directions to the right of a point (east) and above a point (north) positive and to the left of a point (west) and below a point (south) negative. However, you can really do whatever you want, but once you pick a positive and negative direction, you have to stick with your choice through all your work. 3.2
Speed vs. Velocity Speed Velocity v [m/s] Scalar value Vector Defined as how fast. Defined as change in displacement over time. Measured in meters [m/s] Measured in meters [m/s] Can be positive or negative Let’s go out to the track again. This time you are told to run 100 m on the straight portion of the track and then turn around and run back 100 m to where you began. Let’s say it takes you 20 seconds to do this. Aside from the fact you should be training for the Olympics, if you could do this, let’s look at your speed versus your velocity. 100 m
20 s
100 m
– Average velocity [m/s] Δx – Change in displacement [m] t – Time [s] The Δ in the Δx means “change in”. In other words a change indicates the difference between the final value and the initial value. Therefore Δx = xf ‐ xo. (NOTE: I like to use the letter “f” to indicate a final value and the letter “o” to indicate a beginning or initial value). So back to our running 100 m turning around and coming back to the start in 20 s, bet you thought I forgot. As you begin at a certain point, you can call that position the zero (0) position. However, that is the same place you return. So, xo = 0 m xf = 0 m t = 20 s ∆
0
0
20
/ 53 As you can see, since the initial and final positions are the same and displacement is your distance relative to a point, the change in displacement (Δx) is zero (0). This makes the average velocity for the entire trip is zero (0). This, of course, sounds crazy; you did all the running for apparently nothing. This is why in physics we are more often concerned with instantaneous velocities than average velocities. 3.3
Constant Velocity An important velocity that we will come across a lot is known as a constant velocity. This sounds just like it is; a velocity that does not change in terms of magnitude and direction. If I were to graph the motion of something traveling at a constant velocity, the graph would then depend upon how I wanted to graph it. On a velocity – time graph it would look like this: v (m/s)
t (s)
This graph indicates an object is traveling at the same velocity regardless of time. However, what would this look like on a displacement – time graph? As velocity is defined as a change in displacement over time, a constant velocity would indicate that the change in displacement is the same over each time period. This means that for any constant velocity, the slope of the displacement – time graph is a constant slope as shown below. x (m)
Constant positive slope indicates a
constant positive velocity.
t (s)
This indicates an object with a constant positive velocity. An object with a constant negative velocity will have a negative slope. 54 x (m)
Constant negative slope indicates
a constant negative velocity.
t (s)
The steeper the slope, the higher the constant velocity as the change in displacement is greater for a steeper sloped line over a shorter time period. x (m)
High constant velocity
Low constant velocity
t (s)
3.4
Acceleration The third and probably most important part of motion is acceleration. Acceleration is defined as a change in velocity over time. Acceleration, symbolized as (a) is ONLY a vector, and is measured in the unit [m/s2]. If there were no accelerations, physics would be a very easy subject to learn. Of course, if there were no accelerations, then the universe would be a very boring place (and probably not exist). As acceleration is only a vector, it is the direction of acceleration that can be confusing. Acceleration’s direction has nothing to do with what direction an object is actually moving, but what direction the change in velocity is occurring. If velocity is increasing in a certain direction, then the acceleration must be acting in the same direction. If velocity is decreasing in a certain direction, then acceleration must be acting in the opposite direction. Let’s look at how confusing it can be. Let’s say we have a car sitting at rest on a starting line. We will first need to establish positive and negative directions. Every position to the right of the line and any motion towards the right will be considered positive. Every position to the left of the line and any motion towards the left will be considered negative. So you now get in the car, start it up, put it in drive, and hit the gas. You start moving to the right of the starting line. As you are to the right of the line, you have a positive displacement (+x); as you are physically 55 moving to the right, you have a positive velocity (+v); as you are increasing your velocity to the right using the gas pedal, your acceleration is also positive (+a). Start
+x
+v
+a
At this point, you take your foot off the gas pedal and apply the brakes until you come to a stop. Now when you hit the brakes on a car you to not immediately stop, but continue forward at a slower and slower speed until you stop. As you are still physically to the right of the line, you still have a positive displacement (+x) relative to the line; as you are still physically moving to the right, you still have a positive velocity (+v); now as you are decreasing your velocity to the right using the brakes, your acceleration is now negative (‐a). Start
+x
+v
+a
Brakes
+x
+v
-a
Stop
Now that seems easy enough so far. You are now stopped. Let’s put the car in reverse and hit the gas, heading back towards your starting line. As you are still physically to the right of the line, you still have a positive displacement (+x) relative to the line; now you are physically moving to the left, so you now have a negative velocity (‐v); now for the fun part, as you are increasing your velocity to the left using the gas, your acceleration is now negative (‐a). 56 Start
+x
+v
+a
Brakes
+x
+v
-a
+x
-v
-a
Stop
Start
I know, you thought it would be a positive acceleration simply because you were speeding up, but it is negative because you are increasing your velocity in the negative direction, thus your change in velocity is in the negative direction. At this point, you again take your foot off the gas pedal and apply the brakes until you come to a stop back at the starting line. As you are still physically to the right of the line (until you return to the starting line where x = 0), you still have a positive displacement (+x) relative to the line; as you are still physically moving to the left, you still have a negative velocity (‐v); now as you are decreasing your velocity to the left using the brakes, your acceleration is now positive (+a). Start
+x
+v
+a
Brakes
+x
-v
+a
Stop
+x
+v
-a
+x
-v
-a
Stop
Start
Brakes
I know, you may have thought it would be a negative acceleration simply because you were slowing down, but it is positive because you are decreasing your velocity in the negative direction, thus your change in velocity is in the positive direction. Confused, or is it starting to make sense? You are stopped again, but back at the starting line. With the car still in reverse you hit the gas and move to the left of the starting line. As you are now physically to the left of the line, you have a negative displacement (‐x) relative to the line; you are physically moving to the left, so 57 you have a negative velocity (‐v); now for the fun part, as you are increasing your velocity to the left using the gas, your acceleration is negative (‐a). Get it? Start
+x
+v
+a
-x
-v
-a
Brakes
+x
-v
+a
+x
+v
-a
+x
-v
-a
Stop
Start
Brakes
Start
Stop
Let’s see. At this point, you again take your foot off the gas pedal and apply the brakes until you come to a stop. As you are still physically to the left of the line, you still have a negative displacement (‐x) relative to the line; as you are still physically moving to the left, you still have a negative velocity (‐v); now as you are decreasing your velocity to the left using the brakes, your acceleration is now positive (+a). Start
+x
+v
+a
Stop
-x
-v
+a
-x
-v
-a
Brakes
Brakes
+x
-v
+a
+x
+v
-a
+x
-v
-a
Stop
Start
Brakes
Start
Stop
You are stopped again. So you now put the car in drive, and hit the gas. You start moving to the right toward the starting line. As you are still physically to the left of the line, you have a negative displacement (‐x); as you are now physically moving to the right, you have a positive velocity (+v); as you are increasing your velocity to the right using the gas pedal, your acceleration is also positive (+a). 58 Start
Start
Stop
-x
+v
+a
+x
+v
+a
-x
-v
+a
-x
-v
-a
Brakes
Brakes
+x
-v
+a
+x
+v
-a
+x
-v
-a
Stop
Start
Brakes
Start
Stop
Hopefully it is starting to make sense. Last one. At this point, you take your foot off the gas pedal and apply the brakes until you come to a stop back at the starting line. As you are still physically to the left of the line (until you return to the starting line where x = 0), you still have a negative displacement (‐x) relative to the line; as you are still physically moving to the right, you still have a positive velocity (+v); now as you are decreasing your velocity to the right using the brakes, your acceleration is now negative (‐a). Stop
Start
Stop
-x
+v
+a
Brakes
-x
-v
+a
Start
-x
+v
-a
-x
-v
-a
Brakes
+x
+v
+a
Brakes +x
+v
-a
+x
-v
+a
+x
-v
-a
Stop
Start
Brakes
Start
Stop
If you look carefully you should notice a few trends. Displacement vector directions were easy, if you were physically on the right side of the line you had a positive displacement (+x), if you were physically on the left side of the line you had a negative displacement (‐x). Velocity vector directions are also fairly easy, as long as you were physically moving to the right you had a positive velocity (+v) no matter what your position was, as long as you were physically moving to the left you had a negative velocity (‐v) again no matter your position. (Notice that all the velocities on top were positive and all the velocities on the bottom were negative). It is the 59 accelerations that tend to be complicated. However if you look carefully, whenever you were speeding up, your acceleration was in the same direction as your velocity. Whenever you are slowing down, your acceleration was in the opposite direction as your velocity. If we were to graph this car ride on a displacement (x) vs. time (t) graph, we would have something that looks like this: x (m)
3. Stop
x2
4. Gas
x1
2. Brakes
5. Brakes
1.Gas
12. Stop
6. Stop
11. Brakes
7. Gas
- x1
8. Brakes
- x2
t (s)
10. Gas
9. Stop
1. Foot on the gas moving forward (+v) faster, increasing positive slope (+a), displacement is above the axis (+x). 2. Braking, slowing forward motion (+v), decreasing positive slope (‐a) until stopped, displacement is above the axis (+x). 3. Change directions, notice the displacement graph peaks at this point. 4. Foot on the gas moving backward (‐v) faster, increasing negative slope (‐a), displacement is above the axis (+x). 5. Braking, slowing backward motion (‐v), decreasing negative slope (+a), displacement is above the axis (+x) until stopped. 6. Back at original starting point (x = 0). 7. Foot on the gas moving backward (‐v) faster, increasing negative slope (‐a), displacement is below the axis (‐x). 8. Braking, slowing backward motion (‐v), decreasing negative slope (+a), displacement is below the axis (‐x) until stopped. 9. Change directions, notice the displacement graph peaks at this point. 10. Foot on the gas moving forward (+v) faster, increasing positive slope (+a), displacement is below the axis (‐x). 11. Braking, slowing forward motion (+v), decreasing positive slope (‐a) until stopped, displacement is below the axis (‐x). 12. Back at original starting point (x = 0). 60 If we were to graph this car ride on a velocity (v) vs. time (t) graph, we would have something that looks like this: v (m/s)
v
1.Gas
2. Brakes
10. Gas
6. Stop
3. Stop
-v
4. Gas
11. Brakes
9. Stop
7. Gas
5. Brakes
8. Brakes
12. Stop
t (s)
1. Foot on the gas moving forward faster, constant positive slope (+a), velocity is above the axis (+v). 2. Braking, slowing forward motion, constant negative slope (‐a) until stopped, velocity is above the axis (+v). 3. Change directions; notice the velocity is zero at this point. I must stop before I can change directions. 4. Foot on the gas moving backward faster, constant negative slope (‐a), velocity is below the axis (‐v). 5. Braking, slowing backward motion, constant positive slope (+a), velocity is below the axis (‐v) until stopped. 6. Back at original starting point, stopped (v = 0). 7. Foot on the gas moving backward faster, constant negative slope (‐a), velocity is below the axis (‐v). 8. Braking, slowing backward motion, constant positive slope (+a), velocity is below the axis (‐v) until stopped. 9. Change directions; notice the velocity is again zero at this point. I must stop before I can change directions. 10. Foot on the gas moving forward faster, constant positive slope (+a), velocity is above the axis (+v). 11. Braking, slowing forward motion, constant negative slope (‐a) until stopped, velocity is above the axis (+v). 12. Back at original starting point, stopped (v = 0). 61 If we were to graph this car ride on an acceleration (a) vs. time (t) graph, we would have something that looks like this: a (m/s2)
a
1.Gas
5. Brakes
3. Stop
8. Brakes 10. Gas
6. Stop
2. Brakes
-a
4. Gas
7. Gas
1. Foot on the gas moving forward (+v) faster (+a). 2. Braking, slowing forward (+v) motion (‐a). 3. Stopped. 4. Foot on the gas moving backward (‐v) faster (‐a). 5. Braking, slowing backward (‐v) motion (+a). 6. Stopped. 7. Foot on the gas moving backward (‐v) faster (‐a). 8. Braking, slowing backward (‐v) motion (+a). 9. Stopped. 10. Foot on the gas moving forward (+v) faster (+a). 11. Braking, slowing forward (+v) motion (‐a). 12. Stopped. 9. Stop
12. Stop
11. Brakes
t (s)
Note that when we are speeding up, no matter which way, velocity (v) and acceleration (a) have the same sign and are in the same direction. When we are slowing down, again regardless of which way, velocity (v) and acceleration (a) have the opposite sign and are in the opposite direction. One final, ultra important concept, constant velocity as related to acceleration. Acceleration is defined as a change in velocity over time. ∆
As we have already seen, an object with a constant velocity means that the velocity is not changing, therefore Δv = 0. So for a constant linear velocity, the acceleration must always be zero. In the previous car ride, at no time was my car traveling at a constant velocity, thus 62 acceleration was never zero. It is possible to have an acceleration and for a moment have a velocity of zero, we will see that later. 3.5
Kinematic Equations (aka Linear Motion Formulas) Many times in physics we will encounter physical situations in which we know certain values such as a displacement, a velocity, a time, etc. and we wish to determine some unknown value or values. This is where equations or physics formulas become very helpful. To begin with, we will look at the three formulas associated with linear motion. These are very useful formulas as they not only apply to linear motion, but will be seen throughout the year. These formulas are also called the three kinematic equations in case that comes up. So here they are: vf – Final velocity [m/s] x – Displacement [m] vf = vo + at x = vot + ½at2 vo – Initial velocity [m/s] t – Time [s] 2
2
2
vf = vo + 2ax a – Acceleration [m/s ] Each formula has its particular purpose and use. In addition, each is a result of some concept or law of physics. For example, the first formula, vf = vo + at, is based on the definition of acceleration, the change in velocity over time. Here is how it shakes out: ∆
If we expand the Δv, the change in velocity into vf ‐ vo Multiply both sides by time (t) at = vf ‐ vo Add vo to the other side and we get: vf = vo + at So this first formula is really the definition for acceleration rearranged to look a little easier to use. The second formula, x = vot + ½at2, comes from a relationship created by graphing a velocity – time graph that we will see later. The last formula, vf2 = vo2 + 2ax, is derived from a very important law of physics, but I do not want to give that away right now, so we will see it later. However, I like to call this third equation “time unknown” formula as it does not have time (t) in it. There is another equation that also can be employed when accelerations are constant: This can be a useful equation, but I like to avoid it as it relies on average velocities and does not employ acceleration in the equation. 63 3.6
x = 1000 m a = ‐5.0 m/s x = ? A few important concepts to remember 1. Distance is a scalar value indicating how far; displacement is vector indicating how far relative to a specific point. 2. Speed is a scalar value indicating how fast; velocity is a vector value indicating how fast and in which direction. 3. Instantaneous velocity is the velocity of an object at a specific moment in time. 4. The graph of an object on a displacement – time graph with a constant slope indicates that it is traveling at a constant velocity. The steeper the slope, the higher the velocity. 5. Acceleration is defined as the change in velocity of an object over time. Any change is the difference between the final value and the initial value. 6. Any object with a constant linear velocity has an acceleration of zero. 7. When speeding up, no matter which way, velocity (v) and acceleration (a) have the same sign and are in the same direction. 67 8. When slowing down, again regardless of which way, velocity (v) and acceleration (a) have the opposite sign and are in the opposite direction. Questions and Problems 1. What is the term for the change in velocity over time? 2. What is the term for the change in displacement over time? 3. A car drives 500 m, turns around and drives 500 m back to its starting point. (a) What total distance has it traveled? (b) What is the car’s total displacement? 4. If a car is traveling at a constant velocity, what is its acceleration? 5. Explain how a car traveling with a positive velocity could have a negative acceleration. 6. An athlete runs 50 m in 10 s and another 150 m in 40 s. Determine her average velocity. 7. A car travels at a constant velocity of 40 m/s for 5.0 s, determine its displacement during this 5.0 s. 8. A car starts from rest and accelerates at 4.0 m/s2 for 8.0 s. (a) Calculate its velocity at that 8.0 s. (b) Determine its total displacement after 8.0 s. 9. Another car is traveling at 30 m/s when it hits the brakes causing it to stop after 150 m. (a) Calculate its acceleration. (b) Determine how much time it takes to stop. 10. A truck traveling at 15 m/s accelerates at 0.5 m/s2 for 30 s. (a) Determine the displacement of the truck over this 30 s. (b) Determine velocity of the truck at the end of 30 s. 11. A ball rolling at 8.0 m/s rolls up a hill giving it an acceleration of ‐2.2 m/s2. (a) Determine the time it takes the ball to stop. (b) Determine the displacement of the ball up the hill. 12. A bowling ball is placed on a ramp and released. When it reaches the bottom of the ramp it has a measured velocity of 2.0 m/s as it begins to roll across the flat floor. At the end of 6.0 s on the floor it has a measured velocity of 0.4 m/s. (a) Determine the acceleration of the bowling ball along the floor. (b) Determine the displacement of the bowling ball across the floor. 68 13. A runner starts from rest and accelerates at 2.2 m/s2 for 4.0 s. She then runs at a constant velocity for the next 6.0 s. (a) Determine the velocity of the runner at the end of the first 4.0 seconds. (b) Determine the displacement of the runner in the first 4.0 seconds. (c) Determine the displacement of the runner in the next 6.0 seconds. (d) Determine the total displacement of the runner at the end of 10 seconds. 14. A racecar driver is evaluating three cars and wants the one with the greatest acceleration. Car A can accelerate from rest to 33 m/s in 4.2 s; car B can accelerate from rest to 27 m/s in 3.8 s; car C can accelerate from rest to 30 m/s in 4.0 s. Which car should he choose and why? 15. A dragster accelerates from rest to 60 m/s in 4.0 s. A chute then opens giving the dragster an acceleration of ‐2.0 m/s2 until it came to a stop. (a) Determine the total distance the dragster traveled from start to finish. (b) Determine the total time the dragster was in motion. 16. A rock is dropped from rest off an 80 m high cliff. If it hits the ground with a velocity of 40 m/s in the downward direction, (a) determine the acceleration of the rock during the fall. (b) Calculate the time it took the rock to fall the 80 m to the ground. 17. An airplane that takes off and flies at a constant velocity of 250 m/s will arrive at it’s destination in 1 hour. The next day, it takes off initially at 250 m/s and hits a ‐0.12 m/s2 headwind the entire trip. (a) Determine the final velocity of the airplane when it reaches it’s destination. (b) Determine the time it takes the plane to reach it’s destination. (c) By how many minutes is the plane late? 69 Chapter 4 – Acceleration – Velocity – Displacement Graphs 4.1 Introduction If we are given a graph of an object’s accelerations, we can determine and graph its velocities and displacements from that acceleration‐time graph without the use of equations. For example the acceleration‐time graph shown below reveals a lot about the overall motion of an object. Acceleration (a) vs Time (t)
a (m/s2)
3
2
1
0
0
10
20
30
40
50
‐1
t (s)
‐2
‐3
vo = 0 m/s, xo = 0 m
The trick is that we must look at each time interval (10 second space) separately. This is because the acceleration changes at each time interval. Whenever acceleration changes, you have a new situation that must be looked at differently. Equally, if we are given a displacement‐time graph we can determine and graph an object’s velocity, or if we are given a velocity‐time graph, we can determine and graph an object’s acceleration. The velocity‐time graph is the most versatile of the three graphs. No matter what the case, we must always be aware of initial conditions, in other words the initial velocity and displacement of the object prior to any changes in motion. In the example above you will note that the initial velocity (vo) of the object is 10 m/s and the initial position (xo) is 0 m. 70 4.2
Determining Velocities from a Displacement – Time Graph To begin with, let’s review how we can determine the velocity of an object from a displacement‐time graph. What do you mean you have forgotten already? Well, let’s look at the following graph of the displacement of an object over time that starts from rest. Displacement (x) vs. Time (t)
x (m)
40
30
20
10
0
0
10
20
30
‐10
40
50
60
t (s)
‐20
vo = 0 m/s
As this is a displacement vs. time graph and we want velocities, what should we do? Well, the unit of displacement is meters (m) and time is seconds (s) and we want velocity which is in (m/s). As m/s is m divided by s, we must use slopes to determine velocities. Our best approach is to look at the slope of the line in each 10 s interval. In the first 10 s interval, the displacement is from 0 m to 10 m. The slope would be: ∆
∆
10 0
10 0
10
10
m = 1 m/s Ah, now it is all coming back. That was easy. The next 10 s interval (from 10 s to 20 s) has a flat line, which we all remember gives a slope of 0. So this is 0 m/s (aka stopped). For the 20 s to 30 s interval, the slope would be: ∆
∆
71 30
30
10
20
20
10
m = 2 m/s This makes sense as it has a much steeper slope than before. Next, for the 30 s to 40 s interval, it is hard to tell the change in displacement on the y‐axis. However, we can see that the line continues along the same slope until 50 s, so let’s go all the way there from 30 s to 50 s: 0 30
50 30
30
20
m = ‐1.5 m/s This indicates our object is now traveling back towards it’s starting point. Finally, from 50 s to 60 s we have: 10 0
60 50
10
10
m = ‐1.0 m/s Displacement (x) vs. Time (t)
x (m)
40
30
2.0 m/s
20
‐ 1.5 m/s
0 m/s
10
1.0 m/s
0
0
10
20
30
‐10
40
50
60
‐ 1.0 m/s t (s)
‐20
vo = 0 m/s
72 Let’s make a velocity – time graph from our information. As each time interval had a constant slope each one of these was a constant velocity. On a velocity – time graph, a constant velocity must just be a horizontal line. Velocity (v) vs Time (t)
v (m/s)
2.5
2
1.5
1
0.5
0
‐0.5
0
10
20
30
40
50
60
t (s)
‐1
‐1.5
‐2
vo = 0 m/s
At a glance, this graph may not appear to say much, but let’s take a look anyway. For the first 10 s, this object traveled forward at 1.0 m/s. It then stopped for the next 10 s. From 20 s to 30 s it traveled forward at 2.0 m/s. At 30 s it must have switched directions and started going backwards at ‐1.5 m/s (backwards because of the negative velocity) until 50 s where it decreased it’s backward velocity to ‐1.0 m/s. Note how this matches what appears in the displacement – time graph. I have often found that students have trouble with the idea that going from ‐1.5 m/s to ‐1.0 m/s is a decrease in velocity. This is because your math teacher has beaten into your heads that ‐1.0 is bigger than ‐1.5. However, I must remind you again that this is Physics, not math. In Physics, the negative sign is only a symbol for direction. 1.5 is always going to be bigger than 1.0 no matter which way something is traveling, so ‐1.5 m/s is bigger than ‐1.0 m/s in this graph. It is also important to note that we can get accelerations from a velocity – time graph in the same way, by using slopes. 73 4.3
Creating a Velocity–Time Graph from an Acceleration–Time Graph Let’s look now at determining velocities from an acceleration – time graph given the graph below of an object that starts with an initial velocity of 0 m/s and at an initial position of 0 m. Acceleration (a) vs Time (t)
a (m/s2)
3
2
1
0
0
10
20
30
40
‐1
50
t (s)
‐2
‐3
vo = 0 m/s, xo = 0 m
Acceleration is defined as the change in velocity over time. The graph above indicates not only specific accelerations, but how much the velocity is changing during each time interval. To find the changes in velocity, we take the area under each acceleration during each time interval. (Note: The area includes the shaded parts enclosed by the acceleration graph and the x‐axis). Acceleration (a) vs Time (t)
a (m/s2)
3
2
1
+ 20 m/s
+ 10 m/s
0
0
0
10
20
‐1
30 ‐ 10 m/s
40
50
t (s)
‐2
‐3
vo = 0 m/s, xo = 0 m
74 Notice that the area between 10 s and 20 s has an area of 0. This indicates that at this point, the object has a constant velocity. ANY TIME AN OBJECT HAS A LINEAR ACCELERATION OF ZERO, IT HAS A CONSTANT LINEAR VELOCITY. Each one of this areas shown above does not indicate actual velocity, only how much the velocity has changed during each time interval. Given the initial condition that the initial velocity (vo) is zero and the areas from the acceleration‐time graph, we can create a velocity‐
time graph. Velocity (v) vs. Time (t)
v (m/s)
30
20
10
0
0
10
20
30
40
t (s)
‐10
vo = 0 m/s, xo = 0 m
This velocity‐time graph shows that our object started from rest, accelerated, then remained at a constant velocity, then accelerated at a higher rate, and finally slowed down. Our object never stopped because the velocity never reached zero again, so at the end of 40.0 s, our object was traveling at 20 m/s. 4.4
Creating a Displacement–Time Graph from a Velocity–Time Graph This graph can now help us determine displacements during this motion. Velocity is defined as the change in displacement over time. By looking at each time interval again, if we measure the area of each interval, it will give changes in displacement. 75 Velocity (v) vs. Time (t)
v (m/s)
30
+ 250 m
+ 200 m
+ 50 m
20
+ 100 m
10
+ 200 m
+ 100 m
+ 50 m
+ 100 m
0
0
10
20
30
40
t (s)
‐10
vo = 0 m/s, xo = 0 m
In each case, the total area for each time interval does not show actual displacement, but how much the displacement is changing. This information, along with an initial condition of xo = 0 m, allows us to create a displacement‐time graph. (Note: This should be a smooth curve). Displacement (x) vs. Time (t)
600
x (m) 550
500
450
400
350
300
250
200
150
100
50
0
‐50 0
5
10
15
20
25
vo = 0 m/s, xo = 0 m
30
35
40
45
t (s)
76 This graph indicates that the displacement was always increasing, even when our car was slowing down because it continued to move forward. In order for this graph to move downward, the velocity would have to go below the zero mark (become negative). This graph indicates that at the end of 40.0 s, our object traveled 600 m. 4.5
Getting Everything from the Versatile Velocity – Time Graph As I mentioned earlier, the velocity – time graph reveals the most information about the motion of an object. Not only can you determine the velocity of the object at any time (after all, it is a velocity – time graph), but you can determine displacements by using areas or accelerations by using slopes. In addition, the directions and changes in directions of motion can also be established from this graph. So let’s give it a try. Take a look at the following velocity – time graph. Velocity (v) vs. Time (t)
v (m/s)
30
20
10
0
0
5
10
15
20
25
30
‐10
t (s)
‐20
‐30
‐40
xo = 0 m
Wow, look at all the information I am given from this graph. Do you see it? Really, after all this time? OK, let’s look at it bit by bit. 0 – 5 s 1. Started from rest an increased velocity to 20 m/s in the positive direction. 2. Taking the slope of the line ( 4 indicates an acceleration of 4.0 m/s2. 3. Taking the area of the triangle created in this interval, (½ (5) (20) = 50) indicates a displacement of 50 m. 77 5 – 10 s 1. Maintained a constant velocity of 20 m/s. 2. Flat line gives a slope of zero, an acceleration of 0 m/s2, thus a constant velocity. 3. Taking the area of the rectangle formed in this interval ((5) (20) = 100) indicates a displacement of 100 m during this interval; so the total displacement at the end of 10 s is 150 m. 10 – 15 s 1. Velocity decreased from 20 m/s to 0 m/s but is still positive, so the object is still moving forward, just slowing down until it stops at 15 s. 2. Taking the slope of the line ( 4 indicates an acceleration of ‐4.0 m/s2; thus slowing down while moving in the positive direction. 3. Taking the area of the triangle created in this interval, (½ (5) (20) = 50) indicates a displacement of 50 m, and now a total displacement of 200 m. 15 – 20 s 1. Velocity increased from 0 m/s to ‐20 m/s but is now negative, so the object is now moving backwards, but is speeding up in the negative direction. (Remember, 20 is bigger than 0, the negative is for direction). As the velocity has gone from positive to negative, the object has now switched direction of travel. 2. Taking the slope of the line ( 4 indicates an acceleration of ‐4.0 m/s2 this time speeding up in the negative direction. 3. Taking the area of the triangle created in this interval, (½ (5)(‐20) = ‐50) indicates a displacement of ‐50 m, and now a total displacement of 150 m, so it is still in positive location as far as displacement is concerned. 20 – 25 s 1. Velocity increased from ‐20 m/s to ‐30 m/s and is still negative, so the object is still moving backwards and speeding up. (Remember, 30 is bigger than 20, the negative is for direction. These vectors sure are annoying). – 2. Taking the slope of the line ( 2 gives an acceleration of ‐2.0 m/s2; thus still speeding up in the negative direction. 3. Taking the area of the triangle created in this interval, (½ (5)(‐10) = ‐25) plus the rectangle in this interval ((5)(‐20) = ‐100) indicates a displacement of ‐125 m, and now a total displacement of 25 m, so it is still in positive location as far as displacement is concerned. 78 25 – 30 s 1. Velocity decreased from ‐30 m/s to 0 m/s and is still negative, so the object is still moving backwards, but is slowing down in the negative direction until it stopped. (Remember, 0 is less than 30, the negative is for direction). – 6 indicates an acceleration of +6.0 m/s2 2. Taking the slope of the line ( this time slowing down in the negative direction. 3. Taking the area of the triangle created in this interval, (½ (5)(‐30) = ‐75) indicates a displacement of ‐75 m, and now a total displacement of ‐50 m, so it is now in a negative overall location as far as displacement is concerned. 4. This means the object passed its original starting point and is now stopped. Velocity (v) vs. Time (t)
v (m/s)
30
0 m/s2
20
4.0 m/s2
10
‐ 4.0 m/s2
+ 100 m
+ 50 m
+ 50 m
0
0
5
10
15
‐ 50 m
‐10
20
25
‐ 100 m
‐ 4.0 m/s2
‐20
‐ 25 m
‐
‐30
30
‐ 75 m
t (s)
6.0 m/s2
2.0 m/s2
‐40
xo = 0 m
Whew! So to sum up: 1. Object started from rest and accelerated in the positive direction. 2. It then maintained a constant positive velocity. 3. It then slowed down while still moving in the positive direction until it stopped. 4. Next it accelerated in the negative direction (noticed it had to stop before it could change directions) going faster and then even faster. 5. Finally it slowed its negative motion using a positive acceleration until it again stopped. 6. The area under the graph above the zero line is less than the area below thus yielding an overall negative displacement of ‐50 m. 7. That was fun! A few important concepts to remember 1. The slope of a displacement (x) vs. time (t) graph gives exact velocities. 79 2. The area under a velocity (v) vs. time (t) graph indicates changes in displacement. 3. The slope of a velocity (v) vs. time (t) graph gives exact accelerations. 4. The area under an acceleration (a) vs. time (t) graph indicates changes in velocity. Questions and Problems 1. Given the following displacement vs. time graph of an object’s motion,
Displacement (x) vs. Time (t)
x (m)
250
200
150
100
50
0
‐50
0
10
20
30
40
50
t (s)
‐100
‐150
(a) During what time interval(s) is displacement positive?
(b) During what time interval(s) is displacement negative?
(c) During what time interval(s) is velocity positive?
(d) During what time interval(s) is velocity negative?
(e) Determine the velocity between 0 to 10s.
(f) Determine the velocity between 10 to 30s.
(g) Determine the velocity between 30 to 40s.
(h) Determine the velocity between 40 to 50s.
(i) After time t = 0, when did this object return to its original starting position?
80 2. Given the following velocity vs. time graph of an object’s motion,
Velocity (v) vs. Time (t)
v (m/s)
25
20
15
10
5
0
‐5
0
10
20
30
40
50
60
t (s)
‐10
‐15
(a) During what time interval(s) is velocity positive?
(b) During what time interval(s) is velocity negative?
(c) During what time interval(s) is acceleration positive?
(d) During what time interval(s) is acceleration negative?
(e) Determine the acceleration between 0 to 20s.
(f) Determine the acceleration between 20 to 30s.
(g) Determine the acceleration between 40 to 50s.
(h) Determine the acceleration between 50 to 60s.
(i) After time t = 0, at what time(s) did this object stop?
81 Questions 3 – 5 – For each following acceleration (a) vs. time (t) graph, (a) create the velocity (v) vs.
time (t) graph, and (b) create the displacement (x) vs. time (t) graph. Use a computer or graph paper and
note the initial conditions for each graph!!!
3. vo = 0 m/s
xo = 0 m
Acceleration (a) vs Time (t)
a (m/s2)
3
2
1
0
0
10
20
30
40
50
‐1
t (s)
‐2
‐3
4. vo = 0 m/s xo = 0 m
Acceleration (a) vs Time (t)
a (m/s2)
2
1
0
0
‐1
10
20
30
40
50
60
t (s)
‐2
82 5. vo = 10 m/s xo = 0 m
Acceleration (a) vs Time (t)
a (m/s2)
3
2
1
0
0
10
20
30
40
50
60
t (s)
‐1
‐2
6. Given the following table of velocity vs. time:
Velocity vs. Time
t (s)
v (m/s)
0.0
0.0
1.0
5.0
2.0
10.0
3.0
15.0
4.0
20.0
5.0
25.0
6.0
25.0
7.0
25.0
8.0
25.0
(a) Create a velocity – time graph from this data on a computer or graph paper.
(b) Determine the acceleration of the object from 0 – 5.0 s.
(c) Determine the acceleration of the object from 5.0 – 8.0 s.
(d) Determine the total displacement of the object from 0 – 8.0 s.
83 7. Given the following table of velocity vs. time:
Velocity vs. Time
t (s)
v (m/s)
0.0
2.0
1.0
4.0
2.0
6.0
3.0
7.0
4.0
8.0
5.0
9.0
6.0
9.0
7.0
8.0
8.0
4.0
9.0
2.0
10.0
0.0
11.0
-4.0
12.0
-8.0
(a) Create a velocity – time graph from this data on a computer or graph paper.
(b) During what time interval(s) was the object speeding up?
(c) During what time interval(s) was the object slowing down?
(d) During what time interval(s) did the object have an acceleration of 0 m/s2?
(e) At what time(s) did the object change direction?
(f) Determine the acceleration of the object from 0 – 2.0 s.
(g) Determine the acceleration of the object from 10.0 – 12.0 s.
(h) Determine the displacement of the object from 0 – 2.0 s.
(i) Determine the displacement of the object from 10.0 – 12.0 s.
84 8. For the following acceleration (a) vs. time (t) graph, (a) create the velocity (v) vs. time (t)
graph, and (b) create the displacement (x) vs. time (t) graph.
vo = 0 m/s xo = 0 m
Acceleration (a) vs Time (t)
a (m/s2)
3
2
1
0
0
10
20
30
40
50
60
‐1
t (s)
‐2
‐3
85 Chapter 5 – Vertical Linear Motion 5.1 Introduction to Free­Fall We have discussed motion in the horizontal direction. This section looks at motion in the vertical direction, or straight up and down. In particular, we will consider the idea of free‐fall in which the only influence on the object in motion will be gravity. We will consider throwing a ball straight up into the air and allowing it to return to your hand. (We will use the absence of any air resistance to make life easier). 5.2
Vertical Linear Motion Maximum
height
+x
+v
-a
-x
-v
-a
vo
The ball starts with some initial velocity as it leaves your hand. (It must have some velocity (vo) or it would never leave your hand). As the ball rises, its velocity acts in the positive direction (+v) and its displacement becomes positive (+x) from the point where it began. The velocity of the ball begins to slow immediately. The only way for a positive velocity to slow down is due to a negative acceleration (‐a). The acceleration is called gravity (g). (In reality, gravity on Earth has a actual value of ‐9.8 m/s2, but for our purposes and to make life a little easier, we are going to use a value of g = ‐10.0 m/s2. Thus all answers at the end of the text that involve the acceleration of gravity on Earth are based on the value of g = ‐10.0 m/s2). 86 Maximum
height
v = 0 m/s
+x
+v
-a
vo
When the ball reaches the maximum height, its velocity becomes zero, but it is still being accelerated by gravity. (For an object moving through the air, that acceleration is NEVER zero, it is always a = g = ‐10 m/s2). This point must be the maximum height because if the ball is still moving up or down, then it cannot physically be at the maximum height. At the top the ball stops for a very brief moment and then starts back down. As we have seen in linear motion, if an object changes direction, its velocity must be zero at that point. Maximum
height
vo = 0
-x
-v
-a
vf
Starting from the top, the ball has an initial velocity of zero. As it moves downward, it is moving in the negative direction. From the top, it has a negative displacement (‐x) and its velocity increases in the negative direction (‐v). The only way for a negative velocity to increase in the negative direction is to have a negative acceleration (‐a). Again, this is due to gravity (a = g = ‐10 m/s2). GRAVITY ONLY ACTS ONE WAY AND THAT IS DOWNWARD. As the ball falls, it will have some final velocity just before it is caught. 87 It turns out that if the ball travels the same distance up as it does down, then the velocity at which it left your hand will be the same in which it returns, just in the opposite direction. Maximum
height
v = 0 m/s
-x
-v
-a
+x
+v
-a
vo = 20 m/s
vf = -20 m/s
If I were to create acceleration, velocity, and displacement time graphs for this vertical motion, they would look like this: Acceleration (a) vs time (t)
a (m/s2)
1
2
- 10
3
4
t (s)
x (m)
v (m/s)
Velocity (v) vs time (t)
Displacement (x) vs time (t)
20
20
10
- 10
- 20
15
1
2
3
4
10
t (s)
5
1
2
3
4 t (s)
88 The acceleration‐time graph indicates that the ball maintains a constant acceleration at all times of ‐10 m/s2, as it should since it is gravity. The velocity‐time graph shows the velocity begins at 20 m/s, slows to zero (at maximum height), and then gains speed in the opposite direction until it reaches ‐20 m/s at the end. The displacement‐time graph shows that the ball started upward, reached a maximum height of 20 m at 2.0 s, and then came right back down to its starting point (for a total displacement of zero). Any object thrown straight up into the air will look the same on these graphs. The only thing that would change is the numbers. If you throw it harder, the velocity graph will have higher beginning and ending numbers and the displacement graph would go to a larger maximum height. The acceleration graph always looks the same. 5.3
Using Motion Equations in Vertical Motion The linear motion equations we discussed also show this to be true. If I throw something up at 20 m/s, then: vo = 20 m/s vf = vo + at a = g = ‐10 m/s2 0 = 20 + (‐10)t 10t = 20 vf = 0 m/s (at max height) t = ? t = 2.0 s It takes 2.0 seconds to get to the maximum height. This is what the graphs show. Looking at displacement, we have: x = vot + ½ at2 x = (20)(2) + ½ (‐10)(2)2 x = 40 – 20 x = 20 m The ball rises to a maximum height of 20 m after 2.0 s. This also matches the graph. If we look at the situation coming down, we start at the top with a velocity of zero, so: vo = 0 m/s vf = v0 + at 2
a = g = ‐10 m/s ‐20 = 0 + (‐10)t vf = ‐20 m/s t = 2.0 s t = ? So it takes 2.0 s for the ball to fall back from its maximum height, again, like the graphs shows. So whatever happens going up, the same happens coming down. THIS IS DUE TO THE FACT THAT THE ONLY THING ACTING ON AN OBJECT IN THE AIR IS GRAVITY. 89 Example: Let’s look at another situation. We have a skydiver in an airplane that is at a height of 2000 m above the ground. (For the sake of the problem, we will freeze the airplane in mid‐air). The skydiver steps out of the airplane and falls toward ground. vo = 0 m/s
2000 m
Since the skydiver just steps out of the plane, we can say his initial velocity is zero. As the skydiver travels downward, we can say his displacement will be ‐2000 m. Once in the air, the only acceleration he feels is a = g = ‐10 m/s2. Let’s now say that as he falls, his parachute does not open and neither does the back‐up chute. We want to know how fast he will be traveling just before he hits the ground. So let’s look at what we are given: Since time is not known, only one equation works here. vo = 0 m/s a = g = ‐10 m/s2 x = ‐2000 m vf2 = vo2 + 2ax vf = ? vf2 = (0)2 + 2(‐10)(‐2000) vf2 = 40,000 vf = 200 m/s (which is about 450 mph) 5.4
Terminal Velocity Many times skydivers have survived higher falls than 2000 m with chutes not opening. This is because in the problem before, we pretend that air resistance does not exist. In reality, air resistance will slow a person’s fall to about 53.3 m/s (120 mph). 90 If you look at a real velocity‐time graph of the skydiver, it would look like this: Velocity (v) vs time (t)
v (m/s)
vt
t (s)
The skydiver’s velocity would increase at first and then level off at some constant speed known as terminal velocity. Terminal velocity (vt) is the maximum speed an object in free‐fall will reach due to air resistance. What this means is that you continue to fall, but you are no longer accelerating (remember, constant velocity means zero acceleration). All objects, if they fall long enough reach some sort of terminal velocity based on size, shape, and weight of the object. FOR MOST OF WHAT WE DO THIS YEAR, HOWEVER, WE WILL ELIMINATE AIR RESISTANCE IN OUR WORK!!! Example: A student standing on top of a 50 m high building throws a ball downward toward the ground with an initial velocity of 20 m/s. We want to determine how much time it takes the ball to hit the ground. 20 m/s
50 m
t =?
If we look at the givens for this problem we have to be very careful of the vector nature of all of our numbers. (Negative as the initial velocity was downward). vo = ‐20 m/s x = ‐50 m (Negative as the ball ends 50 m south or below where it started). 2
a = g = ‐10 m/s (Of course, the only acceleration while in the air, regardless of the initial speed of the throw). t = ? 91 Now looking at our formulas, the one that seems appropriate is: x = vot + ½at2 So, ‐50 = (‐20)t + ½(‐10)t2 Hey, that is a quadratic equation. I hate those equations! So let’s try another method. I am going to solve for the final velocity (vf) just before it hits the ground. Remember that this is never zero in vertical motion as it is said to occur just before hitting the ground. vo = ‐20 m/s x = ‐50 m a = g = ‐10 m/s2 vf = ? The equation here is: vf2 = vo2 + 2ax vf2 = (‐20)2 + 2(‐10)(‐50) vf2 = 400 + 1000 vf2 = 1400 Now here is where you have to be careful and again consider the vector directions. The square root of any number can be either a positive or negative number. So, 37.4 / √1400
Since the ball is headed downward, its final velocity must be negative, so vf = ‐37.4 m/s This is very important, because now I can find time (t) using vf = vo + at ‐37.4 = ‐20 +(‐10)t ‐17.4 = ‐10t t = 1.74 s (Better than trying to use the quadratic equation in my book). 92 A few important concepts to remember 1. In free‐fall vertical motion, the only acceleration is due to gravity, therefore acceleration is never zero at any point while an object is in free‐fall in the air. 2. In vertical motion, the velocity of an object at its maximum height is always zero. 3. Terminal velocity is the maximum velocity an object in free‐fall will reach due to the effects of air resistance. Questions and Problems 1. A ball is thrown straight up into the air and returns to the thrower. (a) While in the air, at what point is its velocity zero? (b) While in the air, at what point is its acceleration zero? 2. If an arrow is shot straight up in the air at 40 m/s, what is its velocity when it returns to the shooter? 3. A ball is thrown straight up into the air at 30 m/s, (a) determine how much time it takes to reach its maximum height. (b) Calculate the maximum height it reaches. (c) What is the total time the ball is in the air, from the time it is thrown to the time it returns to the thrower? 4. Another ball is dropped from the top of a 100 m building. (a) Calculate the time it takes to fall to the ground. (b) Calculate the velocity of the ball just before it hits the ground. 5. What is the term for the maximum speed an object it free fall will reach due to air resistance? 6. The acceleration of gravity on the moon is 1/6 that of the Earth. If someone on the moon threw a ball up in the air with the same initial velocity as someone doing the same thing on the Earth, (a) which ball would reach a greater maximum height? (b) Which ball would reach its maximum height first (in the shortest time)? (c) Which ball would return to the thrower with the highest final velocity? 7. A person wants to know the depth of a well, so she drops a rock and hears the splash 4.0 s later. Determine the depth of the well. 8. A man standing on top of a 70 m cliff throws a rock downward initially at 20 m/s. (a) Determine the velocity of the rock just before it hits the ground. (b) Determine the time it takes the rock to reach the ground. 93 9. A woman standing on top of a 100 m cliff throws a rock upward at 15 m/s. The rock reaches its maximum height and then falls back past the woman to the bottom of the cliff. (a) Calculate the time it takes the rock to reach its maximum height. (b) Calculate the height to which the rock travels above the top of the cliff. (c) Calculate the time it takes the rock to fall from its maximum height to the bottom of the cliff. (d) Determine the total time the rock is in the air. 10. A ball is thrown up in the air caught 5.0 s later. (a) Determine the initial velocity at which the ball was thrown. (b) Determine the maximum height the ball reached. 11. An astronaut standing on the moon (gmoon = ‐1.62 m/s2) throws a wrench up in the air at 20 m/s. (a) Calculate the maximum height the wrench will reach. (b) Determine the total time it takes before the wrench comes back to her. 12. You are standing on a planet whose gravity is ‐5.0 m/s2. You throw a ball up into the air with an initial velocity of 20 m/s. (a) How fast is it traveling after 1.0 s? (b) 2.0 s? (c) 3.0 s? (d) 4.0 s? (e) Create an acceleration – time graph of the ball’s motion from the time you throw it until it would return to you. (f) Create a velocity – time graph for the same motion. (g) Create a displacement – time graph for the same motion. 94 Chapter 6 – Projectile Motion 6.1 Introduction to Projectile Motion In discussing linear motion, we talked about moving in the horizontal linear direction and in the vertical linear direction. We now look at combining the two motions such as playing a game of catch by throwing a ball back and forth. When doing this, the ball follows an arc shaped path. vy
vo
vx
In this situation, the ball is going up and to the right at the same time at the beginning of its motion. In terms of the vectors that we discussed, it has a velocity in the x‐direction (horizontal) and in the y‐direction (vertical). This causes it to move in the arc path. The important thing to remember is that the horizontal motion is independent of the vertical motion. Consider the motion of a cart with the ball loaded on a vertical spring. Both the cart and the ball are moving forward at the same speed. The ball is launched straight up into the air. Because the ball was already moving forward, it continues to move forward at the same speed as the cart. As it moves upward, the only thing affecting it is gravity. Gravity slows the ball’s upward movement until it reaches its maximum height and then falls back 95 down again. (Thus the horizontal motion and the vertical motion are independent of each other). This allows the ball to land back in the cart. This leads to two important concepts in projectile motion. The horizontal velocity of an object in projectile motion remains constant throughout the motion. This is due to the fact that once the object is in the air, it undergoes no acceleration in the horizontal direction (since nothing is pushing or pulling it in the horizontal direction). Remember, any zero acceleration means a constant velocity in that direction. The only acceleration it does experience is in the vertical direction, due to gravity. In projectile motion, just like in vertical linear motion, at the maximum height of its motion, its vertical velocity is zero. This is due to the affects of gravity, which acts only in the downward direction. In projectile motion, when an object reaches its maximum height, it stops moving up and starts moving down. In order to change directions, it has to stop its upward motion for an instant. Maximum height – ball changes
vertical direction (vy = 0 m/s)
Horizontal velocity of both cart and ball remain constant.
This allows them to stay together in the horizontal direction.
If two balls are placed at the same height and one is shot horizontally out of a cannon and the other is dropped at the same time, they both hit the ground at the same time (if we neglect air resistance). This is due to the fact that they are both affected by gravity in an equal way. So as long as they begin at the same height, they must land on the ground at the same time. 96 t1 = t2
t1
h
t2
This picture shows that the time for the dropped ball to hit the ground is equal to the time for the shot ball to hit the ground. This only works if the shot ball is launched horizontally and from the same height as the dropped ball. If the shot ball is launched at an angle, they will hit at different times. If it is shot upward, the dropped ball hits first. If it is shot downward, the shot ball hits first. 6.2
Shooting the Monkey In the famous shoot the monkey scenario, we see a monkey hanging from a tree a long distance away. We know that as soon as we fire, the monkey will let go of the tree to try to avoid being shot. The question was, where to aim to hit the monkey if it was going to let go of the branch. It turns out that we need to aim directly at the monkey (again, we are pretending there is no air resistance). The shot ball will start upward, but gravity will cause it to follow an arc path. The monkey falls straight downward. They both are only affected by one thing, gravity, which acts the same on both. If we aim directly at the monkey when we fire, the ball and the monkey will intercept at some point in their motion. 97 a=g
a=g
6.3
Horizontal Launch Initial velocity (vo) is in the horizontal direction. There is no acceleration in the horizontal x‐
direction. There is no initial velocity in the vertical y‐direction, however there is an acceleration, gravity (‐10 m/s2). X Y vo 0 vo a ‐10 0 Example: A ball is initially launched horizontally from a point 20 m above the ground at a speed of 50 m/s as shown below. 50 m/s
20 m
t=?
x=?
Determine the time (t) to hit the ground and the horizontal distance (x) the ball travels. 98 Let’s create a Y vs. X chart like we did with vectors and fill in the information that we know. Y
X
vo
a
t
x
vf
We can always fill in the acceleration values as they will never change in projectile motion. The x‐direction is ALWAYS zero and the y‐direction is ALWAYS ‐10 m/s2. Y
X
vo
a
-10 m/s2
0
t
x
vf
As this is a HORIZONTAL launch, meaning it is not launched vertically, the initial velocity in the y‐direction is zero while the initial velocity in the x‐direction is 50 m/s. Y
X
vo
0
50 m/s
a
-10 m/s2
0
t
x
vf
Time we do not yet know, but we do know the vertical displacement as the height of the launch point is 20 m. As it will fall the distance, the displacement in the y‐direction must be ‐20 m. (We do not yet know the displacement in the x‐direction). Y
X
vo
0
50 m/s
a
-10 m/s2
0
t
?
?
?
x
-20 m
vf
As for the final velocities, these occur just before hitting the ground. We do not yet know the final velocity in the y‐direction; however we do know the final velocity in the x‐direction. Since the acceleration in the x‐direction is ZERO, all x‐velocities must be constant. Therefore the final velocity in the x‐direction is the same as the initial velocity in the x‐direction, 50 m/s. 99 Y
0
-10 m/s2
?
-20 m
?
vo
a
t
x
vf
X
50 m/s
0
?
?
50 m/s
So we have filled in our chart with all that we know, let’s see what we can determine. As I have an actual acceleration in the y‐direction, I will focus on that column. This means that I ONLY use numbers in the y‐column and not in the x‐column. Calculate the time based on y‐direction: x = vot + ½at2 (voy = 0) 2
x = ½at Solve for time 2
t = 2.0 s (As time cannot be split, it is the same in both x and y directions). vo
a
t
x
vf
Y
0
-10 m/s2
2s
-20 m
?
X
50 m/s
0
2s
?
50 m/s
Knowing time in the y‐direction will allow us to calculate the final velocity in the y‐direction. Again, we can only use numbers from the y‐column. vf = vo + at vf = 0 + (‐10)(2) vf = ‐20 m/s Y
X
vo
0
50 m/s
a
-10 m/s2
0
t
2s
2s
x
-20 m
?
50 m/s
vf
-20 m/s
100 We can calculate displacement in the x‐direction. As it is the x‐direction, we can only use values from the x‐column and not any from the y‐column. (a = 0) x = vot + ½at2 x = voxt x = (50)(2) x = 100 m Y
X
vo
0
50 m/s
a
-10 m/s2
0
t
2s
2s
-20 m
x
100 m
vf
-20 m/s
50 m/s
We can answer the two initial questions from the beginning, the time (t) in the air and the horizontal distance (x) the ball travels. t = 2.0 s x = 100 m For fun let’s add the following question, at what final velocity does the ball land? The ball will not land straight down in the purely y‐direction, nor will it land flat in the purely x‐direction. If we look in the chart we see that the final velocity has a y‐component (‐20 m/s) and an x‐
component (50 m/s). If we remember our vectors (and I know we all do) when you have two components, to find the resultant, we use the Pythagorean Theorem and then we must also use the Inverse Tangent function to find the angle at which it lands. 50 √20
vf = 53.85 m/s 6.4
tan
tan
θ = 21.8o Angled Launch This is more complex because you have velocity and angle, but just use vector knowledge and motion knowledge. The key, again, is to separate everything into the x‐direction and the y‐
direction. vo
θ
101 In an angled launch, the projectile has an initial velocity (v0) and is launched at an angle (θ). Like any vector, we want to break up the initial velocity into its x‐component and its y‐component. So we still follow our vector rules. voy = vosinθ vox = vocosθ The remaining information remains the same. There is still no acceleration in the x‐direction as nothing affects the motion left or right. The acceleration in the y‐direction is still ‐10 m/s2. The vertical velocity still becomes zero at the maximum height while the horizontal velocity remains constant. Like vertical motion, whatever happens going up, must happen coming back down, so this helps us fill in our chart. Example: We fire a cannon ball at 40 m/s at an angle of 300. We want to determine the maximum height the cannon ball will reach, the total time it is in the air, and the horizontal distance it travels. Start with acceleration, as usual. Y
X
vo
a
-10 m/s2
0
t
x
vf
Break up the initial velocity into its components using the vector rules of Y goes with Sin and X goes with Cos and then (Number)(Function)(Angle): voy = vosinθ = 40sin30 = 20 m/s vox = vocosθ = 40cos30 = 34.6 m/s We can add these to the chart: Y
X
vo
20 m/s
34.6 m/s
a
-10 m/s2
0
t
x
vf
102 The next thing we can fill in is the displacement in the y‐direction. As the ball begins and ends at ground level, its vertical displacement is ZERO. Y
X
vo
20 m/s
34.6 m/s
2
a
-10 m/s
0
t
x
0
vf
Also, as it begins and ends at ground level, the final velocity in the y‐direction must be the same as the initial velocity in the y‐direction except it is negative because it is going down. In the x‐
direction, the final and initial velocities must be the same as there is an acceleration of zero. Y
X
vo
20 m/s
34.6 m/s
a
-10 m/s2
0
t
x
0
vf
-20 m/s
34.6 m/s
Once again, it is easiest to find time in the y‐direction as you have an acceleration. Here you want to use: vf = vo + at ‐20 = 20 + (‐10)t ‐40 = ‐10t t = 4.0 s (Of course, this time is the same in both directions). Y
X
vo
20 m/s
34.6 m/s
a
-10 m/s2
0
t
4s
4s
x
0
vf
-20 m/s
34.6 m/s
To find the total horizontal displacement, we use the x‐direction information: x = vot + ½at2 x = (34.6)(4) + ½(0)(4)2 x = (34.6)(4) x = 138.4 m 103 vo
a
t
x
vf
Y
20 m/s
-10 m/s2
4s
0
-20 m/s
X
34.6 m/s
0
4s
138.4 m
34.6 m/s
As for the other question, determine the maximum height, the ball must reach its maximum height halfway through its motion. This would occur at t = 2.0 s (half the time). Using this time, we can find its displacement in the y‐direction at this moment. As height is a vertical term, we again use only values from the y‐column. x = vot + ½at2 x = (20)(2) + ½(‐10)(2)2 x = 40 + ‐20 x = 20.0 m Other information: If the ball left at 40 m/s, it will land at 40 m/s (or ‐40 m/s as it will be in the opposite direction). If it left at an angle of 300 it will land at an angle to 300. Again, what happens going up, must happen coming down. A few important concepts to remember 1. In projectile motion, the only acceleration occurs in the vertical y‐direction, gravity. 2. In projectile motion, the acceleration in the horizontal x‐direction is always zero. This means that the horizontal velocity is always constant. 3. The vertical y‐velocity of an object in projectile motion is always zero at its maximum height. 4. An object launched horizontally always has an initial vertical y‐velocity of zero. Questions and Problems 1. Neglecting air resistance, what is the only acceleration of an object in projectile motion? 2. If an object is fired horizontally, what is always its initial velocity in the vertical direction? 104 3. (a) Determine the vertical component of the initial velocity (v0y) of an object that is fired at 80 m/s at an angle of 300. (b) Determine the horizontal component of the initial velocity (v0x) of an object that is fired at 80 m/s at an angle of 300. 4. A man running along the top of a 100 m high cliff leaps horizontally off the cliff at a speed of 8.0 m/s. (a) Determine the time it takes him to reach the water below. (b) Calculate the distance from the base of the cliff where he lands in the water. 5. An arrow is fired from a bow at 40 m/s at an angle of 500. (a) Calculate the time it takes to reach the target. (b) Determine the distance to the target. 6. A football leaves the toe of a punter at 20 m/s and at an angle of 400. (a) Determine the total hang time for the football. (b) Calculate how far the football travels from the punter. (c) Determine the maximum height of the football. 7. Dunguaire Castle, located in Kinvara, Ireland, was built in 1520 and has a tower that sits 22.8 m above Galway Bay. A cannon is place in this tower and pointed horizontally toward the bay. If the cannon is fired at 30 m/s, (a) determine the time it takes the cannonball to reach the water. (b) Calculate the distance the cannonball travels from the base of the tower. 8. A 155 mm artillery Howitzer is placed on top of a 60 m high cliff overlooking the water. It is set at an angle of 300 and fires a shell at a speed of 200 m/s. (a) Calculate the time it takes the shell to reach the water. (b) Determine how far from the cliff a ship must be located to be hit by the shell. 9. In a scene at the end of famous movie, a car is driven straight off the edge of the Grand Canyon. If it took the car 12.6 s to land at the bottom of the canyon and it landed a distance of 176.4 m from the base of the canyon wall, (a) determine the velocity of the car when it left the edge of the canyon. (b) Calculate the height of the canyon wall. 10. A cannon is placed on top of a cliff with a height h. The cannon fires a cannonball horizontally with an initial velocity vo causing the cannonball to land a distance x from the base of the cliff after a specific time in the air t. (a) If the height of the cliff is unchanged and the initial launch velocity is doubled to 2vo, how does that specifically affect x and t? (b) If the original initial velocity vo remains and the height of the cliff is change to 4h, how does that specifically affect x and t? 11. Another cannon is placed on the ground and set at a certain angle θ and fires a cannonball with an initial velocity vo causing the cannonball to land a distance x from the cannon after a specific time in the air t. If the angle is unchanged and the initial launch velocity is doubled to 2vo, how does that specifically affect the cannonball’s time in the air t and the landing distance x? 105 12. Ball A is thrown horizontally off the top of a building. At the same moment, a similar Ball B is dropped from the top of the same building. (a) Create a graph of each balls’ accelerations in the horizontal direction and another of each balls’ accelerations in the vertical direction while in the air. (b) Create a graph of each balls’ velocities in the horizontal direction and another of each balls’ velocities in the vertical direction while in the air. 106 Chapter 7 – Forces 7.1 Introduction to Forces Isaac Newton (1642‐1727) was born the year Galileo (considered the father of modern science) died. Many people have tried to establish some mystic connection to this, but basically it is an odd coincidence. At the age of 23, Newton was working at Cambridge University in England when there was an outbreak of plague. The university had to be closed for almost two years. During this time, Newton developed the binomial theorem in math, discovered the law of gravitation, invented calculus (yes, he invented a basically new system of mathematics), and discovered that white light could be separated into the colors of the spectrum. Not bad, try to remember that when you hit 23. Newton was very secretive and often feared criticism. He sometimes would wait for a rival scientist to die before he would publish his own findings. When he did publish, he used geometry rather than the calculus he created to avoid disputes over his work. (Calculus makes doing physics much easier, provided you can understand the Calculus). Newton is considered to be one of the greatest physicists ever to have lived. On his grave is written, “Mortals, congratulate yourself that so great a man lived for the honor of the human race.” Newton is best known for his ideas on forces and motion. In particular, he developed three laws known as Newton’s Laws of Motion. 7.2
Newton’s Laws of Motion 7.2.1
Newton’s 1st Law of Motion – Law of Inertia An object at rest remains at rest, an object in constant straight line motion remains in constant straight line motion, unless acted upon by an external force. Galileo was actually the first person to come to this conclusion, but Newton refined it and made it famous. In fact, none of Newton’s Laws were really his to begin with. He took the work of others and improved on it and published it, thus making it famous. A neat little trick pulled off by many scientists in the past. Consider the idea of inertia. What is it? Galileo describe inertia as a property of matter that opposes changes in motion. This might not mean much to you, so let’s use the following definition; inertia is the tendency for an object to remain at rest or in constant motion. A large rock at rest will stay that way and is very difficult to move or start into motion. A huge truck going down the highway is hard to stop moving or bring to rest. If you ride the Tower of Terror at Disney’s MGM Park, you come out of your seat when the tower first drops because your body wants to remain where it is while the floor drops out from under it. It turns out that 107 inertia is directly related to mass. The more mass something has the less it is likely to change its state of rest or motion. The less mass something has, the easier it becomes to change its state of rest or motion. The famous penny and the hoop trick – a circular plastic hoop is placed on top of a glass flask with a very small opening. A stack of pennies this then placed on top of the hoop directly over the opening of the flask. Now, how does one get all the pennies in the glass without touching the pennies or the glass? Pennies
Loop
Being made of plastic, the hoop is very flexible. By pulling the loop straight out really fast, the loop bends and separates from the pennies. Because the pennies are at rest in the beginning, their inertia keeps them at rest for a moment while the loop gets out of the way. Then gravity takes over and pulls the pennies straight down into the glass. Neat trick isn’t it. By the way, magicians use this same principle when they perform the famous “Pull the tablecloth out from under the dishes” trick, but we will see that later. Notice how this directly relates to Newton’s 1st Law of Motion. The pennies were at rest and stay at rest. When the loop is pulled away, they are acted upon by an external force, in this case, gravity. This sets the pennies in motion. The pennies remain in motion until acted upon by another external force, in this case, the bottom of the glass, which stops their motion. Another good example of Newton’s 1st Law is to consider motion in space. If I were to throw something in space, it would continue to travel on a straight line at the speed it was thrown, forever, or until something acted upon it. 108 7.2.2
Newton’s 2nd Law of Motion Newton’s 2nd Law of Motion shows the relationship between force, mass, and acceleration. The acceleration of an object is directly proportional to the amount of force applied to the object and inversely proportional to the object’s mass. Basically this was really just a formula: F = ma F – Force [N] m – Mass [kg] 2
a – Acceleration [m/s ] This little equation (F = ma) is one of the most important in all of physics. It is the true foundation of most of physics. The unit of force [N] is called the Newton after old Isaac himself. Let’s look at each part. Force (F) is defined as any influence which can change the velocity of an object. In simpler terms, a force is a push or a pull. Gravity is a force, magnets provide force, if you get punched in the face you certainly experience a force, etc. A force can start something in motion, stop it from moving, or change its direction (that last one is important to remember). Force is a vector, which means the direction in which a force is acting makes a difference. Force is measured in the unit called the Newton [N], where 1.0 N = 1.0 kgm/s2, so it still uses the basic units of mechanics. (In the English system, force is measured in pounds (lbs)). This equation is viewed by how things affect acceleration. If acceleration is directly proportional to the force applied, that means that as the force applied on a mass increases, its acceleration increases by the same amount. So if I triple the force on a mass, the acceleration also triples. If acceleration is inversely proportional to the amount of mass, that means if force is held constant and I increase the amount of mass, the acceleration will decrease by the same amount. So if I have a constant force available and I triple the mass, the acceleration will decrease by 1/3. Example: If I have an object that is 25 kg in mass and I apply a 50 N force to it, (a) determine its acceleration. m = 25 kg F = ma F = 50 N 50 = 25a a = ? a = 2 m/s2 109 If this object starts from rest and I apply that force for 10.0 seconds, (b) how fast will it be moving, and (c) how far will I have moved it? (a) vf = vo + at (b) x = vot + ½ at2 vo = 0 a = 2.0 m/s2 x = (0)(10) + ½ (2)(10)2 vf = 0 + (2)(10) t = 10.0 s vf = 20 m/s x = 100 m So I can use F = ma to find the acceleration and then use the acceleration to find velocities and displacements. I can also do the reverse, using linear motion I can find acceleration and then find the force applied. Example: A car with a mass of 1000 kg is traveling at 30 m/s when it hits the brakes. It comes to a stop in 5.0 s. What is the force required to stop the car in 5.0 s? m = 1000 kg Find acceleration first. Then find force. vo = 30 m/s vf = vo + at F = ma 0 = 30 + a(5) F = 1000(‐6) vf = 0 m/s t = 5.0 s ‐30 = 5a F = ? a = ‐6.0 m/s2 F = ‐6000 N Notice that a force can be either positive or negative, because it is a vector. To stop something from moving forward in the positive direction requires a negative force. 7.2.3
Weight One of the most common forces we will deal with during mechanics is the force called weight. WEIGHT IS A FORCE IT IS NOT A MASS!!!! Weight is the force of gravity on an object’s mass. It is calculated by: W = mg W – Weight [N] m – Mass [kg] g – Acceleration of gravity (g = 10 m/s2 on Earth) So a person with a mass of 80 kg would have a weight of: W = mg W = (80)(10) W= 800 N During linear motion, we used g = ‐10 m/s2 because of the vector aspect of the acceleration of gravity. When we use weight and for most other problems we encounter in physics, we will use 110 a gravity of g = + 10 m/s2 (on or near Earth). (ONLY WHILE USING LINEAR MOTION FORMULAS WILL g = ‐10 m/s2). 7.2.4
Net Force More than one force may act on an object at a time. For example, while moving a crate, one person may push on it while another pulls on it. F1 = 40 N
m = 30 kg
F2 = 20 N
The total force on the crate is the vector sum of all the forces acting on the table. So Net Force is the vector some of all forces acting on an object. In the above situation, Fnet = F1 + F2 = ma (The sum (Σ) of all forces acting on object equals mass x acceleration). ΣF = ma If the first person pushes with a force of 20 N, the second pulls with a force of 20 N, and the crate has a mass of 30 kg, (a) what is the net force on the table? (b) What is the acceleration of the table? (c) Starting from rest, how far will they be able to move the table in 10.0 s? (b) Fnet = 60 N (c) a = 2.0 m/s2 (a) F1 = 40 N m = 30 kg t = 10.0 s F2 = 20 N Fnet = ΣF = F1 + F2 F = ma vo = 0 m/s Fnet = 40 + 20 60 = (30)a x = vot + ½ at2 x = (0)(10) + ½ (2)(10)2 Fnet = 60 N a = 2.0 m/s2 x = 100 m Since forces are vectors, they can be added when acting together (as above), and subtracted when acting against each other. 111 Example: 100 N
60 N
F1
20 kg
F2
If a 60 N (F1) force acts on a 20 kg box to the right and a 100 N (F2) force acts on it to the left, the net force is 40 N to the left. Or, Fnet = ΣF = F1 – F2 = 60 – 100 = ‐40 N This means that the box will end up accelerating to the left, or in the negative direction. Again, the acceleration can be found by using: Fnet = ma ‐40 = 20a a = ‐2 m/s2. NET FORCE AND ACCELERATION MUST ALWAYS BE IN THE SAME VECTOR DIRECTION. 7.2.5 Newton’s 3rd Law of Motion –Law of Action‐Reaction For every action, there is an equal and opposite reaction. What this means is that if I push on something with a certain force, that object must push back on me with the same force, but in the opposite direction. Keep in mind that we are talking about two different objects acting in different directions at the same time. For example, if an ice skater pushes against the wall, we know that the wall must push back against her for her to start moving. If you press your hand against the corner of the table, your hand deforms and you can feel pain (if you push hard enough). This means that the table must be pushing back against your hand. The harder you push against the table, the harder the table pushes back against you. If you were standing on a skateboard holding a large rock and you threw the rock one direction, you would move in the other direction, indicating that the rock must have applied a force on you. So if I apply a force of 20 N to a table, the table applies a force of ‐20 N back on me (same magnitude, opposite direction). Now I know what you are thinking, if I throw a baseball I must apply a force to it. When I do this, the baseball takes off very fast and I don’t move at all, so how could the ball apply any force back to me. Well, it does, so there. Just kidding, you have to consider all the forces involved, all the masses involved, and Newton’s 2nd Law. 112 Example: Let’s say you have a mass of 60 kg while the baseball only has a mass of 0.6 kg. If you apply a 5.0 N force to the ball then the ball applies a 5.0 N force back to you. Now, let’s do Newton’s 2nd Law for the ball AND for you. You Ball F = ma F = ma 5 = 0.6a 5 = 60a a = 8.33 m/s2 a = 0.0833 m/s2 Since you have a mass 100 times greater than the ball, you experience an acceleration 100 times less than that of the ball. So the ball takes off, but you don’t. You may be thinking, well I do have some acceleration yet I still don’t go anywhere. That is because we did not do the net force on you correctly. You see you experience a force of 5.0 N trying to push you backwards, but your feet are firmly planted on the Earth which provides an additional force called friction (that we will hit on later) that helps to lower the net force acting on you, giving you an even lower acceleration than 0.0833 m/s2. (By the way, if you did this in outer space, you would physically begin to move backwards. Working is outer space is tricky business, you really need to know your Physics). So when thinking about Newton’s 3rd Law, always remember that the forces acting between two objects are equal, but in the opposite direction. However also keep F = ma in mind and that if an equal force acts on two objects, one with a big mass (M) and one with a little mass (m), the big mass (M) will have a small acceleration (a), the small mass (m) will have a big acceleration (A). F = ‐ F Ma = ‐ mA 7.3
Free Body Diagrams One of the best ways to understand how forces act on objects is through the use of free body diagrams. A free body diagram means that you take the object and draw it in free space. You then draw all the forces acting on the object in the direction in which they are acting. 113 F1 = 40 N
m = 30 kg
F2 = 20 N
Looking at the picture above, we want to determine the net force and the acceleration of the 30 kg crate. In addition, we want to understand how all the forces are acting on the crate. So we draw a free‐body diagram of the crate. There are some basic rules to drawing any free‐body diagram that you have to follow. 1. Draw the object that you want to analyze in free space. So if you are looking at the forces on a car, you draw the car in free space; if you are looking at the forces on a person, draw the person; if you are looking at the forces acting on a crate (a physics teachers love to use crates for some reason, maybe we all really wanted to be movers, draw the crate in free space. 2. Then draw all the forces acting on the crate in the direction they are acting by using arrows. Now it does not matter which side you draw the arrows as long as they are pointing in the correct direction. Also, you can have as many arrows as necessary on the same side of a free‐body diagram and in the same direction. So if there are two forces acting to the left on an object, you draw two arrows pointing to the left; if there are 50 forces acting to the left, you draw 50 arrows pointing to the left. 3. Label the forces. This means label each arrow with either letters to represent the force or numbers if they are given or known. Most of the time it is letters and the numbers come later. 4. Whichever direction the object is actually moving, call that the positive direction. So if it is moving left, call left the positive direction, if it is moving up, call up the positive direction, even if it is moving down, call down the positive direction. This way, you will never have to worry about whether acceleration is positive or negative (like we had to do in linear motion) as it will work out correctly in the end. 114 There are some types of forces that you should know, as they are quite common. Weight (W) – this is the force of gravity which acts on an object’s mass. As we have mentioned before, weight is the product of mass and gravity, or W = mg. As the letter W is used later on for Work and for the unit the Watt, I recommend that you always label the weight force as mg. This makes things a little easier down the line and you always know that weight is calculated as mg. Normal Force (FN) – a force that acts perpendicular to two surfaces in contact. When an object sits on the ground, its weight acts downward on the ground, so the ground must push back upward on the object or it would sink. This upward force by the ground is called the Normal Force. (The word “Normal” in mathematics means “perpendicular to”). So anything laying flat has a normal force acting straight up as shown in Figure 1 below. However, if something is resting on a slope, the normal force is perpendicular to the slope, so up at an angle as shown in Figure 2 below. FN FN Figure 2 Figure 1 NOTE: Many physics teachers label the normal force as N instead of FN. I used to do that myself, because that was what I was taught. I like FN better as using N to represent normal force confuses everyone with the unit Newton [N]. Whatever you prefer to use is fine, but throughout this text you will FN used to represent the normal force acting on an object. Tension (T) – This is a special force associated with ropes or strings. Whenever a force is applied to a rope or string, it can only be a pulling force. This is because if you try to push on a rope, the rope simply folds up or collapses, so there is no real force to speak of. If you encounter any situation that involves a rope or string attached to an object, draw an arrow pulling away from the object and label it T for tension force. Let’s get back to or original problem. F1 = 40 N
m = 30 kg
F2 = 20 N
115 We have two people moving a 30 kg crate where one is pushing forward from the back with a 40 N force and the other is pulling forward from the front with a 20 N force, thus being lazier than his partner. We want to determine the acceleration of the crate. (Now I know this seems overly simple and that most people can do this without the benefit of a free‐
body diagram, but we need to start somewhere). We first draw a free‐body diagram by following the three rules: Draw the crate alone in free space, draw arrows to indicate the forces acting on the crate, and label the forces. Finally, since the movers are moving the crate to the right, we will say that all forces acting to the right are positive forces which make all forces to the left negative. NOTE: The acting on the crate is important as the crate also creates reaction forces on the movers (Newton’s 3rd Law), but we are not analyzing the movers motion, so we do not show those forces. In addition, as the crate is not moving up or down, we will use the default convention that up is the positive direction and down is the negative direction. y
FN
x
F1
F2
mg
In the free‐body diagram above, F1 represents the mover pushing from the back, F2 represents the mover pulling on the front, mg represents the weight of the crate (or if you prefer, the force of gravity acting on the crate), and FN represents the Normal force provided by the floor pushing up on the crate. Force is a vector. In the situation above, we have two forces acting horizontally (in the x‐
direction) and vertically (in the y‐direction). We treat each direction separately, just like in projectile motion. So we want to find the net force in each direction. We do this by adding the vectors in each direction separately. Now the way physics works, we like to view an object as only having the ability to move in one direction at any given time. Thus an object can only be said to accelerate in a single direction, not in two separate directions. This way, one direction has an acceleration while the other does not (so one direction always equals ma while the other must equal zero). 116 This gives us the following: Σ Fx = F1 + F2 = ma Σ Fy = FN – mg = 0 In the x‐direction, the net force is equal to ma because the crate moves in that direction. In the y‐direction, the net force is equal to zero because it does not move in that direction. The FN in the y‐direction again stands for the Normal Force – force perpendicular to the point of contact between two surfaces. Since we want the acceleration and it is only occurring in the x‐direction, we only look at the x‐direction. Σ Fx = F1 + F2 = ma Σ Fx = 20 + 10 = (60)a 60 = 30a a = 2.0 m/s2 OK, so you probably could have done that without that entire free‐body diagram stuff, but what if I had something that appears to be a little more complicated. For example, let’s say you have a 600 kg elevator that is accelerating downward at 0.5 m/s2 as shown below. We want to determine the tension provided by the cable that will allow this to happen. 600 kg
0.5 m/s2
HA! Not so simple as you may think. First let’s draw our free‐body diagram and label our forces. In this case there are only two forces, weight (mg) which only ever acts downward and the tension (T) from the cable (rope) that must pull upward, unless you want a freely falling elevator which would not be fun for the riders. 117 x
T
y
mg
Since the elevator is physically moving downward, we will say that down is positive and up is negative in terms of vector directions. So if we sum our forces in both the x‐direction and the y‐
direction, there is a ma in the y‐direction (direction of motion) and an acceleration of zero in the x‐direction. ΣFx = 0 (There are no forces at all in the x‐direction.) ΣFy = mg – T = ma (Remember, positive is down, so mg is positive and T negative.) mg – T = ma (600)(10) – T = (600)(0.5) 6000 – T = 300 T = 5700 N Now think about this, the elevator cable would have to provide 6000 N of tension (equal to the weight of the elevator) for it to hold still. In order for the elevator to move downward, the cable must reduce its tension to less than the weight and gravity does the rest to move it downward. How much tension do you think it would take to accelerate the same elevator upward at a rate of 0.5 m/s2? (Answer: 6300 N, think about it). 7.4
Equilibrium When a system is in equilibrium, all forces are balanced. This does not necessarily mean that the object is at rest. Newton said that being at rest and being in constant velocity motion are basically the same as they both have a = 0 m/s2. So being at rest or at a constant velocity means both situations are in equilibrium. It is like having a plant hanging from the ceiling, the plant is at equilibrium because the tension and the weight balance each other. Of course, that is easy to understand. The other example is a car traveling on the highway at a steady 80 km/hr. If it is steady at that rate, that means that the forward force provided by the engine is equal to the 118 backward forces of friction and air resistance. These forces are balanced so the car still moves, it just does not accelerate (speed up or slow down) while they are balanced. So for any equilibrium, a = 0 m/s2, therefore: ΣF = 0 (Sum of all forces equals zero is the primary requirement for equilibrium). F2 = 20 N
F1 = 20 N
m = 30 kg
Looking at the picture above, the free‐body diagram would look like this: y
FN
x
F1
m
F2
mg
Summing the forces in both directions gives: ΣFx = F1 – F2 = ma ΣFy = FN – mg = 0 Looking at this then F1 ‐ F2 = ma 20 – 20 = 30a 0 = 30a a = 0 m/s2 (F1 = F2) Also, FN – mg = 0 and thus FN = mg, which is pretty obvious. So the entire system is in equilibrium. Equilibrium is a nice situation to have as you always know that the acceleration is equal to zero. It also does not matter what direction is called positive. 119 Here are a couple of trickier free‐body diagrams when you have more than one object. In each case, you treat each object as its own free body diagram. m1 = 20 kg
m1 = 30 kg
F = 200 N
Our friend is going to use a force of 200 N to drag two crates across the floor. The crates are tied together with a rope. One crate has a mass of 20 kg and the other a mass of 30 kg. The free‐body diagrams look like this: FN1
FN2
y
x
m1
m1g
T
T
m2
F
m2g
Those two tensions (T) must be equal and opposite (they are part of the same rope) and must be less than the value of F in order for the two crates to accelerate forward. As both crates move right, positive for both is considered to the right. 120 m1 = 4.0 kg
m2 = 1.0 kg
Here I have two objects attached together by a rope. One has a mass of 4.0 kg and is sitting on a frictionless table at rest. The second has a mass of 1.0 kg and is hanging freely at the other end of the rope which is passed over a frictionless pulley. I let go of the masses. The one on the table slides to the right while the other one falls. The free‐body diagrams look like this: FN
y
m1
T
x
T
m2
x
y
m1g
m2g
Note that in the x‐direction for the second mass, there are no forces to add up. I have also decided that since the second mass is moving downward, I am going to call down the positive y‐direction. (I can do that!!) In addition, these tensions (T) are once again equal and opposite as they are part of the same rope and must obey Newton’s 3rd Law. Isn’t this fun? 121 One last example, where we can now put our knowledge of forces and free‐body diagrams together with our knowledge of vectors. F = 200 N
m = 50 kg
θ = 30o
A mover drags a 50 kg crate forward by applying a 200 N force at an angle of 300. We want to determine (a) the acceleration (a) of the crate and (b) the normal force (FN) on the crate. First we draw a free‐body diagram. FN
F
θ = 30o
mg
We then break up the pulling force into its x and y components the same way we do with all vectors. (You do remember your vector rules, right?) Fx = Fcosθ = 200cos(30) = 173.2 N Fy = Fsinθ = 200sin(30) = 100 N FN
Fy
Fx
mg
Now we can create the net force equations. Σ Fx = Fx = ma Σ Fy = FN +Fy – mg = 0 122 To determine the acceleration (a), the y‐direction does not matter for this part. Σ Fx = Fcosθ = ma 173.2 = (50)a a = 3.46 m/s2 To get the value of the normal force (FN) we use the forces in the y‐direction and ignore the forces in the x‐direction. Σ Fy = FN +Fy – mg = 0 FN +100 – (50)(10) = 0 FN = 400 N (Amazing how it all comes together!) 7.5
Pressure A common term we are all accustom to hearing is called “pressure.” Pressure comes in many forms, though I am certain my students often use it in terms of the “pressure” of taking a test. However the term “pressure” in physics is related to the idea of force. A classic example of the idea of pressure is the “bed of nails” trick. I saw this demonstrated once where a person lay on a large bed of nails. A large cinder block was then placed on his chest and his assistant then hit the block with a sledge hammer (I am certain there was a lot of practicing so that he knew how to hit the block and not the instructor’s head). Anyway, the block broke into many pieces and the instructor got up without a single puncture in his skin. So how is this possible? First let’s begin with a definition of pressure: Pressure – the amount of force applied over a certain area. This leads to the basic formula for pressure: P – Pressure [Pa] or [N/m2] Pa ‐ Pascal F – Force [N] A – Area [m2] So as you can see, pressure is directly proportional to the amount of force applied and inversely propottional to the area over which the force is applied. So as force increases, so does the pressure by an equal amount. So if you double the force with no change in area, the pressure also doubles. As the area over which the force is applied increases, the pressure decreases by 123 an equal amount. So if the area over which a force is distributed doubles, the pressure is cut in half. Example: A 50 kg ballerina is standing on one foot. Her foot has an area of 0.014 m2. (a) Calculate the pressure on her one foot. (b) She then pushes herself up onto her toes which have a combined area of 0.001 m2. Determine the pressure on her toes. (a) The amount of force is equal to her weight. F = mg 50 10
0.014
P = 35,714.3 Pa (b) Now the area changes, but her weight does not. 50 10
0.001
P = 50,000 Pa By going up on her toes, she has increased the pressure by 1.4 times. So let’s look at the “bed of nails”. However we will use a balloon as a demonstration with two different situations. First we will place our balloon on a single nail and then place a 1.0 kg mass on top of it. 1.0 kg The weight applied on the tip of the nail is around 10 N. The area of the tip of a single nail is somewhere around 3.14 x 10‐6 m2. So the pressure would be: 10
,
,
3.14 10
Wow, that is way more than our ballerina on her toes, so certainly enough to break a balloon. 124 Now let’s put it on a bed of 200 nails. 1.0 kg This does not change the weight, but increases the area to 6.28 x 10‐4 m2. The new pressure is now: 10
,
6.28 10
So that is how the instructor on the bed of nails is not in any real danger. By distributing his weight over hundreds of nails, the pressure applied is greatly reduced. One of the aspects you will discuss in Chemistry is the idea of pressure as related to a gas. When you blow up a balloon, the air in the balloon applies pressure to keep it inflated. Thus the balloon is in a state of equilibrium. Of course, Newton’s Laws basically indicate that if the air in the balloon is providing pressure on the inside, there must be a counterpressure on the outside of the balloon, otherwise it would continue to expand outward. That outside pressure comes from the air around the balloon. It is hard to imagine the air that surrounds us really applies any significant force or pressures on our bodies. However, there are many kilometers of air on top of us and it all begins to add up. To give you and idea, what is known as atmospheric pressure, the pressure of all the air above sea level, is 101.325 kPa ( or 101,325 Pa of pressure). That is a significant amount of pressure on our bodies. As humans, we have evolved in this type of pressure. So if you went into outer space, where the pressure is zero and you took off your spacesuit, your internal body pressure which is calibrated for Earth would basically cause you to explode. Now imagine what happens to a person who goes diving. They not only have the air pressure above them, but now the pressure of the water they are underneath. Diving too deep, the pressure could be so great as to cause an implosion. This is basically an inward collapse. On a diver, this increase in pressure actually compresses the nitrogen gas molecules in your body the deeper the dive. As the diver rises, these nitrogen molecules begin to expand with less pressure. A slow expansion is not dangerous, this is why divers, as they rise, must stop about every 10 m and pause to give the nitrogen time to slowly expand. If they race to the surface, the nitrogen will rapidly expand and cause the “bends” which is when these expanding bubbles of nitrogen damage joints, arteries, lungs, etc. 125 Example: A 4.0 kg paint can is sitting on a table. If the can has a radius of 8.0 cm, determine the pressure it applies to the table. The amount of force is equal to the weight. The area is that of a circle. F = mg A = πr2 40 10
.08
P = 1,989.4 Pa A few important concepts to remember 1. A force is defined as any influence which can change the velocity of an object. 2. Newton’s 1st Law of Motion states that an object at rest remains at rest, and object in constant straight line motion remains in constant straight line motion, unless acted upon by an external force. 3. Inertia is the tendency for an object to remain at rest or in motion. 4. Newton’s 2nd Law of Motion states that the acceleration of an object is directly proportional to the amount of force applied to the object and inversely proportional to the object’s mass. As force increases, acceleration increases, as mass increases, acceleration decreases. 5. Acceleration is always in the same direction as the net force. 6. Newton’s 3rd Law of Motion states that for every action, there is an equal and opposite reaction. 7. Weight is the force of gravity acting on an object’s mass. 8. A normal force is a force created due to two surfaces in contact and acts perpendicular to the point of contact. 9. Tension is a pulling force associated with ropes or strings. 10. The net force on an object is the sum of all forces acting on an object. 126 11. Pressure is the amount of force applied over a certain area. Pressure is directly proportional to force applied and inversely proportional to the area over which the force is applied. Questions and Problems 1. What is the term for the tendency of an object to remain at rest or in motion? 2. What is the term for the sum of all the forces acting on an object? 3. According to Newton’s 2nd Law of Motion (F=ma), if the mass of an object remains constant and the force is doubled, specifically how is the acceleration changed? 4. According to Newton’s 2nd Law of Motion (F=ma), if the force on an object remains constant and the mass is tripled, specifically how is the acceleration changed? 5. The acceleration of gravity on Earth is approximately 10 m/s2, on the moon it is 1.6 m/s2, and on Jupiter it is about 25 m/s2. Calculate the weight of a 50 kg person on (a) Earth, (b) the moon, and (c) Jupiter. 6. A 60 kg crate is pushed with a horizontal force of 20 N. (a) Calculate its acceleration. (b) If it started from rest, determine how far it is moved in 10 seconds. 7. A quarterback accelerates a ball from rest to 20 m/s in 0.4 s. (a) Calculate the acceleration of the football. (b) Calculate how much force the quarterback applied to the ball if it had a mass of 0.6 kg. 8. An 800 kg dragster starts from rest and accelerates to 40 m/s in the first 20 m. (a) Calculate the acceleration of the dragster. (b) Determine the amount of force is used to accelerate the dragster. If the driver of the dragster has a mass of 80 kg, (c) calculate the weight of the driver. (d) As the driver would experience the same acceleration as the dragster, determine the amount of force exerted on the driver and compare this to his weight. 9. A 0.15‐kg baseball is moving at 30 m/s when it strikes a catcher’s glove. The glove recoils 8.0 cm while stopping the ball. (a) Determine how much force the glove applies to the ball. (b) Determine how much force the ball applies to the glove. 10. A 4000 kg truck traveling at 15 m/s brakes and comes to a stop in 30 s. Calculate how much force was required by the brakes to stop the truck. 11. A car traveling at 15 m/s crashes into a wall. The car crushes 20 cm while coming to a stop. (a) Assuing the driver undergoes the same accerleration as the car, determine how much force the 80 kg driver experiences. (b) If the car had crumple zones and instead stopped in a distance of 100 cm, calculate how much force the same driver now experiences. 127 12. A pitcher can throw a 100‐mph (44 m/s) fastball from rest in 0.2 s. If the ball has a mass of 0.15 kg, determine how much force he applies to the ball. 13. Draw the appropriate free‐body diagram for each picture below (assume no friction or air resistance). (NOTE: The first rule is to draw the object in free space, so do not draw on these pictures!) F1
(a)
F2
m
(b)
m
F
m1
(d)
(c)
m2
m2
m1
(e)
m
F
14. A 200 kg crate is pushed horizontally along the floor with a 50 N force. As it is pushed, it experiences a resisting force against it of 10 N. (a) Draw a free‐body diagram of the crate as it is described. (b) Determine the net force acting on the crate. (c) Calculate the acceleration of the crate. (d) Determine the weight of the crate. (e) Determine the amount of normal force acting on the crate. 15. A crane is used to lift a 1000 kg steel beam. (a) Draw a free‐body diagram of the steel beam as it is being lifted. (b) If the crane’s cable can withstand a maximum tension of 15,000 N, determine the maximum acceleration at which the crane can lift the beam. (c) Using this maximum acceleration, determine the time it takes the crane to lift this beam from rest to a height of 60 m. 128 16. Another crane is being used to lower a 200 kg barrel of tools from a 40 m high platform. (a) Draw a free‐body diagram of the barrel as it is being lowered. (b) If, starting from rest, it takes the crane 20 s to lower the tools from 40 m to the ground, determine the acceleration of the barrel. (c) Calculate the tension provided by the crane’s cable to lower the barrel at the acceleration found in (b). 17. (a) If the force applied to a certain area is tripled, how does that specifically affect the pressure. (b) If the area over which a constant force is applied is cut in half, how does that specifically affect the pressure. (c) If the radius of a circular area over which a constant force is applied is doubled, how does that specifically affect the pressure. 18. Determine the pressure a 2.0 kg brick can apply when (a) lying on its largest side (20 cm x 9.0 cm), (b) lying on its edge (20 cm x 5.0 cm), and (c) standing on its end (9.0 cm x 5.0 cm). 19. If the pressure on the atmosphere is 101.325 kPa and the area of the top of your head is about 0.05 m2, determine the mass of the air that sits on top of your head. 20. A 6.0 kg cylinder can apply 4,000 Pa of pressure. Determine the radius of the cylinder. 129 Chapter 8 – Friction 8.1 Introduction to Friction Friction is a force, much like any other force, except that it always acts in a specific direction. Friction is the force which resists motion or intended motion between two surfaces in contact. That means that the direction of the force of friction is always opposite motion or intended motion. F
m
FN
F
f
m
mg
If I apply a force on a crate to move it to the right, friction will oppose that motion and act to the left. (Note: In this text I use f to indicate friction forces to distinguish it from other forces; others may use Ff, again whatever you prefer). Friction is a very useful force in many cases. Nails would not hold in a wall without friction, you could not stop a car without friction, and you could not walk without friction. In other cases, however, friction is a force we could do without. About half of the power in a car’s engine is used just to overcome friction in its moving parts. Friction causes a loss in energy along electric power lines. (Friction interferes with most of our labs). 8.2
Forces of Friction There are two types of friction forces. The first is called static friction. Static friction (fs) is the force that prevents motion from beginning, what I refer to as intended motion. It is sometimes called starting friction. One must apply a certain amount of force to overcome static friction to begin moving something. If you apply a force less than the static friction force, the object intends to move, however the object will not move. The second type of friction is called kinetic friction. Kinetic friction (fk) is the force which resists actual motion. If you apply a force to an 130 object equal to the kinetic friction force, the object will move at a constant velocity (acceleration of zero). The magnitude of any friction force is based on two things, (1) the types of surfaces in contact, and (2) the magnitude of the normal force acting on the object. The formulas used to calculate friction are: fs = μsFN – static friction fk = μkFN – kinetic friction μs and μk are called the coefficients of static friction and kinetic friction respectively. They are based on the types of materials in contact. For example, a certain type of wood moving on a certain type of stone has a static coefficient of μs = 0.5 and a kinetic coefficient of μk = 0.4. (Note: there are no units associated with coefficients of friction.) Other examples: μk = 0.2 wood on wood μs = 0.5 steel on steel μs = 0.78 μk = 0.58 steel on steel (with oil) μs = 0.15 μk = 0.06 rubber on concrete μs = 0.8 μk = 0.65 rubber on concrete (wet) μs = 0.6 μk = 0.4 Notice the coefficients are between 0 and 1, which is where most, but not all exist. You will also notice that the coefficients of steel on steel drop dramatically when oil is added. That is the primary reason why oil is used on car engines and many other machines, to reduce the amount of friction. Rubber on concrete also drops when the concrete is wet; this is why you are told to allow for more distance when stopping on a wet road versus a dry road. The FN in the equation fs = μsFN and fk = μkFN represents the value of the Normal Force acting on the object. Let us look at the situation of pushing a wood crate on a stone floor. F
wood
m = 50 kg
m
stone
If the crate has a mass of 50 kg, I want to determine how much force is required to get it to start moving (μs = 0.5, μk = 0.4). First I want to look at the free body diagram of the crate. 131 FN
m
fs
F
mg
In this diagram, the weight (mg) pulls down, the Normal force (FN) pushes up from the contact with the floor, and the pushing force (F) acts to the right. Since the pushing force is to the right, the friction force (fs) must oppose that force, so it acts to the left. This is the static friction force (fs) because the object is not in motion. If I want to write out the net force for this situation it would be: ΣFx = F – fs = ma ΣFy = FN – mg = 0 (Note: normal force and weight are not trying to move the crate) Since the crate is not yet moving, the acceleration must be zero. This gives us: ΣFx = F – fs = 0 thus F = fs This tells us that I must push with enough force to overcome the static friction force (fs). I can calculate the static friction force using: fs = μsFN Since it is wood on stone, μs = 0.5. But I also need the value of FN. If I look at the free body diagram, the normal force (FN) and the weight (mg) balance each other out. I know this because the crate does not move up or down. If this is true, then: ΣFy = FN – mg = 0 FN = mg Since the mass of the crate is 50 kg and g = 10 m/s2, then FN = (50)(10) FN = 500 N Now we apply this to the formula for static friction. fs = μsFN fs = (0.5)(500) fs = 250 N This means that I need to push with a force (F) that is greater than 250 N, so F > 250 N to get this to start moving. 132 Once I have the crate moving, I want to know how much force it takes to keep it moving at a constant velocity. Again I look at the free body diagram. FN
m
fk
F
mg
In this case, none of the forces have changed except friction. Friction has changed from static friction (fs) to kinetic friction (fk) because it is moving. If I write the net force for this situation, I have: ΣFx = F – fk = ma ΣFy = FN – mg = 0 However, if I have a constant velocity, the acceleration is zero. So I get: ΣFx = F – fk = 0 or F – fk = 0 so F = fk This says that to push something along the floor at a constant speed, I have to push it with a force (F) equal to the force of friction (fk). Most people have a hard time with that idea, how can the forces be equal but be moving in one direction. The key is that they are moving at a constant velocity, so they are not accelerating (moving faster or slower). If the pushing force (F) was greater than friction (fk) (F > fk) then the crate would move faster and not at a constant speed. If friction (fk) was greater than the pushing force (F) (F < fk) the crate would slow down and not move at a constant speed. So as long as the crate is in motion, and that motion is at a constant velocity, F = fk. fk = μkFN (Note: The normal force has not changed because of motion, but the coefficient of friction has changed to μk = 0.4). ΣFy = FN – mg = 0 FN = mg = (50)(10) = 500 N fk = μkFN fk = (0.4)(500) fk = 200 N So I must push with a force of 200 N to keep the crate moving across the floor at a constant velocity. If I push with more than 200 N, it will accelerate to a faster velocity. If I push with less 133 than 200 N, it will slow down. You will also notice that it took a force greater than 250 N to start the crate moving, but only 200 N to keep it moving. Again, that is because the static friction force (fs) is greater than the kinetic friction force (fk). Example: Let’s say that I do push this same 50 kg crate with a force of F = 300 N. Starting from rest, I want to know how far I move it in 10 s. My net force told me that: ΣFx = F – fk = ma ΣFy = FN – mg = 0 I calculated fk = 200 N. If my pushing force is F = 300 N and my crate is 50 kg in mass, then I can say that If vo = 0 m/s and t = 10 s F – fk = ma 300 – 200 = (50)a then I can use: 100 = 50a x = vot + ½ at2 a = 2.0 m/s2 x = (0)(10) + ½ (2)(10)2 x = 100 m Different example: A 500 kg car is traveling down the road on a sunny day at 30 m/s. Suddenly a cow walks out into the road 80 m in front of the car. The driver slams on the brakes. If μk = 0.6 between the tires and the road, is the cow safe? To answer this, we need to determine how far the car travels before it comes to a stop. 30 m/s
500 kg
80 m
μk = 0.6
So we are given: m = 500 kg vo = 30 m/s vf = 0 m/s We need to determine displacement (x). In order to do that we need the acceleration (a) of the car as it is braking. 134 If we look at the free body diagram of the car while it is braking we get: 30 m/s
500 kg
80 m
μk = 0.6
FN
fk
mg
The forces acting on the car are its weight (mg), the normal force (FN), and the kinetic friction (fk) that is stopping the car. There is no forward force because the driver is not pressing on the accelerator, he is braking. (Note: The 30 m/s is a velocity, not a force or acceleration). Our net force then becomes: ΣFx = ‐fk = ma ΣFy = FN – mg = 0 fk = μkFN FN = mg so, FN = (500)(10) = 5000 N fk = (.6)(5000) = 3000 N ‐fk = ma ‐3000 = (500)a a = ‐6 m/s2 If vo = 30 m/s and vf = 0 m/s and we want displacement (x), we can use vf2 = vo2 + 2ax 02 = 302 + 2(‐6)x 0 = 900 ‐12x ‐900 = ‐12x x = 75.0 m Since the cow was 80 m away, the cow is safe. Hurray for the cow!!!!! Stopped
80 m
135 Example: Let’s say it was a rainy day, which drops the coefficient of friction to μk = 0.5 (not really that different, right?). Everything stays the same except the calculations. FN = mg = (500)(10) = 5000 N (Still the same). fk = μkFN = (0.5)(5000) = 2500 N (Different force of friction because of different μk). ‐fk = ma ‐2500 = (500)a a = ‐5 m/s2 (Lower acceleration due to less friction). If vo = 30 m/s and vf = 0 m/s and we want displacement (x), we can again use vf2 = vo2 + 2ax 02 = 302 + 2(‐5)x 0 = 900 ‐10x ‐900 = ‐10x x = 90.0 m (Probably don’t want to see that picture). Since the cow was 80 m away, the cow is toast and so is your car. Poor cow, also try explaining that one to Mom or Dad!! Note that the rain increased your stopping distance by 15 m (that is over 45 ft to you English people). This is one of the major reasons why when you learn to drive you are told to keep a larger distance between you and the car in front of you when it is raining as you will not be able to stop as quickly. Let’s look at just the free‐body diagrams around some more complex situations. A previous example: m2
m1
F
μk
Our friend is going to use a force (F) to drag two crates across the floor. The crates are tied together with a rope. One crate has a mass (m1) and the other a mass (m2). There is a coefficient of kinetic friction (μk) between the floor and the crates. I want to understand all the forces with a free body diagram of each crate. 136 y
FN1
FN2
x
m1
m2
T
T
f1
F
f2
m1g
m2g
Notice the same tension (T) on the rope acts on both masses. Since the rope is attached to both masses, the tension on the rope must be the same for each mass, but in the opposite direction. However, each mass has its own unique frictional force (f1 and f2) acting on it. Let’s look at another example we saw before. m1
μk
m2
I have two objects attached together by a rope. One has a mass (m1) and is sitting on a table at rest. There is a coefficient of kinetic friction (μk) between the m1 and the table. The second has a mass (m2) and is hanging freely at the other end of the rope which is passed over a frictionless pulley. I let go of the masses. The one on the table slides to the right while the other one falls. I want to understand all the forces with a free body diagram of each crate. FN1
y
m1
x
T
x
T
y
f1
m1g
m2
m2g
137 Using free body diagrams allows you to understand how several complex forces act on any object. More complex F = 200 N
m = 50 kg
θ = 30o
μk = 0.4
A mover drags a 50 kg crate by applying a 200 N force at an angle of 300. The μk = 0.4 between the crate and the floor. We want to determine the acceleration of the crate. First we draw a free‐body diagram. FN
F
θ = 30o
fk
mg
The weight (mg) and the normal force (FN) both act in the y‐direction. The kinetic friction force (fk) acts in the x‐direction, but the pulling force (F) acts at an angle. So we need to break up the pulling force into its x and y components the same way we do with all vectors with angles. Fx = Fcosθ = 200cos(30) = 173.2 N Fy = Fsinθ = 200sin(30) = 100.0 N FN
Fx
fk
mg
Fy
138 Create the net force equations. Σ Fx = Fx – fk = ma Σ Fy = FN +Fy – mg = 0 The y‐component does not matter for this part. Σ Fx = Fcosθ – fk = ma Σ Fy = FN +Fy – mg = 0 FN = mg – Fy FN = mg – Fsinθ = (50)(10) – 100 = 400 N Fcosθ – fk = ma Fcosθ – μkFN = ma Fcosθ – μk(450) = ma 173.2 – (0.4)(400) = (50)a 173.2 – 160 = 50a 17.2 = 50a a = 0.344 m/s2 A few important concepts to remember 1. Friction is a force that opposes motion or intended motion between two surfaces in contact. 2. Static friction resists motion from starting; kinetic friction resists actual motion. Static friction is always greater than kinetic friction. 3. Friction is based on the materials in contact and the amount of normal force. 4. A braking car does not have a forward force as the gas pedal is not in use; it only has the force of friction acting to slow it down. 5. If an object is in equilibrium (a = 0 m/ss) while a force is applied to move it, the force applied must be equal to the static friction force if it is not moving, or the kinetic friction force if it is moving at a constant velocity. Questions and Problems 1. What is the term for the force that resists starting motion? 2. What is the term for the force that resists actual motion? 3. What two things is the force of friction based on? 139 4. Explain how an object can be at rest on the ground when a single horizontal pushing force is applied to it. 5. Explain how someone can push a sofa across the room with a continuous horizontal force and yet the sofa travels at a constant velocity across the floor. 6. Explain why a car traveling down a flat road immediately begins to slow down when the driver takes her foot off the gas pedal. 7. Draw the free body diagram for each object, include the forces of friction. (a)
m1
m2
F
(b)
m
μk
μk
F
8. If a mover is pushing a large crate across the floor at a constant velocity with a force of 200 N, how much frictional force does the crate experience? 9. A 50 kg crate is pushed with 500 N of force. It experiences a frictional force of 300 N. (a) Calculate the acceleration of the crate. (b) If the crate starts from rest, calculate how far it will move in 10 s. 10. A 50 kg stone slab is on a wood floor (μs = 0.5, μk = 0.4). Calculate (a) the maximum static friction force and (b) the maximum kinetic friction force. 11. In a strong man contest, the competitor must pull a 2,000 kg truck from rest. If there is a μs = 0.8 between the truck and the ground, calculate the amount of force the man must apply to start the truck moving. 12. A driver in a 1,000 kg car traveling at 40 m/s sees our favorite cow wander out into the road. The driver slams on the brakes and skids to a stop. If μk = 0.8 between the tires and the road, calculate how far the car skids before stopping. 13. A 60 kg metal trunk full of clothes is sitting on your wood floor (μs = 0.6, μk = 0.5) and you are asked to move it to the other side of the room. (a) Draw a free‐body diagram of the trunk when you start to push it. (b) Calculate the amount of force you must apply to start the trunk moving. (c) Once you have it moving, calculate the amount of force you must apply to keep it sliding across the floor at a constant velocity. 140 14. A 20 kg plastic sled is sitting on the snowy ground (μs = 0.3, μk = 0.1). (a) Draw a free‐
body diagram of the sled when you start to pull it. (b) Calculate the amount of force you must apply to start the sled moving. (c) Once you have it moving, calculate the amount of force you must apply to accelerate it across the snow at 0.5 m/s2. 15. A 200 kg steel crate is sitting on a concrete floor. (a) If it takes 1500 N of force to start it moving, determine the static coefficient of friction (μs) between the crate and the floor. (b) If it takes 1200 N of force to keep it moving at a constant velocity, determine the kinetic coefficient of friction (μk) between the crate and the floor. 16. Using the following setup: m1 = 20 kg
m1
m2 = 10 kg
μs = 0.4
μs
m2
(a) Draw a free body diagram for each mass. (b) Calculate the static friction force (fs) for m1. (c) Calculate the weight of m2. (d) Is there enough mass on m2 to start m1 moving? (Justify your answer). 141 Chapter 9 – Work and Energy 9.1 Work Forces can produce changes. They set things in motion, alter the path of an object’s motion, or bring a moving object to rest. Work is a measure of how much change a force produces. Work – the amount of change a force can cause. If I push against the wall, it does not move. So I have not caused any real change in the wall, therefore I have done no work on the wall. Yet if I apply the same force to a ball, it does move. Since I have affected a change in the ball, I have done work on the ball. Some people would argue that if I push on a wall for a while, I will start to get tired, so I must be doing some sort of work. They would be close to being correct. Based on Newton’s 3rd Law (Action‐Reaction) as I push on the wall with a force, it pushes back on me with the same force. That force from the wall causes the cells in my muscles to stretch and contract. So there is a change affected on me. I do not do any work on the wall, but the wall does do work on me (specifically on my muscles). So when you “work out” you gain strength not so much because of forces you are applying to any weights, but from the reaction forces those weights are applying to your muscles. The actual definition of work is the magnitude of the force in the direction of the displacement multiplied by the magnitude of the displacement. This is just a fancy way of giving a formula for work. W = Fx W – Work [J] F – Force [N] x – Displacement [m] Work is measured in the unit called the Joule [J] after James Joule who discovered the Law of Conservation of Energy which we will discuss later. (The formula is often written as W = Fxcosθ in the cases when the force is not applied on the same line as the displacement, but we will try not to do that here). 142 F
m
x
Example: Let’s say our favorite mover pushes a crate with mass (m) of 50 kg across a frictionless floor with a force (F) of 100 N and moves it a distance (x) of 5 m. W = Fx W = (100)(5) W= 500 J of work done to move the crate. Not too hard. Keep in mind that work only occurs while the force is being applied. For example, if I use 50 N of force to throw a ball 100 m, I have not actually done 5000 J of work. (It seems like it if I use W = Fx = (50)(100) = 5000 J). If I throw a ball 100 m, my hand is not in contact with the ball during that 100 m, so I was not applying a force that entire time. The force occurs while the ball is in my hand. So my arm may only move the ball 1.0 m before I let it go. The work in this case would be: W = Fx W = (50)(1.0) W = 50 J Now, let’s combine everything we have learned so far in terms of motion, force, and work. Example: I pick up a ball with a mass of 0.5 kg and throw it from rest to a speed of 20 m/s by moving my arm 2.0 m. (a) Calculate the acceleration of the ball. (b) Calculate the force of my arm on the ball. (c) Calculate the work done by my arm on the ball. m = 0.5 kg (a) vf2 = vo2 + 2ax (b) F = ma (c) W = Fx 202 = 02 + 2(a)(2) F = (0.5)(100) W = (50)(2) vo = 0 m/s vf = 20 m/s 400 = 0 + 4a x = 2.0 m F = 50 N W = 100 J a = 100 m/s2 143 9.2
Work in the Vertical All this work is horizontal work. Just like in horizontal motion, the acceleration can vary depending on the situation. What about work in the vertical direction? In this direction, we mostly have one acceleration that we are concerned about, gravity (g = 10 m/s2). Consider the idea of lifting an object off the ground to a certain height at a constant velocity. F
h
m
If you look at the free‐body diagram once it is off the ground you have: F
mg
ΣFx = 0 (There are no forces at all in the x‐direction). ΣFy = F – mg = ma If we assume that we lift with a constant velocity, then F – mg = 0 (no acceleration) and therefore F = mg. The displacement (x) is equal to the height (h) to which we lift the object. So: W = Fx (F = mg and x = h) Wg = mgh This is known as work against gravity or vertical work. If I were to lower the same object from the same height, I would do the same amount of work. Force is a vector and displacement is also a vector. Work, however, is a scalar value, so it does not have a specific direction. Example: I want to lift a 20 kg box to a shelf that is 2.0 m high. How much work is done? Wg = mgh Wg = (20)(10)(2.0) Wg = 400 J Not too difficult. 144 9.3
Energy Energy – the ability to do work. If something has energy, it can exert a force on an object. Thus energy can do work. It turns out, however, that you can do work and transform it into energy. This concept is sometimes referred to as the Work – Energy Theorem – Energy can do work and work can transform into energy. Energy is measured in the same unit as work (Joule). There are three categories of energy. Kinetic Energy (KE) – Energy due to motion. This means that any moving object automatically has kinetic energy. Potential Energy (PE) – Energy due to position. Objects in certain positions have available energy. This is a little more complex. Rest Energy (E) – Energy due to mass. This means that all masses have energy stored in themselves just because they have mass. Rest energy was discovered and made somewhat famous by Einstein and his equation: E = mc2 This is not his theory of relativity. It is his equation for rest energy. He basically said that anything that has mass can be converted into energy. This equation led to nuclear fission and fusion. Rest energy applies to an area called Modern Physics which may be a little beyond what we will study this year. 9.3.1 Kinetic Energy Kinetic energy is energy due to motion. We have seen that when we throw a ball, we do work on the ball. In addition, we also give the ball kinetic energy because we put it into motion. Under the idea of the Work‐Energy Theorem, however much work I do on the ball equals the amount of kinetic energy the ball has. So if I do 100 J of work to throw the ball, it will have 100 J of kinetic energy. Knowing the kinetic energy will allow us to find velocities more quickly. KE = ½ mv2 Example: Let’s say I again throw my 0.5 kg ball with a velocity of 20 m/s. KE = ½ mv2 KE = ½ (0.5)(20)2 KE = 100 J 145 Notice this 100 J was equal to the work I did earlier to throw the ball. So the work I do creates the kinetic energy. Thus W = KE (sometimes shown as W = ΔKE). If the ball were to be caught and brought to a stop, all the KE would be gone. The KE will end up doing 100 J of work on the catcher. If the catcher’s hand were to recoil 10.0 cm upon catching the ball, we can determine how much force the ball applied to the catcher. KE = W = 100 J W = Fx x = 10 cm = 0.1 m 100 = F(0.1) F = 1000 N So W = KE = W Look at this from another direction. Let’s say I have a 4.0 kg ball with a KE of 200 J and I want to know its velocity. KE = ½ mv2 200 = ½ (4.0)v2 200 = (2)v2 100 = v2 (Take the square root of both sides). v = 10 m/s 9.3.2 Potential Energy Potential energy is energy due to position. By position, what we are going to look at is height above something. If I hold an object at a certain height, it has the ability to do work if I were to let it go. If I increase the height, it has a greater ability to do work. If I increase the mass it also has a greater ability to do work. This is known as Gravitational Potential Energy (PEg). In the Work – Energy Theorem above, we said that work and energy were equal. When I lift an object to a certain height, I have done a certain amount of work against gravity (Wg = mgh). At that height, the work I have done has now given that object potential energy. That PE is equal to that work. Wg = PEg (sometimes shown as W = ΔPEg). PEg = mgh 146 Potential energy is always measured relative to some point. For example: m =2.0 kg
h1 = 10 cm
h2 = 1 m
h3 = 10 m
A person standing on top of a building is holding a 2.0 kg box 10 cm above a table. The box is also 1.0 m above the roof and 10 m above the ground. The box then has three different potential energies based on each height. PEg = mgh PE1 = mgh1 = (2)(10)(.1) = 2.0 J PE2 = mgh2 = (2)(10)(1) = 20 J PE3 = mgh3 = (2)(10)(10) = 200 J So if this person were to drop the box, it would do 2.0 J of work on the table, 20 J of work on the roof, and 200 J of work on the ground. Think of it as landing on a nail at each location. The nail on the ground would be driven in a deeper distance due to the ability of the box to do more work. 9.4
Conservation of Energy The idea of a conservation principle in science is that in an isolated system (think of it as a box that does not let anything in or out) a certain quantity will always remain constant, regardless of what changes occur within the system. For example, in chemistry you learn the Law of Conservation of Mass which says the total mass of the product of a chemical reaction is equal to the total mass of the original substances that went into the reaction. In other words, if you burn wood, the total mass of the ashes and smoke at the end would equal the total mass of wood in the beginning. Thus the total mass of the system always remains constant. 147 In physics, the first conservation law we will look at is the Law of Conservation of Energy. Law of Conservation of Energy – the total energy in an isolated system always remains constant. This means that however much energy a system starts with, it must end with that same amount of energy, regardless of what goes on within the system. The energy may change form and often does, but in whatever form or forms it exists, its total magnitude must remain constant. Example: We want to drop a 2.0 kg ball from a height of 50.0 m. (Neglect air resistance). m = 2.0 kg
h = 50 m
I want to look at a table of several values as the ball falls. I will measure its height (h), its gravitational potential energy (PEg), its kinetic energy (KE), and its total energy (TE). The total energy (TE) is equal to the sum of all its energies, in this case TE = PEg + KE at any point. I will put these values in a table for comparison. At 50 m with a mass of 2.0 kg, PEg = mgh PEg = (2.0)(10)(50) PEg = 1000 J The ball starts from rest so, KE = ½ mv2 KE = ½ (2.0)(0)2 KE = 0 J Now the total energy can be determined at the beginning to be, TE = PEg + KE TE = 1000 + 0 TE = 1000 J 148 I will put these values in my table. h (m)
PEg (J)
KE (J)
TE = PEg + KE (J)
50
0
1000
1000
The ball is released and begins to fall. I will now look at what happens as it passes 40 m above the ground. At 40 m with a mass of 2.0 kg, PEg = mgh = (2.0)(10)(40) = 800 J To determine the KE at this point (because the ball is moving) we need the velocity (as KE = ½ mv2). To find it, we will use motion. We know the ball was released from rest, and has only fallen 10 m (from 50 m to 40 m), and the acceleration must be due to gravity. vf2 = vo2 + 2ax vo = 0 m/s 2
a = g = 10 m/s vf2 = 02 + (2)(10)(10) (vf = 14.14 m/s) x = 10 m vf2 = 200 I am going to leave the final velocity at this point in squared form (200) because the formula KE = ½ mv2 has velocity squared in it. So, KE = ½ mv2 = ½ (2.0)(200) = 200 J If the PEg = 800 J and the KE = 200 J, then the TE = 1000 J h (m)
PEg (J)
KE (J)
TE = PEg + KE (J)
50
1000
0
1000
40
800
200
1000
If we were to continue our calculations every 10 m, we would find that the PEg decreases due to the loss of height. At the same time, the ball’s velocity increases as it falls which causes the KE to continue to increase. However, the TE remains the same value (constant) the entire time. h (m)
PEg (J)
KE (J)
TE = PEg + KE (J)
50
1000
0
1000
40
800
200
1000
30
600
400
1000
25
500
500
1000
20
400
600
1000
10
200
800
1000
0
0
1000
1000
149 Just before the ball hits the ground, its entire PEg is gone because it does not have any height. However, KE is at a maximum because the ball reaches its highest speed right before it hits the ground. Throughout the entire motion, the total energy remained 1000 J, so no energy was gained or lost in the fall, it was just transferred from PEg to KE. Thus, the total energy is conserved. (Note: At 25 m, or halfway down, both the PEg and KE were equal for an instance). When the ball hits the ground, it will have the ability to do 1000 J of work. PEg = KE = W The PEg at the beginning became KE at the end, which becomes work when it hits the ground. We can now apply conservation of energy to situations we have seen before. For example, let’s say that we throw a ball straight up into the air at 30 m/s and we want to know how high it will rise. When we learned about vertical motion, we had to consider the initial velocity, the final velocity at the maximum height, time, and the acceleration of gravity. Now we can approach this using energy. At the beginning, the only energy the ball has is KE because it is at its lowest point and only has velocity. At the maximum height, the only energy the ball will have is PEg because it comes to a stop at the maximum height. Energy is all PEg
vo = 30 m/s
Energy is all KE
Using the Law of Conservation of Energy, the total energy at the bottom which is entirely KE must equal the total energy at the maximum height which is entirely PEg. So, TE = TE KE = PEg If KE = ½ mv2 and PEg = mgh, then I can say that 2
½ mv = mgh (Note that mass is on both sides of the equation, so it cancels). ½ v2 = gh ½ (30)2 = (10)h ½ (900) = 10h 450 = 10h h = 45 m 150 The mass does not affect the problem in this situation. This makes sense because we never used mass while doing vertical motion. If we were to use vertical motion equations, we would determine the same result. vo = 30 m/s vf2 = vo2 + 2ax 2
a = g = ‐10 m/s 02 = (30)2 + (2)(‐10)(x) 0 = 900 – 20x vf = 0 m/s x = ? 20x = 900 x = 45 m 9.5
Pendulum A pendulum can basically be anything that is attached to a string or rope and allowed to oscillate back and forth. The oscillation of a pendulum is very regular, which is why it is often used to keep time. When a pendulum is pulled back to a certain point and released from rest, that point will be the highest possible point for it to return. This is based on the Law of Conservation of Energy. A
h1
C
h2
D
B
In the picture above, Point A is the release point. It is a certain height (h1) above the bottom of the swing. (It does not matter how far above the ground the pendulum is, only how far above the bottom of its swing). At Point A, the pendulum has all PEg. It has no KE because it is not moving. When it is released, it swings down to point B where all of its energy is KE. This is because it is moving and it is at the lowest point of its swing. When it swings up to point C on the other side, all the energy returns to PEg as the pendulum comes to a stop before it goes back the other way. So, the PEg at point A equals the KE at point B, which equals the PEg at point C. This being true, then the height at point A (h1) must equal the height at point C (h2). The pendulum will continue to swing back and forth between those points, and can never get higher than point A or point C. (Again, we neglect air resistance and friction. In reality, these would take energy from the system and cause it to swing lower each time). What about point D, somewhere between points A and B. At this point, the pendulum is moving and it is also above the lowest point in its swing. If it is moving, it has KE, if it has height, 151 it has PEg. So at point D, the total energy of the pendulum is made of both KE and PEg. That total energy still equals the PEg at points A and C, and the KE at point B. Example: Let’s say I pull a pendulum with a mass of 2.0 kg back to a height of 5.0 m above the bottom of its swing. I want to determine (a) its PEg at point A, (b) its KE at point B, and (c) its velocity at point B. A
2.0 kg
5.0 m
C
2.0 m
(a) TE = PEg = mgh PEg = (2)(10)(5) PEg = 100 J B
(b) TE = PEg = KE 100 J = KE (c) TE = KE = ½ mv2 100 = ½ (2)v2 100 = v2 10 m/s = v If point C is only 2.0 m above the lowest point, what is the velocity of the pendulum at that point? We look at this in terms of the total energy. The total energy at any point must be 100 J, as was determined above. The total energy at point C consists of both KE and PEg because at that point it has both height and a velocity. So we can say that: TE = PEg + KE = 100 J (Substitute formulas) 2
mgh + ½ mv = 100 (2)(10)(2) + ½(2)v2 = 100 (Notice that of the TE of 100 J, 40 J of it is PEg). 40 + ½(2)v2 = 100 2
v = 60 v = 7.7 m/s This makes sense because the velocity is greater than zero at the highest point, but less than the 10 m/s at the lowest point. 9.6
Roller Coasters The roller coaster is another good example of conservation of energy. A typical roller coaster car is pulled up to the top of the first hill by a chain. Thus the chain does a certain amount of work to give the car PEg at the top of the first hill. That first hill is significant because that hill must be the highest hill of the ride. At this point, all of the car’s energy is PEg, and that is all the energy it will have. No other part of the ride may be higher than that first hill because the car 152 will not have enough energy to get over a higher hill. In fact, most parts of the ride get lower and lower because the car loses energy to friction all the ride. They make the ride twist and turn and make the loops and turns tighter and sharper to keep your body constantly changing directions. These changing directions create accelerations on your body which give you the thrills. Example: In the picture below, a 500 kg car is pulled to the top of a 100 m hill and released from rest at point A on the ride. m= 500 kg
A
C
h1 = 100m
h2 = 60 m
B
At point A, the total energy (TE) of the car is all PEg because it has height, but no velocity. When the car reaches point B, the TE is now all KE because the car has velocity, but no height. At point C, the car has both velocity and height, so its TE is both PEg and KE. I want to determine (a) the work to pull the car to point A, (b) the PEg at point A, (c) the KE at point B, (d) the velocity at point B, and (e) the velocity at point C. (a) W = mgh (b) The work done on the car gives it PEg, so W = (500)(10)(100) W = PEg = TE W = 500,000 J PEg = 500,000 J (c) Since the TE = PEg at point A and the TE = KE at point B, then PEg at point A must equal the KE at point B. So, TE = PEg = KE = 500,000 J (d) TE = KE = ½ mv2 500,000 = ½ (500)v2 500,000 = 250v2 2000 = v2 v = 44.72 m/s 153 (e) At point C, the car must have both PEg and KE because it has height and must be moving. (You know it must be moving because the hill at point C is lower than the hill at point A). Therefore, the TE = PEg + KE = 500,000 J TE = PEg + KE = 500,000 J (h = 60 m at point C) TE = mgh + ½ mv2 2
500,000 = (500)(10)(60) + ½ (500)v 500,000 = 300,000 + 250v2 (Of the 500,000 J of TE, 300,000 is PEg) (200,000 J left over must be KE) 200,000 = 250v2 2
800 = v v = 28.3 m/s It is always important to understand the type(s) of energy you have at any given point in an object’s motion. If the object has height, but no motion, its TE is all PEg. If an object has motion, but no height, its TE is all KE. If it has both motion and height, its TE is PEg + KE. 9.7
Elastic Potential Energy Anything that stretches or can be compressed is said to be elastic. A spring or rubber bands are good examples of elastic materials. If I have a spring, it has a certain shape and strength. When I hang a mass from the spring, it stretches due to the force of the weight acting on the spring. x
m
How far the spring stretches depends on how much force (in this case weight) is applied, and how strong the spring is. The strength of the spring is known as the elastic constant (k). k – Elastic constant [N/m] 154 If a spring has an elastic constant of 100 N/m, it means that it would take 100 N of force to stretch the spring 1.0 m. Many springs, rubber bands, and other elastic materials have very low spring constants and are easy to stretch or compress. The springs in a car’s shock absorber have very high spring constants (k = 10,000 N/m) to withstand the rigors of supporting a car as it drives. (Although, the stronger the spring, the less it will move, and the more you will feel the bumps in the road). To stretch a spring, you must apply a force for a certain distance to get it to stretch. F
x
If I apply a force (F) for a certain distance (x), then I have done work (W = Fx). The work I do to stretch the spring gives it elastic or spring potential energy (PEs). So the work and the PEs must be equal. W = PEs = TE If I release the spring, it will move, changing the PEs to KE. We determine PEs by: PEs = ½ kx2 So, if I have a spring with k = 100 N/m and I stretch it 20 cm, then I can determine my PEs. PEs = ½ kx2 PEs = ½ (100)(0.2)2 PEs = 2.0 J Let’s look at a more complex situation. 155 Example: A 500 kg elevator starts at rest at a height of 50 m above a spring platform (point A) as shown below. The cables holding the elevator break, causing it to fall and hit a spring with an elastic constant (k = 50,000) N/m). The spring compresses and stops the elevator (point C). A
m
h
B
k
x
m
C
We want to determine (a) The PEg at point A, (b) the velocity (v) of the elevator just before it hits the platform at point B, and (c) how far the spring platform compresses (x) at point C. (a) TE = PEg = mgh PEg = (500)(10)(50) PEg = 250,000 J (b) The TE at point A was entirely PEg because the elevator had height, but no velocity, and the spring platform was not involved. At point B, the TE becomes all KE because it will no longer have height, but will be in motion. In addition, at point B, the spring platform has not been compressed. So, TEA = PEg = TEB = KE = 250,000 J KE = ½ mv2 250,000 = ½ (500)v2 250,000 = 250v2 1000 = v2 v = 31.6 m/s (c) When the spring platform is compressed, the TE all becomes PEs at point C. At this point there is no motion, so no KE, and no height, so no PEg. Therefore, TEC = PEs PEs = ½ kx2 250,000 = ½ (50,000)x2 250,000 = 25,000 x2 10 = x2 x = 3.16 m 156 If I consider any point between points A and B, the TE = PEg + KE. This is because as the elevator falls, it loses height and thus loses PEg. However, it is gaining velocity, so it gains KE. Whatever it loses in PEg it must gain in KE. That way it conserves energy and TE remains constant. If I consider any point between points B and C, the TE = PEs + KE. This is because as the elevator compresses the spring platform, it loses velocity and thus loses KE. However, it is gaining elastic energy, so it gains PEs. Whatever it loses in KE it must gain in PEs. That way it conserves energy and TE remains constant. 9.8
Non­conservative Forces A force is said to be conservative if the work done by the force does not take energy from the system. For example, gravity is a conservative force. As you lift and object or allow it to fall, the work and energy are equal. This is because gravity does not add or subtract energy from the system. A force is called a non‐conservative force if it does take energy from the system. By taking energy from the system, we mean that the amount you begin with is not the same as what you end with in regards to the system itself. Friction is a good example of a non‐conservative force. As friction acts on a moving object, it takes energy away and converts it into heat energy. In this case the friction does negative work to take away energy. TE = PEg
m
h
x
TE = KE
A crate starts from rest at the top of a ramp and slides to the bottom. At the top, all of its energy is PEg. At the bottom, all of its energy is KE just before it hits the ground. If there was no friction, then PEg = KE (TE = TE), due to the Law of Conservation of energy. If we were to actually record its mass and velocity at the bottom of the ramp and calculate the KE, we would actually find PEg > KE. If the PEg at the top of the ramp is greater than the KE at the bottom of the ramp, then the remaining energy had to go somewhere. In this case, friction turns some of the PEg into heat energy. The lost energy would be the change in the total energy of the system (ΔTE), or the difference between the final total energy and the initial total energy. Example: Let’s say that the mass (m) of the crate is 20 kg and the height (h) of the ramp is 5.0 m. The crate is released from rest and we then measure the velocity (v) to be 6.0 m/s at the bottom of the ramp. (a) Calculate the PEg at the top of the ramp. (b) Calculate the KE at the bottom of the ramp. (c) Determine how much energy was lost (Elost) to friction. 157 (a) TEo = PEg = mgh PEg = (20)(10)(5) PEg = 1000 J (b) TEf = KE = ½ mv2 KE = ½ (20)(6)2 KE = 360 J (c) Elost = ΔTE Elost = TEf – TEo Elost = KEf – PEo Elost = 360 –1000 Elost = ‐640 J 9.8.1 Air Resistance Air resistance is another non‐conservative force. Also, when objects strike other objects, there is energy lost in the collision. A dropped ball will not bounce back to its original height because when it hits the ground, some of its energy is converted into other energies such as heat and sound (and sometimes light). Example: A ball with mass (m) of 2.0 kg is dropped from an initial height (h1) of 10 m. It bounces on the ground and then rises to a height (h2) of 6.0 m as shown below. m
h1
h2
(a) Calculate the PEg at the first height. (b) Calculate the PEg at the second height. (c) Determine the energy lost (Elost) due to the impact with the ground. (a) TEo = PEg = mgh PEg = (2)(10)(10) PEg = 200 J 158 (b) TEf = PEg = mgh PEg = (2)(10)(6) PEg = 120 J (c) Elost = ΔTE = TEf – TEo Elost = PEg(b) – PEg(a) Elost = 120 – 200 Elost = ‐80 J Any time the change in total energy (ΔTE) is a negative value, it indicates that energy is lost from the system. 9.9
Power We often hear about how much power a car has or the power of a boxer’s punch. These types of power are known as mechanical power (as opposed to electrical or nuclear power). Power – the rate at which work is done. Another way of saying this is the rate at which energy is changed from one form to another. Whenever we talk about a rate, it involves something divided by time. So if power is the rate at which work is done, then: P – Power [W] W – Work [J] t – Time [s] Power is measured in Watts [W] after James Watt who improved the steam engine. We often hear the term horsepower (HP) instead of power. 1 HP = 746 W (An interesting bit of trivia, but not that important to us.) Example: Let’s say we have our mover again, and he applies 100 N of force to a 50 kg crate to push it 10 m across the floor. 159 100 N
50 kg
10 m
If it takes him 20 s to move the crate, how much power does he use? Need work (W) W = Fx = (100)(10) = 1000 J 1000
20
P = 50 W Example: How about our mover who lifts the 20 kg crate to a height of 2.0 m? Let’s say she can do that in 5.0 s. How much power does he use? 2.0 m
20 kg
Again we need work, but this time it is work against gravity. W = mgh W = (20)(10)(2) W = 400 J 400
5
P = 80 W 160 Imagine you have two people with the same mass who run up the same flight of stairs. As they carry the same mass up the same height, they do an equal amount of work. However, if one person gets to the top of the stairs before the other, that person has a greater power output as she did the same work in less time. A few important concepts to remember 1. Work is the amount of change a force can cause. It is the product of force and displacement. 2. Vertical work is often known as work against gravity. 3. Energy is the ability to do work. 4. Energies fall into three categories: kinetic energy, potential energy, and rest energy. 5. Kinetic energy is energy due to motion, potential energy is energy due to position, and rest energy is energy due to mass. 6. The Work – Energy Theorem indicates that work and energy are interchangeable and that energy can do a certain amount of work and work can produce a certain amount of energy. 7. The Law of Conservation of Energy states that the total energy of an isolated system always remains constant. Energy can transform from one type to another, but the total amount remains constant. 8. Gravity is a conservative force because it does not cause a loss of energy from a system. Friction and air resistance are non‐conservative forces because they do cause energy to leave the system in the form of other energies such as heat, sound, light, etc. 9. Power is the rate at which work is done. Questions and Problems 1. What is the term for the amount of change a force can cause? 2. What is the term for the ability to do work? 3. List the three types of energies. 161 4. Given the following picture: A
B
C
As the pendulum swings from its highest position at point A to its lowest position at point C, (a) what type(s) of energy does it have at point A? (b) What type(s) of energy does it have at point B? (c) What type of energy does it have at point C? 5. Given an object with a constant mass, (a) if I double its height, how does that specifically affect its PEg? (b) If I double its velocity, how does that specifically affect its KE? 6. A 50 kg crate is pushed with a constant 100 N force 10 m across the floor. (a) Calculate the work done on the crate. (b) Determine how much KE is given to the crate. (c) Determine the velocity of the crate at 10 m. 7. (a) Calculate the work done to lift a 30 kg crate to a height of 5 m. (b) Determine how much PEg the crate has at 5 m. 8. A bow fires a 0.2 kg arrow from rest to 45 m/s by applying a constant force over a distance of 0.4 m to the arrow. Calculate the work done by the bow on the arrow. 9. Given the following graph of the amount of Force (F) applied to a 4.0 kg object over a given displacement (x), Force (F) vs Displacement (x)
F (N)
14
12
10
8
6
4
2
0
0
2
4
6
8
10
12
x (m)
(a) determine the work done on this object from 0 m to 10 m. (b) If the 4.0 kg object started from rest, calculate it’s final velocity at the end of 10 m. 162 10. A pendulum with a 2.0 kg mass is pulled back to a height of 1.8 m above its lowest point. (a) Calculate its PEg at this point. (b) The pendulum is released; determine its KE at its lowest point. (c) Calculate the velocity of the pendulum at its lowest point. 11. Using the following picture: m = 100 kg
A
C
h1 = 50 m
h2 = 20 m
B
A 100 kg car is pulled to the top of a 50 m hill. (a) Calculate the work done to pull the car to point A. (b) Determine the PEg of the car at point A. (c) The car is released from rest, calculate its velocity at point B, the lowest point. (d) Calculate the velocity of the car at point C on top of the 20 m hill. 12. What type of energy exists in a stretched or compressed spring? 13. What is an example of a conservative force? What is an example of a non‐conservative force? 14. Given a stretched rubber band with a certain elastic constant. If I double the distance it is stretched, what specifically happens to its PEs? 15. What is the term for the rate at which work is done? 16. A spring (k = 300 N/m) is compressed 20 cm. Calculate its elastic potential energy (PEs). 163 17. Given the following graph of the amount of Force (F) applied to a spring stretching it a given displacement (x) from equilibrium, Force (F) vs Displacement (x)
F (N)
30
25
20
15
10
5
0
0
0.1
0.2
0.3
0.4
0.5
0.6
x (m)
(a) determine the elastic (spring) constant for this spring. Determine the maximum PEs of this spring at 0.5 m. 18. Given the following picture: h
x
m
k
A 300 g ball is placed on a spring (k = 2,000 N/m) which is compressed 10 cm from its rest position. (a) Calculate the PEs of the spring at this point. (b) The spring is released and fires the ball straight up in the air. Determine the KE of the ball just as it leaves the spring. (c) Calculate the velocity of the ball as it leaves the spring. (d) The ball rises up to its maximum height. Determine the PEg of the ball at this point. (e) Calculate the maximum height of the ball. 19. A drill can do 3000 J of work in 30 s. Calculate its power output. 20. A 50 kg girl climbs up a 4.0 m high ladder in 8.0 s. (a) Calculate the work she does in climbing the ladder. (b) Calculate her power output. 164 21. A tennis ball machine can shoot a 300 g tennis ball from rest to 50 m/s in 0.4 s. Determine the power delivered to the tennis ball by the machine. 22. It is discovered that as an object slides down a ramp, the KE of the object at the bottom of the ramp does not equal the PEg of the object at the top of the ramp. The Law of Conservation of Energy says that the total energy must remain constant. Where does this “lost” energy go? 23. A 40 kg sled starts at the top of an 8 m high hill. (a) Calculate the sled’s PEg. It slides down the hill, experiencing friction along the way. If the sled has a velocity of 9.0 m/s at the bottom of the hill, (b) calculate the sled’s KE at the bottom of the hill. (c) Determine the energy lost due to the friction on the hill. 24. A 200 g hockey puck is initially hit at a velocity of 12.0 m/s. After sliding 20 m along the ice it is traveling at 8.0 m/s. (a) Calculate the acceleration of the hockey puck along the ice. (b) Determine the energy lost by the hockey puck over the 20 m. 165 Chapter 10 – Momentum and Collisions 10.1 Introduction to Momentum If I throw a baseball to someone and that person throws it back to me, it is fairly easy to catch (provided that other person is not a major league pitcher). However, if I throw a baseball towards someone and they hit a line drive back at me with a baseball bat, it becomes harder to catch. The difference in these situations is the velocity of the ball as it returns to me. In a second situation, a baseball is thrown to me and then a 7.25 kg (16 lb) shot put is thrown to me at the same velocity. The shot put is much harder to catch than the baseball. In this situation, the difference lies in the mass of each object that is thrown to me. 10.2 Linear Momentum Each situation described above is related to the object’s momentum. In this case, we are going to look at what is called linear momentum, or the momentum of an object along a straight line. The linear momentum of any object is the product of its mass and its linear velocity. p = mv p – Momentum [kgm/s] or [Ns] m – mass [kg] v – velocity [m/s] Momentum is a vector due to velocity (v) being a vector, so direction is very important to the idea of momentum. Example: If I have a car with a mass (m) of 500 kg traveling at a velocity (v) of 20 m/s, then its momentum would be: p = mv p = (500)(20) p = 10,000 Ns (If v = ‐20 m/s, then p = ‐10,000 Ns, direction matters). This simple equation shows that if I increase the mass or velocity of an object, I will increase its momentum. (Keep in mind that momentum is a vector, so direction makes a big difference). You can also see that momentum must be related to something else we have discussed, kinetic energy. p = mv KE = ½mv2 If an object has momentum, it must have KE. If an object has KE, it must have momentum. Keep in mind, though, they are not equal. In fact, if I double an objects velocity, the momentum of 166 the object also doubles, but the KE increases by 4 times (due to the squared nature of velocity in kinetic energy). 10.3 Impulse Most people understand the basic idea of momentum. Once I get something moving, it often continues to move under its own momentum. It is important to realize that momentum is not a force. However, to give something momentum or to change or take momentum away requires a force. If I apply a small force to an object for a short period of time, it will have a small momentum. If I apply that same force continuously for a long period of time, its velocity and therefore its momentum will increase. This method of applying forces for certain times is known as giving an object an impulse. Impulse – a force applied for a certain period of time. I = Ft I – Impulse [Ns] or [kgm/s] F – Force [N] t – Time [s] If I give something at rest an impulse, it will set it in motion. If I apply an impulse to a moving object, I will speed it up, slow it down, or change its direction. In any case, I will cause a change in the object’s momentum. This led to what is known at the Impulse – Momentum Theorem. (Note that both impulse and momentum have the same units, making them interchangeable). Impulse – Momentum Theorem – any impulse will cause a change in an object’s momentum. This means that: I = Δp = Δmv Let’s take a closer look at this relationship. If: I = Δmv then: Ft = m(vf – vo) Remember, any change is the final value minus the initial value, and mass usually does not change. Now if I divide both sides by t, I get Hey, we have seen that before it is just Newton’s 2nd Law of Motion!!! F = ma So the Impulse – Momentum Theorem is really a restatement of Newton’s 2nd Law of Motion. By the way, don’t forget the importance of vectors and direction in anything involving Impulse, Force, and Momentum. 167 10.4 Conservation of Momentum Law of Conservation of Momentum states that the total momentum of a system of interacting objects always remains constant if no external forces act upon the system. Considering two interacting objects leads to the formula for Conservation of Momentum: m1v1o + m2v2o = m1v1f + m2v2f Each individual object has its own initial momentum before they interact with each other. After the interaction, they each have their own final momentum. The equal sign (=) in that formula is the point where the two objects physically interact with each other. Interactions come in many forms. We will look at when two objects start together and then separate (called a recoil) and when two objects start off apart and come together (called a collision). The Law of Conservation of Momentum and its big and somewhat intimidating equation does not appear out of nowhere. Let’s explore where this equation came from. Imagine two objects, each with their own mass and velocity (and thus their own momenta) that are moving towards each other. v2o
v1o
m1
m2
These two objects hit each other. Now according to Newton’s 3rd Law of Motion (Action‐
Reaction), upon impact they apply equal, yet opposite forces to each other. So as shown below, mass one applies a force on mass two to the right (F1) and mass two applies an equal but opposite (negative) force on mass one to the left (‐F2). F1
-F2
m1
m2
So according to Newton’s 3rd Law of Motion: F1 = ‐ F2 After the collision, the two objects bounce off each other and head away from each other in opposite directions, each with a certain final velocity and thus each with a certain final momenta. 168 v1f
m1
v2f
m2
During the collision, they had to be in contact with each other for the exact same amount of time. Making the following relationship: You can see that this is the formula for Impulse (I = Ft), so F1t = ‐ F2t I1 = ‐I2 they apply equal and opposite impulses to each other. According to the Impulse‐Momentum Theorem, any impulse is equal to a change in the object’s momentum (I = Δmv), so I1 = ‐I2 Δmv1 = ‐Δmv2 If we expand the change part out, final minus initial we get m1v1f – m1v1o = ‐(m2v2f – m2v2o) or, m1v1f – m1v1o = ‐m2v2f + m2v2o Now let’s put the initial terms together on one side and final terms together on the other side and get m1v1o + m2v2o = m1v1f + m2v2f So you can see that the Law of Conservation of Momentum is really just Newton’s 3rd Law of Motion restated. That Newton, he is just everywhere. Now for some examples. No matter how many objects interact with each other, the total momentum of all the objects added together must remain the same. Example: Let’s look at a rifle firing a bullet. A rifle with a mass of 5.0 kg fires a 5.0 g bullet from rest to a velocity of 300 m/s. This causes the rifle to recoil. What is the recoil velocity of the rifle? (The rifle must recoil because of Action – Reaction and Conservation of Momentum). mB = m2 = 5.0 g = 0.005 kg mR = m1 = 5.0 kg v1o = 0 m/s v2o = 0 m/s v1f = ? v2f = 300 m/s The initial velocities of the bullet and rifle are both zero because both were at rest before the point at which the two interact with each other when the gun is then fired. The key to applying the Law of Conservation of Momentum is to consider what happens just before and immediately after objects interact with each other. 169 m1v1o + m2v2o = m1v1f + m2v2f (5)(0) + (0.005)(0) = (5)v1f + (0.005)(300) 0 + 0 = 5v1f + 1.5 ‐1.5 = 5v1f v1f = ‐0.3 m/s It makes sense that the velocity of the rifle is both very small and in the negative direction. As the much smaller bullet fires forward at a high velocity, the much larger rifle recoils in the opposite direction with a much smaller velocity. Example: A man is floating downstream in a boat at 5.0 m/s. He picks up a 10 kg package and throws it forward at 10 m/s. If the man and the boat have a combined total mass of 500 kg, what is the combined velocity after he throws the package? m2 = 10 kg m1 = 500 kg v2o = 5.0 m/s v1o = 5.0 m/s v1f = ? v2f = 10 m/s m1v1o + m2v2o = m1v1f + m2v2f (500)(5) + (10)(5) = (500)v1f + (10)(10) Both package and boat are moving together. 2500 + 50 = 500v1f + 100 2550 = 500v1f +100 2450 = 500v1f v1f = 4.9 m/s The man and the boat still continue forward, but at a slightly slower velocity. 10.5 Introduction to Collisions Momentum is conserved in all interactions of objects. When two objects come together and apply a force to each other, this is known as a collision. The total momentum of the two objects before they collide must equal the total momentum of the two objects after they collide. The collision tends to redistribute the momenta among the objects. However, in any collision, there is usually energy that is lost to heat, sound, light, etc. So energy tends to leave the system during a collision. To sum up, all collisions conserve momentum, but not all collisions conserve energy. Any collision falls into one of three categories: elastic, inelastic, or perfectly inelastic. 10.5.1 Elastic Collision An elastic collision occurs when two or more objects collide and rebound off each other and there is no loss of energy in the collision. The result is that each object may have a different velocity (and thus a different momentum and different KE) after the collision. However, the 170 total momentum and the total energy of all the objects before and after the collision are the same. An elastic collision conserves both momentum and energy. In reality, there can be no true elastic collision since all collisions tend to produce heat and thus lose some energy in the form of heat. A game of pool is a close representation of an elastic collision and a good use of conservation of momentum. In the collision between the balls on the table, very little energy is lost and it is possible to easily transfer energy and momentum from one ball to another. When solving for an elastic collision, we still apply conservation of momentum. Let’s look at some examples: A
v1o
m1
B
v1o
m1
C
m1
m2
v1o
m2
m2
v2o = 0
v2o = 0
v2o = 0
v1f
m1
m1
v1f
m2
v1f = 0
m1
v2f
m2
m2
v2f
v2f = v1o
80 cm
2.0 kg
It slides down and makes a perfectly inelastic collision with a 2.0 kg block at rest. (a) Calculate the potential energy of the 4.0 kg block at the top of the ramp. (b) Determine the velocity of the 4.0 kg block at the bottom of the ramp before it collides with the 2.0 kg block. (c) Calculate the final velocity of the two blocks immediately after the collision. 176 21. During a hockey game, a 100 kg defenseman skating east at 12 m/s applies a viscous head‐on check to a 90 kg player handling the puck who is skating west at 5.0 m/s. The two skaters become stuck together. Determine the magnitude and direction of their final velocity. 22. Given the following graph of a force applied to a 20 kg object for a certain time,
Force (F) vs. Time (t)
F (N)
35
30
25
20
15
10
5
0
0
2
4
6
8
10
12
t (s)
(a) Determine the impulse applied to the object during the 12 s indicated. (b) Determine the change in velocity of the object. 177 Chapter 11 – Circular Motion 11.1 Introduction to Angular Motion Up to this point, we have largely dealt with motion in a straight line (linear motion, momentum, and forces). Even when we discussed projectile motion, we would break up the curved motion into its two linear parts, one in the horizontal x‐direction and the other in the vertical y‐
direction. However, not everything travels in a straight line. In fact, more things travel in some sort of curved motion than in straight line motion. Running around a track, a car rounding a curve, an airplane following the curvature of the Earth, and the motion of all the planets and stars, are all examples of circular motion. Observe two points on a spinning object. One point (A) is located on the outer edge of the object and the other (B) is closer to the center. A
B
B
A
Both points must rotate at the same speed as the object they are resting upon. However, point A travels a longer distance along the circle than point B. If it takes the same time for each point to travel their respective distances and Point A travels a longer distance, then logically Point A must be traveling faster than Point B. So how can two points travel at the same speed and at different speeds at the same time? The answer lies in the types of speed. Any object that is rotating has two velocities at any given moment. One velocity is what we have been studying, linear velocity (v). The second velocity is new and is called an angular velocity (ω). Any circle is divided up into 3600, but what is a “degree”? It could be said that a degree is 1/360th of a circle, and that is correct, but it does not mean that the “degree” is a unit. A circle is also divided up into 2π radians. This comes from the idea that the circumference of a circle is found by multiplying the radius times 2π (C = 2πr). In linear motion, we learned that as you move from one point to another, you have a certain linear displacement (x) relative to where you began. In circular motion, as something rotates from one point to another, it is said to rotate through a certain angle (θ), or to have an angular displacement, given by the Greek letter theta (θ) and measured in a unit called rad (for the radian). Now the “rad” is not a real unit any more than the “degree”, but we have to call it something. 178 11.2 Angular Displacement A
B
xθB
B
xA
A
So as Point A rotates to its position and Point B rotates to its position, they both have the same angular displacement (θ). However, Point A has a greater linear displacement (xA) than Point B (xB). In mathematics, this distance is often referred to as the arc length. If Point A makes one complete rotation, its total linear displacement is zero because it has returned to its starting point. However, its angular displacement is equal to 2π rad. (Note: another more common form of angular displacement is the number of revolutions a rotating object has made. 1 rev = 2π rad). It turns out there is a direct relationship with the linear distance and the angular distance traveled by a point on a circle (the arc length). That relationship is based on the radius of the circle and is given by: x = rθ x – Linear distance [m] r – Radius of circle [m] θ – Angular distance [rad] Example: Let’s say the distance from the center of the circle to Point A is 0.8 m and the distance from the center of the circle to Point B is 0.6 m. The two points make ¼ of a revolution. Determine the angular distance (θA and θB) and the linear distance (xA and xB) of each point. A
B
0.8 m
0.6 m
θ
xB
B
The both have to travel the same angular distance of ¼ of a revolution: θA = θB = ¼(2π) = 1.57 rad. xA
A
r
New path
v
ω
So the motion of the object is changed from angular motion to linear motion. The linear (tangential) velocity (v) came from the angular velocity (ω). Again the relationship between the two velocities is based on v = rω. So any two objects rotating with the same angular velocity (ω) but at different radii will have different linear velocities (v). The greater the radius, the greater the linear velocity. 11.5 Centripetal Force If I spin an object attached to a rope over my head, what keeps the object from flying away? (Answer: the tension (T) on the rope). If a car rounds a turn, what keeps it from sliding off the road? (Answer: the friction (f) between the road and the tires). If you ride the gravitron ride at an amusement park that spins you around so fast that you stick to the side of the wall, what keeps you from flying out of the ride? (Answer: the normal force (FN) of the wall pushing back against you). What keeps the moon orbiting the Earth in a circular path and not flying away? (Answer: the force of gravity (Fg)). In each case, in which direction is each force acting? 181 v
v
v
v
v
v
T
v
f
FN
v
Fg
In all these cases, each force is directed toward the center of its circular motion and perpendicular to its linear (tangential) velocity. In order for an object to travel in a circular or curved path, there must be a force acting on the object directed toward the center of its circular or curved path. This force is called a centripetal force. The word centripetal means “center‐seeking”. The centripetal force (Fc) is always perpendicular to the direction of the linear velocity (v) of the object. There is no circular motion without a centripetal force. You may have heard of a force called centrifugal force. Many people believe that it is centrifugal forces which keep things moving in a circle. The word centrifugal means “center‐fleeing”. In reality, there is no such thing as a centrifugal force; it is not a true force in physics. What people mistake for centrifugal forces is actually inertia. For example, when you are driving down the road and your make a sharp left turn, you feel your body sliding to the right. Most people think that a centrifugal force is pushing your body to the right. In reality, your inertia is still continuing to carry your body forward along the original line of motion based on Newton’s 1st Law of Motion. However, as your body continues to move forward under its own inertia, the right side of the car (or the seat) comes up to meet you. This gives the appearance that you are sliding to the right when in reality it is the side of the car making a left that comes up to meet you. So remember, there is no such thing as a centrifugal force. 182 To determine the centripetal force on an object based on its linear velocity (v): Fc – centripetal force [N] m – mass [kg] v – linear (tangential) velocity [m/s] Let’s look at a couple of examples. First, let’s say I am spinning a 0.5 kg mass on a 2.0 m rope at a rate of 10 m/s. Determine the centripetal force (Fc) acting on the mass. 0.5 10
2
Fc = 25 N Not too hard. Now let’s say I am spinning that same mass on the same rope at 10 rad/s and I want to determine the Fc acting on the mass. This is not a linear velocity (v). We can convert the angular velocity into a linear velocity by the relationship: v = rω If we were to substitute this equation into the equation, it would look like this: (substitute v = rω) (We can cancel one of the r’s on top with the one on the bottom). Fc = mrω2 This is another formula for Fc using angular velocities (ω). So if ω = 10 rad/s, m = 0.5 kg, and r = 2.0 m, then Fc = mrω2 Fc = (0.5)(2.0)(10)2 Fc = 100 N We now have two ways to find Fc based on linear or angular velocities. 183 11.6 Centripetal Acceleration If a centripetal force is a force acting on a mass directed toward the center of its circular motion, then there also exists an acceleration directed toward the center of the circular motion. (This is based on Newton’s 2nd Law of Motion, F = ma). It turns out that if I increase the velocity of an object in circular motion I will increase both its centripetal acceleration and its centripetal force. This leads to the formulas related to centripetal acceleration. Since , if Newton’s second law states F = ma then Fc = mac. Therefore, whatever is not mass (m) must be acceleration (a): ac – Centripetal acceleration [m/s2] v – Linear (tangential) velocity of object at any instance [m/s] r – Radius of the circular path [m] If Fc = mrω2 and Fc = mac, then ac – Centripetal acceleration [m/s2] ω – Angular velocity of object at any instance [rad/s] r – Radius of the circular path [m] This brings up a very important, new concept; an object in circular motion can have a constant velocity (tangential or angular) and now have a non‐zero acceleration, the centripetal acceleration (ac). In fact, all circular motion must have a non‐zero centripetal acceleration. 11.7 Vertical Circular Motion The circular motion we have discussed so far is in the horizontal plane of motion. Let us now focus on circular motion in the vertical plane. What is the difference? In the horizontal plane, gravity acts uniformly on the object in motion at all points as it rotates. So gravity does not speed up or slow down the motion. mg
mg
mg
mg
In the vertical plane, gravity also acts in the downward direction on the object at all times as it rotates. However, as the object moves downward in the vertical circle, it is moving in the direction of gravity. As the object moves upward in the vertical circle, it is moving opposite the direction of gravity. 184 v
v
g
g
v&g
v
g
So as the object moves downward, gravity increases its velocity as they are in the same direction. As the object moves upward, gravity decreases its velocity as they are in the opposite direction. More importantly, let’s look at it in terms of energies and conservation of energy. vtop
TE = PEg + KE
r
h = 2r
vbottom
TE = KE
At the top of a vertical circle, the object has motion and height, so its total energy (TE) consists of both PEg and KE. At the bottom of the vertical circle, the object has motion, but is at the lowest point, so its total energy (TE) is all KE. 2.0 kg 4.0 m/s
TE = PEg + KE 50 cm h = 2r = 1.0 m vbottom TE = KE 185 Example: If an object moving in the vertical circle (as shown above) is attached to a rope with a radius of 50 cm and has a mass of 2.0 kg and has a velocity at the top of the circle of 4.0 m/s, we can determine its velocity at the bottom of the vertical circle by using conservation of energy. m = 2.0 kg TEtop = TEbottom r = 50 cm = 0.5 m (PEg + KE)top = KEbottom vtop = 4.0 m/s (mgh + ½mv2)top = (½mv2)bottom h = 2r = 1.0m (2)(10)(1.0) + (.5)(2)(4)2 = (.5)(2)vb2 36 = vb2 vbottom = ? vbottom = 6 m/s So this indicates that the velocity increases due to gravity, as it should. (Note: we could have done this without mass because mass appears in all parts of the energy equations and would cancel out). Now if the velocities are different at different points on the vertical circle, then the centripetal forces must also be different on each part of the vertical circle. Let’s look at the free body diagram of the object at the top and bottom of the vertical circle. vtop
r
mg
Fc = T + mg
T
T
Fc = T - mg
vbottom
mg
At the top of the vertical circle, both the tension on the rope (T) and the weight of the object (mg) point toward the center of the circle. This means that both contribute to the centripetal force (Fc), thus Fc = T + mg. At the bottom of the vertical circle, the tension (T) points towards the center of the circle and the weight (mg) points away from the center of the circle. This means that one contributes to the centripetal force and the other takes away from the centripetal force, thus Fc = T ‐ mg. If we were to solve these two relationships in terms of the tension force (T), we would have: Ttop = Fc – mg Tbot = Fc + mg 186 However, tension is not the only possible force that could be involved in vertical circular motion; it could be any type of force. So to find the force (F), whatever it may be, at either the top or bottom points on a vertical circle, we will use: Ftop = Fc – mg Fbot = Fc + mg So let’s take our previous example of an object moving in the vertical circle attached to a rope with a radius of 50 cm and has a mass of 2.0 kg and has a velocity at the top of 4.0 m/s. We have already determined that it has a velocity of 6.0 m/s at the bottom of the circle. We want to determine the tension on the rope at (a) the top of the circle, and (b) at the bottom of the circle. 2.0 kg
4.0 m/s
50 cm
6.0 m/s
Fbot = Fc + mg Ftop = Fc – mg m = 2.0 kg Ttop = Fc – mg Tbot = Fc + mg r = 50 cm = 0.5 m Ttop = mv2/r – mg Tbot = mv2/r + mg vtop = 4.0 m/s Ttop = (2)(4)2/.5 – (2)(10) Tbot = (2)(6)2/.5 + (2)(10) vbot = 6.0 m/s Ttop = 64 – 20 Tbot = 144 + 20 TTop = ? Tbot = ? Tbot = 164 N Ttop = 44 N A big difference!! Different example: A pilot decides to take his fighter plane into a dive and then pull upward out of the dive. He is traveling at 200 m/s when he pulls out of the dive into a vertical circle with a radius of 1 km. (a) Determine the force his 80 kg body experiences. (b) How many times the force of gravity on his body (i.e. his weight) is this force? m = 80 kg Fbot = Fc + mg W = mg 2
v = 200 m/s Fbot = mv /r + mg W = (80)(10) 2
r = 1 km = 1000 m Fbot = (80)(200) /(1000) + (80)(10) W = 800 N Fbot = ? Fbot = 4000 N 4000/800 = 5 times 187 This means the pilot feels a force 5 times his weight, or 5 times the force of gravity. This would be known as a +5‐g maneuver because he feels a force 5 times his weight. This is a dangerous maneuver because that means he would have a harder time lifting his arms and maneuvering the aircraft. More importantly, this would also cause the blood in his brain and body to sink downward. This would result in a “blackout” for the pilot due to lose of blood from the brain, rendering the pilot unconscious. To prevent this from happening, fighter pilots wear pressure suits that force pressure to different parts of the body to keep blood from rushing one way or the other. In case you were wondering, the human body can only withstand about a 6‐g maneuver before blacking out. 12‐g can be fatal as it will cause rupturing to the internal organs. However, pressure suits have been designed to help the human body withstand enormous accelerations up to 25‐g (take off speed of the now sadly retired space shuttles). Bet you didn’t know that! 11.8 Critical Speed If I swing something in a vertical circle, there is a certain speed I must maintain in order for the object to completely make it through the top of the circle. If the object moves slower than that speed, it will fall out of the vertical circle. This minimum speed required to maintain a vertical circle is called critical speed. Critical speed can only be determined at the top of a vertical circle. This is because the top is the only point with the lowest possible velocity. It is based on whatever force that is causing the vertical circle dropping to zero. Ftop = Fc – mg 0
Mass cancels out on both sides. rg So going back to my 2.0 kg object tied to a 50 cm rope, I can determine its critical speed. .
. / Any slower and the object will fall out of the circle at the top. 188 Example: I am building a roller coaster loop that has a radius of 20 m. I want to determine the minimum height the hill before the loop has to be for the car to safely make through the loop. TE1 = PEg
vcrit
r
h1
TE2 = PEg + KE
h2 = 2r
I know I need enough height to reach the critical speed at the top of the loop. If I use conservation of energy, my total energy (TE1) is all PEg on top of the hill. At the top of the loop, I will have height, and I better have motion or my riders will fall out of the loop and I will be in big trouble, so my total energy (TE2) at the top of the loop is PEg and KE. TE1 = TE2 (Conservation of energy) (PEg)1 = (PEg + KE)2 masses all cancel out mgh1 = mgh2 + ½mv2 gh1 = gh2 + ½v2 gh1 = g(2r) + ½(
)2 where h2 = 2r and v = gh1 = 2gr + ½gr gravity now cancels out h1 = 2r + ½r h1 = 2.5r h1 = (2.5)(20) h1 = 50 m So we need at least a 50 m hill before the loop for the car to make it through if there is no friction or air resistance. Critical speed can also be a maximum speed if one is on the outside edge of a vertical circle. For example, as a roller coaster car goes over the top of a hill, it can go very slow and still go over just fine. However, if it went too fast, it could leave the track (if it did not have under track wheels). So in this case, critical speed is a maximum value; if you go faster than the critical speed you then leave the circle. This is often what is done on a roller coaster, the car is made to exceed the critical speed of a small hill so the rider feels like he or she is coming out of the seat as they are beginning to leave the circle. Good thing there are lap bars or the riders would fly out of the car. 189 Example: If I have a roller coaster that is going over the top of a 20 m radius hill, the critical speed would be: vcrit = 14.1 m/s Any faster and the car will jump out of the circle at the top. A few important concepts to remember 1. An object in circular motion has two velocities at the same time, an angular velocity and a linear (tangential) velocity. 2. All points on a rotation object have the same angular velocity, regardless of location. 3. As a point on a rotating object gets farther from the center of rotation, its linear velocity increases. 4. An object in circular motion always has an acceleration (centripetal acceleration) toward the center of its motion, even if it is traveling at a constant velocity. 5. An object spinning faster or slower has three accelerations, an angular acceleration (α), a tangential acceleration (at), and a centripetal acceleration (ac). 6. A centripetal force is a force that causes an object to move in a circle and is always directed toward the center of the circle. 7. In a vertical circle the velocity of an object is always highest at the bottom and lowest at the top. 8. In a vertical circle the forces acting on an object are always highest at the bottom and lowest at the top. 9. Critical speed is the speed required for an object to maintain a vertical circle and only occurs at the top of the vertical circle. Questions and Problems 1. Two corks are attached to a string, one at the end and the other halfway up the string. The corks are rotated on the string together. (a) Which cork has the greater angular velocity, the one in the middle or the one on the end? (b) Why? (c) Which cork has the greater linear velocity, the one in the middle or the one on the end? (d) Why? 190 2. Given the following two circles, the radius of the first is 60 cm and the radius of the second is 80 cm. If point A makes 2.5 revs and point B makes 4.0 revs, determine (a) the angular displacement, (b) the linear displacement of each circle. (1 rev = 2π rad) A
r = 60 cm
B
r = 80 cm
θ = 2.5 rev
θ = 4.0 rev
Fg
Fg
m2
r
Note that the force of gravity on the larger object is equal and opposite to the force of gravity on the smaller object, as it should be according to Newton’s 3rd Law of Motion. The Universal Gravitational Constant is an interesting value. Newton recognized that in order for his law to work, there was some mathematical constant that was required that had to be applicable everywhere in the universe. He had no idea what that number was or how to find it. However, he was able to set up proportionalities in his calculations that would divide the unknown constant out, so he could work without it. Nearly a century later, Lord Henry Cavendish was able to accurately determine the Universal Gravitational Constant (G) using a 194 devise called a torsional balance and a beam of light. His measurement was so accurate at the time that it was not improved on for another 100 years. Cavendish was a rich eccentric who made many scientific discoveries, most of which were not known until after he died. He preferred to discover things for his own knowledge and did not care about sharing them with the rest of the world. Example: Let’s look at Newton’s Law of Gravity using the legend of the apple and the Earth. Let’s say that an apple has a mass of 0.5 kg. The mass of the Earth is 5.98 x 1024 kg. The distance used to calculate gravity is the distance between the center of each mass, so from the center of the apple to the center of the Earth. The distance from the surface of the Earth to its center is 6.38 x 106 m (so an extra meter or so above the surface that the apple may be is irrelevant). 6.67 10
0.5 5.98 10
6.38 10
Fg = 4.92 N (Force at which the Earth pulls the apple and the apple pulls the Earth). Using the idea of Newton’s 2nd Law of Motion (F = ma) we can determine the acceleration of each mass due to the force each applies to the other. First the Earth: FE = mEaE 4.92 = (5.98 x 1024)aE aE = 8.23 x 10‐25 m/s2 (Very small, and that is a good thing). Now the apple: FA = mAaA 4.92 = (0.5)aA aA = 9.8 m/s2 Hey, that is gravity on the surface of the Earth! Go figure! So every time something is pulled downward by the Earth, the Earth is accelerated upward, but at such a small amount that it is not noticeable. 195 12.3 Gravitational Field Strength (aka What You Earthlings Call Gravity) Using this idea of gravitational force, we can determine the acceleration of gravity at any point above the surface of the Earth or any astronomical object. Fg = mg In this case, m1 and m are both the same mass, such as the apple. Also, m2 is the mass of the astronomical body such as the Earth. m and m1 cancels out since they are the same mass. m2 – Mass of the astronomical body such as Earth [kg] r – Distance from the center of the body to any point [m] Example: We can determine gravity on the moon. Mass of the moon is 7.35 x 1022 kg and the radius of the moon is 1.74 x 106 m. To determine gravity on the moon: 6.67 10
7.35 10
1.74 10
.
.
vorbit = 7453. 3 m/s 12.6 Introduction to Kepler’s Laws Up until the 16th century and even for a while afterward, the Earth was considered to be not only the center of our Solar System, but the center of the entire universe. This idea goes back to Pythagoras, Plato, and Aristotle. It was first published under what was called the Geocentric Theory (Earth‐centered) in the year 2 C.E. by a Greek astronomer Claudius Ptolemy. Ptolemy is given the credit for this idea because he first published it, even though it came hundreds of years after Pythagoras. This continued to be the accepted theory for the next 1400 years. So for all this time, people believed that the Sun, moon, all the planets, and even the stars orbited around the Earth. In 1543, an astronomer named Nicolas Copernicus published the first Heliocentric Theory (Sun‐
centered), which said that the planets, including the Earth, orbited around the Sun. Copernicus was a very religious man and was tormented by the fact that what he was writing went against the accepted teachings of the church. Fortunately for Copernicus, he died the same year his works were published, though not before receiving large criticisms from the church. Other scientists began to embrace and teach the Copernican model against the wishes of the church. Giordano Bruno was actually burned at the stake by the Holy Inquisition for teaching the fact that the Earth revolved around the Sun. Galileo was also brought before the Inquisition for teaching the Copernican system and was forced to recant his teachings in public. He was saved by being good friends with an intellectual Cardinal that would often meet with Galileo to discuss science and religion. Galileo, however, was later excommunicated and placed under house arrest for the remainder of his life by the church. This was largely due to a technicality in religious law that the Pope and the Cardinals who did not approve of Galileo used to arrest him. In the late 20th century, the church reopened Galileo’s case and proposed lifting his 200 excommunication. In the end, they admitted that the arrest had been a mistake but did not do much more than that. Anyway, after Copernicus came a wealthy nobleman named Tycho Brahe. Tycho loved astronomy and put together the most comprehensive astronomical catalog known to that day. Tycho, being a wealthy patron, took in a young apprentice named Johannes Kepler. Kepler learned quickly and advanced Tycho’s work more than any other person. He then went on to formulate his own laws governing planetary motion. These three laws became known as Kepler’s Laws of Planetary Motion. 12.6.1 Kepler’s 1st Law of Planetary Motion – Law of Elliptical Orbits All planets (and other bodies) move in elliptical orbits with the Sun at one of the focal points of the orbit. Any ellipse has two focal points. The Sun exists at one focal point of a planet’s orbit, the other is just space. aphelion
.
.
perihelion
Planet
Sun
The point in a planet’s orbit when it is closest to the Sun is called Perihelion. The point where it is farthest from the Sun is called aphelion. These points were a significant discovery because it was previously thought that all orbits were perfect circles. This helped to explain some of the unusual observations of planets like Mars. 12.6.2 Kepler’s 2nd Law of Planetary Motion – Law of Equal Areas A line drawn from the Sun to any planet sweeps out equal areas in equal time intervals. Planet
t1
Sun
A1
A2
t2
If t1 = t2 then A1 = A2
When a planet is close to the Sun it will move faster in its orbit than when it is farther away. In the time it takes the planet to travel a certain distance (t1) it will sweep out an area of space 201 (A1). Later in its orbit, if we wait a certain time (t2) for it to travel a certain distance, it will sweep out another area (A2). If the time periods are the same (t1 = t2) then the areas must also be equal (A1 = A2). This also follows the principle of the Law of Conservation of Angular Momentum. 12.6.3 Kepler’s 3rd Law of Planetary Motion – Law of Harmony Kepler’s 3rd law was a method of determining the time it takes one object to orbit around another. It turns out that the object that is doing the orbiting is irrelevant. For example, any object that is the same distance from our Sun as the Earth would take the same Earth year to orbit around the Sun. Also, any object the same distance from the Earth as the moon would take the same time to orbit the Earth as the moon does. This became very important when we started placing satellites in space. We need to know the exact distance for a satellite to be from the Earth to orbit the Earth a specific number of times each day. There is one specific distance at which a satellite will orbit the Earth at the same rate at which the Earth rotates. Thus the satellite remains in the same position above the same point on the Earth at all times. This orbit is called geosynchronous orbit. Kepler’s 3rd Law is actually just an equation for period (T), the time to complete one orbit. Determining the period of an orbit, assumed to be both constant and circular for this argument, is based on its orbital velocity (vorbit). If we assume a constant velocity, then x = vot + ½ at2 (a = 0 m/s2 due to constant velocity) x = vt The distance (x) would equal the circumference of the orbit (2πr) with r being the radius of its orbit. The time (t) would be the period (T). This would give you 2
Solving for period (T) you get, 202 Using the previous example of an orbiting satellite 800 km above the surface of the Earth (mE = 5.98 x 1024 kg, rE = 6.38 x 106 m), we can determine the period of its orbit. (We already found out that vorbit = 7453.3 m/s and r = 7.18 x 106 m). 2
7.18 10
7453.3
T = 6052.8 s (1.68 hrs) A few important concepts to remember 1. The four fundamental forces in the universe in order of strongest to weakest are the strong nuclear force, the electrostatic force, the weak nuclear force, and the gravitational force. 2. A single mass will create a gravitational field that extends from the mass and gets weaker with distance. 3. The gravitational force requires two masses and increases as mass increases and decreases as distance increases. 4. Escape velocity is the minimum required velocity for an object to break away from a planet or other body’s gravitational pull. 5. Orbital velocity is the speed at which one object orbits another. 6. Kepler’s 1st Law refers to the elliptical nature of a planet’s orbit. 7. Kepler’s 2nd Law, the law of equal areas, relates the motion of a planet at different points in its orbit around the Sun. 8. Kepler’s 3rd Law is used to determine the period of the orbit of one object around another. 203 Questions and Problems NOTE: There is a table of astronomical data at the end of this problem section. 1. (a) As the mass of an object is doubled, what happens to the gravitational force it exerts on another mass? (b) As the distance separating two objects is tripled, what happens to the gravitational force between them? 2. Describe Kepler’s 1st Law of Planetary Motion. 3. Describe Kepler’s 2nd Law of Planetary Motion. 4. Who first discovered the actual value of the Universal Gravitational Constant? 5. (a) Describe the Geocentric Theory. (b) Who first published it? (c) Describe the Heliocentric Theory. (d) Who first published this theory? 6. (a) At what point in a planet’s orbit around the Sun is it traveling the fastest? (b) What is the name of this location? (c) At what point in a planet’s orbit around the Sun is it traveling the slowest? (d) What is the name of this location? 7. A 50 kg student is sitting 2.0 m from a 60 kg student. Calculate the gravitational force attracting them. 8. (a) Determine the acceleration of gravity (g‐field) at the surface of the moon. (b) Determine the acceleration of gravity at the surface of Mars. (c) Calculate the weight of a 70 kg person standing on the surface of the Earth, the moon, and Mars. 9. A 2,000 kg satellite is at an orbit of 600 km above the surface of the Earth. Calculate the gravitational force between the Earth and the satellite. 10. Titan, the largest moon of Saturn and one of the few with an actual atmosphere, has a mass of 1.35 x 1023 kg and a radius of 2.58 x 106 m (making it larger than our own moon, Pluto, and nearly as large as Mercury). In 2005, the 300 kg Huygens probe was dropped on the surface of Titan where it sent back pictures of this methane clouded moon. (a) Calculate the acceleration of gravity at the surface of Titan. (b) Calculate the weight of the probe on the surface of Earth and on the surface of Titan. 11. The International Space Station (ISS) orbits at a distance of 350 km above the surface of the Earth. (a) Determine the gravitational field that astronauts onboard the ISS would experience. (b) Calculate the orbital velocity of the ISS. (c) Determine the period of the ISS orbit around Earth. 204 12. A neutron star has a mass equal to that of the Sun, but a radius equal to that of the Earth. Calculate the gravitational field at the surface of a neutron star. (Yes, it is that big!) 13. The New Horizons spacecraft was launched on January 19, 2006. It will reach Pluto around July 14, 2014 (only 8 years, it took the Voyager crafts about 12 years to go that far). Upon reaching Pluto it will orbit within 10,000 km of this dwarf planet. (a) Calculate its orbital velocity around Pluto. (b) Determine its orbital period around Pluto at this same distance. 14. The 600 kg Messenger spacecraft settled into orbit around Mercury in March, 2011 at a distance of 200 km about the surface of the planet. (a) Calculate the gravitational force of Mercury on the Messenger spacecraft while in orbit. (b) Determine the orbital velocity of the Messenger spacecraft around Mercury. (c) Calculate the orbital period of Messenger around Mercury. Solar System Data Name Mass (kg) Average Radius (m) Average Distance from Sun (m) 23
6
Mercury 3.30 x 10
2.44 x 10
5.79 x 1010
Venus 4.87 x 1024
6.05 x 106
1.08 x 1011
Earth 5.98 x 1024
6.38 x 106
1.496 x 1011
Mars 6.42 x 1023
3.40 x 106
2.28 x 1011
27
7
Jupiter 1.90 x 10
7.15 x 10
7.78 x 1011
Saturn 5.69 x 1026
6.03 x 107
1.43 x 1012
Uranus 8.68 x 1025
2.56 x 107
2.87 x 1012
Neptune 1.02 x 1026
2.48 x 107
4.50 x 1012
Moon 7.40 x 1022
1.70 x 106
‐‐‐
30
8
Sun 1.99 x10
6.96 x 10 ‐‐‐ 22
6
Pluto (dwarf planet) 1.25 x 10
1.20 x 10
5.91 x 1012
r
r
Axis of rotation
F
Axis of rotation
No matter what the scenario, though, to achieve the highest torque, the angle (θ) between the force and the line to the axis of rotation needs to be 90o. Anything lower will result in a lower amount of torque. For example, let’s look at the best way to close a door. The axis of rotation of a door is located at the hinges. The hinges are what allow the door to rotate open or closed. In the three pictures below we apply a force (a) at hinges, then (b) halfway down the door from the hinges, (c) at the farthest point from the hinges, and finally (d) at an angle other than 90o. 900
900
300
900
(a)
(b)
(c)
(d)
Examples: Let’s say that in each case, the force applied is 10 N. The distances (r) for each part will be (a) 2.0 cm, (b) 60 cm, (c) 1.2 m, and (d) 1.2 m along with the given angles. Determine the torque for each case. (a)
(b) (c) (d) 10 0.02 sin 90 10 0.6 sin 90 10 1.2 sin 90 10 1.2 sin 30 = 0.02 Nm = 6.0 Nm = 12.0 Nm = 6.0 Nm Notice that as you get farther from the hinges you can apply more torque and close the door more easily. However note that applying your force at a 30o angle at the end of the door produced the same toque as applying a 90o force at the middle of the door, so the angle is also important. Next time you watch the movie Jurassic Park, watch the scene when the velociraptor is trying to get into the control room. The one guy is pushing hard against one end of the door when the woman comes to help by pressing her back against the door. He tells her to try to reach the gun, but she says she cannot because she would have to move away from the door. Yet if you look carefully, she is pushing very near the hinges, thus providing little to no help at all in 207 keeping the door closed, so she may as well just have gone for the gun. Ah, movie makers and their lack of proper physics applications. 13.3 Torque and Equilibrium Most of us have been on a seesaw at some point in our lives. One person sits at each end of the seesaw and goes up and down. A little trick we sometimes try to do is to balance the seesaw perfectly horizontal. In some cases we can do it, but in others we cannot. When the seesaw is balanced, it is said to be in equilibrium. As a reminder, the rule for equilibrium is that the sum of all forces acting on an object in equilibrium must be equal to zero (and thus no acceleration). We all remember: ∑
This still applies to rotational mechanics except there is a second condition required for equilibrium, and that is that the sum of all torques acting on an object must also equal zero, or: ∑
Example: Let’s say we have a 50 kg seesaw that is balanced in the middle where it can pivot on a triangular support. A 20 kg crate is placed 80 cm from the pivot point, as shown below. 80 cm
20 kg
50 kg
m1
m3
m1 = 20 kg
m2 = 30 kg
m3 = 50 kg
Weight (mg) will act downward on each object and there will be a normal force (FN) acting upward on the center of the seesaw from its contact with the triangular support. m1g
r1
m2g
m3g
r2
m1 = 20 kg
m2 = 30 kg
m3 = 50 kg
FN
r1 = 80 cm
Note that both FN and m3g act right at the center of mass of the seesaw which is also the location of the pivot point for the seesaw. This means that neither FN nor m3g provide any torque as the distance between each force and the axis of rotation is zero. So the only torques provided are by m1g and m2g. So to determine where to place m2 (in other words to determine r2) we apply the rule for equilibrium: ∑
If we look at these two forces and their relationship to the pivot point, we see that m1g will cause the seesaw to rotate counterclockwise around the pivot point and m2g will cause the seesaw to rotate clockwise around the pivot point. As we often found it useful to define positive and negative directions in force analysis, we need to do the same thing for torques. So we need to decide which direction of rotation (clockwise or counterclockwise) should be considered a positive torque and which should be negative. In the end, it really does not matter, but I always like to consider all counterclockwise torques to be positive and all clockwise torques to be negative. 209 Keeping in mind that the formula for torque is τ = Frsinθ and that all of the angles given here are 90o, the two torques are: Counterclockwise m1g
torque
r1
m2g
m3g
Clockwise torque
r2
-τ2
τ1
FN
FN and m3g provide no torque as they act
on the center of mass of the system
τ1 = m1gr1 and τ2 = ‐m2gr2 (Note that τ2 is negative because it is a clockwise torque.) Now we can apply ∑
. ∑
∑
0 ∑
0 What this really means, and is probably pretty obvious is that m1gr1 = m2gr2 (Gravity can cancel out if you want.) (20)(10)(0.8) = (30)(10)r2 r2 = 0.533 m So we need to put our 30 kg crate 53.3 cm from the pivot point to balance the seesaw. Now to determine the normal force (FN) we apply the other requirement for equilibrium: ∑
(We must consider all the forces involved, the distance do not matter.) m1g
m3g
m2g
m1 = 20 kg
m2 = 30 kg
m3 = 50 kg
FN
∑
r1 = 80 cm
Remember, it has to hold the seesaw, too. 210 Therefore, FN = m1g + m2g + m3g FN = (20)(10) + (30)(10) + (50)(10) FN = 1000 N Now I know some of you are thinking, “I could have figured all of this out without all that work!”, but what if there are more than two objects? Let’s look at a couple of different situations. Example 1: A 40 kg crate is placed on a beam 20 cm from the pivot point. Another 20 kg crate is placed 60 cm from the pivot point and on the same side as the 40 kg crate. Determine where a 50 kg crate must be placed (r3) to balance the beam. r1
r2
m2
m1 = 40 kg
r3
m1
r1 = 20 cm
m3
m2 = 20 kg
r2 = 60 cm
m3 = 50 kg
r2
m1g
r1
r3
m1 = 40 kg
m3g
r1 = 20 cm
m2 = 20 kg
mBg
τ2
τ1
FN
-τ3
r2 = 60 cm
m3 = 50 kg
As we are only concerned about balancing the beam, we can ignore the weight of the beam itself (mBg) and the normal force (FN) because they occur at the pivot point. ∑
∑
0 ∑
0 (Both m1g and m2g are counterclockwise torques.) m1gr1 + m2gr2 = m3gr3 (40)(10)(0.2) + (20)(10)(0.6) = (50)(10)r3 r3 = 0.4 m 211 Example 2: A 60 kg crate is placed on a beam 40 cm from the pivot point. Another 20 kg crate is placed 80 cm from the pivot point and on the same side as the 60 kg crate. Determine how much mass (m3) must be placed 50 cm on the other side to balance the beam. r2
m2
m1 = 60 kg
r1
r3
m1
r1 = 40 cm
m3
m2 = 20 kg
r2 = 80 cm
r3 = 50 cm
This is really no different than before, there is just a different unknown, so the free‐body diagram looks the same. m2g
r2
m1g
r1
r3
m1 = 60 kg
m3g
r1 = 40 cm
mBg
τ2
τ1
FN
m2 = 20 kg
-τ3
r2 = 80 cm
r3 = 50 cm
Again, we are only concerned about balancing the beam, so we can ignore the weight of the beam itself (mBg) and the normal force (FN) because they occur at the pivot point. ∑
∑
0 ∑
0 (Both m1g and m2g are counterclockwise torques.) m1gr1 + m2gr2 = m3gr3 (60)(10)(0.4) + (20)(10)(0.8) = m3(10)(0.5) m3 = 80 kg So again, this is not so bad. Let’s look at one more, slightly tricky example. 212 Example 3: A 50 kg crate is placed 40 cm on the left side of a beam from the pivot point. Then a 60 kg crate is placed 20 cm to the right of the pivot point. Determine where (r3) to place a 30 kg crate to balance the beam. The tricky part here is that we do not know on which side to place the third crate. So right now all we have is: m1 = 50 kg
r1
m3
r2
m1
r1 = 40 cm
m2
m2 = 60 kg
r2 = 20 cm
m3 = 30 kg
r3 = ?
If we just sum the torques of the known values we have: ∑
0 m1gr1 = m2gr2 (50)(10)(0.4) = (60)(10)(0.2) 200 = 120 This means to balance out, we need another 80 Nm of torque on the right side, or the side of m2. 80 = (30)(10)r3 r3 = 0.267 m So the 30 kg crate needs to be placed 26.7 cm from the pivot point on the right side or the same side as the 60 kg crate. 13.4 Torque and Rotational Acceleration As with forces, it would be nice if everything was always in equilibrium, but that is not the way the universe works. Of course that is a good thing, otherwise the universe would not be what it is. Anyway, back in Chapter 7 on Forces, we discovered that when an object experienced a net force other than zero, it accelerated. In rotational mechanics, if an object experiences a net torque other than zero, it will experience an angular acceleration. For example, if I have a bicycle wheel with a certain mass (m) at rest that is free to rotate around its center axis (r) and I apply a force (F) to the outer edge of the wheel, the wheel will begin to rotate with an angular velocity. As it started from rest and now has an angular velocity, it must have experienced an angular acceleration (α). 213 F
α
r
m
r
m
If we look at this from the perspective of Newton’s 2nd Law (F = ma) we can get a picture of how this relates torque and angular acceleration. F = ma where a is the tangential acceleration (at) as the force and acceleration act tangent to the circle. F = mat if I multiply both sides by the radius (r), then Fr = mrat from rotational motion we know that at = rα so we can substitute in and get Fr = mr(rα) Fr = mr2α and since τ = Frsinθ and the angle here is 90o, this leaves us with 2
τ = mr α So there is a relationship between torque and angular acceleration. However, that mr2 relationship within that equation is rather important. This is known as the moment of inertia of the wheel. 214 13.5 Moment of Inertia We have seen the word “inertia” before as a tendency for an object to remain at rest or in motion. In rotational mechanics, the moment of inertia is known as an object’s resistance to rotation. The greater the moment of inertia of an object, the more it will resist rotation. The moment of inertia (I) of an object is based on its mass and its shape. This gives each shape a specific equation for its moment of inertia. Moments of inertia for common, uniform shapes m
Thin Hoop
r
I = mr2
Solid Disk or Cylinder
r
m
Solid Sphere
m
r
Long Thin Rod Rotating Around Center
m
l
Long Thin Rod Rotating Around End
m
l
215 Looking at these shapes, a thin hoop has a higher resistance to rotation than a solid disk if they had the same mass and radius. A long, thin rod of a certain mass (m) and length (l) spinning around its center has a very low resistance to rotation, especially compared to an equal rod in terms of mass and length that is rotating around one end of the rod. The more mass distributed away from the axis of rotation, the greater the resistance to rotation and thus the greater the moment of inertica. Moment of inertia is measured in kgm2. I [kgm2] As mentioned before, a relationship between torque and angular acceleration is: τ = mr2α That relationship was applied to a bicycle wheel which is basically a thin hoop. Given that the moment of inertia of a thin hoop is given by I = mr2, we have a new relationship between torque and angular acceleration. τ = Iα In this relationship, the angular acceleration (α) is directly proportional to the amount of torque (τ) applied and inversely proportional to the moment or inertia (I) of the object. For example, if I applied the same torque to a thin hoop (I = mr2) as I did to a solid disk (I = ½mr2), as the solid disk has half the moment of inertia as the thin hoop, the solid disk will have twice the angular acceleration as the thin hoop. Example: Given a thin hoop with a mass (m) of 2.0 kg and a radius (r) of 40 cm and a solid disk with the same mass and radius, I apply an 8.0 N force (F) to each one at the outer edge and tangent to each one. (a) Determine the torque (τ) applied to each shape. (b) Determine the moment of inertia (I) of each shape. (c) Determine the angular acceleration (α) of each shape. Thin Hoop Solid Disk (a)
τ = (8.0)(0.4)sin 90 τ = (8.0)(0.4)sin 90 τ = 3.2 Nm τ = 3.2 Nm They experience the exact same torque. (b) I = mr2 I = ½mr2 I = (2.0)(0.4)2 I = ½(2.0)(0.4)2 2
I = 0.32 kgm I = 0.16 kgm2 The thin hoop has a higher moment of inertia. (c) τ = Iα τ = Iα 3.2 = (0.32)α 3.2 = (0.16)α 2
α = 10 rad/s α = 20 rad/s2 The solid disk has a higher angular acceleration. 216 13.6 Rotational Kinetic Energy When any object is given motion, it automatically has kinetic energy due to that motion. The kinetic energy associated with this motion is called translational kinetic energy. Translational kinetic energy means that the energy is moving in one direction. m
v
KET
However, when an object spins, it is moving in a circular direction. As this is still motion, the spinning object must have kinetic energy, even if it is spinning in place. This spinning energy is called rotational kinetic energy and is related to the kinetic energy we have been using so far. Let’s start with the idea of a bicycle wheel (thin hoop) that is spinning around its center axis (r). The wheel has a certain mass (m) and any point on the end of the wheel has a certain constant linear (tangential) velocity (v). v
ω
r
m
r
m
KER
If it is spinning, it must also have a constant angular velocity (ω). If we look at kinetic energy, We can relate v to ω by v = rω and substitute, This gives us, If we recall, the moment of inertia for a thin hoop is I = mr2, so This is the equation for rotational kinetic energy. As with most of rotational mechanics, the moment of inertia of an object is important to the amount of rotational kinetic energy it has. So if an object is actually rolling down the street, it has two types of kinetic energy, translational kinetic energy (the one we are used to) and rotational kinetic energy. So for any rolling object, KE = KET + KER 217 ω
r
v
m
KET + KER
Example: Given a solid disk with mass (m) of 4.0 kg and radius (r) of 20 cm that is rolling down the hall with a linear velocity (v) of 5.0 m/s, (a) determine its translational kinetic energy, (b) determine its rotational kinetic energy, and (c) determine the total kinetic energy of the disk. (a)
4.0 5.0 KET = 50 J (b)
For this part, we need both I and ω. Moment of inertia for a solid disk. 4.0 0.2 I = 0.08 kgm2 v = rω 5.0 = (0.2)ω ω = 25 rad/s Relationship between v and ω. Now, 1
0.08 25 2
KER = 25 J (c) KE = KET + KER KE = 50 + 25 KE = 75 J Rotational KE is measured in Joules like all energy. 218 Notice that by rolling, the disk has an extra 25 J of energy that it would not have if it were just sliding. 13.7 Angular Momentum We have looked at rotational motion, rotational forces, and rotational energy. That leaves the idea of momentum. Just as an object with motion will have both kinetic energy and momentum, a rotating object will have rotational kinetic energy and rotational or angular momentum (L). I am not sure why L is used to represent angular momentum, but then I did not know why linear momentum is represented with a p. The relationship for angular momentum is: L = Iω L – Angular momentum [kgm2/s] I – Moment of inertia [kgm2] ω – Angular velocity [rad/s] Example: Given a solid sphere with mass (m) of 2.0 kg and radius (r) of 10 cm that is rolling down the hall with a linear velocity (v) of 5.0 m/s, determine its angular momentum (L). L = Iω To solve this we need both I and ω. Moment of inertia for a solid sphere. 2.0 0.1 I = 0.008 kgm2 v = rω 5.0 = (0.1)ω ω = 50 rad/s L = Iω L = (0.008)(50) L = 0.4 kgm2/s Relationship between v and ω. 219 13.8 Conservation of Angular Momentum If you have ever watched a figure skater go into a spin, she will start spinning slower and then with very little apparent effort will suddenly spin faster and faster. If you watch carefully, you will notice that when she begins her spin, her arms and legs are stretched far from her body. She then brings her arms and legs closer to her body and she begins to spin faster. The closer her arms and legs come to her body, the faster she spins. Then if she extends her arms and legs away from her body, her spinning slows down. Why does this happen? Well, the skater has a set mass, so we can’t change that. When she begins her spin, she has a certain moment of inertia (I) dictated by her shape and a certain angular velocity (ω). By bringing in her arms and legs, she is changing her shape or radius (r) and thus changing her moment of inertia (her resistance to rotation). This in turn changes her angular velocity. As her radius decreases, her moment of inertia decreases, but her angular velocity increases. In addition, to stop herself she extends her arms back out, increasing her radius, increasing her moment of inertia, and decreasing her angular velocity. The reason is that in her spin, she must obey the Law of Conservation of Angular Momentum which states that the total angular momentum of a system always remains constant so long as no external torques act on the system. Therefore: Lo = Lf Example: Let’s consider the figure skater to be a long, thin rod rotating around its center axis (some of these men and women skaters look like this anyway). Given that the skater has a mass (m) of 50 kg and a length (l) of 1.6 m with her arms stretched out. She begins spinning at an angular velocity (ω) of 10 rad/s. She then pulls her arms into her side thus reducing her length to 0.5 m. (a) Determine her initial and final moments of inertia, and (b) determine her final angular velocity. (a)
Long rod rotating around center axis. 50 1.6 50 0.5 2
Io = 10.67 kgm If = 1.042 kgm2 (b) Apply conservation of angular momentum L = Iω Lo = Lf Iωo = Iωf (10.67)(10) = (1.042)ωf ωf = 102.4 rad/s Certainly spinning much faster. 220 A few important concepts to remember 1. A torque is a kind of turning force related to the amount of force applied, the distance from the applied force and the axis of rotation, and the angle between the applied force and the line to the axis of rotation. 2. For a system to be in equilibrium, ∑
and ∑
on the system. 3. The center of mass of a system is the point where all the mass of an object is concentrated and where all forces are considered to act upon. 4. The moment of inertia of an object is its resistance to rotation and is based on its mass and shape. The more mass distributed away from the axis of rotation, the greater the resistance to rotation and the greater the moment of inertia. 5. The angular acceleration of an object is directly proportional to the amount of torque applied to the object (if torque increases, angular acceleration increases) and inversely proportional to the moment of inertia of the object (if the moment of inertia increases, angular acceleration decreases). 6. A rolling object will have both translational kinetic energy and rotational kinetic energy. 7. Rotational kinetic energy and angular momentum are both based on the moment of inertia of an object and its angular velocity. 8. The Law of Conservation of Angular Momentum states that the total angular momentum of a system always remains constant so long as no external torques act on the system. Questions and Problems 1. What are the two conditions for an object to be in equilibrium? 2. (a) At what angle must a force be applied to yield the maximum torque? (b) At what angle must a force be applied to yield no torque? 3. Define center of mass. 4. A student applies 15 N of force to a door at a point 80 cm from the hinges. If the force is applied perpendicular to the door, calculate the amount of torque he applies to the door. 221 5. A 40 kg student needs to change a flat tire. She has a 60 cm long tire iron. Calculate the amount of torque she can apply to the lug nuts if she stands perpendicularly on the end of the tire iron. 6. A worker is capable of applying 30 N of force to a 15 cm long wrench. Determine the torque she can apply to a bolt at an angle of (a) 300, (b) 450, (c) 600, and (d) 900. 7. Given the picture below, a 4.0 kg block is placed on a 20 kg board at a point 90 cm from the balance point. 4.0 kg
90 cm 20 kg
(a) Determine where a 9.0 kg block must be placed on the other side for the system to be in equilibrium. (b) Determine the normal force provided by the balancing point for the system to be in equilibrium once the 9.0 kg block is placed on the balance. 8. Given the picture below, a 60 kg crate is placed on a 100 kg board at a point 50 cm from the balance point. A 40 kg crate is placed on the same side, 70 cm from the balance point. 40 kg 60 kg 50 cm 100 kg 70 cm (a) Determine the necessary mass of a crate that must be placed 40 cm on the other side for the system to be in equilibrium. (b) Once this crate is added, determine the force provided by the support to hold up the system. 222 9. Given the picture below, a 50 kg crate is placed on a board at a point 30 cm from the balance point. A 40 kg crate is placed 60 cm from the balance point on the other side. 50 kg 40 kg
30 cm 60 cm
Electron
Cloud
p
n
e
It was once believed that all these particles mingled together like an alphabet soup. Later, the theory of the atom was revised with the protons and neutrons in the nucleus and the electrons in orbits around the nucleus much like the way that planets orbit a star. It is now understood that the electrons move in what is known as an electron cloud around the nucleus. This configuration is essential to not only keeping an atom stable, but allowing it to combine with other atoms or to become an ion (an atom that is charged, not neutral). 225 Each one of these particles has a physical mass. The proton and the neutron have the larger masses (mp = 1.673 x 10‐27 kg, mn = 1.675 x 10‐27 kg) while the electron has a much smaller mass (me = 9.11 x 10‐31 kg). Thus the mass of an individual atom (called the atomic mass) is based on the number of protons and neutrons in its nucleus. In addition, each of these particles has what is known as an electrical charge (q), measured in a unit called the Coulomb [C], which is named for Charles Coulomb who established a law of electrical forces. Originally electrical charges were referred to as positive (+) and negative (‐) (or neutral (0) for no charge). This would make the atom look like this: Electron
Cloud
Nucleus
0
+
+
0
-
These terms (positive and negative) were first coined by Ben Franklin in his studies of electricity. Later came an understanding of what an electric charge was, how it worked and then actual values were determined for the charge of these particles: qp = 1.60 x 10‐19 C qe = ‐ 1.6 x 10‐19 C qn = 0 C Notice that the proton and electron have the same charge, just opposite signs; the neutron is electrically neutral and has no charge. (The charge of an electron was first determined by Robert Millikan in 1909). In addition, the mass of a neutron is equal to the mass of a proton plus the mass of an electron (and the mass of another particle called an antineutrino. Neutrons can decay or break up into these other particles. 14.3 Conservation of Charge Objects in nature are typically electrically neutral, meaning they have an equal balance of protons and electrons. In order for something to become electrically charged, it must have an excess of electrons or an excess of protons. However, protons do not move from one place to another as they are fixed in the nucleus. A charged object is generally due to the flow of electrons. So a negatively charged object has an excess of electrons and a positively charged 226 object has a deficiency of electrons. When two objects are rubbed together, like the balloon against someone’s head, electrons move off the person’s hair (making it positively changed) and flow on the balloon (making it negatively charged). This method of charging is called charging by friction. The number of electrons leaving the person’s hair is equal to the number of electrons that flow onto the balloon. Law of Conservation of Charge – the total charge for a system always remains constant (electrons lost = electrons gained). The reason that the balloon will now stick to the wall has to do with another property of electric charges. Opposite charges attract and like charges repel. The wall is electrically neutral and the balloon is negatively charged. When the balloon is brought near the wall, the positive charges within the wall are attracted towards the balloon and the negative charges within the wall are repelled farther into the wall away from the balloon. +
+
+
+
-
+
-
- -
-
+
- + - - -
+
+
+
- - - +
- +
- -
+
-
This is one of the primary reasons why the electromagnetic force is considered to be more powerful than the gravitational force. If you consider the balloon in the above example, let’s look at the free body diagram on the balloon. Since the balloon has mass, gravity (mg) must act on it pulling it downward. In addition, there must be a normal force (FN) acting to the right as the balloon and the wall are two surfaces in contact. FN
mg
However, the balloon does not move to the right nor does it fall downward, so what are the other forces? The upward force acting against gravity in this case would be static friction (fs) which prevents the balloon from moving downward. That now leaves the force acting to the left. This is the electrostatic force (FE). 227 fs
FN
FE
mg
14.4 Electric Force The attraction of the opposite charges holds the balloon to the wall with a physical force strong enough to overcome the pull of gravity on the balloon. This force must be due to the electrical interaction of those very small positive and negative charges. In 1785, Charles Coulomb established a law of electrical forces between charged objects. Coulomb’s Law – The electrical force between two charged objects is directly proportional to the product of their charge and inversely proportional to the distance separating the charges. This law sounds a lot like Newton’s Law of Gravitation, and in fact the equation for electric force looks a lot like the equation for gravitational force. FE – Electric force [N] q – Charge [C] r – distance separating charges [m] k = 9.0 x 109 Nm2/C2 (Coulomb’s constant) We have discussed how the gravitational force is actually weaker than the electrical force, so let’s look at an example of both. Two protons are placed a distance of 2.0 m apart, calculate the electrical force and the gravitational force between the two protons. mp = 1.673 x 10‐27 kg qp = 1.60 x 10‐19 C 9.0 x 10 1.60 x 10
1.60 x 10
2
FE = 5.76 x 10‐29 N 228 6.67 x 10
1.673 x 10
2
1.673 x 10
Fg = 4.67 x 10‐65 N As you can see, there is quite a big difference in forces. The electric force is 1.23 x 1036 times stronger than the gravitational force. In fact, this force is so strong that the repulsive force of like charges at the atomic level is what really keeps you from falling through the ground to the center of the Earth as it is far more powerful than Earth’s gravity. In Chemistry, you will learn about chemical bonding, the way elements combine together to form compounds such as Sodium (Na) bonding with Chlorine (Cl) to form Sodium Chloride (NaCl), also known as salt. The types of chemical bonding known as ionic bonding and covalent bonding rely on the electrostatic forces between the elements to bond them together. At this level, the gravitational force is of little consequence, so we will ignore it for the most part. Gravitational forces and electrical forces do act in similar ways, the more mass, the greater the gravitational force, the more charge, the greater the electrical force. In addition, as masses or charges get farther away from each other, the forces they exert decrease by a squared value. The one difference is that gravitational forces only pull masses towards each other, electrical forces either pull charges together (if oppositely charged) or push them away (if like charged). It is possible to force an electrically neutral object to become charged through a method called charging by conduction. Much like heat or the flow of water, charge will naturally flow from a high concentration of charge to a low concentration of charge. So if a highly charged object comes in contact with an object that is neutral (that is a charge of zero), charge will flow from that high concentration to the neutral object, causing it to be charged as shown in the two cases below. Charging by Conduction
Neutral
- +
+ - + - + - +
+ - + - +
Neutral
Negatively Charged
---------
+ -+ - + - +
+ - + - +
+++++++++
Both Positively Charged
Both Negatively Charged
-- +
- - - +
- - + +
- - + +
- - +
Positively Charged
+
- - - - - - +++
+
+ +- + + + -
+++++----
229 A third method of charging a neural object is called charging by induction. Unlike charging by friction or conduction, this method requires no physical contact. First, a charged object is brought close to the neutral object. This causes the neutral object to become polarized so that one side is positive and the other is negative as shown below on the left. If the netural object is then grounded, some of the charge on the grounded side leaves the neutral object and heads to the ground. When the ground is removed, the object now has an excess of charge and is thus no longer netural, as shown below on the right. It has taken on the opposite charge of the rod. Charging by Induction
Neutral
Negatively
Charged
- +
+ - + - + - +
+ - + - +
---
---------
Ground
---------
Now induced positive charge
Netural object polarized
- +
+
+
+
+
+
+
+
+
+
+ +
+
+
+
+
+
+
+ +
---------
---------
14.5 Electric Field So what causes this attraction and repulsion of charges? The answer lies in a charges electric field. What is a field? In Physics, a field is a region in space in which a certain quantity has a definite value at every point. For example, in a room there exists a temperature field. At every point in the room there is a definite value for the temperature, and that value may differ in different parts of the room. 18 oC
20 oC
20 oC
35 oC
Fireplace
30 oC
25 oC
In a room with a window and a fireplace, the room is the entire field and the temperatures are different at different locations. Another example of a field the gravitational field discussed in Chapter 12. There is a gravitational field (g) that surrounds the Earth (and any large mass). The 230 strength of that field is different at the surface of the Earth, at the top of Mt. Everest, and as you get into outer space. An electric field (E‐field) exists around any charged object, just like a gravitational field (g‐field) exists around any mass. This electric field radiates out from the charged object and its strength is based on the amount of charge, just like a planet’s gravitational field is based on the amount of mass the planet has. It is sometimes difficult to visualize an electric field, so we have developed a way of representing an electric field by using electric field lines. Electric field lines are based on the electric field being a vector and three (3) rules. 14.6 Rules of Electric Field Lines 1. Electric field lines must begin on a positive charge and/or end on a negative charge. For a single charge, the lines begin or end at infinity. +
-
+
-
2. The number of field lines is proportional to the amount of charge. +
(+q)
3. No two field lines can cross. +
2x Greater charge (+2q)
231 Example: Two positive charges of different values near each other would look like this. +q
+2q
The charges repel, and the one on the right is twice the charge of the one on the left. Determining the strength of the electric field at any point is similar to determining the strength of a gravitational field at any point near a mass. E – Electric Field Strength [N/C] k = 9.0 x 109 Nm2/C2 r – distance from charge [m] Example: Calculate the strength of an electric field 10 cm from a 2.0 x 10‐20 C charge. . . .
E = 1.8 x 10‐8 N/C Notice that the equation for the electric force and the equation for the electric field are very similar. In fact, an electric force cannot exist without a second charge entering the electric field created by the first charge. This is very similar to the idea that a gravitational force cannot exist without two masses, one that sets the gravitational field and the second that enters the field. therefore: FE = Eq This second equation for electric force is useful only if the electric field strength is known along with the value of a charge placed within the electric field. It is also useful for what is known as electrically charged parallel plates. When two oppositely charged plates are set parallel to each other and a certain distance apart, they create a uniform (constant) electric field between them. This means that the electric field strength is the same everywhere between the two plates, unlike a single charge where the strength changes with distance. Considering the above 232 equation (FE = Eq) if the E‐field is constant everywhere between the plates, then the electric force acting on a single charge (q) will also be constant everywhere between the plates. +
+
+ +q
+
+
+
+
E
FE
-
Note that the direction of the electric force (FE) is in the same direction as the direction of the electric field (E). Thus the direction of the acceleration is also in the same direction. That is always true for a positive charge in an electric field. For a negative charge, the direction electric force and acceleration is opposite the direction of the electric field. +
+
+
+
+
+
+
E
FE
-q
-
Example: A set of parallel plates has a uniform electric field (E) of 5.0 x 104 N/C between the plates. If I place a positive charge (q) of 4.0 x 10‐15 C with mass (m) 2.0 x 10‐20 kg next to one plate with a second plate placed 20 cm away, (a) determine the electric force (FE) on the charge. (b) If the charge starts from rest, determine the acceleration (a) of the charge. (c) Determine the final velocity (vf) of the charge at the other plate. (a) FE = Eq FE = (5.0 x 104)(4.0 x 10‐15) FE = 2.0 x 10‐10 N (b) F = ma 2.0 x 10‐10 = (2.0 x 10‐20)a a = 1.0 x 1010 m/s2 (c) vf2 = vo2 + 2ax vf2 = 02 + 2(1.0 x 1010)(.2) vf2 = 4.0 x 109 vf = 6.32 x 104 m/s You mean we still have to use F = ma and linear motion while doing electricity? Of course, all physics is related to all topics. 233 14.7 Electric Potential Energy We have already seen many similarities between gravity and electricity in terms of fields and forces. Fg = mg Force on a mass within a specific gravitational field FE = Eq Force on a charge within an specific electric field If a mass is placed in a constant gravitational field and that mass is raised to a certain height, work has been done to give it height. At that final height, the mass now has gravitational potential energy that is equal to the work done to place it there. m
PEg = mgh
h
h
W = mgh
KE = ½mv2
m
m
If the object is released, that potential energy is transferred into kinetic energy as the object is pulled to the ground. This is similar to the idea of electric potential energy. If I have two oppositely charged parallel plates, they create a uniform electric field. If I place a positive charge next to the positively charged plate, it is repelled by that plate and attracted to the negatively charged plate by providing an electric force on the charge. This means that as the charge is forced along the distance between the two plates, some amount of work is done on the charge to move it from one plate to the other. d
+
+
+ +q
+
+
+
+
W = Fx = FEd FE
FE = Eq E
-
W = Eqd (W = qEd) If the charge is held at the one plate, it is similar to a mass being held at some height, they both have the potential to start moving. Therefore, the charge will have some electric potential energy (PEE) just like the mass has gravitational potential energy (PEg). PEE = qEd (Formula for electric potential energy) 234 Similarly, if that charge is released, it gains velocity as it crosses the distance between the plates. So the electric potential energy changes into kinetic energy. d
+
+
+
+ +q
+ PE
E
+
+
-
E
d
+
+
+
+
+
+
+
E
+q
KE
-
Example: A charge of 4.0 x 10‐15 C with a mass of 2.0 x 10‐20 kg next to one plate with a second plate placed 20 cm away and a uniform electric field of 5.0 x 104 N/C between the plates. (a) Determine the electric potential energy of the charge. (b) Determine the velocity of the charge if it is released from rest and reaches the other plate. (a) PEE = qEd PEE = (4.0 x 10‐15)(5.0 x 104)(.2) PEE = 4.0 x 10‐11 J (b) PEE = KE (Conservation of energy still applies) 2
KE= ½mv 4.0 x 10‐11 = ½(2.0 x 10‐20)v2 v2 = 4.0 x 109 v = 6.32 x 104 m/s Wow, it all works out the same!!! A few important concepts to remember 1. An atom is made of three particles, the proton, the neutron and the electron. Each particle has a specific mass and a specific charge. Protons are positively charged, electrons are negatively charged, and neutrons have no charge. 235 2. The Law of Conservation of Charge indicates that the total charge of a system always remains constant. 3. There are three ways to charge an object: (1) charging by friction, (2) charging by conduction, and (3) charging by induction. 4. Like charges will repel each other, opposite charges will attract each other. 5. A single electric charge will create an electric field around it. The strength of this field decreases with distance. 6. Two electric charges are required to create an electric force. This force increases if the amount of charge increases and this force decreases if the distance between the charges increases. 7. Electric field lines are used to represent the strength of charges and how they interact with other charges. 8. The electric field between charged parallel plates is constant. A charge placed anywhere between the plates will experience a constant force and a constant acceleration. 9. As a charge moves from one charged parallel plate to another, it’s electric potential energy decreases, its kinetic energy increases, and its velocity increases. 10. The electrostatic force is similar to the gravitational force in that they both provide “action at a distance”, they both require two objects to create a force, and they both get weaker as distance increases. Questions and Problems mproton = 1.673 x 10‐27 kg, melectron = 9.11 x 10‐31 kg qproton = 1.60 x10‐19 C, qelectron = ‐1.6 x 10‐19 C 1. Describe why rubbing a balloon against someone’s head allows you to stick the balloon to the wall. 2. (a) What are the three methods by which an object can become charged. (b) Describe how charging by conduction is different than charging by induction? 3. (a) Describe two similarities between the gravitational force and the electric force. (NOTE: The fact that both formulas have a constant or does not count). (b) Describe two differences between the gravitational force and the electric force. (NOTE: The fact that both formulas have different constants does not count). 236 4. (a) Draw the electric field lines around two negative charges that are near each other if one charge is three times as large as the other charge. (b) Draw the electric field lines between two oppositely charged parallel plates. (a)
(b)
-q
+
+
+
+
+
+
+
- 3q
-
5. Given the following electric fields, determine the amount of each charge. (b)
(a)
6. Given the following picture, a positive charge is placed at point A and released: +
+
+
+A
+
+
+
B
C
D -
(a) At which point(s) is the potential energy of the positive charge the greatest? (b) At which point(s) is the velocity of the positive charge the greatest? (c) At which point(s) is the force on the positive charge the greatest? Why? 7. A single charge of +5.0 x 10‐15 C is sitting in space. (a) Calculate the strength of the electric field at a distance of 20 cm from the charge. (b) A second charge is ‐2.0 x 10‐15 C is placed a distance of 20 cm from the first charge, determine the electric force between the two charges. (c) Is this an attracting force or a repelling force? Why? 237 8. Two charges of ‐4.0 nC and ‐2.0 nC, shown below, are separated by 0.3 mm of distance. (a) Draw the electric field around the two charges. (b) Calculate the electric force acting on the two charges. (c) Is this an attracting force or a repelling force? Why? - 4.0 nC
- 2.0 nC
0.3 mm
9. Older televisions worked by accelerating electrons through a constant electric field that would then strike the screen and form an image. Based on the mass and charge of an electron, (a) calculate the electric force applied on an electron in a television that is using a 2.0 x 105 N/C electric field. (b) Determine the acceleration of the electron. 10. A single charge of 4.0 x 10‐15 C with a mass of 6.0 x 10‐22 kg is placed next to the positive side of a set of parallel plates. The negative plate is placed 20 cm from the positive plate. There is a 2,000 N/C uniform electric field across the plates. (a) Calculate the electric force on the charge. (b) Determine the acceleration of the charge. (c) If the charge is released from rest, calculate its velocity as it reaches the negative plate. 11. A single charge of 5.0 x 10‐16 C with a mass of 4.0 x 10‐23 kg is placed next to the positive side of a set of parallel plates. There is a constant electric field of 8,000 N/C between the plates and they are 10 cm apart. (a) Determine the electric potential energy of the charge. (b) If the charge is released and moves to the opposite plate, determine the kinetic energy of the charge as it reaches the opposite plate. (c) Calculate the velocity of the charge as it reaches the other plate. 12. When Robert Millikan first discovered the charge of an electron, he did it with his famous Oil‐Drop Experiment (you will hear all about it in chemistry). Oil drops were sprayed into a special container and charged by friction. Gravity then caused them to fall through a small hole into an area that had uniformly charged parallel plates that were horizontal to each other as shown below. The electric field was adjusted until the upward electric force on the charged oil drop was in equilibrium with the downward gravitational force (weight) on the oil drop. + + + + + + +
-q
‐ ‐ ‐ ‐ ‐ ‐ ‐
(a) Draw a free body diagram of the oil drop in its equilibrium state. (b) If the mass of the oil drop is 0.2 mg, determine the weight of the oil drop. (c) If the uniform electric field between the two plates is 4.0 x 108 N/C, calculate the amount of charge (q) on the oil drop. 238 13. A 2.0 x 10‐15 C positive charge with a mass of 8.0 x 10‐24 kg is traveling straight and horizontally at 6.0 x 106 m/s when it passes between a set of horizontal parallel plates as shown below. The plates have a constant electric field of 3.0 x 104 N/C acting downward. + + + + + + +
E
+q
‐ ‐ ‐ ‐ ‐ ‐ ‐
(a) Draw the path the charge would travel as it passes between the plates. (b) Determine the electric force that will act upon the charge while between the plates. In what direction will this force act on the charge? (c) Determine the horizontal and vertical accelerations of the charge while between the plates. (Hint: Doesn’t this look like projectile motion, specifically a horizontal launch, but now gravity is not the force?) (d) If it takes 4.0 x 10‐7 s for the charge to pass horizontally through the plates, calculate the vertical displacement of the charge as it passes between the plates. (e) If a negative charge was now sent between the plates, explain any difference from the positive charge motion that would occur. + + + + + + +
E
-q
‐ ‐ ‐ ‐ ‐ ‐ ‐
14. A 0.08 N force is required to move a 50 μC charge with a mass of 4.0 x 10‐18 kg from rest over a distance of 60 mm in a uniform electric field. (a) Determine the strength of the electric field. (b) Calculate the velocity of the charge after 60 mm. 15. A charge experiences a force of 2.0 x 10‐12 N at some distance from a second charge that generates a 4,000 N/C electric field at the location of the first charge. (a) Calculate the strength of the first charge. (b) If the second charge is 2.0 x 10‐16 C, determine the distance between the two charges. 239 Chapter 15 – Electric Circuits 15.1 Introduction to Electric Circuits Every electrical device is made up of an electrical circuit. These circuits are used to collect, route, store, and use electrical energy. Before we can study complete circuits, we must understand the basic components of a simple circuit. There are three basic parts: current, voltage, and resistance. We will represent them in a circuit using the following symbols: +
-
V
V (voltage)
I (current)
+
Simple
Circuit
I
R (resistance)
R
What is known as a simple circuit is a circuit that contains a single voltage source (V), as single current (I), and a single resistance (R). What are these three things? Well, let’s find out. 15.2 Current (I) A circuit must carry electricity from one point to another; this is done by moving electric charge. Current (I) is defined as the rate at which charge flows. Current is different from static charge in that static charge only flows on the surface of an object while current flows through an object. This is also why current electricity is far more dangerous than static electricity. I – Current [A] q – Charge [C] t – time [s] Current is the actual flow of electrons, which in reality flows from the negative point to the positive. This is known as true current flow. However, when Ben Franklin came up with the ideas of positive and negative charge, he assumed charge would flow from positive to negative. This is known as conventional current. We will use conventional current because it basically works for our purposes. Current is measured in the unit [A] for Amperes or just Amps for short. 240 15.3 Voltage (V) In circuits, there must be a source of current (in other words, a source of energy). This is known as an Electromotive Force (EMF). An EMF is any device that can transform non‐electrical energy into electricity and then force that electricity to flow through a circuit. Examples of an EMF include a battery, a generator, or a turbine. This EMF creates a potential difference (known as voltage) that allows the charge (current) to move from the high potential to the low potential. The EMF basically pushes the charge along, so the higher the voltage, the more charge it is able to push. Thus a 12 V battery can drive more electricity than a 1.5 V battery. Voltage is measured in the unit of [V] for Volts after Alessandro Volta, the inventor of the battery. 15.4 Resistance (R) All circuits contain some sort of resistance, something that slows the flow of current. Resistance is like a form of electrical friction in that it can take away electrical energy and convert it into heat energy. The resistance of a substance is based on four (4) specific things: (1) length, (2) cross‐sectional area, (3) material, and (4) temperature. The resistance of an object like a wire can be found with the formula: R – Resistance [Ω] L – length [m] A – cross‐sectional area [m2] ρ – resistivity [Ωm] (A constant based on material) As the length of a resistor increases, so does its resistance, making resistance directly proportional to length. So if you double length, you double R. As the cross‐sectional area of a resistor increases (in other words, it gets wider) the resistance decreases, making resistance inversely proportional to area. So if you double cross‐sectional area of a resistor, R decreases by ½. Resistance is measured in the unit [Ω] for Ohms. The resistivity (ρ) is based on the type of material the resistor is made of and is measured in [Ωm] for Ohm‐meter. This is usually found in a table of materials. Finally, the lower the temperature, the lower the resistance of a wire. Example: Let’s say we have a copper wire (ρ = 1.70 x 10‐8 Ωm) that is 20 cm long and has a radius of 0.2 mm. To determine its resistance, we must consider that the cross‐sectional shape of a wire is a circle. . .
. R = 0.027 Ω 241 Objects can be classified by their electrical resistance. There are four categories of electrical resistance. Conductors – any object that easily allows the flow of electricity. Most things will conduct some form of electricity, but metals are usually the best conductors. Insulators – any object that resists the flow of electricity. Many things can be used as insulators, but rubber and plastics tend to be the best. Semiconductors – materials that regulate the flow and direction of electricity. These were important materials as they made the computer age possible. The most common form of semiconductors comes from silicon and is used in making computer chips. Superconductors – materials with no electrical resistance. A modern discovery, it was found that when a regular wire was cooled to 7 K (‐266 oC), the molecular motion slowed down so much that it did not interfere with the flow of electric current. Remember, temperature can affect resistance. From the idea of superconductors comes a fourth item that can affect resistance and that is temperature. As temperature goes down, so does the resistance of a conductor. 15.5 Ohm’s Law The three parts of a circuit can all be related together under Ohm’s Law. Ohm’s Law is not a true law of nature like Newton’s Laws or Conservation of Energy. Ohm’s Law is an experimental law that works in most situations, but not all. As far as circuits are concerned, however, it works just fine. Ohm’s Law is essentially just an equation that relates voltage, current, and resistance all together. V = IR V – Voltage [V] I – Current [A] R – Resistance [Ω] It is pretty easy to use as shown in the three examples below. If you are given any two, you can find the third. (1) V = 10 V (2) I = 3 A (3) V = 20 V R = 5 Ω R = 5 Ω I = 5 A I = ? V = ? R = ? V = IR V = IR V = IR 10 = I(5) V = (3)(5) 20 = (5)R I = 2 A V = 15 V R = 4 Ω (Not too difficult). 242 We can now combine all things so far into one situation. Let’s say we have 0.5 C of charge passing through a 10 Ω resistor every 0.2 s. Determine the voltage across the wire. V = IR First need to find current (I). .
.
I = 2.5 A Now find Voltage (V). V = IR = (2.5)(10) V = 25 V 15.6 Joule’s Law Along with Ohm’s Law comes Joule’s Law which is related to electrical power. We discussed mechanical power as the rate at which work is done. Electrical power is very similar, but relates power to voltage, current, and resistance. P = IV P – Power [W] I – Current [A] V – Voltage [V] We can combine Ohm’s Law and Joule’s Law together to get a second power equation. P = IV V = IR P = I(IR) P = I2R We can combine with Ohm’s Law by solving for current and substituting into Joule’s Law for a third power equation. V = IR so P = IV V
V P
R
243 We can use any of the equations for power based on what is given with one exception. If a question ever asks to determine the power lost, we must use: Plost = I2R (Power lost is based on current and resistance, not voltage). Example: A 100 W light bulb is plugged into a 120 V source P = 100 W V = 120 V I = ? R = ? P = IV V = IR 100 = I(120) 120 = (0.83)R I = 0.83 A R = 144. 5 Ω 15.7 Cost of Electricity So how much does electricity cost? The power company likes to charge by their own units and not by standard units. (I guess they don’t hire picky physics teachers). On your power bill may \$ .
be something that says . This means that they charge you \$0.15 (fifteen cents) per kilowatt of power that you use per hour. Example: Let’s say your TV uses 200 W (0.2 kW) of power and you watch it for 4 hours, which is way too much TV. To solve this, we need to go back to dimensional analysis to get rid of kW and hr and leave the unit of \$. \$ .
V (voltage)
I (current)
Simple
Circuit
I
R (resistance)
R
Again, what is known as a simple circuit is a circuit that contains a single voltage source (V), as single current (I), and a single resistance (R). What we will look at here is how to deal with circuits that contain more than one resistance within the circuit. For example, the lights in a classroom are all wired together in a specific way. If one bulb goes out, the others do not, however I can turn off one group of lights all at once, while the others stay on, with the flip of a switch. Each individual light is a resistance. How they are all hooked together and how the current flows through each is all part of DC (direct current) circuit analysis. Let’s make a quick distinction between DC (direct current) and AC (alternating current). Every electrical device runs on DC, this is current running in one direction. AC indicates that a current flows in one direction and then switches and flows in the opposite direction. AC is necessary for the transmission of electricity over long distances. The current in the United States electrical system alternates directions in a 60 cycle rate (60 times per second). If, for example, a fan were hooked to AC instead of DC, the fan blades would constantly switch direction in which they rotated and would not do us any good. When we plug things into the outlets, special devices convert the AC electricity into DC to run our devices. We will focus only on DC circuits. Circuits are required to obey two major principles of physics, Conservation of Charge and Conservation of Energy. These two laws control the flow and use of electricity. However, when analyzing a circuit we will apply Ohm’s Law and Joule’s Law for any calculations. 248 16.2 Conservation of Charge in a Circuit When any device (resistance) is attached to a source or electricity (voltage) a current will flow out of the voltage source, through the resistance, and back to the voltage source. Again we will use the idea of conventional current in that the current will flow from the positive to the negative. This means that whatever current leaves the positive end of the voltage source must return to the negative end in order to conserve charge. The total current must remain constant from beginning to end. I
V
Both currents (I)
are the same.
R
I
When a current encounters a point where it can go two different directions, the current will divide itself up and part will go one way and the rest the other way(s). A current can split many times within a single circuit, but in the end, all the divided currents must join back together before returning to the voltage source in order to conserve charge. In the example below, I1 is the current that leaves the voltage source. It encounters a split and divides into I2 and I3. I2 moves down one path through resistance R1 and I3 continues on until it encounters another split where it divides into I4 and I5. V
I5
I3
I1
I2
I4
R1
R2
R3
I4 will travel its path through R2 while I5 travels around its path through R3. However, at this point, I4 and I5 encounter the same connection point and will now rejoin each other. Since they both split from I3, by conservation of charge, they rejoin and become I3 again. This I3 will then meet up with I2, join together and become I1 again. I1 then returns to the voltage source. We have obeyed Conservation of Charge in the entire circuit. 249 I5
I3
I1
I2
I4
R1
R2
V
I1
R3
I3
I5
Currents will often travel the “path of least resistance”. This means that more current will flow through an area of lower resistance, though some current always flows through all parts of a circuit. Also, in order for current to flow from the positive to the negative of a voltage source, it must have a complete path from beginning to end. If a break in the circuit occurs and there is no way for the current to go from the positive all the way back to the negative, the current flow will stop. There is a certain condition called a short circuit. This tends to be when something of very little resistance (like a bare wire) is placed across a resistance. Normal Circuit
I1
I
I2
Short Circuit
V
R
V
I3
R
I3 >>> I2
I
I1
In a short circuit, two things happen; first the overall resistance of the circuit becomes very low causing a very large current (I1) to leave the voltage source. Second, the majority of that current (I3) then bypasses the resistance leaving only a very small current (I2) for the resistance. This current is too small to provide power to run the device, so you have short circuited the device and it will not work. This can be dangerous because of the high value of I1. 16.3 Conservation of Energy in a Circuit Any device that is attached to a voltage source will use energy supplied by that source. The idea behind conservation of energy within a circuit is that all of the voltage provided by the source must be used in the circuit. If a single light bulb is powered by a 6.0 V battery, then that one light bulb will use all 6.0 V. If four (4) light bulbs are attached to that 6.0 V battery, then they will each use a certain amount of the voltage, but in the end, all 6.0 V will be used. The secret to this lies in Ohm’s Law (V = IR). As a current (I) passes through a resistance (R), it will “drop” voltage (V) at that point. The term voltage drop will indicate that portion of the voltage that is 250 used by an individual resistance. The actual amount of voltage dropped will depend on the total available voltage, the amount of resistance each object has, and the arrangement of the resistances within the circuit itself. The key to understanding all of this will be Ohm’s Law. To understand circuit analysis, we will look at three types of circuits, series circuits, parallel circuits and complex circuits. Our goal in each case is to understand the current flow and voltage drops through all parts of the circuit, particularly through any resistance. 16.4 Series Circuit A series circuit is a circuit that contains two or more resistors attached in a way that all resistors in the circuit get the same current. Typically this is easy to see as the resistors are attached in a row, one after the other. However, despite each resistor getting the same current, they will have different voltage drops based on their individual resistance. A typical series circuit may look like this: R1
I1
V
R2
R3
Notice that the current (I1) must pass through all three resistors; therefore they are all in series with each other. In dealing with circuit analysis, our first goal will always be to reduce our given circuit to a simple circuit. I1
I1
R1
V
V
Req
R2
Becomes
R3
251 In this process, all the resistors in the circuit are reduced to a single resistor known as the equivalent resistance (Req). The equivalent resistance (Req) is equal to all the resistors in the circuit. In a series circuit, determining the Req is pretty easy. Example: Given the following series circuit, determine (a) Req, (b) the current (I1), and (c) the voltage drop on each resistor (VR): 4Ω
I1
40 V
10 Ω
6Ω
(a)
R
4
10
6 Req = 20 Ω Draw the simple circuit with one resistance, Req. I1
40 V
20 Ω
(b) Using Ohm’s Law, determine I1. V = IR 40 = I(20) I1 = 2.0 A 252 Now take this current (I1) back to the original circuit. 4Ω
2A
40 V
10 Ω
6Ω
(c) To determine the voltage drop on each resistor, using Ohm’s Law, apply the 2.0 A current to each resistor as it must pass through each resistor in series. V4Ω = IR V10Ω = IR V6Ω = IR V4Ω = (2)(4) V10Ω = (2)(10) V6Ω = (2)(6) V10Ω = 20 V V6Ω = 12 V V4Ω = 8 V 2A
4Ω
8V
40 V
20V
10 Ω
6Ω
12 V
Notice that when you add the three voltage drops together that they add up to 40 V, so you obeyed conservation of energy. Also, if I were to attach more resistors to the series circuit, the Req would go up which would cause the current (I) to go down. Adding more resistors in series is like increasing the length of a resistor, it causes overall resistance to increase. A series circuit is usually easy to deal with. 253 16.5 Parallel Circuit A parallel circuit is a circuit that contains two or more resistors attached in a way that all resistors in the circuit get the same voltage drop. This can be a little harder to see, but the best way to recognize resistors in parallel is that they come AFTER a point where the current can split into two or more different currents. However, despite each resistor getting the same voltage drop, they will have different currents based on their individual resistance. A typical parallel circuit may look like this: I5
I3
I1
V
I2
I4
R1
R2
R3
Each resistor has a distinctly different current passing through it. However, each resistor is also directly connected to the voltage source. I know that is a little hard to see, but if we remove any one resistor, the others still have a complete, unbroken path to and from the voltage source. When thinking of a parallel circuit, think about a power strip. When I plug something into a wall outlet, that outlet provides 120 V to whatever I plugged in. Everything we plug into the wall is made to run on 120 V. So when I plug a power strip into a wall outlet, it is like I am making a bigger wall outlet. That means that each device that I plug into that power strip needs 120 V. Therefore, each plug in the power strip is set up in parallel. That way everything I plug in gets 120 V. So in parallel, it is like they are sharing the voltage drops; they do not divide it up like in series circuit. Now the more items I plug into the power strip, the more current I will need to run everything, so you have to be careful with a large parallel circuit as it will generate a large current flow. Once again, to analyze a circuit, we first want to reduce it to a simple circuit. V
I1
I5
I3
I1
I2
I4
R1
R2
Becomes
R3
V
Req
254 In this process, all the resistors in the circuit are again reduced to a single resistor known as the equivalent resistance (Req). In a parallel circuit, determining the Req is a little different. Example: Given the following parallel circuit, determine (a) Req, (b) all the indicated currents, and (c) the voltage drop on each resistor (VR): I5
I3
I1
54V
I2
I4
9Ω
6Ω
18 Ω
(a)
Find the common denominator, in this case 18. Add the fractions. Then invert your answer to get Req. Req = 3 Ω Draw the simple circuit with one resistance, Req. I1
54 V
3Ω
(b) Using Ohm’s Law, determine I1. V = IR 54 = I1(3) I1 = 18 A 255 That just gives us I1, but what about the other currents. In parallel, to get those, we need the voltage drops first. (c) This is the part that always trips students up. As all three resistors are in parallel and attached to the 54 V source, all three resistors drop 54 V. V9Ω = 54 V V6Ω = 54 V V18Ω = 54 V I5
I3
18 A
54V
I2
I4
9Ω
54V
6Ω
54V
18 Ω
54V
This does not violate conservation of energy as they are sharing the voltage together, not splitting it up. We can get the rest of the currents using Ohm’s Law and Conservation of Charge. V9Ω = I2R V6Ω = I4R V18Ω = I5R 54 = I2(9) 54 = I4(6) 54 = I5(18) I2 = 6 A I4 = 9 A I5 = 3 A 54V 3 A I3 18 A 6 A 9 A 9 Ω 54V 6 Ω 54V 18 Ω 54V That just leaves I3. Since I1 split into I2 and I3, then by Conservation of Charge: I1 = I2 + I3 18 = 6 + I3 I3 = 12 A (Note that I3 splits into 9 A and 3 A, which add up to 12 A.) 54V
3A
12 A
18 A
6A
9A
9Ω
54V
6Ω
54V
18 Ω
54V
256 We have obeyed conservation of charge because the 3 A and 9 A will rejoin to become 12 A which will rejoin with the 6 A to become 18 A again. Also, if I were to attach more resistors to the parallel circuit, the Req would go down which would cause the main current (I1) to go up. Adding more resistors in parallel is like increasing the area of a resistor, it causes overall resistance to decrease. 16.6 Complex Circuits Now the fun really begins. Most circuits are neither only series nor only parallel, but a complex combination of series and parallel parts. When analyzing a complex circuit, we will always begin with the same goal and that is to reduce the circuit to a simple circuit and find Req. After that, we will work our way backwards to find currents and voltage drops. To do these things, we will simplify the circuit in small parts at a time and then redraw the circuit in a simpler form until we have reached the simple circuit. This is tricky at first, but once you see the pattern it becomes fairly easy to do, so pay close attention. Example: Given the following complex circuit, determine (a) Req, (b) all the indicated currents, and (c) the voltage drop on each resistor (VR): I2
I1
6Ω
1Ω
60 V
I3
6Ω
5Ω
I4
4Ω
I5
12 Ω
3Ω
If we look at this circuit, it looks like one big loop with some small loops along the way. So it looks a lot like a basic series circuit if we could get rid of those two small loops, so let’s do that. If we look carefully, the 6 Ω and 6 Ω have two different currents (I2 and I3), so they must be in parallel with each other (they also immediately follow a split in the circuit). In addition, the 4 Ω and 12 Ω have two different currents (I4 and I5), so they must be in parallel with each other (they also immediately follow a split in the circuit). What I am going to do is to squeeze those 257 two sets of resistors together to make one equivalent resistance for each set and they redraw the circuit. As they are each a parallel set, I need to use my parallel resistance equation: For the 6 Ω and 6 Ω For the 4 Ω and 12 Ω Flip them both over. Req = 3 Ω These are like mini‐Req’s. Req = 3 Ω Now redraw the circuit with these new values in place of the small loops. I1
1Ω
3Ω
60 V
5Ω
3Ω
3Ω
Notice that I did not change any of the other resistors and all the currents except I1 are gone. I can now see that I have a basic series circuit that I can reduce to a simple circuit using: Req = 15 Ω This is the Req for the whole circuit. 258 The simple circuit now looks like this: I1
60 V
15 Ω
I can apply Ohm’s Law to determine I1 for the simple circuit. V = I1Req 60 = I1(15) I1 = 4 A 4A
60 V
15 Ω
I bring this 4 A back to the previous circuit (not the original circuit). 4A
3Ω
1Ω
60 V
5Ω
3Ω
3Ω
As this is a basic series circuit, I take that 4 A around the circuit and determine the voltage drops on each resistor in the circuit using Ohm’s Law (V = IR). V1Ω = IR V5Ω = IR V3Ω = IR V3Ω = IR V3Ω = IR V1Ω = (4)(1) V3Ω = (4)(3) V5Ω = (4)(5) V3Ω = (4)(3) V3Ω = (4)(3) V1Ω = 4 V V3Ω = 12 V V5Ω = 20 V V3Ω = 12 V V3Ω = 12 V 259 A quick check: if we add all of these voltage drops up, we get 60 V, so we are doing well. I now know all currents and voltage drops on this circuit. 4A
1 Ω 4V
3Ω
12V
20V
60 V
5Ω
3Ω
12V
3 Ω 12V
I carry all of this information back to the main circuit. The tricky part is dealing with those small loops from the original circuit. On the original circuit, the 6 Ω and 6 Ω that were in parallel with each other reduced to a 3Ω. Since that 3Ω resistor was just shown to drop 12 V then the 6 Ω and 6 Ω each drop 12 V. (I know, it sounds crazy, but parallel resistors must share voltage, so they always drop the same amount). That means that on the original circuit, the 4 Ω and 12 Ω that were in parallel with each other reduced to a 3Ω. Since that 3Ω resistor was just shown to drop 12 V then the 4 Ω and 12 Ω each drop 12 V. 260 So we get something like this: 4A
1Ω
I2
4V
60 V
I3
6Ω
12V
20V
5Ω
6 Ω 12V
I4
12V
4Ω
3Ω
I5
12V
12 Ω
12V
I know what you are thinking, if I add up all of these voltage drops I get more than 60 V, so I have violated conservation of energy. WRONG!! The parallel loops are sharing their voltage drops, so you can’t add them separately; they are like one voltage drop, so we have not violated anything, so there! We can find the missing currents using the voltage drops and Ohm’s Law (V = IR). V6Ω = I2R V6Ω = I3R V4Ω = I4R V12Ω = I5R 12 = I2(6) 12 = I3(6) 12 = I4(4) 12 = I5(12) I3 = 2 A I4 = 3 A I5 = 1 A I2 = 2 A 261 4A
1Ω
2A
4V
60 V
6Ω
12V
20V
2A
5Ω
6 Ω 12V
3A
12V
4Ω
3Ω
1A
12V
12 Ω
12V
Notice how these answers make sense using conservation of charge, I1 which is 4 A splits into I2 and I3 which together add up to 4 A. They then rejoin and split again into I4 and I5 which also add up to 4 A. So it all makes sense. Our final answers are: Req = 15 Ω I3 = 2 A I1 = 4 A I2 = 2 A I4 = 3 A I5 = 1 A V1Ω = 4 V V6Ω = 12 V V6Ω = 12 V V5Ω = 20 V V4Ω = 12 V V12Ω = 12 V V3Ω = 12 V Yay we did it! Let me throw one more question in here, (d) determine the total power to operate the circuit. 262 To determine power, we use Joule’s Law, but which version and what numbers do we use? The answer lies with the simple circuit. 4A
60 V
15 Ω
This circuit represents the TOTAL circuit in its simplest form, so we use these values and then any version of Joule’s Law. P = IV P = I2R P = (4)(60) P = 240 W P = (4)2(15) P = 240 W P = 240 W Let’s try another one. Example: Given the following complex circuit, determine (a) Req, (b) all the indicated currents, (c) the voltage drop on each resistor (VR), and (d) the total power. I1
2Ω
I3
6Ω
I2
60 V
10 Ω
8Ω
1Ω
3Ω
If I look at this circuit, it looks a lot like a basic parallel circuit, yet it appears to have some series parts to it, so I will reduce a series component first. Resistors in series must have the same current running through each. The only resistors that can clearly get the same current (I3) are the 6 Ω, 1 Ω, and 3Ω. So I will reduce that part first and redraw the circuit with those three resistors squeezed into a single resistor. As they are in series I will use the series equation first. Req = 10 Ω For that part, I redraw the circuit. 263 2Ω
I1
I3
I2
60 V
10 Ω
10 Ω
8Ω
When I look at the circuit, the next reduction that makes sense is that the 10 Ω and 10 Ω that follow a split in current must be in parallel, so I will reduce that pair and squeeze them together into one resistor. As the 10 Ω and 10 Ω are in parallel, I use the parallel equation. Req = 5 Ω I1
For that part, I redraw the circuit. 2Ω
60 V
5Ω
8Ω
What remains is a basic series circuit that I can reduce to a simple circuit. This is the Req for the entire circuit. Req = 15 Ω 264 I now draw the simple circuit. I1
60 V
15 Ω
I can then apply Ohm’s Law to determine I1 for the simple circuit. V = I1Req 60 = I1(15) I1 = 4 A 4A
60 V
15 Ω
Next, I bring this 4 A back to the previous circuit (not the original circuit). 4A
2Ω
60 V
5Ω
8Ω
As this is a basic series circuit, I take that 4 A around the circuit and determine the voltage drops on each resistor in the circuit using Ohm’s Law (V = IR). V5Ω = IR V8Ω = IR V2Ω = IR V1Ω = (4)(2) V3Ω = (4)(5) V5Ω = (4)(8) V3Ω = 20 V V5Ω = 32 V V1Ω = 8 V 265 A quick check: if we add all of these voltage drops up, we get 60 V, so we are doing well. I know all currents and voltage drops on this circuit. 4A
2 Ω 8V
60 V
20V
5Ω
8Ω
32V
I now carry all of this information back to the previous circuit (still not the main circuit in this case). The tricky part is dealing with those small loops from the original circuit. On this circuit, the 10 Ω and 10 Ω that were in parallel with each other reduced to a 5 Ω. Since that 5 Ω resistor was just shown to drop 20 V, then the 10 Ω and 10 Ω each drop 20 V. So we get something like this: 4A
2Ω
8V
I3
I2
60 V
20V
10 Ω
8Ω
20V
10 Ω
32V
Next, we can find the missing currents using the voltage drops and Ohm’s Law (V = IR). V10Ω = I2R V10Ω = I3R 20 = I3(10) 20 = I2(10) I2 = 2 A I3 = 2 A Note that these add up to I1 = 4 A. 266 I carry all of this information back to the main circuit. 4A
2Ω
8V
2A
6Ω
2A
60 V
10 Ω
8Ω
32V
20V
1Ω
3Ω
I am only missing the voltage drops on the 6 Ω, 1 Ω, and 3 Ω. They all get the current I3 = 2 A. So I take that 2 A around that part of the circuit and determine the voltage drops on each resistor using Ohm’s Law (V = IR). V6Ω = IR V1Ω = IR V3Ω = IR V1Ω = (2)(1) V3Ω = (2)(3) V6Ω = (2)(6) V6Ω = 12 V V1Ω = 2 V V3Ω = 6 V Notice that they add up to 20V, which is the total voltage that the 10 Ω dropped, so we must be doing something correct. 267 So we have: 2Ω
4A
8V
2A
6Ω
12V
2A
60 V
10 Ω
8Ω
20V
2V
1Ω
3Ω
32V
6V
Our final answers are: Req = 15 Ω I1 = 4 A I2 = 2 A I3 = 2 A V2Ω = 8 V V10Ω = 20 V V8Ω = 32 V V6Ω = 12 V V1Ω = 2 V V3Ω = 6 V Total Power comes from the simple circuit. 4A
60 V
P = IV P = (4)(60) P = 240 W 15 Ω
Not so bad once you understand the method. 268 A few important concepts to remember: 1.
2.
3.
4.
5.
A simple circuit contains a single voltage, current, and resistance. In a series circuit, each resistor in the circuit gets the same current, but drops different voltages. As you add more resistors to a circuit in series, the overall resistance increases and the current decreases. Adding more resistors in series is like increasing the length in resistance. In a parallel circuit, each resistor in the circuit drops the same voltage, but uses different currents. As you add more resistors to a circuit in parallel, the overall resistance decreases and the current increases. Adding more resistors in parallel is like increasing the area in resistance. Questions and Problems 1. Describe the difference between a series circuit and a parallel circuit. 2. Explain a “short circuit”. Why is it so dangerous? 3. (a) How does increasing the number of resistors in a series circuit affect the total resistance of the circuit? (b) Given the situation in (a) and a constant voltage source, how does this change affect the total current within the circuit? 4. (a) How does increasing the number of resistors in a parallel circuit affect the total resistance of the circuit? (b) Given the situation in (a) and a constant voltage source, how does this change affect the total current within the circuit? 5. Explain why you would rather have your house wired in parallel circuits rather than in series circuits. 6. Explain the two principles of physics that govern DC circuit analysis and how they relate to a circuit. 7. Given three resistors of 10 Ω, 20 Ω, and 60 Ω, and a 90 V source, (a) arrange them in a circuit that will result in the lowest possible current leaving the 90 V source. (b) Determine the equivalent resistance and total current for your circuit in (a). Given three resistors of 10 Ω, 20 Ω, and 60 Ω, and a 90 V source, (c) arrange them in a circuit that 269 will result in the highest possible current leaving the 90 V source. (d) Determine the equivalent resistance and total current for your circuit in (c). 8. Given the following circuit, find (a) Req, (b) the total current (I) leaving the EMF, (c) the voltage drop on each resistor (VD). 10 Ω
I
3Ω
70 V
6Ω
12 Ω
4Ω
9. Given the following circuit, find (a) Req, (b) the total current (I) leaving the EMF, (c) the voltage drop on each resistor (VD). I
12 Ω
144V
36 Ω
18 Ω
6Ω
270 Problems 10 – 14: For each circuit, find (a) Req, (b) all indicated currents (I), (c) the voltage drop on each resistor (VD), and (d) the total power used in each circuit. 10. 7Ω
I1
I3
I2
20 V
4Ω
1Ω
2Ω
5Ω
11. I2
8Ω
2Ω
I1
120 V
I3
8Ω
5Ω
I4
3Ω
I5
6Ω
7Ω
271 12. 2Ω
I1
I3
6Ω
I2
40 V
1Ω
4Ω
3Ω
2Ω
5Ω
13. I1
I4 4 Ω
8Ω
I2
2Ω
4Ω
80 V
I5
I3
8Ω
12 Ω
1Ω
2Ω
I6
2Ω
I7
5Ω
272 14. 7Ω
I1
3Ω
I3
I5
4Ω
I4
I2
60 V
10 Ω
10 Ω
3Ω
2Ω
1Ω
5Ω
273 Chapter 17 – Magnetism 17.1 Introduction to Magnetism Magnetism was known to the ancient Greeks as early as 800 B.C.E. It was found that certain stones, now called magnetite, attracted small pieces of iron. Magnetite was found in large quantities in a town called Magnesia, so that is the origin of the term magnetism (or magnet). In ancient days, sailors would take a piece of magnetite and tie it to a string. The part of the rock that faced north was painted to form the earliest compass. We have all played around with magnets of all different shapes and sizes. All magnets, regardless of their shape have what are known as magnetic poles. We will use the idea of a bar magnet as it is the easiest to understand. These magnetic poles are referred to as north poles and south poles. This is because it was found that when a bar magnet was suspended from a string or floated in a liquid, one end pointed toward the north pole of the Earth and the other towards the south pole of the Earth. All magnets have both poles, there is no evidence of a “mono‐pole” (single poled magnet), though it is predicted to exist. So even if I break a bar magnet in half, each part will now have north and south poles. N
S
N
S
N
S
The interesting part is that all magnets point towards the Earth’s magnetic poles which are not located at the actual geographic poles on the Earth. The Earth’s south magnetic pole is located about 1200 miles south of the geographic North Pole, somewhere in Canada. The Earth’s north magnetic pole is located about 2000 miles north of the geographic South Pole in the Pacific Ocean. For some unexplained reason, the Earth’s magnetic poles have reversed many times. The polarity has lasted as short as 50,000 years before changing to as long as several million years before changing. The last known change was about 780,000 years ago. So we may all need to buy new compasses one day soon. Of course, GPS has rendered compasses obsolete to everyone except maybe the Boy Scouts. It is found that if you bring two magnets near each other, they will either attract or repel each other depending on their orientation to each other. Magnetism and electricity are directly related to each other and share similar properties. One of the first similarities is how magnets interact. For any two magnets, like poles repel and opposite poles attract, (much like charge interaction). It turns out that in order for something to actually be magnetic, the electric charges within the substance must move in a very specific way. The charges must be aligned and move together in a very set pattern. 274 -
-
-
-
-
Non-magnet – charges move randomly
Magnet – charges move together
17.2 Magnetic Fields When we studied gravitation, we learned that a gravitational field (g) exists around a single mass. When we studied electrostatics, we learned that an electric field (E) exists around a single charge or a group of interacting charges. This is always true whether a charge is at rest or in motion. However, if a charge is in motion, a second field is created that encircles the moving charge. E
+q
B
+q
v
This is known as a magnetic field (B). A MAGNETIC FIELD IS THE RESULT OF MOVING CHARGE!! All magnetism is the result of moving charges, very important concept discovered by a man named Orestead who demonstrated that a current running through a wire would change the direction of a compass needle. The strength of the magnetic field is based on the amount of charge and the speed of the charge. Magnetic field strength is given by the letter B and measured in Tesla [T] after Nicola Tesla. 17.3 Magnetic Field Lines Just as we can represent electric fields with lines, we can do the same for magnetic fields. There are similar rules: 1. Magnetic field lines must begin on a north pole and end on a south pole. 2. For a single moving charge or a current in a wire, magnetic field lines move in circles around the charge or around the wire. 3. No field lines may cross. 275 N
S
N
B
N
B
B
N
S
B
S
B
S
B
+q
v
I
17.4 Magnetic Force Like gravity and electricity, magnets have the ability to create physical forces. Also, like gravity and electricity, a single magnet cannot create a force, only a magnetic field (B). It takes two interacting magnetic fields to create a force, just like gravitational forces require two masses and electrical forces require two charges. A single moving charge creates a magnetic field. If that charge enters a second magnetic field such as passing near a magnet, a magnetic force (Fm) acts on the charge. This gives the equation: Fm = Bqv Fm – Magnetic force [N] B – Magnetic field strength [T] that the charge is passing through q – Charge [C] v – velocity [m/s] Example: Let’s say we have a 5.0 x 10‐15 C charge moving at 2.0 x 104 m/s through a 0.5 T magnetic field. To determine the amount of force the charge experiences, Fm = Bqv = (0.5)(5.0 x 10‐15)(2.0 x 104) Fm = 5.0 x 10‐11 N (very large force for charge) 276 To prove that moving electric charges created magnetic fields, a man named Orestead created an experiment using a compass, a voltage source, and a current carrying wire. When the wire was placed over the compass and connected to the voltage source, the compass needle changed directions due to the strong magnetic field created by the current running through the wire. If the direction of the current is reversed, the compass needle moved the other direction. (The Earth’s magnetic field has an average strength of around 5.0 x 10‐5 T, which is strong, but not as strong as a regular magnet of about 0.1 T). 17.5 Magnetic Fields and the Right Hand Rule As we saw with Orestead’s experiment, if I changed the direction of the current flow through the wire, the compass needle changed direction. This indicated that the magnetic field around the wire must have changed direction. It also indicated that a magnetic force must exist in order to move the compass needle one way or another. There is a simple and often useful method for determining the direction of magnetic fields and magnetic forces called the Right Hand Rule. It is called the Right Hand Rule because a person’s right hand is used to indicate magnetic field line direction, charge or current flow, and magnetic force directions. With your right hand fully open and your thumb pointing 90 degrees from your fingers, your thumb indicates the direction of motion of a positive charge or conventional current, your fingers indicate the direction of a magnetic field, and the palm of your hand pushes in the direction of the magnetic force. Pictur Picture courtesy of commons.wikimedia.com. As magnetic fields, charge or current movement, and forces act in three dimensions, we have methods for indicating these things on paper. Anything moving up or down the page, or left or right on the pager is indicated with arrows in the appropriate directions. 277 left
right
up
down
However, something that moves perpendicular to the page is different. We will use a dot (.) to represent an arrow coming out of the page and an x (x) to represent an arrow going into the page. .
x
Out of page
Into page
Think of it as the tip of an
arrow coming out of the page
toward you or the cross
feathers of an arrow moving
into the page away from you.
Example: We have a charge moving to the right through a magnetic field that is going into the page as shown below. Which way is the magnetic force acting? Bin
x x x x x x x x
x x x x x x x x
x x x x x x x x
x x x x x x x x
v
Using the right hand rule, your thumb points to the right (v) and your fingers point into the page or away from you (B). Therefore the palm of your hand is pushing down the page. So the direction of the force on the charge is up the page. Fm
Bin
x x x x x x x x
x x x x x x x x
x x x x x x x x
x x x x x x x x
v
The right hand rule works for any of the three parts as long as you know two of them. 278 Example: If a charge is traveling with a velocity (v) down the page and there is a magnetic force (Fm) acting on the charge coming out of the page, as shown below, what is the direction of the magnetic field (B)? .
Fm (out of the page)
v
Using the right hand rule, your thumb points down the page (v) and the palm of your hand is pushing out of the page toward your face (Fm). This makes your fingers point to the right (B). So the direction of the magnetic field around the charge is to the right. .
Fm (out of the page)
B
Example: Let’s look at when velocity is not known. If a charge is moving through a magnetic field (B) that is pointed up the page and there is a magnetic force (Fm) acting on the charge to the right, what is the direction of the charge velocity (v)? B
Fm
Using the right hand rule, your fingers point up the page (B) and the palm of your hand is pushing to the right (Fm). Therefore your thumb ends up pointing into the page (vin). So the direction of the velocity of the charge is into the page (x). B
vin
x Fm
279 Let’s put it all together. Example: A 6.0 nC charge traveling downward (‐y direction) at 4.0 x 106 m/s experiences a force of 8.0 x 10‐3 N into the page (‐z direction) as shown below. B=?
x Fin
v
Determine the magnitude and direction of the magnetic field. For the calculation, using the equation for magnetic force: Fm = Bqv 8.0 10
6.0 10
4.0 10
B = 0.33 T (This is the magnitude, now apply the right‐hand rule for direction). B
B field direction to the left (-x direction)
x Fin
v
280 17.6 Magnetism and Circular Motion A charge moving through a magnetic field has a force that will act on it perpendicular to the direction of motion. This force will change the direction of motion which changes the direction of the charge velocity. By the Right‐Hand‐Rule, the direction of the force will change which once again alters the direction of motion of the charge as shown below. x x x x x x x x x x x x
q
v
x x x x x x x x x x x x
v
Fm
v
x x x x x x x x x
Fm
x x x x x x x x x
Fm
x x x x x x x x x
Fm
Fm
v
x x x x x x x x x
q
v
x x x x x x x x x
x x x
x x x Bin
x x x
x x x
x x x
Since the magnetic force (Fm) always acts towards the center of the circular path of the charge, it must act like a centripetal force (Fc). So, Fm = Fc. Fm Bqv Example: A charge (q) of 8.0 x 10‐16 C with a mass (m) of 3.0 x 10‐22 kg is moving through a 5.0 T magnetic field (B) with a velocity (v) of 4.0 x 105 m/s. Determine the radius (r) of its circular path. Fm = Fc Cancel one of the velocities (v) on each side 3.0 10
4.0 10
5 8.0 10
r = 0.03 m (3 cm) 281 17.7 Magnetic Force on a Current Carrying Wire A current passing through a wire must generate a magnetic field (Bw) that surrounds the wire based on the direction of the current. If you apply the right hand rule in a different way, you can determine what the magnetic field around the wire looks like. In this case, point your thumb in the current (I) direction and then curl your fingers into a circle, this reveals the direction of the magnetic field lines (Bw) around the wire. Bw
I
If the current carrying wire (with its own magnetic field, Bw) is then placed inside a secondary magnetic field (B), the interacting fields create a magnetic force (Fm) on the wire based on the current (I), the magnetic field it is in (B) and the length of the wire that is in the field (l). l
X X X X X X X X X X X X X X X X
I
X X X X X X X X X X X X X X X X
Bin
Formula for magnetic force on a current carrying wire: Fm = BIl
Example: Given a current passing (I) through a 40 cm wire (l) that is in a 0.2 T magnetic field (B), if the wire has a mass (m) of 10 g, determine the amount of current required to float the wire in the magnetic field. Using a free‐body diagram, the weight of the wire (mg) pulls it down and the magnetic force (Fm) lifts it up. To float, the two forces must be equal or in equilibrium. 282 Fm
X X X X X X X X X X X X X X X X
I?
Fm = mg (for equilibrium)
X X X X X X X X X X X X X X X X
mg
Bin
Fm = mg
BIl = mg
0.01 10
0.2 0.4
I = 1.25 A A few important concepts to remember 1. All magnetism is the result of moving electric charge. 2. All magnets have two poles, N‐Pole and S‐Pole. 3. Like poles repel, opposite poles attract. 4. A single magnet produces a magnetic field; it takes two magnetic fields to produce a magnetic force. 5. The magnetic field around a single moving charge or a current carrying wire travels in a circle. 6. A charge will travel in a circular path within a magnetic field. 7. The magnetic force is similar to the electrostatic and gravitational forces. 8. A current in a wire will generate a magnetic field and thus will experience a magnetic force when placed inside another magnetic field. 283 Questions and Problems 1. Given the direction of the velocity of a moving charge (v) and the direction of the magnetic field (B), determine the direction of the magnetic force (Fm). (a) B
(b)
v
x x x x x x
x x x x x x
v
Bin
x x x x x x x
x x x x x x x
v
(e)
(d)
. . . . . . . . .
. . . . . . . . .
(c)
Bout
v
(f)
.
x
vin
B
. . . . . . . . .
. . . . . . . . .
B
vout
B
2. Given the direction of the velocity of a moving charge (v) and the direction of the magnetic force (Fm) on the charge, determine the direction of the magnetic field lines (B). Fm
(a)
(b)
(c)
x
v
v
Fout
.
vin
Fm
(d)
(e)
(f)
v
x
Fin
v
Fm
Fm
.
vout
284 3. Given the direction of the magnetic field lines (B) and the direction of the magnetic force (Fm) on a charge, determine the direction of the velocity of the moving charge (v). (a) B
(b)
x x x x x x
x x x x x x
x
Fin
Bin
(d)
x x x x x x
x x x x x x
(e)
.
B
Fout
. . . . . .
. . . . . .
(c)
Fm
. . . . . .
. . . .Fm. .
Fm
(f)
Fin
x
B
B
4. Draw the magnetic field lines between each set of magnetic poles. (a)
(b)
N
N
Bout
(c)
S
N
S
S
5. If you were given an unmarked bar magnet, explain how you could determine which end was the north pole and which end was the south pole. 6. (a) Determine the magnetic force on a 3.0 x 10‐16 C charge moving to the right at a velocity of 5.0 x 105 m/s in a 4.0 T magnetic field that is moving into the page. (b) What is the direction of this magnetic force? 7. (a) Determine the magnetic force acting on a 60 cm long wire with a 0.5 A current moving up the page in a 0.2 T magnetic field moving to the right. (b) What is the direction of this magnetic force? 8. A 5.0 x 10‐16 C charge with a mass of 8.0 x 10‐22 kg is moving at 2.0 x 105 m/s in a 3.0 T magnetic field. Calculate the radius of its path of motion. 285 9. A 20 cm long wire carrying 0.6 A of current experiences a 0.1 N magnetic force when placed in a magnetic field. Determine the strength of that magnetic field. 10. A Z‐boson particle (a subatomic particle related to the weak nuclear force) has a mass of 3.1 x 10‐25 kg and a charge of 1.6 x 10‐19 C. One is traveling in a 2.0 m radius circle in an 8.0 T magnetic field. Determine the velocity of the Z‐boson particle. 11. A 60 cm wire is placed in a 2.0 T magnetic field directed out of the page. A 0.4 A current is sent to the left through the wire. (a) Calculate the magnetic force on the wire. (b) If the wire has a mass of 0.02 kg, determine the weight of the wire. (c) Draw a free‐body diagram of this scenario. (d) Based on the weight of the wire, determine how much current is now required to make the same 60 cm wire float in equilibrium in the 2.0 T magnetic field. 12. A 6.0 x 10‐22 kg particle with a charge of 4.0 x 10‐15 C is traveling in a 0.2 T magnetic field at 7.5 x 106 m/s. (a) Determine the acceleration of the particle. (b) What type of acceleration is this? 13. A 4.0 nC positive charge with a mass of 2.0 x 10‐20 kg, is traveling at 5.0 x 106 m/s to the right (+x direction) near the surface of the Earth. At this location, there is a 100 N/C electric field acting to the left (‐x direction) and a magnetic field of 5.0 x 10‐5 T acting upward (+y direction). (a) Determine the magnitude and direction of the gravitational force on this charge if it is near the surface of Earth. (b) Determine the magnitude and direction of the electric force acting on this charge. (c) Determine the magnitude and direction of the magnetic force acting on this charge. (d) Draw a diagram, one for each situation, of (a), (b), and (c). Be certain to include the charge, its velocity, the appropriate field, and the force vectors. 286 Chapter 18 – Waves and Vibrations 18.1 Simple Harmonic Motion, Periodic Motion, and Springs There are many things in the universe which vibrate or oscillate: pendulums, guitar strings, springs, etc. Even at the atomic level things vibrate. Waves are related to vibrations. Earthquakes, water waves, and sounds are all produced by some sort of vibration. Even objects that appear to be rigid, such as a building, have waves that pass through them. If a mass is attached to the end of a spring and the spring is hung so it is stretched by the mass, the spring stretches a distance (x) relative to the amount of hanging mass (m) and the elastic constant of the spring (k). They are related by Hooke’s Law, which is just an equation: Fs = ‐kx Fs – Spring (elastic) force [N] k – Elastic constant [N/m] x – Displacement from equilibrium [m] The negative sign indicates that the spring force always opposes any action on the spring. In other words, if you pull a spring to the left, the spring force will automatically act to the right. If you compress a spring upward, the spring force will be downward. Spring force is always a reaction force, which is the point of the negative. As the negative sign is more of a concept, it will not be used in any calculations that follow. Looking at the situation below, an unstretched spring is in equilibrium. When a mass is hung from the spring it stretches a limited distance to a new equilibrium position. At this point, the spring force (Fs) is equal to the weight (mg) of the mass attached to it. k
k
Fs = kx1
m
x1
m
mg
287 Example: If a 0.5 kg mass is hung from a spring with an elastic constant of 40 N/m, how far will it stretch? The spring force is balanced by the weight, so they must be equal. (Again, drop the negative for calculations). Fs = kx = mg (40)x = (0.5)(10) x = 0.125 m If I pull down on the mass slightly (x2) from this new equilibrium position, it will then begin to oscillate (move up and down) around its equilibrium point. k
k
k
-x2
m
x2
m
m
A graph of its displacement (x2) versus time (t) would look like this: x (m)
x2
T
A
t (s)
- x2
Any type of motion that can be graphed in this manner is said to be Simple Harmonic Motion. Simple Harmonic Motion (SHM) only occurs in situations where the force required to bring an object back to equilibrium is proportional to the displacement of the object from equilibrium, such as in a spring. This should be distinguished from a similar thing called Periodic Motion. Periodic Motion is when an object repeats the same cycle of motion at regular intervals. Both SHM and Periodic motion have things called period and frequency. 288 So what’s the difference? Well, all SHM is also Periodic Motion, so that is easy. However, not all Periodic Motion is SHM. For example, an oscillating spring is both Simple Harmonic and Periodic. However, a bouncing ball, a water wave, or a planet in orbit around the Sun are all examples of periodic motion (they repeat the same cycle in a set time period) but are not SHM (they do not oscillate around an equilibrium point). Hope that clears that up. Anyway, back to the graph of SHM. x (m)
x2
T
A
t (s)
- x2
The graph has three specific parts, amplitude (A), period (T), and frequency (f). Amplitude (A) – the maximum displacement from equilibrium, measured in meters [m]. Period (T) – the time to execute one complete cycle of motion (i.e. return to the starting point) measured in seconds [s]. Frequency (f) – the number of cycles completed per time period, measured in Hertz [Hz]. Amplitude is basically displacement of the mass from equilibrium. However, period and frequency have specific formulas. (NOTE: Period and frequency exist in periodic motion, too). The formula for the period of a mass oscillating on a spring is: Ts – Period of spring [s] m – mass [kg] k – Elastic constant [N/m] f – Frequency [Hz] T – Period [s] 289 Example: A 0.2 kg mass is hung from a 50 N/m spring. The mass is pulled back 4.0 cm and allowed to oscillate. Determine the (a) amplitude, (b) period, and (c) frequency of its motion. A = x = 0.04 m (4.0 cm is as far as the spring can and will stretch). 2
0.2
50
Ts = 0.4 s 1
0.4
f = 2.5 Hz 18.2 Pendulums A pendulum is another device that exhibits simple harmonic motion (and periodic motion). It is similar to a spring in that it has a mass that oscillates around an equilibrium point. However, a pendulum does not have an elastic constant, and as we learned with spring potential energy (PEs = ½kx2) it has nothing to do with mass. It turns out that the motion, period, and frequency of a pendulum is due to the length of the pendulum and the strength of the gravitational field around the pendulum. Pendulums of specific lengths have very steady and specific periods. This is why they are used in clocks. Galileo first discovered this while watching a chandelier in a church rock back and forth. He timed its motion using his pulse rate as a time keeper. He later experimented with pendulums of different length and mass and determined a relationship for the period of a pendulum. The formula for the period of a pendulum: Tp – Period [s] l – Length [m] g – gravity [m/s2] 290 Example: Determine the period (on Earth) of any pendulum that is 50 cm long. 2
Tp = 1.4 s 0.5
10
Regardless of the mass attached at the end. 18.3 Wave Motion in Mechanical Waves Periodic motion is related to wave motion, but what exactly is a wave? First, there are two types of waves, mechanical waves and electromagnetic waves. If you drop a stone on the water, water waves move out from the center point where the stone hit the water. If an object were floating on the water, it would move up and down as the wave passed by, but would not move forward or backward. The water itself moves up and down but is also not carried forward with the wave. What is happening is that the kinetic energy of the stone is transferred to the water, and that energy is then carried through the water to another place. The water is considered a medium, a material through which the wave energy can travel. This type of wave is a mechanical wave because all mechanical waves require a medium through which to travel. A mechanical wave cannot travel through the vacuum of space. We will look at electromagnetic waves later. There are two types of mechanical waves, a traveling wave and a standing wave. 18.3.1 Traveling Waves If I attach one end of a slinky to a wall and hold the other, I can give the slinky a little flip. This creates a bump (called a pulse) that travels down the slinky, hits the wall, and comes back. This is known as a traveling wave. There are two types of traveling waves. The first is called a transverse wave because the disturbance in the medium moves perpendicular to the direction of the wave velocity. A water wave is a typical transverse wave as the “bump” created by the wave moves up and down as the wave itself moves forward. v
The second type of traveling wave is called a longitudinal wave because the disturbance in the medium moves parallel to the direction of the wave velocity. This is a little harder to understand. If I take a large slinky and squeeze the coils together at one end, the squeezed part, called the compression, will move through the slinky compressing parts that it moves through and stretching out the parts it passes, called the rarefaction. 291 rarefaction
v
compression
In each case, the wave has a specific amplitude (A), period (T), and frequency (f). However, these waves have something else ‐ wavelength (λ) and velocity (v). The wavelength (λ) is the distance between the peak of one wave and the peak of the next. A (m)
x2
λ
A
x (m)
- x2
It is very similar to the idea of period, only it is a measurement of length [m], not time. However, it is clear to see that on the same graph of a wave’s motion, both wavelength (λ) and period (T) correspond to the same location. x (m)
x2
λ (m)
A
T (s)
t (s)
- x2
If a certain distance is covered over a certain time , which would give units of (m/s) then this must somehow related to a wave velocity . That is nice, but since there is a more common relationship for wave velocity that includes wavelength and frequency. v = λf v – Velocity [m/s] λ ‐ Wavelength [m] f – Frequency [Hz] 292 Example: If you have a wave that is 50 cm from one peak to the next and has a period of 0.1 s, determine (a) frequency, and (b) velocity of the wave. v = λf = (0.5)(10) = 5 m/s As waves travel, they often encounter barriers, changes in media, or other waves. Depending on the shape of the wave, the shape of the barrier, the change in medium, and other factors, waves can undergo four processes. These are reflection, diffraction, refraction, and interference. 18.3.2 Reflection When a wave encounters a solid barrier that it cannot get past, it will bounce off, or reflect off the barrier. Again, a wave is a form of energy, just like a falling ball. When that ball hits the ground, it tends to bounce off the ground, losing some of its energy. When a wave strikes a barrier, it will also lose some of its energy, but that will not change its velocity, frequency, or wavelength as long as the medium has not changed. The reflection of the wave will depend on (1) the shape of the wave, (2) the shape of the barrier, and (3) the angle at which it strikes the barrier. To begin with, let’s consider waves with a flat wave front shape. If a flat wave front strikes a flat barrier head‐on, it will reflect back in the opposite direction with no change in shape as shown below. Barrier
Barrier
Reflected wave
Incoming wave
If the barrier is angled, then the wave will reflect off the barrier so that it runs now parallel to the barrier itself. The greater the angle of the barrier, the greater the reflection of the wave. Barrier
Barrier
Incoming wave
Reflected wave
293 More interesting is when the barrier is curved and a flat wave strikes the barrier. Depending on which way the curve faces the barrier produces different results. If the barrier is curved inward (concave), then the flat wave fronts are reflected at different times. However, the curve forces each point on the wave front to be reflected towards a specific point called the focus or focal point of the curve. This then forms circular wave fronts that radiate out from the focal point. Focal point
Concave Barrier
Concave Barrier
Reflected wave
Incoming wave
Concave Barrier
Focal point
New circular waves
That focal point is an interesting location. If I were able to originate a circular wave at that focal point, when it hit the concave barrier, all points on the wave would hit at once and match the barrier. This would then cause a flat wave front to reflect off the concave barrier. Concave Barrier
Concave Barrier
Focal point
Incoming circular waves
Flat reflected waves
294 If the curved barrier is turned around (convex) and it encounters flat wave fronts, the reflected wave will be curved and move away from the barrier. If you were to trace the reflected waves backwards (make smaller curves) you would find that they appear to originate at the focal point. Convex Barrier
Convex Barrier
Focal point
Incoming wave
18.3.3 Diffraction When a wave encounters a barrier, but it has the ability to get past it, either around the side or through an opening, the part of the barrier that the wave encounters will cause the wave to bend and change directions. Think of it like friction on the wave by the barrier in which the barrier will slow that part of the wave while the untouched part of the wave will maintain its original speed. For example, when a flat wave encounters a barrier where it has the ability to get past one side of the barrier, it will bend at that point, but not at points farther from that side of the barrier, producing a curved shape at the point of contact with the barrier. Barrier
Barrier
Diffracted wave
Incoming wave
295 If the flat wave encounters an opening within the barrier, it will be diffracted on both sides of the opening, so it will slow down on both sides while the center continues to move on at its original speed. This will form a curved wave front after the opening. Opening in Barrier
Opening in Barrier
Diffracted wave
Incoming wave
18.3.4 Refraction Refraction is the bending of a wave due to a change in the medium. This is different than diffraction which is just the bending of a wave around a barrier. When the medium through which a wave is traveling is changed, the velocity and the wavelength of the wave will also change, the frequency however remains constant. This change in velocity of the wave can cause it to bend. A classic example occurs at the beach when an ocean wave moves from deep water to shallow water. This is a change in the medium and will often cause the wave to change directions. One way to look at it again is like friction, the shallower area starts to drag the wave and slow it down. If one part of the beach is shallower than another, the shallowest part will slow the wave more than the deeper part causing it to bend. Deep water
Shallow
Water
Refracted wave
Incoming wave
296 If the shallow shape is a convex, the waves will be bent in on each other, so they will converge toward each other. Convex shape
Converging refracted wave
Incoming wave
If the shallow shape is a concave, the waves will be bent outward, so they will diverge away from each other. Concave shape
Diverging refracted wave
Incoming wave
We will see this similar effect when we study optics. 18.3.5 Interference Sound and other waves have the ability to interfere with each other in good ways and bad ways. When two or more waves pass through the same region in space, they have the ability to interfere, particularly if they are the same frequency. If the two waves meet in phase, they undergo constructive interference in which they will add to each other making a larger amplitude wave with the same frequency. Becomes
297 However, if they meet out of phase, they undergo destructive interference and cancel each other out (perfectly if they are the same amplitude and frequency). Becomes
18.4 Sound If one plucks a guitar string or blows through a reed or strikes a drum head, a sound is formed. In each case, something is vibrating. All sound waves are produced by something vibrating. (So if a tree does fall in the woods, it does make a sound, even if nobody is there to hear it). Consider a tuning fork, when it is struck it vibrates back and forth causing the air around it to be compressed. The compression of air then moves away from the tuning fork like ripples in the water where a stone is dropped. The air then carries the sound energy from one place to another. Sound requires a medium through which to travel, making it a mechanical wave. Sound waves cannot travel through the vacuum of space, so all those movies where explosions make sounds in space are quite wrong. The speed at which a sound wave travels depends on the medium through which it is traveling and is greatly affected by density and temperature. Sound travels faster in more dense materials such as solids than in less dense materials such as gases. This is because in a solid, the molecules are more closely packed and can transfer the energy faster. vs (air) = 340 m/s vs (water) = 1490 m/s vs (aluminum) = 5100 m/s Although the speed of sound is different in different media, it travels at a constant speed within any medium. 298 Example: You see a lightning flash off in the distance and 3.0 seconds later you hear the thunder that was generated by the lightning. So how far away was the lightning? vs = 340 m/s x = vot + ½at2 (since constant velocity, a = 0 m/s2) t = 3.0 s x = vt x = (340)(3) v = 1,020 m (Just over 1.0 km away, so you better get inside soon). Sound waves can fall into different three categories based on the frequency of the sound wave. Infrasonic – Low frequency sounds of less than 20 Hz. Audible – Range in which the human ear is sensitive (20 Hz – 20,000 Hz) Ultrasonic – High frequency sounds of more than 20,000 Hz Many animals can detect infrasonic sounds as these tend to precede earthquakes. Others can detect ultrasonic sounds such as dog whistles. 18.5 Intensity and the Decibel Scale Since sound is a form of energy, it has physical power. The sound energy passes through an area in space at a certain rate. The greater the energy or the smaller the area affects what is known as the intensity of the sound. Intensity is basically how loud the sound is. I – Intensity [W/m2] P – Power [W] 2
A – Area [m ] The faintest sound we can hear must have an intensity of 1.0 x 10‐12 W/m2. This is known as the Threshold of Hearing. The loudest sound that the human ear can tolerate before it can be damaged is about 1.0 W/m2. This is known as the Threshold of Pain. However, we do not often hear of sound being measured in W/m2. Sound is usually measured on a scale called the Decibel Scale. The Decibel Scale gets its name from Alexander Graham Bell. At the Threshold of Hearing, I = 1.0 x 10‐12 W/m2, so on the Decibel Scale: β = 0 dB At the Threshold of Pain, I = 1.0 W/m2, so on the Decibel Scale: β = 120 dB 299 Some other levels to compare on the Decibel Scale: Airplane – 150 dB Subway – 100 db Conversation – 60 dB Whisper – 30 dB Good Rock Concert – 110 – 120 dB Prolonged exposure to high decibel levels can lead to hearing damage. 18.6 Doppler Effect When a sound is made, it spreads out in all directions, which is why everyone can hear a sound, not just the people who are directly in line with it. So a stationary object would create sound waves that look like a rock dropped in a pond. All observers of this sound hear the same sound at the same frequency (f). The intensity (loudness) of the sound will depend on how far from the source of the sound each observer is standing. Constant f , λ, and v
If you listen to a car blowing a horn as it moves towards you or away from you, the intensity of the sound grows or shrinks. However, the pitch (how high or low of a frequency) of the sound also appears to change. This is because the frequency of the sound does appear to change for an observer. This change in observed frequency due to motion of the source of the sound or the observer of the sound is called the Doppler Effect. This has to do with the motion of the source or observer and the speed and direction of the sound wave. The sound waves created by a moving source travel at a specific speed (aka the speed of sound (vs)). As the source moves faster (v), it starts to catch‐up with its own sound waves. This makes the sound waves bunch closer together which decreases the distance between each wave. This has the effect of decreasing the wavelength (λ) of the sound in front of the moving source. 300 v
vs
According to the formula for wave velocity v = λf, if wavelength (λ) decreases, then the frequency (f) must increase for the velocity (v) to remain constant. Therefore, as a moving source approaches an observer, the observed frequency increases. However as the sound passes the observer and moves away, the trailing sound waves are moving farther apart, causing the wavelength (λ) to increase. If λ increases then f must decrease for the velocity (v) to remain constant. This makes the observed frequency decrease as the sound moves away. λ increasing
λ decreasing
v
vs
Observed f decreasing
Observed f increasing
18.7 Speeds Relative to Sound Waves from a stationary source travel out from the source in circular patterns much like ripples on the water when you drop a stone in the water. When the source of the sound starts to move, it starts to catch up with its own sound waves. As long as its sound waves remain out in front, it is said to be traveling at subsonic speed, slower than the speed of sound. At this point, the sound will reach observers before the source. v
vs
Subsonic Speed
Eventually, a source may be able to travel fast enough to catch its own sound waves. It is now traveling at the speed of sound or what is known as sonic speed. At this point, the sound wave 301 is traveling with the source, so any observer will not hear the sound until the source is next to the observer. So you would not hear the source coming until it is right next to you. v
vs
Sonic Speed
With the modern age, we gained the technology to travel at speeds faster than sound. This is called supersonic speed. At this point, the sound waves made by the source are trailing behind the source. So the source of the sound could pass an observer and the observer would not hear it until later when the sound wave comes past. There is a second effect of traveling supersonic speed, and that is the sonic boom. When you travel faster than sound, you create a large pressure wave in the air called a mach cone. This compresses the air so fast that it creates that loud sound. Mach cone
v
vs
Supersonic Speed
Thunder is an example of a sonic boom. A lightning flash superheats the air causing it to compress and form the high pressure sound wave. This pressure wave is strong enough to rattle buildings and shatter windows. There is a ratio of speeds versus the speed of sound called the Mach number. M – Mach number (no units) v – speed of object [m/s] vs – speed of sound [m/s] M < 1 – subsonic speed (M = 0.7 = 0.7 x speed of sound) M = 1 = sonic speed M > 1 – supersonic speed (M = 1.5 = 1.5 x speed of sound) 302 The fastest speed on record for an aircraft is the SR‐71 spy plane that could travel at a speed of M = 3.3. This aircraft was used to overfly the Soviet Union during the Cold War to take pictures. Rockets and missiles travel even faster; the Space Shuttle would hit a speed of about M = 25 during liftoff. 18.8 Standing Waves Many musical instruments create a special kind of sound wave known as a standing wave. This is a wave that oscillates within the medium and does not travel anywhere. However, the result of this standing wave is a sound wave that does travel away from the standing wave. 18.8.1 Standing Waves in Strings If I pluck a guitar string or rub a bow across a violin string, it vibrates in a specific pattern giving a certain note (frequency). I can change that note by either plucking a different string (which is either heavier of lighter than the other), or by pressing my finger on the string (changing its length) and then plucking it, or by changing the tension on the string. So the three things that affect the note (frequency) of a string are: length of string, tension, and mass of string. As the length of a string increases, its frequency decreases; as the tension on a string increases, its frequency increases; as the mass of a string increases, its frequency decreases. I can get more than one note out of the same string by pressing my finger down on the string at a certain point. This has the affect of creating what is known as a node. A node is the stationary point of a standing wave. The large vibrating point on a standing wave is called an antinode. antinode
nodes
f1
antinodes
nodes
f2
As you can see, a string must have at least 2 nodes, but can have more. A string must have at least one antinode, but can also have more. The number of antinodes affects the frequency or note of the sound. The string in the first picture with one antinode will generate what is known as the fundamental frequency for the string. This indicates the lowest frequency of sound or the lowest note. 303 This note is found by: f1 – Fundamental frequency of string [Hz] v – Speed of vibrating string [m/s] l – Length of the string [m] The reason for the 2 in the denominator of this equation is that the fundamental frequency of a string produces ½ of a wavelength (λ). Example: A string vibrating at 100 m/s that is 20 cm long, the frequency of the sound produced is .
= 250 Hz If this string is made to vibrate at a different rate to set up two antinodes, or three antinodes, then the sound produced will be a harmonic of that fundamental note. This is found by: fn = nf1 (n = 1, 2, 3, 4…) Example: If the fundamental frequency (f1) was 250 Hz, then the 2nd harmonic (f2) is: f2 = 2f1 = 2(250) = 500 Hz. The third harmonic (f3) is: f3 = 3f1 = 3(825) = 750 Hz. 18.8.2 Standing Waves in Open Tubes Standing waves are also found in wind instruments. The best example is a pipe organ. As the air in a pipe organ is vibrated, it sets up a standing wave inside the pipe. However, a pipe can either be closed at one end and open at the other, or open at both ends. An open pipe is open at both ends. Because it is open at both ends, an antinode must exist at each end. Like a string, the length is a very important factor to the note that comes out of the pipe. Also like a string, the open tube has a fundamental frequency and harmonics depending in this case on the number of nodes. 304 nodes
node
antinodes
f1
antinodes
f2
The fundamental frequency of an open tube is found the same way as the string: f1 – Fundamental frequency of open tube [Hz] v – Speed of sound [m/s] l – Length of the open tube [m] You will notice a “2” in the denominator again because the fundamental frequency produced in an open tube is ½ of a wavelength (λ). The harmonics are also found the same way: fn = nf1 (n = 1, 2, 3, 4…) Example: Find the 3rd harmonic of a 30 cm long open tube using a speed of sound of 330 m/s: f4 = 3f1 = 3
3
.
f4 = 1650 Hz 18.8.3 Standing Waves in Closed Tubes If one end of a tube is open and the other is closed, then a node is always created at the closed end and an antinode at the open end. This is similar to when you blow across the top of a water bottle it makes a certain note. If you drink some of the water and blow again, you get a different note because you have changed the length of the air column in the tube. This effect of changing the note produces and limits the number of harmonics to the odd harmonics only. This limits the notes that are produced to f1, f3, f5, etc. In other words, a closed tube only produces the odd harmonics, never the even harmonics. 305 node
antinode
f1
antinodes
node
This also changes the equation for the frequency of a closed tube: f3
The harmonics are found the same way: fn = nf1 (n = 1, 3, 5, 7…) Only odd harmonics. Example: If our 30 cm long tube is closed and I want the 4th harmonic, it does not exist. If I want to find the 3rd harmonic (using a speed of sound of 330 m/s): f3 = 3f1 = 3
3
.
f3 = 825 Hz 18.9 Resonance Specific sound frequencies are often produced by a process called resonance. When an object resonates, it is vibrating at what is known as its natural frequency. The natural frequency of any object is based on many characteristics including material, shape, size, etc. Resonance tends to produce the maximum amplitude a vibration can have. If two materials are similar, the vibrations of one can cause the other to resonate or vibrate without any physical contact. For example, two tuning forks with the same natural frequency are placed near each other. If one is struck, the second one will start to vibrate. This is because the sound waves from the one tuning fork travel at the same frequency as the resonant frequency of the second tuning fork. When these waves hit the second turning fork, it causes it to resonate because the sound frequency matches the natural frequency of the tuning fork. If the tuning forks were of different frequencies, this would not work. The first would send out sound waves that would strike the second, but would not produce vibrations in the second tuning fork because the sound frequency would not match the natural frequency of the second tuning fork. 306 18.10 Electromagnetic Waves So far, we have been discussing mechanical waves, waves that require a medium in order to travel from one place to another. A second type of wave, called an electromagnetic wave (EM), does not require a medium to travel from one point to another. The theory behind electromagnetic waves was first developed by James Clerk Maxwell. Maxwell was the first to suggest a direct relationship between the invisible electric field and the invisible magnetic field. He did this through a “thought experiment”, one that is done on paper but not in a lab. He did this with the idea of charged parallel plates and a current carrying wire. Maxwell knew that a current carrying wire had a magnetic field (B) around it and that charged parallel plates had an electric field (E) between them. But how could the current pass through the empty region between the two parallel plates to return to the other end of the voltage source used to charge the plates? He reasoned that somehow, the electric field (E) carried with it the magnetic field (B) across the plates. This allowed both the magnetic field and the current to continue through the circuit. So in this empty region between the two parallel plates, was something he called electromagnetism, the combination of electric and magnetic fields that could transport energy through empty space. (See the figure below). B
........
I
+
B
........
B
........
E
xxxxxx
xxxxxx
I
xxxxxx
I
+
-
I
V
From Maxwell’s thought experiment, he developed some mathematical laws relating to this idea of electromagnetism. While doing his work around what would become electromagnetic waves, he calculated the ratio of the electric field strength (E) to the magnetic field strength (B) in a vacuum and came up with approximately 3.0 x 108 m/s, which is the speed of light. This was a very important discovery, and lead to the primary rule about electromagnetic waves: All electromagnetic waves travel at the speed of light in a vacuum. Later on, Heinrich Hertz proved Maxwell correct. He set up an electric coil attached to a voltage source on one side of his lab and another that was not attached to anything on the other side of the lab. When the sparks jumped between the ends of one coil, they suddenly appeared to 307 jump between the ends of the coil at the other end of the lab with no physical connection. So energy had to be traveling across the lab from one coil to the other. Sparks
High V
Invisible
EM waves
Though Hertz did not realize it, he had just created a transmitter and a receiver. Later, an Italian named Marconi read about Hertz’s experiment and invented the radio. Since an EM wave is a wave, it has the same characteristics as a mechanical wave except that it does not require a medium to travel. It does, however, have amplitude (A), period (T), frequency (f), and velocity (v). Therefore, it follows wave velocity formula, v = λf, however for all EM waves, v = c, so: c = λf (for all EM waves, where c = 3.0 x 108 m/s) An EM wave is a traveling wave and is also considered a transverse wave. 308 18.11 EM Spectrum Since EM waves travel at the speed of light and they are waves, c = λf. With c being a constant, wavelength and frequency can be changed. As one goes up, the other goes down. This creates a large range of possible EM waves. This range of electromagnetic waves is known as the Electromagnetic Spectrum. The EM spectrum has a very large range which can be divided into three parts based on energy levels.
Picture courtesy of "Electromagnetic radiation." New World Encyclopedia. 25 Feb 2009, 23:06 UTC. 23 Jun 2010, 14:00 Low Energy Waves ELF AM FM, TV Radar, Microwave Medium Energy Waves IR Visible UV High Energy Waves UV X – Ray Gamma Ray f (Hz) λ (m) < 1.0 x 104 3
3 540 x 10 – 1600 x 10 88 x 106 – 108 x 106 3.0 x 1011 > 30,000 555 – 187.7 3.4 – 2.7 .001 f (Hz) λ (m) 3.0 x 1011 – 4 x 1014 4.0 x 1014 – 7.5 x 1014 7.5 x 1014 – 5 x 1015 1.0 x 10‐3 – 7.0 x 10‐7 7.0 x 10‐7 – 4.0 x 10‐7 4.0 x 10‐7 – 6.0 x 10‐8 f (Hz) 5.0 x 1015 3.0 x 1016 – 3.0 x 1021 3.0 x 1018 – 3.0 x 1022 λ (m) 6.0 x 10‐8 1.0 x 10‐8 – 1.0 x 10‐13 1.0 x 10‐10 – 1.0 x 10 ‐14 309 18.11.1
Medium Energy Waves – Light Waves Infrared (IR) – Light produced by hot objects. Anything that gives off heat will also give off infrared light. As this light range is outside of the range in which the human eye can detect, we do not see anything. Some animals, particularly desert snakes and pit vipers (rattlesnake), can see into the infrared. This allows them to hunt at night by being able to see the light given off a warm animal against the cold background. 310 Visible Light – Light that can be detected by the human eye. Visible light consists of all the color and white light. There are six specific frequencies of light that form the basis of all colors of light. They are Red, Orange, Yellow, Green, Blue, and Violet (ROYGBV). (You will notice that there is no Indigo (I) because it is not a true member of the color spectrum). The combination of all these colors at once forms white light. Black is the complete absence of all color or all light. All other colors of light are a combination of one or more of these colors. (NOTE: Colors of light are different from paint colors called “pigments”). The visible light range is the smallest range on the entire electromagnetic spectrum. Ultraviolet (UV) – Light produced by stars, it has a higher energy than other light and is outside the human eye range. (There are some species, particularly bees that see ultraviolet light). Ultraviolet has enough energy to penetrate the skin and stimulate the production of a “tanning” pigment in most skin called melanin. This pigment darkens our skin to try to prevent any more ultraviolet light from penetrating our skin. So a tan is our skin trying to defend itself from the Sun. (Both the UVA and the more dangerous UVB rays are at this level). Ultraviolet light has had a big role in human evolution when it comes to permanent skin color. Humans who evolved in parts of the world with a lot of sunshine all year (Africa, Middle East, Mediterranean, Pacific Islands, Caribbean Islands, etc.) have evolved with darker skin to fight off the constant ultraviolet light. Humans who evolved in parts of the world with less sunlight (England, Scandinavia, Canada, Siberia, etc.) have evolved with lighter skin as they do not need as much protection. Can’t tan very well? Blame your ancestors. 18.11.3
We will use this:
316 19.2 Reflection When light that is traveling through a medium encounters a second medium, it must strike the boundary that lies between the two media. Depending on the characteristics of the second medium, some or all of the light will be reflected off of the boundary the same way a water wave will reflect off a barrier. If the boundary is smooth, then all parallel light rays striking the boundary will be reflected off in the same direction parallel to each other, as shown below on the left. If, however, the boundary is rough, those parallel rays will be reflected in different directions (called dispersion) as shown. For example, on a normal day, we do not see any glare of sunlight coming off the dark asphalt of the road. However, when it rains, the water fills in the broken surface making it smooth. When the sun hits the road, we get a glare as all the sunlight is reflected into our eyes rather than being dispersed in all directions. Smooth boundary reflection
Rough boundary dispersion
We will deal mostly with smooth boundaries in this unit. The incoming light ray that strikes the boundary is called the incident ray (incident meaning “to happen upon or coming in contact with”). The ray that bounces off the boundary is called the reflected ray. Together they form the Law of Reflection. Law of Reflection – The angle of incidence equals the angle of reflection. (Hard one, isn’t it)? θ1 – Angle of incidence
θ1
θ1'
θ1′ – Angle of reflection
Boundary
Normal line
The formula for the Law of Reflection says: θ1 = θ1′ 317 NOTE: All angles in light are based upon the angle formed by the light ray and a line drawn “normal” (perpendicular) to the boundary. This is very important to remember. Basically, the Law of Reflection is an exercise in Geometry (and I know you love Geometry). Example: Given the two mirrors below, they are placed at an angle of 1200 to each other. A beam of light strikes Mirror 1 at an angle of incidence of 650 to the normal line for that mirror. It bounces off the first mirror and strikes the second mirror. We want to determine the angle of reflection off Mirror 2 (again relative to a normal line to that mirror). θ?
650
Mirror 2
1200
Mirror 1
Using the Law of Reflection and a little Geometry, we can find the answer. The reflected angle off of Mirror 1 must be 650 due to the Law of reflection. Then since a normal line must be 900 to the boundary, the angle with respect to the boundary of Mirror 1 must be 250 as indicated below. θ?
650
650
0
250 120
Mirror 1
Mirror 2
The shape formed by the reflected ray and the two mirrors is a triangle. We now know two of the angles of the triangle. As all three angles of a triangle must total up to 1800, the other angle of the triangle must be 350. Again since the ray that strikes the normal line (900) to Mirror 2 then the angle of incidence must be 550 to the normal. θ?
550
650
650
350
0
0
25 120
Mirror 1
Mirror 2
We now use the Law of Reflection again with an angle of incidence of 550 so the angle of reflection (the unknown angle) must also be 550. 318 550
550
650
650
350
0
25 1200
Mirror 1
Mirror 2
That is about all there is to it. And you thought you would never use Geometry for anything. 19.3 Refraction When a light ray in one medium strikes a transparent boundary with another medium, part of the light is still reflected. (This is why you can see a faded reflection of yourself in a window). The remainder of the light passes through the boundary and into the second medium. The light that enters the new medium is bent away from its original path so that it is now traveling a different direction. This is similar to when a water wave moves from deep water to shallow water its direction may change. (NOTE: This only occurs when the light strikes the boundary at some angle to the normal, if light strikes the boundary directly along the normal line, it will not bend). This bending of a light ray due to a change in medium is called refraction. The angle at which the light refracts depends upon the properties of the two media. Medium 1
Medium 2
θ1 – Angle of Incidence
θ1 θ1'
θ2
θ1' – Angle of Reflection
θ2 – Angle of Refraction
Original path
Actual refracted path
This refraction is what makes a straw in a glass of water appear to be broken. The straw is still the same, but the light that comes off of the straw is refracted by the water and the glass making it travel a different path and appear to be broken. One of the reasons that the light bends is because the velocity of light changes as it enters different media. Light is an electromagnetic wave, therefore it travels at the speed of light (duh) in a vacuum, or at 3.0 x 108 m/s. The constant, c. (c = 3.0 x 108 m/s) represents the speed of light in a vacuum. However, upon entering a certain type of glass, light slows to 2.0 x 108 m/s. It is different still for water. Once light returns back to a vacuum, its velocity returns to 3.0 x 108 m/s which is a pretty cool phenomenon. Therefore the speed of light in any medium other than a vacuum will ALWAYS be less than 3.0 x 108 m/s. (NOTE: We will treat air and a vacuum as the 319 same thing since they are close enough). We can determine the speed of light in a medium based on the medium’s index of refraction. Index of refraction (n) – The ratio of the speed of light in a vacuum to the speed of light within a certain medium. n – Index of refraction c – Speed of light in a vacuum (3.0 x 108 m/s) v – Speed of light within the medium [m/s] With this in mind, the smallest possible index of refraction for any material is n = 1, and that is for a vacuum. All other indices of refraction must be greater than 1. (NOTE: The index of refraction has no units and is generally found in a book of materials). Some Common Index of Refractions: Vacuum – 1.0 Air ~ 1.0 Water – 1.333 Ethanol – 1.36 Crown Glass – 1.52 Quartz – 1.54 Flint Glass – 1.62 Diamond – 2.42 Ice – 1.309 Example: The index of refraction for water is 1.333; determine the speed of light in water. 1.333
3.0 10
3.0 10
1.333
v = 2.25 x 108 m/s Remember for any light, 3.0 10 / and 1. Knowing the index of refraction of a medium will allow us to determine the angle of refraction (θ2) within the medium. This is done with Snell’s Law of Refraction. Snell’s Law of Refraction n1sinθ1 = n2sinθ2 320 Example: Given a light ray in air (n = 1) it strikes water (n = 1.333) at an angle of incidence (θ1) of 350. (a) Determine the angle of refraction (θ2) within the water. (b) The light then passes out of the water and back into the air, determine the final angle (θ3) of the light in the air. θ1
air (n = 1)
water (n = 1.333)
θ2
θ3
air (n = 1)
(a) n1sinθ1 = n2sinθ2 (1) sin 35 = (1.333) sin θ2 0.57 = (1.333) sin θ2 (Divide both sides by 1.333) 0.428 = sin θ2 (You will now have to take the sin‐1 of 0.428) θ2 = sin‐1(0.428) θ2 = 25.30 (b) Since 25.30 (θ2) is the angle of refraction going from the air to water, it now becomes the angle of incidence (θ1) going from water (now n1) to air (now n2). (This is due to the Geometric law of alternate interior angles). The angle of refraction (θ2) is now θ3 as it occurs when the light comes out of the water. n1sinθ1 = n2sinθ2 (1.333) sin 25.3 = (1) sin θ3 (Divide both sides by 1) 0.57 = (1) sin θ3 0.57 = sin θ3 (You will now have to take the sin‐1 of 0.57) ‐1
θ3 = sin (0.57) θ3 = 350 (Same as original angle of incidence. Coincidence? I don’t think so). 321 My colleague, Seth Karpinski, uses a great analogy for why the light bends in the different directions. Imagine you are sent out to mow the lawn. As you push the lawnmower along the concrete driveway it moves very easily. Yet when you hit the grass it moves much more slowly. So imagine that you approach the grass with the front wheels at an angle as shown below. concrete
grass
The wheel that hits the grass is suddenly slowed by the grass while the wheel that is still on the concrete continues to move as before. This will cause the lawnmower to suddently pivot towards the grass, changing the angle of motion. concrete
grass
Notice this looks just like when the light went from the fast moving air to the slower moving water. Now when the lawnmower reaches another area of concrete like a neighbor’s driveway, the reverse pivot will happen. The wheel that first hits the concrete starts to move faster than the one that is still on the grass. (Of course, your parents may ask why you are cutting the grass at these strange angles, but at least you are getting the job done). concrete
grass
322 Back to the air to water to air situation. Upon returning back to the air, the light ray returns to its original angle of direction, however, it has been shifted slightly from its original path as shown below. 350
air (n = 1)
25.30
water (n = 1.333)
25.30
air (n = 1)
35
0
Original path
If you ever go to an aquarium which has water and very thick glass, when you look at a fish at an angle, you are not seeing it exactly where it is, but offset from its actual postion. This is why it took a lot of practice for ancient spear fishermen to hit a fish under the water when they were spearing from above the water as the fish was not really where they saw it. NOTE: If light strikes a boundary directly along the normal line (θ1 = 00) so there is no refraction (θ2 = 00). air (n = 1)
θ 1 = 00
water (n = 1.333)
sin 00 = 0
θ 2 = 00
air (n = 1)
NO REFRACTION!!
Like any wave, as light changes medium its velocity changes, but its FREQUENCY NEVER CHANGES. Considering the wave formula v = λf, if the frequency is always constant and the velocity of light changes, then the wavelength must also change. They are in fact directly proportional, as velocity decreases, so does wavelength. Thus Snell’s Law for Refraction and the Index of Refraction can be combined to determine the wavelength of light within a different medium. n1λ1 = n2λ2 323 Example: Given a light ray with a wavelength of 600 nm (600 x 10‐9 m) in a vacuum, it enters a piece of quartz (n = 1.54). Determine (a) the speed of light in the quartz, (b) the wavelength of the light in the quartz, and (c) the frequency of the light. (a) 3.0 10
1.54
v = 1.95 x 108 m/s (b) n1λ1 = n2λ2 (1)(600 x 10‐9) = (1.54)λ2 (λ2 = 390 x 10‐9 m = 390 nm) λ2 = 3.90 x 10‐7 m (c) This can be done two ways with the same formula, just be careful to keep the correct values together. In the vacuum In the quartz v = λf v = λf 3.0 x 108 = (600 x 10‐9)f 1.95 x 108 = (3.90 x 10‐7)f f = 5.0 x 1014 Hz (Answer same for both as f is always constant) 19.4 Critical Angle (θc) When light travels from a low index of refraction to a higher index of refraction (in other words n1 < n2), the light will always bend TOWARDS THE NORMAL. No matter how big the angle of incidence is, there will always be an angle of refraction until you hit 900. In other words, when n1 < n2, for every θ1 there is a definite θ2. n1 < n2
n1
n2
θ1
θ2 n1
n2
n1
θ1
θ1
θ2
n2
θ2
324 However, when light travels from a high index of refraction to a lower index of refraction (in other words n1 > n2), the light will always bend AWAY FROM THE NORMAL. This means at some point, the angle of incidence will be so great that the light ray merely refracts along the boundary and never enters the second medium. n1 > n2 θ1 n1 n2 n1 n2 θ2 n1 θ1 θ1 = θc θ2 n2 No θ2 At this point, the angle of incidence is said to have reached the critical angle (θc). The critical angle is the maximum angle at which refraction will occur, traveling from a high index of refraction to a low index of refraction. At this angle, light will refract along the boundary between the two media. It ONLY occurs when traveling from a high index of refraction to a low index of refraction. It is found by: where n2 < n1. Example: Determine the critical angle between air (n = 1) and water (n = 1.333). NOTE: No matter what, the smaller number must always be n2, or you will get an error answer on your calculator. 1
sin
1.333
θc = 48.60 Once the angle of incidence exceeds the critical angle (θ1 >θc), in the above example, if the angle of incidence (θ1) is greater than 48.60, there is only total internal reflection. This means that the light ray is completely reflected back into its original medium. So even though the boundary is transparent, the light ray will not cross the boundary, but will be completely reflected back into the original medium. n1 > n2
n1
n2
n1
θ1 = θc
θ1 > θc
No θ2
Total internal reflection
n2
325 19.5 Introduction to Color When we talked about electromagnetic waves we discussed a portion of the EM Spectrum called visible light. This was the range that the human eye could detect. It ran from 4.0 x 1014 Hz to 7.5 x 1014 Hz (or 750 nm to 400 nm). The longest wavelength (750 nm) corresponds to the color red while the shortest (400 nm) corresponds with violet. Visible white light is when all the colors of the visible spectrum are incident upon the eye at once. The color black is merely the complete absence of any color or light. White light can be broken into six basic colors that make up the visible spectrum, Red, Orange, Yellow, Green, Blue, and Violet (ROYGBV). Sorry, there is no Indigo as there is no specific wavelength that separates it within the spectrum from violet. The color of an object and the way we see it depends on one of three processes: emission, reflection, and transmission. 19.5.1 Emission Any object whose atoms have been excited by the absorption of energy will emit light as those excited atoms turn around and give off the energy to return to their natural state. For example, when iron is heated until it glows, the added heat energy causes the electrons to give off light in colors of red, yellow, or even white. The sun emits light (but the moon does not), fire is the emission of light and color. Blue flame emits blue light
19.5.2 Reflection Most color is seen through reflection. For example, a red rose is red because when white light strikes it, something about its natural chemical make‐up causes it to absorb all of the light except the red wavelength of light. This red wavelength bounces off the rose and we see it as red. Colors
absorbed
R
O
Y
G
B
V
White
Light
Only red
reflected
326 If the red rose was placed in a room with only green light, there would be no red to reflect and it would absorb the green, so it would appear black to us, no color. Anything that is white will reflect any color of light. So if a white rose is placed in a green lit room, it will appear to be green as white reflects all colors. Anything that is black absorbs all colors and will always appear black under any light. 19.5.3 Transmission This occurs with transparent materials that only let certain colors pass through. At theaters and in concerts, it would be difficult to have multi‐colored light bulbs all over the place to provide different light colors. So they use what are called “gels” that are color filters. If they want a violet light on the stage, they put a violet gel in front of a white light. The gel then only allows the color violet to pass through the filter, producing a violet light on stage. White
Light
R
O
Y
G
B
V
Only violet
transmitted
Colors absorbed
A few important concepts to remember 1. Light can act as a particle (photon) or a wave. 2. When light strikes a boundary between two media, all or part of it is reflected. 3. When light changes medium it is refracted unless it strikes the medium directly along the normal line to the medium. 4. As light changes medium its velocity and wavelength change, but its frequency remains constant. 5. The critical angle for a boundary between two media only occurs when light travels from a high index of refraction to a low index of refraction. The critical angle will then be the maximum angle at which refraction will occur between the two media. 6. When light strikes a boundary at an angle greater than the critical angle, it will experience total internal reflection, no refraction will occur. 327 7. Color, in terms of light, can be produced by three processes: emission, reflection, and transmission. Questions and Problems NOTE: There is a list of index of refractions at the end of this section. 1. (a) In the particle theory of light, what are the particles called? (b) What are four things that light can do that indicates that it acts like a wave? 2. Determine the speed of light in (a) crown glass, (b) quartz, and (c) flint glass. 3. A light ray generated by excited sodium atoms has a wavelength of 589 nm in a vacuum. (a) Calculate its frequency. (b) Determine its velocity in ethanol. (c) Determine its wavelength in ethanol. 4. The speed of light in a substance is found to be 2.29 x 108 m/s. (a) Calculate its index of refraction (n). (b) What is this substance? 5. A light ray in air strikes a glass window (n = 1.5) at an angle of 35o to the normal, calculate the angles of (a) reflection, and (b) refraction. 6. (a) A diver shines a light up from under the water at an angle of 28o to the normal; determine the angle it refracts into the air. (b) The boat captain shines another light from the air so that it refracts in the water at an angle of 35o to the normal; determine the angle of incidence of the light in the air. 7. Calculate the critical angle between (a) quartz and air, (b) crown glass and air, and (c) flint glass and water. 8. A 500 nm light ray in a vacuum strikes a boundary with quartz at an angle of 30o with the normal. (a) Calculate the angle of refraction in the quartz. (b) Determine the wavelength of the light in the quartz. If instead, the light strikes a boundary with a diamond at the same initial angle of 30o with the normal; determine (c) the angle of refraction in the diamond. (d) Determine the wavelength of the light in the diamond. 9. A light ray has a wavelength of 450 nm in Plexiglas. (a) Determine the speed of light in Plexiglas. (b) Determine the frequency of the light in Plexiglas. (c) If the light leaves the Plexiglas and enters a vacuum, calculate the wavelength of the light in the vacuum. (d) Determine the frequency of the light ray in the vacuum. 328 10. A light ray with a wavelength of 600 nm is in a vacuum. It strikes a boundary with water at an angle of 20o with the normal. (a) Calculate the angle of reflection. (b) Determine the angle of refraction. (c) Calculate the wavelength of the light in the water. (d) Determine the speed of the light in the water. (e) Determine the frequency of the light. (f) Calculate the critical angle of the boundary. 11. (a) What happens to the wavelength of a light ray as it moves from air to water? (b) What happens to its frequency? 12. What occurs beyond the critical angle? 13. Describe how color is formed through the process of (a) reflection, (b) emission, and (c) transmission. 14. A person wearing a blue shirt walks through three different rooms, what color does her shirt appear to be if the room is lit with (a) white light, (b) yellow light, and (c) blue light? 15. A light beam travels from air to flint glass to water back to air. If the light beam strikes the air‐flint glass boundary as shown below, draw the path of the light beam as it passes through the glass, then the water, and back out to the air. Clearly indicate which direction the light is refracted at each boundary. air (n = 1)
flint glass (n = 1.62)
water (n = 1.333)
air (n = 1)
16. A material is found to have a critical angle of 47.30 with air. (a) Determine its index of refraction. (b) What is this material? Index of Refractions: Vacuum – 1.0 Air ~ 1.0 Water – 1.333 Ethanol – 1.36 Crown Glass – 1.52 Quartz – 1.54 Flint Glass – 1.62 Diamond – 2.42 Ice – 1.31 Plexiglas – 1.51 Sodium Chloride – 1.53 329 Chapter 20 – Optics 20.1 Introduction to Optics How do we actually see things? The human eye is a highly developed organ designed to give us the ability to see things. Not everything. As we saw back in electromagnetic waves, there is only a small part of the entire EM spectrum that the human eye can detect. That is probably a good thing because if we could see the entire spectrum, we would never really be able to determine what we are actually looking at. There are animals that have evolved very specialized capabilities with their eyes. Pit vipers, rattlesnakes for example, have sensors that allow them to “see” infrared light given off by other animals. This is often why they hunt at night when the background air is much cooler, allowing them to see the infrared better. Bees can see into the ultraviolet spectrum. It was often wondered why bees only went to certain flowers. It was later discovered that under UV light, flowers have different patterns than what we actually see. But again, how do we actually see things. One of the most important aspects of the human eye is that it contains an optical device called a converging lens. Optics is the study of the behavior and properties of light and comes from the Greek word for “appearance”. Optical devices include telescopes, binoculars, microscopes, eyeglasses, contact lens, etc. In this section, we will focus on two general optical devices, mirrors and lenses. Each works based on principles of waves and light that we have discussed. Mirrors obey the Law of Reflection and lenses obey the Law of Refraction. Along the way we will learn a method called Ray Diagrams to understand how these devices produce images. 20.2 Introduction to Mirrors Mirrors obey the Law of Reflection in that when light strikes a mirror, it does not pass through the mirror, but bounces off of it like a barrier. Most mirrors are actually just plain glass that has a silver reflective material coated on one side. Contrary to popular belief, anyone can see through a two‐way mirror as the reflective coating is very thin. When the police use a two way mirror, the observation room must be very dark or those on the other side will see through the mirror. The darkness helps the reflection of light. There are two types of basic mirrors, plane mirrors and curved mirrors. 330 20.3 Plane Mirror This is your ordinary, everyday mirror that most of us are used to seeing each morning as we get ready for school or work. The word plane means “flat”, so this is any flat mirror. When we discuss mirrors we must look at them as if we had the ability to “go inside” the mirror. So if we consider the plane mirror on the next page, we are looking at it as if the mirror were turned sideways. Plane Mirror
In front of Mirror
Inside Mirror
On the left side is anywhere in front of the mirror, so where you would be standing while brushing your teeth (you do brush your teeth, right?). This is the real world and we will call it real space. Now when we look in the mirror, our faces do not appear to be pressed against the glass. If we are standing 50 cm in front of the mirror, it appears as if our image is 50 cm “inside” the mirror. In other words the mirror shows depth, even though it may only be 1.0 cm in actual depth. So the right side of the mirror will be viewed as this depth within the mirror itself. As this is not real space we will call this side of the mirror virtual space (imaginary space). Plane Mirror
In front of Mirror
Inside Mirror
Real space
Virtual space
To help us with understanding distances and heights and other information, we find it useful to draw a line through the center of an optical device. This line that bisects the optic is called the principal axis. Plane Mirror
Real space
Virtual space
Principal axis
331 We will represent objects, such as you, as a simple arrow placed at some point in front of the mirror and denoted with the letter O for “object” (aren’t we clever). Plane Mirror
Real space
Virtual space
O
Principal axis
Imagine this, you are standing in front of a plane mirror with a flashlight held next to your head and pointed straight at the mirror. The ray of light goes straight toward the mirror. Real space
Plane Mirror
Virtual space
O
Principal axis
When you look in the mirror, what you see appears to be someone who looks an awful lot like you shining a beam of light straight back at you. Your light is a real light ray; the one shining back at you is a virtual light ray as it exists in virtual space. I like to use solid lines for real light rays and dashed lines to indicate virtual light rays. Real space
Plane Mirror
Virtual space
O
Principal axis
Now let’s say that you decide to tilt your flashlight downward at an angle towards the mirror. The light ray strikes the mirror at an angle and, following the Law of Reflection, must bounce off 332 the mirror at the same angle. This causes the real light from your flashlight to appear on the floor. Real space
Plane Mirror
Virtual space
O
Principal axis
If we look in the mirror, what we see is again someone who looks a lot like you pointing a flashlight towards the spot on the floor where we actually see that real light. If we trace the reflected light ray backwards into virtual space, we find it intersects our other virtual light ray from before. Real space
Plane Mirror
Virtual space
O
Principal axis
At the point where these two virtual light rays meet we find you. Actually it is the image of you. So we will use another arrow to represent the image that is formed in the mirror and label it I for image. Real space
Plane Mirror
Virtual space
O
I
Principal axis
There are a few things to note about this image (I) that is formed. First, this is called a virtual image. The reason is that it is formed in virtual space and is the result of virtual light rays coming together. Virtual images cannot be projected onto a screen, so your image remains there in the mirror. Another thing to note is the distances relative to the mirror itself. The 333 object stands a certain distance in front of the mirror that we will label as the object distance (s). The image appears to also be a certain distance from the mirror that we will label as the image distance (s’). Plane Mirror
O
s′
s
I
Principal axis
These distances must be the same, therefore s = s’. That is why if you are standing 50 cm in front of a mirror, your image appears 50 cm “inside” the mirror. The next relationship of note is the height of the object which label as the object height (h) and the height of the image which we will label as the image height (h’). All heights in optics are measured with respect to the principal axis. Plane Mirror
O
s
s′
h
h′
Principal axis
In a plane mirror, your image never appears to be taller or shorter than you are, thus h = h’. The ratio of image height to object height is often referred to as magnification (M). This gives the relationship: . For all plane mirrors, M = 1. Finally, a somewhat unusual aspect of all mirrors, if you raise your right hand, your image appears to raise its left hand. This is known as right‐left reversal. All mirrors reverse your image. The word AMBULANCE is often written because the rear‐view mirror in a car is a plane mirror. So just in case you didn’t notice the loud siren and flashing lights, when you looked in the rear‐view mirror, the word AMBULANCE could be properly read, as it has been reversed. Therefore, all plane mirrors have the same characteristics: 1. Produce virtual, upright images 2. Object distance = Image distance (s = s′) 3. Object height = Image height (h = h′) 4. M = 1 5. Right‐left reversal 334 20.4 Introduction to Curved Mirrors There are two types of curved mirrors, concave mirrors and convex mirrors. These are the types of mirrors that you often encounter at a funhouse. They make you look taller or shorter or they twist your image around. These “mirrors” are also found when you look at a spoon, depending on which side you look at. 20.4.1 Concave Mirrors A concave mirror is an inward curving mirror. Again we will consider the mirror as if we were looking at it from the side with real space being on the left and virtual space (inside the mirror) on the right. In addition, we will bisect the concave mirror with a principal axis. Concave Mirror
Virtual space
Real space
Principal axis
As the mirror is curved like an arc, it has a specific radius of curvature. If the curved arc were to be made into a complete circle, there would exist a center to that circle that is located at the same point as what we will call the center of curvature (C) for the mirror. In addition, with respect to light and waves, any curved barrier has a location known as its focus or focal point (f). For all curved mirrors, the focal point is located halfway between the center of curvature and the mirror. Thus f = ½C or C = 2f. Concave Mirror
Virtual space
Real space
C
f
Principal axis
C = 2f
335 As light from a distant source strikes the curved mirror, it is reflected based on how it strikes the mirror. If the light rays are traveling parallel to the principal axis, then when they hit the mirror, they will be reflected towards the focal point (f). One reason is that all points on the mirror are curved inward, thus reflecting the light inward. This was the same reason why a straight front water wave, upon striking an inward curved barrier, formed circular waves that originated at the focal point of the barrier. So if we have multiple, parallel light rays that strike the concave mirror (either above or below the principal axis): Concave Mirror
C
f
Principal axis
All the reflected light rays will pass through the focal point (f). Concave Mirror
C
f
Principal axis
So what happens when an object is placed in front of a concave mirror? What kind of image is formed in the mirror? Well, that has a lot to do with exactly where the object is standing. There are two ways we can determine information about the image formed in a concave mirror, geometrically and mathematically. Both work very well. We will first look at the geometric method which involves a process called Ray Diagrams. 20.4.2 Ray Diagrams for Curved Mirrors Ray Diagrams is a method by which we draw light rays using specific rules to determine the location, height, magnification, and orientation of an image. In addition, we can determine if an image is real or virtual. The physical difference between a real image and a virtual image is that a real image can be projected onto a screen. In the case of a mirror and the process of Ray 336 Diagrams, a real image will be located in real space (left side or in front of the mirror) and a virtual image will be located in virtual space (right side or “inside” the mirror). There are several methods to drawing Ray Diagrams, but I like to keep it to two simple rules. 1. Draw a ray from the top of the object through the center (C). 2. Draw a second ray from the top of the object parallel to the principal axis to the mirror, and then from this point on the mirror through the focal point (f). So C and f are always going to be your important points. We will see how we can apply these two rules to all of optics. Example 1: let’s say we have an object (O) located in front of a concave mirror at a location beyond C as shown below. Concave Mirror
O
C
f
I apply my first rule and draw a ray from the top of the object through C. Concave Mirror
O
C
f
337 I then apply my second rule and draw a ray from the top of the object parallel to the principal axis and straight across to the mirror. Concave Mirror
O
C
f
Since this is a concave mirror and it is curved toward the focal point, this light ray is then reflected down through f. Concave Mirror
O
C
f
In Ray Diagrams, an image (I) is formed either where real light rays intersect, or virtual light rays intersect. The image is drawn from the principal axis to the point where the light rays intersect. Therefore in this case, the image (I) is formed as shown below. Concave Mirror
O
C
f
I
338 If we look carefully at this image, we can see that it is located in front of the mirror, or in real space. This means that it is a real image. We can also see that, unlike the object, it is upside down or inverted. If we were to measure the height of the image versus the height of the object, we would find that the image is smaller or reduced. If we recall, magnification was the ratio of image height (h’) to object height (h). Since h’ < h then this gives us M < 1. So the type of image formed is a real, inverted, reduced (M <1) image. Example 2: What if the object (O) were located right at C as shown below? Concave Mirror
O
C
f
If we apply our first rule, then we must draw a line from the top of the object straight down through C. Concave Mirror
O
C
f
339 I then apply my second rule and draw a ray from the top of the object parallel to the principal axis and straight across to the mirror and it is then reflected down through f. Concave Mirror
O
C
f
The point at which these rays intersect is directly below C, so the image (I) is drawn from the principal axis to the point of intersection and is thus right below the object. Concave Mirror
O
C
f
I
This image is once again in real space and is also upside down. In addition, if we measured it carefully, we would find that the image height and the object height are equal. So this image formed is real, inverted, and equal (M = 1). Example 3: The object (O) is located between C and f. Concave Mirror
O
C
f
340 We apply our two rules and get: Concave Mirror
O
C
f
The image (I) is now formed in real space, but out beyond the object and beyond C. Concave Mirror
O
C
f
I
This image is real, inverted, and enlarged (M > 1). Example 4: The object (O) is located at f. Concave Mirror
O
C
f
341 This one is interesting because when we apply our Ray Diagram rules, we get: Concave Mirror
O
C
f
The two rays form parallel lines that will never intersect, therefore NO IMAGE is formed. Example 5: The object (O) is located between f and the mirror. Concave Mirror
O
C
f
Let’s apply our two rules: Concave Mirror
O
C
f
These rays do not appear to intersect, yet they are NOT parallel. That means they must intersect somewhere. If we trace the rays backwards (to the right) we find that they do intersect, but “inside” the mirror, or in virtual space. 342 Concave Mirror
O
C
f
The image (I) that is formed is located where these virtual light rays intersect and is again drawn from the principal axis to the point of intersection. I
Concave Mirror
O
C
f
Since the image (I) is formed in virtual space, it is a virtual image. In addition, it is now right‐
side up or upright. The image height is clearly much larger than the object height indicating it is an enlarged image. So the type of image formed here is a virtual, upright, enlarged (M > 1) image. 20.4.3 The Math Way for Mirrors Wow, wasn’t that fun. So that is the geometric method. It works great, but of course lacks mathematical accuracy. So let’s look at the math method. As was mentioned with the plane mirror, an object is located a certain distance from the mirror called the object distance (s). The image appears a certain distance from the mirror called the image distance (s’). For a curved mirror, there is another important distance and that is the distance from the focal point to the mirror known as the focal length (f). All three are related by the optic equation: 343 In addition, we use the magnification equation, but there is a second one related to s and s': For a concave mirror, f is always a positive number as it is always in real space. Then there are some other rules to remember that will apply to all optics. They are: If s’ is positive, it indicates that a real image is formed. If s’ is negative, it indicates that a virtual image is formed. If M is positive, it indicates that an upright image is formed. If M is negative, it indicates that an inverted image is formed. Example: A 30 cm tall object is placed 20 cm in front of a concave mirror with a focal length of 15 cm. (a) Determine the location of the image. (b) Determine the magnification of the image. (c) Determine the height of the image. (d) Describe the image (real or virtual, upright or inverted, enlarged, equal, or reduced). The givens in this case are: f = 15 cm (positive as it is a concave mirror) s = 20 cm (object distance) h = 30 cm (object height) (a) Apply the optic equation: (Note: you do not need to convert units here.) Subtract and then invert your answer to get s′. (1/0.0166) s’ = 60 cm Since s′ is a positive number, than means it is a real image. (b) Apply the first part of the magnification equation (Note: Magnification still has no units): 60
20
M = ‐ 3 Since M is negative, this indicates that the image is inverted. (c) Apply the second part of the magnification equation: h′ = ‐ 90 cm Since h′ is greater than h, the image is enlarged. 344 This matches the Ray Diagram for this situation where if f = 15 cm then C = 2f = 30 cm: Concave Mirror
O
30
60 cm 45 cm
30 cm
15 cm
90
I
Type:_REAL______
Orientation:____INVERTED___ Size:___ENLARGED____
20.4.4 Convex Mirrors A convex mirror is an outward curving mirror. Again we will consider the mirror as if we were looking at it from the side with real space being on the left and virtual space (inside the mirror) on the right. Convex Mirror
Real space
Virtual space
Principal axis
As the mirror is curved this way, both the center of its curvature (C) and its focal point (f) are located to the right of the mirror in virtual space. Convex Mirror
Real space
Principal axis
Virtual space
f
C
C = 2f
345 As light from a distant source strikes the curved mirror, it is again reflected based on how it strikes the mirror. If the light rays are traveling parallel to the principal axis, then when they hit the mirror, they will be reflected away from the focal point (f). One reason is that all points on the mirror are curved outward, thus reflecting the light outward. This was the same reason why a straight front water wave, upon striking an outward curved barrier, formed circular waves that curved away from the barrier. Therefore, if we have multiple, parallel light rays that strike the convex mirror (either above or below the principal axis): Convex Mirror
Real space
Virtual space
Principal axis
f
C
All the reflected light rays will be directed away from the focal point (f). As this focal point lies in virtual space, these reflected rays appear to originate from the focal point as shown below. Convex Mirror
Real space
Virtual space
Principal axis
f
C
What happens when an object is placed in front of a convex mirror? What kind of image is formed in the mirror? Again we will look at this idea through the Ray Diagrams method and the mathematical method. 346 Example 1: Let’s start again with an object (O) located in front of the convex mirror at a location far from the mirror. Now since C and f are in virtual space and we are going to place our object in real space, we need some reference points. In real space, we will use f to represent the same distance from the mirror in real space as the focal point (f) is from the mirror in virtual space. NOTE: We will use 2f to represent the same distance from the mirror in real space as the center point (C) is from the mirror in virtual space (convenient since C = 2f). Convex Mirror
O
2f
f
f
C
I apply my first rule and draw a ray from the top of the object through C, which is all the way back in virtual space. Convex Mirror
O
2f
f
f
C
347 I then apply my second rule and draw a ray from the top of the object parallel to the principal axis and straight across to the mirror. Convex Mirror
O
2f
f
f
C
Since this is a convex mirror and it is curved away from the focal point, this light ray is then reflected as if it originated at f and then bounces away from the mirror as shown below. Convex Mirror
O
2f
f
f
C
In Ray Diagrams, an image is formed either where real light rays intersect, or virtual light rays intersect. The image is drawn from the principal axis to the point where, in this case, the virtual light rays intersect. Therefore in this case, the image (I) is formed as shown below. Convex Mirror
O
I
2f
f
f
C
348 If we look carefully at this image, we can see that it is located inside of the mirror, or in virtual space. This means that it is a virtual image. We can also see that, like the object, it is right side up or upright. If we were to measure the height of the image versus the height of the object, we would find that the image is smaller or reduced. If we recall, magnification was the ratio of image height (h') to object height (h). Since h’ < h then this gives us M < 1. So the type of image formed is a virtual, upright, reduced (M <1) image. Example 2: What if the object (O) were located right at 2f (the equivalent distance as C is from the mirror, but in real space) as shown below? Convex Mirror
O
2f
f
f
C
We apply our two rules and get: Convex Mirror
O
I
2f
f
f
C
Hey, that looks a lot like what we had in the first example except that the image (I) is formed slightly closer to the mirror. The type of image formed is a virtual, upright, reduced (M <1) image. Well, it turns out that a convex mirror is only capable of producing one type of image and that is a virtual, upright, reduced (M <1) image. That is VERY IMPORTANT to remember. Also note that the image is always formed between f and the mirror. 349 Let’s look at the math method. As was mentioned with the concave mirror, an object is located a certain distance from the mirror called the object distance (s). The image appears a certain distance from the mirror called the image distance (s'). Again for any curved mirror, there is another important distance and that is the distance from the focal point to the mirror known as the focal length (f). All three are related by the optic equation and the magnification equations: For a convex mirror, f is always a negative number as it is always in virtual space. Then we apply the same rules to optics as before. They are: If s’ is positive, it indicates that a real image is formed. If s’ is negative, it indicates that a virtual image is formed. If M is positive, it indicates that an upright image is formed. If M is negative, it indicates that an inverted image is formed. Example: A 30 cm tall object is placed 20 cm in front of a convex mirror with a focal length of 15 cm. (a) Determine the location of the image. (b) Determine the magnification of the image. (c) Determine the height of the image. (d) Describe the image (real or virtual, upright or inverted, enlarged, equal, or reduced). The givens in this case are: f = ‐15 cm (negative as it is a convex mirror) s = 20 cm (object distance) h = 30 cm (object height) (a) Apply the optic equation: (Note: you do not need to convert units here.) s’ = ‐8.57 cm Subtract and then invert your answer to get s’. (1/‐0.1167) Since s’ is a negative number, than means it is a virtual image. 350 (b) Apply the first part of the magnification equation (Note: Magnification still has no units): 8.57
20
M = 0.43 Since M is positive, this indicates that the image is upright. (c) Apply the second part of the magnification equation: .
h′ = 12.86 cm Since h′ is less than h, the image is reduced. This matches the Ray Diagram for this situation where if f = 15 cm then C = 2f = 30 cm: Convex Mirror
O
I
30 cm
Type:_VIRTUAL____
15 cm
Orientation:____UPRIGHT___
-15 cm
-30 cm
Size:___REDUCED____
The convex mirror is commonly used in stores and is also the mirror that is located on the right side of a car. We have all seen this mirror and the words OBJECTS IN THE MIRROR ARE CLOSER THAN THEY APPEAR. The reason for using a convex mirror is to allow someone to see more of the road or area and to ensure that all the images are always upright, though they will always be reduced. 20.5 Introduction to Lenses The second major optic to discuss is called a lens. Unlike a mirror, a lens will allow light to pass directly through it. This will cause the light to change medium as it passes from the air through the lens and back out again. Therefore a lens must obey the Law of Refraction and bend the light ray as it passes through. How the light is bent depends upon the shape and thickness of the lens. We will focus on two types of lenses, converging lenses and diverging lenses. 351 20.5.1 Converging Lens A converging lens (also called a convex lens) is an outward curving lens. As it is transparent, light can pass through the lens from either direction. Converging lenses are typically found in magnifying glasses, telescopes, microscopes, and is the type of lens within the human eye. Anyone who has played with a magnifying glass knows that it does not matter which side one looks through to use the magnifying glass. Since it is curved and light can pass through it in either direction, a converging lens has two focal points, one on each side of the lens. Unlike a mirror, it is difficult to say that there is a virtual space with regards to a lens as one can find real space on both sides of the lens. This makes the difference between real and virtual images a little harder to understand. So we will apply a sort of general rule to images with regards to a lens. We will once again locate objects in the real space on the right side of a lens. However, unlike a mirror, all real images will form on the right side of the lens as light should pass through the lens. This means that if any image forms on the left side (i.e. the same side as the object) it will be considered a virtual image. I know, confused already, but hopefully we will see how this works through some examples. A typical converging lens looks like this: Converging Lens
Principal axis
Secondary axis
f
C
f
We will once again bisect the optics with a principal axis. In addition, for lenses, we will bisect them the other way with a secondary axis. As you can see, the lens has focal points (f) on both sides. Instead of a center of curvature like the curved mirrors had, the center for a lens (C) will conveniently be located at the center of the lens where the principal axis and secondary axis cross each other, as show in the previous picture. As light from a distant source enters the converging lens, it is refracted as it passes through the lens. If the light rays are traveling parallel to the principal axis, then when they enter the lens, they will be refracted towards the focal point (f) on the other side of the lens. Thus a converging lens will converge light rays or bring them together. This was the same reason why a straight front water wave, passing over an outward curved shape, will bend in on itself. So if we have multiple, parallel light rays that enter the converging lens (either above or below the principal axis): 352 Converging Lens
Principal axis
Secondary axis
f
C
f
The light rays are converged towards the focal point. Converging Lens
Principal axis
Secondary axis
f
C
f
This is, of course, why you can use a magnifying glass to focus sunlight from a great distance to a very strong and fiery point. 20.5.2 Ray Diagrams for Lenses Ray Diagrams can be used to locate images for lenses. The physical difference between a real image and a virtual image is still that a real image can be projected onto a screen. In the case of a lens and the process of Ray Diagrams, a real image will be located to the right of the lens and a virtual image will be located to the left of the lens. Once again there are several methods to drawing Ray Diagrams, but I like to keep the same two simple rules we used with mirrors. 1. Draw a ray from the top of the object through the center (C) which is now in the center of the lens where the principal axis and the secondary axis cross. 2. Draw a second ray from the top of the object parallel to the principal axis to the lens, and then from this point on the lens through the appropriate focal point (f). So C and f are still going to be your important points. The tricky part is knowing which focal point (f) to use. For the converging lens, the appropriate focal point (f) will always be on the side of the lens opposite of the side where the object is located. 353 Example 1: let’s say we have an object (O) located in front of a converging lens at a location beyond 2f as shown below. Converging Lens
O
2f
f
f
2f
I apply my first rule and draw a ray from the top of the object through C which will be at the center of the lens. Converging Lens
O
2f
f
f
2f
I then apply my second rule and draw a ray from the top of the object parallel to the principal axis and straight across to the lens. Converging Lens
O
2f
f
f
2f
354 Since this is a converging lens and will converge or bring light rays together, this light ray is then refracted down through the focal point (f) on the right side of the lens. This will bring the two light rays together. Converging Lens
O
2f
f
f
2f
The image (I) is again formed either where real light rays intersect, or virtual light rays intersect. The image is drawn from the principal axis to the point where the light rays intersect. Therefore in this case, the image (I) is formed as shown below. Converging Lens
O
2f
f
f
2f
I
If we look carefully at this image, we can see that it is located on the right side of the lens. This means that it is a real image. We can also see that, unlike the object, it is upside down or inverted. If we were to measure the height of the image versus the height of the object, we would find that the image is smaller or reduced. If we recall, magnification was the ratio of image height (h’) to object height (h). Since h’ < h then this gives us M < 1. So the type of image formed is a real, inverted, reduced (M <1) image. 355 Example 2: What if the object (O) were located right at 2f as shown below? Converging Lens
O
2f
f
f
2f
We apply our two rules and get: Converging Lens
O
2f
f
f
2f
I
The image (I) is located to the right of the lens and exactly at 2f. This image is once again real and is also upside down. In addition, if we measured it carefully, we would find that the image height and the object height are equal. The image formed is real, inverted, and equal (M = 1). Example 3: The object (O) is located between f and 2f. Converging Lens
O
2f
f
f
2f
356 We apply our two rules and get: Converging Lens
O
2f
f
f
2f
I
This image is real, inverted, and enlarged (M > 1). Example 4: The object (O) is located at f. Converging Lens
O
2f
f
f
2f
This one is interesting because when we apply our Ray Diagram rules, we get: Converging Lens
O
2f
f
f
2f
The two rays form parallel lines that will never intersect, therefore NO IMAGE is formed. 357 Example 5: The object (O) is located between f and the lens. Converging Lens
O
2f
f
f
2f
Let’s apply our two rules: Converging Lens
O
2f
f
f
2f
These rays do not appear to intersect, yet they are NOT parallel. That means they must intersect somewhere. If we trace the rays backwards (to the left) we find that they do intersect, but on the left side of the lens. Converging Lens
2f
O
f
f
2f
358 The image (I) that is formed is located where these virtual light rays intersect and is again drawn from the principal axis to the point of intersection. I
Converging Lens
O
2f
f
f
2f
Since the image (I) is formed on the left side of the lens (same side as the object), it is a virtual image. In addition, it is now right‐side up or upright. The image height is clearly much larger than the object height indicating it is an enlarged image. So the type of image formed here is a virtual, upright, enlarged (M > 1) image. An interesting comparison is between the concave mirror and the converging lens. If you look at each of the five (5) examples for both, they are identical. This has to do with the fact that both the concave mirror and the converging lens will bring light rays together. 20.5.3 The Math Way for Lenses Still having fun? So again that was the geometric method. It works great, but of course lacks mathematical accuracy. So let’s look at the math method. As was mentioned with the mirrors, an object is located a certain distance from the lens called the object distance (s). The image appears a certain distance from the lens called the image distance (s'). For a converging lens, there is another important distance and that is the distance from the focal point to the lens known as the focal length (f). All three are related by the same optic equations: 359 For a converging lens, f is always a positive number. Then we apply the same rules to optics as before. They are: If s’ is positive, it indicates that a real image is formed. If s’ is negative, it indicates that a virtual image is formed. If M is positive, it indicates that an upright image is formed. If M is negative, it indicates that an inverted image is formed. Example: A 30 cm tall object is placed 25 cm in front of a converging with a focal length of 10 cm. (a) Determine the location of the image. (b) Determine the magnification of the image. (c) Determine the height of the image. (d) Describe the image (real or virtual, upright or inverted, enlarged, equal, or reduced). The givens in this case are: f = 10 cm (positive as it is a converging lens) s = 25 cm (object distance) h = 30 cm (object height) (a) Apply the optic equation: (Note: you do not need to convert units here.) s′ = 16.67 cm Subtract and then invert your answer to get s'. (1/0.06) Since s′ is a positive number, than means it is a real image. (b) Apply the first part of the magnification equation (Note: Magnification still has no units): 16.67
25
M = ‐ 0.67 Since M is negative, this indicates that the image is inverted. (c) Apply the second part of the magnification equation: .
h′ = ‐ 20 cm Since h′ is less than h, the image is reduced. 360 This matches the Ray Diagram for this situation where if f = 10 cm then: Converging Lens
O
20 cm
10 cm
10 cm
20 cm
I
Type:_REAL______
Orientation:____INVERTED___ Size:___REDUCED____
20.5.4 Diverging Lenses A diverging lens (also called a concave lens) is an inward curving lens. Like the converging lens it is transparent, light can pass through the lens from either direction. Diverging lenses are typically found in flashlights and eyeglasses. Anyone who is nearsighted is more familiar with a diverging lens than they may know. Since it is curved and light can pass through it in either direction, a diverging lens has two focal points, one on each side of the lens. We will once again locate objects in the real space on the right side of the lens. Like the converging lens, real images will form on the right side of the lens and virtual images will form on the left side of the lens. A typical diverging lens looks like this: Diverging Lens
C
f
f
We will once again bisect the lens with a principal axis and a secondary axis. As you can see, the lens has focal points (f) on both sides. The center for a lens (C) will still conveniently be located at the center of the lens where the principal axis and secondary axis cross each other. 361 As light from a distant source enters the diverging lens, it is refracted as it passes through the lens. If the light rays are traveling parallel to the principal axis, then when they enter the lens, they will be refracted away from the focal point (f) on the other side of the lens. Thus a diverging lens will diverge light rays or spread them apart. This was the same reason why a straight front water wave, passing over an inward curved shape, will bend outward. So if we have multiple, parallel light rays that enter the diverging lens (either above or below the principal axis): Secondary axis
Diverging Lens
C
f
f
Principal axis
The light is then diverged away from the focal point on the right or it appears to originate from the focal point on the left. Secondary axis
Diverging Lens
f
C
f
Principal axis
This is why a diverging lens is often used in a flashlight; it will spread the light apart giving you a larger field of light. So what happens when an object is placed in front of a diverging lens? What kind of image is formed? Again we will look at this idea through the Ray Diagrams method and the mathematical method. 362 Example 1: let’s start again with an object (O) located in front of the diverging lens at a location far from the lens, beyond 2f. Diverging Lens
O
2f
f
f
2f
I apply my first rule and draw a ray from the top of the object through C which is in the center of the lens. Diverging Lens
O
2f
f
f
2f
I then apply my second rule and draw a ray from the top of the object parallel to the principal axis and straight across to the lens. Diverging Lens
O
2f
f
f
2f
Since this is a diverging lens this light ray is then refracted away from the other light ray or as if it originated at the focal point (f) on the left side and then diverges from the other light ray as shown below. 363 Diverging Lens
O
2f
f
f
2f
Once again an image is formed either where real light rays intersect, or virtual light rays intersect. In this case a real light ray intersects with a virtual light ray. As long as at least one light ray is a virtual light ray, then a virtual image must be formed. The image is drawn from the principal axis to the point where, in this case, the real and virtual light rays intersect. Therefore in this case, the image (I) is formed as shown below. Diverging Lens
O
I
2f
f
f
2f
If we look carefully at this image, we can see that it is located on the left side of the lens. This means that it is a virtual image. We can also see that, like the object, it is right‐side up or upright. If we were to measure the height of the image versus the height of the object, we would find that the image is smaller or reduced. If we recall, magnification was the ratio of image height (h’) to object height (h). Since h’ < h then this gives us M < 1. So the type of image formed is a virtual, upright, reduced (M <1) image. 364 Example 2: What if the object (O) were located right at 2f as shown below? Diverging Lens
O
2f
f
f
2f
We apply our two rules and get: Diverging Lens
O
I
2f
f
f
2f
Hey, that looks a lot like what we had in the first example except that the image (I) is formed slightly closer to the lens. The type of image formed is a virtual, upright, reduced (M <1) image. Well, it turns out that a diverging lens is only capable of producing one type of image and that is a virtual, upright, reduced (M <1) image. That is VERY IMPORTANT to remember. Also note that the image is always formed between f and the lens. An interesting comparison is between the convex mirror and the diverging lens. If you look at each of the examples for both, they are identical. This has to do with the fact that both the convex mirror and the diverging lens will spread light rays apart. Let’s look at the math method. As was mentioned with the mirrors and converging lens, an object is located a certain distance from the lens called the object distance (s). The image appears a certain distance from the lens called the image distance (s'). For a diverging lens, there is another important distance and that is the distance from the focal point to the lens known as the focal length (f). All three are related by the same optic equations: 365 For a diverging lens, f is always a negative number. Then we apply the same rules to optics as before. They are: If s’ is positive, it indicates that a real image is formed. If s’ is negative, it indicates that a virtual image is formed. If M is positive, it indicates that an upright image is formed. If M is negative, it indicates that an inverted image is formed. Example: A 30 cm tall object is placed 25 cm in front of a diverging with a focal length of 10 cm. (a) Determine the location of the image. (b) Determine the magnification of the image. (c) Determine the height of the image. (d) Describe the image (real or virtual, upright or inverted, enlarged, equal, or reduced). The givens in this case are: f = ‐10 cm (negative as it is a diverging lens) s = 25 cm (object distance) h = 30 cm (object height) Apply the optic equation: (Note: you do not need to convert units here.) Subtract and then invert your answer to get s'. (1/‐0.14) s′ = ‐7.14 cm Since s′ is a negative number, than means it is a virtual image. Apply the first part of the magnification equation (Note: Magnification still has no units): 7.14
25
M = 0.29 Since M is positive, this indicates that the image is upright. Apply the second part of the magnification equation: .
h′ = 8.57 cm Since h′ is less than h, the image is reduced. 366 This matches the Ray Diagram for this situation where if f = 10 cm then: Diverging Lens
O
I
20 cm
Type:_VIRTUAL______
10 cm
10 cm
Orientation:____UPRIGHT___
20 cm
Size:___REDUCED____
So that is the story of mirrors, lens, and Ray Diagrams. 20.6 Human Eye As was mentioned previously, the human eye is a specialized organ designed for seeing and processing images. It contains a converging lens, which is important as it is necessary for the eye to produce real images, otherwise the brain would have nothing to analyze. There are many other important parts to the human eye, so let’s take a look (no pun intended). Picture in public domain 367 The human eye has a number of important parts, some of which we will define here. ¾ Cornea – a transparent window at the front of the eye that allows light to enter the eye. It aids with focusing of images and provides a large amount of refraction of light. In LASIK surgery, a laser is used to reshape the cornea to improve sight. ¾ Aqueous humor – a clear fluid located behind the cornea that provides nutrients to the lens and cornea. The lens and cornea contain no blood vessels, so this liquid is important to their use. ¾ Iris – a circular, colored portion of the eye that controls the size of the pupil thus regulating the amount of light entering the eye. This is the location of our eye color and it will grow and shrink depending on the need for light within the eye. ¾ Pupil – an opening in the iris that determines the amount of light that enters the eye. The pupil does not change size; the iris regulates the size of the pupil. Contrary to what you may think, the pupil is not black, it is clear; the black comes from the back of the eye. ¾ Lens – a transparent, converging lens that focuses light on the back portion of the eye called the retina. ¾ Suspensory ligament – tissue that holds the lens in place. ¾ Ciliary muscle – muscle tissue that can be used to alter the shape of the lens for focusing. When you squint, you are using these muscles. Also, as you change from looking at something close up to looking at something far away, these muscles help refocus the lens. ¾ Vitreous humor – a clear, jelly‐like material that fills the eye and is important for refraction. ¾ Retina – a layer of nerves along the wall of the eye, this area receives light and images and transmits them to the optic nerve. When we look at an object, a real image of the object is formed on the retina. The retina contains special cells called rods and cones. The rods are very sensitive to light and are what allow us to see black and white and in very low light. The cones provide our ability to see color and in bright light. ¾ Sclera – the tough, white part of the eye. ¾ Fovea – area of the retina that provides the sharpest and clearest vision. Light hitting the fovea will provide the retina with the sharpest image. 368 ¾ Optic disc – point where the optic nerve connects to the retina. This is also the location of our “blind spot”. If light enters the eye at a certain angle, it will fall directly on the optic disc, thus not hitting the retina. So any object located at this angle cannot be seen. ¾ Optic nerve – the connection between the eye and the brain, this is the pathway for nerve impulses to travel back and forth to the brain. All of these parts work together to all us to see objects either close up or far away. Some of us, either from the very beginning of life or later on, develop problems with our sight. We have found many ways of using optics to correct these problems. 20.6.1 Myopia One of the most common eye conditions is called myopia, or nearsightedness. If you are nearsighted, this means that you see things very well when they are near to you, but you cannot see things far away. This is because of how light is focused on your retina. In an eye with 20/20 vision, light is focused perfectly on to the retina. However, if you are nearsighted, the light is focused too early and then becomes unfocused when it hits the retina. So to correct myopia, we place a diverging lens in front of the eye. This can be in the form of glasses, contact lens, or laser surgery to reshape the cornea into a more diverging shape. The diverging lens then spreads apart the light before it enters the eye so that it can then be focused properly on to the retina. 369 20.6.2 Hyperopia Another common eye conditions is called hyperopia, or farsightedness. If you are farsighted, this means that you see things very well when they are far from to you, but you cannot see things well close up. This is because the light is focused too late and then is unfocused when it hits the retina. So to correct hyperopia, we place a converging lens in front of the eye. This again can be in the form of glasses, or a contact lens. The converging lens then brings the light together before it enters the eye so that it can then be focused properly on to the retina. A few important concepts to remember 1. Plane mirrors only produce virtual, upright, equal (M = 1) images. 370 2. All mirrors have right‐left reversal. 3. Mirrors obey the Law of Reflection; lenses obey the Law of Refraction. 4. Concave mirrors and converging lens bring light together and are capable of producing a variety of images including real or virtual, inverted or upright. 5. Convex mirrors and diverging lens spread light apart and ONLY produce virtual, upright, reduced (M < 1) images. 6. All real images are inverted, all virtual images are upright. 7. The human eye uses a converging lens. 8. Correct myopia (nearsightedness) with a diverging lens; correct hyperopia (farsightedness) with a converging lens. Questions and Problems 1. Why are the letters on the front of an AMBULANCE often spelled backwards? 2. (a) Describe the difference between a real image and a virtual image. (b) Explain why the film in a movie projector must be loaded upside down. 3. What is the only type of image a plane mirror can produce? (3 parts to answer: type, orientation, and size) 4. What is the only type of image a convex mirror can produce? (3 parts to answer: type, orientation, and size) 5. What types of images can a converging lens produce? (Note: multiple combination of answers) 6. What is the only type of image a diverging lens can produce? (3 parts to answer: type, orientation, and size) 7. (a) Where must an object be located to produce a real image in a concave mirror? (b) In a convex mirror? (c) In a converging lens? (d) In a diverging lens? 8. (a) Where must an object be located to produce a virtual image in a concave mirror? (b) In a convex mirror? (c) In a converging lens? (d) In a diverging lens? 371 9. (a) Where must an object be located to produce an image of equal size in a concave mirror? (b) What is the magnification at this point? 10. (a) Where must an object be located to produce a reduced image in size in a converging lens mirror? (b) What is the magnification at this point? 11. Explain why the words “Objects in mirror are closer than they appear” are printed on the right side mirror of a car. 12. Two plane mirrors are placed at a right angle to each other as shown below. A light beam strikes the first mirror at an angle of 250 to the normal. Determine the angle (θ) at which it reflects off the second mirror. θ
250
13. Draw the ray diagrams for the following situations; in each case describe the image in terms of type (real or virtual), orientation (upright or inverted), and size (reduced, equal, or enlarged): (a) Concave Mirror, object located beyond C.
O
C
f
372 (b) Convex Mirror, object located at 2f.
O
2f
f
f
C
(c) Converging Lens, object located between f and 2f.
O
2f
f
f
2f
(d) Diverging Lens, object located between f and lens.
O
2f
f
f
2f
373 14. A 6.0 cm high object is placed 8.0 cm in front of a concave mirror with a focal length of 5.0 cm. (a) Determine the location of the image. (b) Determine the magnification of the image. (c) Determine the height of the image. (d) Indicate the type, orientation, and size of the image. (e) Draw the ray diagram. Concave Mirror
O
10.0 cm
5.0 cm
15. A 6.0 cm high object is placed 8.0 cm in front of a convex mirror with a focal length of 5.0 cm. (a) Determine the location of the image. (b) Determine the magnification of the image. (c) Determine the height of the image. (d) Indicate the type, orientation, and size of the image. (e) Draw the ray diagram. Convex Mirror
O
10.0 cm
5.0 cm
-10.0 cm
- 5.0 cm
374 16. A 6.0 cm high object is placed 15.0 cm in front of a converging lens with a focal length of 5.0 cm. (a) Determine the location of the image. (b) Determine the magnification of the image. (c) Determine the height of the image. (d) Indicate the type, orientation, and size of the image. (e) Draw the ray diagram. Converging Lens
O
15.0 cm
10.0 cm
5.0 cm
5.0 cm
10.0 cm
17. A 6.0 cm high object is placed 15.0 cm in front of a diverging lens with a focal length of 5.0 cm. (a) Determine the location of the image. (b) Determine the magnification of the image. (c) Determine the height of the image. (d) Indicate the type, orientation, and size of the image. (e) Draw the ray diagram. Diverging Lens
O
15.0 cm
10.0 cm
5.0 cm
5.0 cm
10.0 cm
10.0 cm
5.0 cm
Real, Inverted, Enlarged
I
15. (a) ‐3.08 cm (b) 0.38 (c) 2.3 cm O
I
10.0 cm
5.0 cm
- 5.0 cm
-10.0 cm
Virtual, Upright, Reduced
389 16. (a) 7.5 cm (b) ‐0.5 (c) ‐3.0 cm O
15.0 cm
10.0 cm
5.0 cm
5.0 cm
10.0 cm
I
Real, Inverted, Reduced
17. (a) ‐3.75 cm (b) 0.25 (c) 1.5 cm O
I
15.0 cm
10.0 cm
5.0 cm
5.0 cm
10.0 cm
Virtual, Upright, Reduced
390 ```