Uniform Circular Motion October 2014 Circular Motion • A is towards the center. • V is tangential to the motion • Speed is constant, V changes • A force directed towards the center is what causes the acceleration (e.g. gravity, a string) • If the force is removed, the ball will continue in a straight line at the speed it had. Check your understanding … • When a wheel rotates about a fixed axis, do all the points on the wheel have the same tangential speed? Yes! Do they all have the same velocity? No! Check your understanding … Circular Motion Equations • Vt= 2πr/T • ac = vt2/r Where Vt = tangental velocity R = radius T = period (time required to make one complete circle) Ac = centripetal acceleration Circular Motion– We do • The radius of a spacecraft orbiting earth is 6.67 x 106 m. If it orbits earth in 5292 seconds, what is the velocity of the spacecraft? Vt= 2πr/T = 2*π*6.67X106m / 5292 sec = 7919 m/s Circular Motion– We do Jimmie Johnson is driving his #48 Lowe’s NASCAR around a bend that has a radius of 70 meters. It takes him 30 seconds to travel the track. What was the centripetal acceleration of Jimmie John’s #48 Lowe’s NASCAR? Strategy: find vt, then find ac. Vt= 2πr/T = 2π*70m/30sec = 14.66m/s ac = vt2/r = (14.66 m/s)2 / 70 m = 3 m/s2 Centripetal Acceleration – You do a. A girl sits on a tire that is attached to an overhanging tree limb by a rope. The girl’s father pushes her so that her centripetal acceleration is 3.0 m/s2. If the length of the rope is 2.1 m, what is the girl’s tangential speed? b. A boy swings a yo-yo horizontally above his head so that the yo-yo has a centripetal acceleration of 1.5m/s2. If the yo-yo’s tangential speed is 11m/s, what is the length of the yo-yo? c. Correct the following statement: The racing car rounds the turn at a constant velocity of 145 km/h. Quadratic Equation Quadratic equations, such as x = vit + ½ a t2 can be tricky to solve. 3 strategies 1. Factoring (doesn’t usually work well with physics) 2. Quadratic Equation solution (always works, but you need to memorize) 3. Trial & Error / Plugging in values (can be useful for multiple choice) We will talk about #2 and #3. Quadratic Equation Solution • First, put your equation into this form: 0 = at2 + bt + c for example, x = vit + ½ at2 becomes 0 = ½ at2 +vit – x so , a=½a b = vi c = -x Then, use this equation to solve for t Quadratic Equation Solution – We do A ball is launched upward with a speed of 15 m/s from an intial height of 5 m. What are the two approximate times that the object will be located at the height of 10 m above the ground? Quadratic Equation Solution – We do A ball is launched upward with a speed of 15 m/s from an intial height of 5 m. What are the two approximate times that the object will be located at the height of 10 m above the ground? 0 = ½ at2 +vit – x so , a = ½ a = ½ * - 9.8 m/s2 = -4.9 m/s b = vi = 15 m/s c = -x = -10 m tt = 0.5 sec and 1 sec Plugging in Numbers • Sometimes, you can just plug in numbers into the equation to find the solution. • Works best with multiple choice. Plugging in Numbers • Sometimes, you can just plug in numbers into the equation to find the solution. • Works best with multiple choice. Example – WE DO • A car and truck start from the same position. The car has a constant velocity of 20 m/s. The truck has an initial velocity of zero, but accelerates 3 m/s. At approximately what time does the truck overtake the car? A. 6.6 s B. 10.2 s C. 13.4 s D. 15.1 S Plugging in Numbers • Sometimes, you can just plug in numbers into the equation to find the solution. • Works best with multiple choice. Example – WE DO • A car and truck start from the same position. The car has a constant velocity of 20 m/s. The truck has an initial velocity of zero, but accelerates 3 m/s. At approximately what time does the truck overtake the car? A. 6.6 s B. 10.2 s C. 13.4 s D. 15.1 S Use x = vit + ½ at2 for each car and compare x. At which answer choice does the truck final reach the car? Plugging in Numbers • Sometimes, you can just plug in numbers into the equation to find the solution. • Works best with multiple choice. Example – WE DO • A car and truck start from the same position. The car has a constant velocity of 20 m/s. The truck has an initial velocity of zero, but accelerates 3 m/s. At approximately what time does the truck overtake the car? A. 6.6 s B. 10.2 s C. 13.4 s D. 15.1 S Use x = vit + ½ at2 for each car and compare x. At which answer choice does the truck final reach the car? Quadratic Equation – You do • A ball is dropped from 15 ft. At what time will the ball be 5 ft from the ground? • Car 1 and Car 2 start at the same position. Car 1 has an initial velocity of 10 m/s and an acceleration of 5 m/s2. Car 2 has an initial velocity of 15 m/s and an acceleration of 2 m/s2. At one time will car 2 overtake car one? A. 1.2 sec B. 2.5s C. 3.4 s D. 4.2 s Exit ticket & Homework Exit ticket: write facts about circular motion Homework: Circular acceleration problems Review problems Extra credit: Write 3 challenging physics problems and solve them on another sheet of paper. Upcoming: CFA Wed / Thur next week. Exam right after break.

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