1. Purdue University, Physics 220
Feb. 2
Lecture 7 missed
Chapter 5
Feb. 4
Recitation 4
HW4
Feb. 7
Lecture 7
Feb. 9
Lecture 8
Feb. 11
Recitation 5
Feb. 14
Lecture 9
Chapter 6
Feb. 16
Lecture 10
Exam 1
Feb. 18
Recitation 6
HW5 /HW6
Feb. 21
Lecture 11
Chapter 7
Lecture 7
Purdue University, Physics 220

2. Purdue University, Physics 220
UNIMPORTABLE: #1D987FFA,2
#1DB9EC48,2
#2238DDC7,4
#DB602B9,2
#DD17CA0,4
#DF9EC18,2
#E316E51,4
Lecture 7
Purdue University, Physics 220

3. Purdue University, Physics 220
Lecture 07
Circular Motion
Lecture 7
Purdue University, Physics 220

4. Examples of Circular Motion
Rotate ruler 90 degrees ask students how far rotated. Motivate meters bad unit, use angles. Also w/ record player
Lecture 7
Purdue University, Physics 220

5. Uniform Circular Motion
Assume constant speed
The direction of the velocity is continually changing
The vector is always tangent to the circle
Uniform circular motion assumes constant speed
period of the motion: T = 2r/v [s]

6. Purdue University, Physics 220
Angular Variables
The motion of objects moving in circular (or nearly circular) paths, is often described by angles measured in radians rather than degrees.
The angle  in radians, is defined as:
If s = r the angle is 1 rad
If s = 2r (the circumference of the circle) the angle is  rad. (In other words, 360° = 2 rad.)
Rotate ruler 90 degrees ask students how far rotated. Motivate meters bad unit, use angles. Also w/ record player
Lecture 7
Purdue University, Physics 220

7. Purdue University, Physics 220
Circular Motion
Angular displacement Dq = q2-q1
How far it has rotated
Angular velocity wav = Dq/Dt
How fast it is rotating
Units: radians/second
(2p = 1 revolution)
Period = 1/frequency
T = 1/f = 2p / w
Time to complete 1 revolution
Rotate ruler 90 degrees ask students how far rotated. Motivate meters bad unit, use angles. Also w/ record player
Lecture 7
Purdue University, Physics 220

8. Purdue University, Physics 220
Circular to Linear
Displacement Ds = r Dq (q in radians)
Speed |v| = |Ds/Dt| = r |Dq/Dt| = r|w
|v| = 2rf
|v| = 2r/T
Direction of v is tangent to circle
Lecture 7
Purdue University, Physics 220

9. Purdue University, Physics 220
iClicker
Bonnie sits on the outer rim of a merry-go-round with radius 3 meters, and Klyde sits midway between the center and the rim. The merry-go-round makes one complete revolution every two seconds.
Klyde’s speed is:
Klyde
Bonnie
A) the same as Bonnie’s
B) twice Bonnie’s
C) half Bonnie’s
Bonnie travels 2 p R in 2 seconds vB = 2 p R / 2 = 9.42 m/s
Klyde travels 2 p (R/2) in 2 seconds vK = 2 p (R/2) / 2 = 4.71 m/s
Lecture 7
Purdue University, Physics 220

10. Purdue University, Physics 220
Examples
The wheel of a car has a radius of 0.29 m and is being rotated at 830 revolution per minute (rpm) on a tire-balancing machine. Determine the speed (in m/s) at which the outer edge of the wheel is moving:
The speed could be obtained by |v| = 2r/T
A CD spins with an angular frequency 20 radians/second. What is the linear speed 6 cm from the center of the CD?
v = r w = 0.06  20 = 1.2 m/s
Lecture 7
Purdue University, Physics 220

11. Purdue University, Physics 220
Circular Motion
A ball is going around in a circle attached to a string. If the string breaks at the instant shown, which path will the ball follow? 1 2 3 v 4 5
Use this to motivate circular motion involves acceleration.
Answer: 2
Lecture 7
Purdue University, Physics 220

12. Centripetal Acceleration
Magnitude of the velocity vector is constant, but direction is constantly changing
At any instant of time, the direction of the instantaneous velocity is tangent to the path
Therefore: nonzero acceleration
Lecture 7
Purdue University, Physics 220

13. Uniform Circular Motion
Circular motion with constant speed R
centripetal acceleration
Recall: v = w R v a
Instantaneous velocity is tangent to circle
Instantaneous acceleration is radially inward
There must be a force to provide the acceleration
Lecture 7
Purdue University, Physics 220

14. Roller Coaster Example
What is the minimum speed you must have at the top of a 20 meter diameter roller coaster loop, to keep the wheels on the track.
y-direction: F = ma
N + mg = m a
Let N = 0, just touching
mg = m a
mg = m v2/R
g = v2 / R
v = sqrt(g*R) = 10 m/s N y mg
Lecture 7
Purdue University, Physics 220

15. Purdue University, Physics 220
Unbanked Curve
What force accelerates a car around a turn on a level road at constant speed?
A) it is not accelerating
B) the road on the tires
C) the tires on the road
D) the engine on the tires
Lecture 7
Purdue University, Physics 220

16. Purdue University, Physics 220
Unbanked Curve
What is the maximum velocity a car can go around an unbanked curve in a circle without slipping?
The maximum velocity to go around an un-banked curve depends only on s (for a given r)
Dry road: s=0.9
Icy road: s=0.1
Lecture 7
Purdue University, Physics 220

17. Purdue University, Physics 220
Banked Curve
A car drives around a curve with radius 410 m at a speed of 32 m/s. The road is banked at 5.0°. The mass of the car is 1400 kg.
A) What is the frictional force on the car?
B) At what speed could you drive around this curve so that
the force of friction is zero?
Work out on overhead. Use all the standard steps: draw figure, list what is given using symbols, draw a FBD, write 2nd law (component form), write equations that result, give answer (can skip algebra). Algebra is rather complicated and is in the answer book. No need to spend time on that.
Lecture 7
Purdue University, Physics 220

18. Purdue University, Physics 220
Banked Curve  x y
y-direction
(1)
(2) N W f
x-direction
Lecture 7
Purdue University, Physics 220

19. Purdue University, Physics 220
Banked Curve
2 equations and 2 unknown we can solve for N in (1) and substitute in (2)
Lecture 7
Purdue University, Physics 220

20. Purdue University, Physics 220
Banked Curve
A car drives around a curve with radius 410 m at a speed of 32 m/s. The road is banked at 5.0°. The mass of the car is 1400 kg.
A) What is the frictional force on the car?
B) At what speed could you drive around this curve so that
the force of friction is zero?
Like an airplane
Work out on overhead. Use all the standard steps: draw figure, list what is given using symbols, draw a FBD, write 2nd law (component form), write equations that result, give answer (can skip algebra). Algebra is rather complicated and is in the answer book. No need to spend time on that.
Lecture 7
Purdue University, Physics 220

21. Purdue University, Physics 220
iClicker
Suppose you are driving through a valley whose bottom has a circular shape. If your mass is m, what is the magnitude of the normal force N exerted on you by the car seat as you drive past the bottom of the hill.
A) N < mg
B) N = mg
C) N > mg v mg N R
a=v2/R
correct
Since there is centripetal acceleration, the normal force is greater than simply mg
SF = ma
N - mg = mv2/R
N = mg + mv2/R
Lecture 7
Purdue University, Physics 220