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```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.
• 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!
Circular Motion Equations
• Vt= 2πr/T
• ac = vt2/r
Where
Vt = tangental velocity
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 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.
• 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
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?
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?
• 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.
``` # 2.1a Uniform Motion Grahps.notebook October 12, 2014 v =   d 