1. Physics 221 Chapter 10

2. Problem 1 . . . Angela’s new bike
The radius of the wheel is 30 cm and the speed v= 5 m/s. What is the rpm (revolutions per minute) ?

3. Solution 1 . . . Angela’s rpm r = radius circumference = 2  r
f = revolutions per second
v = d/t
v = 2  f r
5 = (2 )(f)(0.3)
f = 2.6 revolutions per second or 159 rpm

4. What is a Radian?
A “radian” is about 60 degrees which is 1/6 of the circle (360 degrees).
To be EXACT, the “radian pie” has an arc equal to the radius.

5. Problem 2 What EXACTLY is a Radian?

6. Solution 2 What EXACTLY is a Radian?
If each pie has an “arc” of r, then there must be 2  radians in a 3600 circle.
2  radians = 3600
6.28 radians = 3600
1 radian = 57.30

7. Angular Velocity = radians / time

8. Problem 3 . . . Angular Velocity
The radius of the wheel is 30 cm. and the (linear) velocity, v, is 5 m/s. What is the angular velocity?

9. Solution 3 . . . Angular Velocity
We know from problem 1 that :
f = 2.6 rev/s
But 1 rev = 2  radians
So  =  / t
 =(2.6)(2 ) /(1 s)
 = 16.3 rad/s

10. v = r  V and  Linear (m/s) Angular (rad/s) V  d / t  / t
2  r f / t  f / t
v = r 

11. a = r  a and  Linear (m/s2) Angular (rad/s2) a 
( Vf - Vi ) / t ( f - i ) / t
a = r 

12. Problem Your CD player
A 120 mm CD spins up at a uniform rate from rest to 530 rpm in 3 seconds. Calculate its:
(a) angular acceleration
(b) linear acceleration

13. Solution 4 . . . CD player  = ( f - i ) / t
 = (530 x 2  / ) / 3
 = 18.5 rad/s2
a = r 
a = 0.06 x 18.5
a = 1.1 m/s2

14. Problem CD Music
To make the music play at a uniform rate, it is necessary to spin the CD at a constant linear velocity (CLV). Compared to the angular velocity of the CD when playing a song on the inner track, the angular velocity when playing a song on the outer track is
A. more
B. less
C. same

15. Solution CD Music
v = r 
When r increases,  must decrease in order for v to stay constant. Correct answer B
Note: Think of track races. Runners on the outside track travel a greater distance for the same number of revolutions!

16. Angular Analogs
d 
v 
a 

17. Problem 6 . . . Angular Analogs
d = Vi t + 1/2 a t ?

18. Solution 6 . . . Angular Analogs
d = Vi t + 1/2 a t  = i t + 1/2  t2

19. Problem Red Corvette
The tires of a car make 65 revolutions as the car reduces its speed uniformly from 100 km/h to 50 km/h. The tires have a diameter of 0.8 m. At this rate, how much more time is required for it to stop?

20.  = - 4.4 rad/s2 f = i +  t Solution 7 . . . Corvette
100 km/h = 27.8 m/s = 69.5 rad/s since v = r 
Similarly 50 km/h = 34.8 rad/s
(f)2 = (i)2 + 2  
(34.8)2 = (69.5)2 + (2)()(65)(6.28)
 = rad/s2
f = i +  t
0 = t
t = 7.9 s

21. Torque Torque means the “turning effect” of a force.
SAME force applied to both. Which one will turn easier?

22. Torque = distance x force
 = r x F
Easy!

23. Which one is easier to turn?
Torque
Which one is easier to turn?

24. Torque . . . The Rest of the Story!
 = r F sin 
Easy! 

25. Problem 8 . . . Inertia Experiment
SAME force applied to m and M. Which one accelerates more?

26. Solution 8 . . . Inertia Experiment
Since F = ma, the smaller mass will accelerate more

27. Problem 9 Moment of Inertia Experiment
SAME force applied to all. Which one will undergo the greatest angular acceleration?

28. Solution 9 Moment of Inertia Experiment
This one will undergo the greatest angular acceleration.

29. What is Moment of Inertia?
F = m a
Force = mass x ( linear ) acceleration
 = I 
Torque = moment of inertia x angular acceleration

30. I = mr2
The moment of inertia of a particle of mass m spinning at a distance r is I = mr2
For the same torque, the smaller the moment of inertia, the greater the angular acceleration.

31. All about Sarah Hughes . . .
Click me!

32. Problem Sarah Hughes
Will her mass change when she pulls her arms in?
Will her moment of inertia change?

33. Solution Sarah Hughes
Mass does not change when she pulls her arms in but her moment of inertia decreases.

34. Problem Guessing Game
A ball, hoop, and disc have the same mass. Arrange in order of decreasing I
A. hoop, disc, ball
B. hoop, ball, disc
C. ball, disc, hoop
D. disc, hoop, ball

35. Solution 11 . . . Guessing Game
I (moment of inertia) depends on the distribution of mass. The farther the mass is from the axis of rotation, the greater is the moment of inertia.
I = MR I = 1/2 MR I = 2 /5 MR2
hoop disc ball

36. Problem K.E. of Rotation
What is the formula for the kinetic energy of rotation?
A. 1/2 mv2
B. 1/2 m2
C. 1/2 I2
D. I 

37. Solution 12 . . . K.E. of Rotation
The analog of v is 
The analog of m is I
The K.E. of rotation is 1/2 I2

38. Problem Long, thin rod
Calculate the moment of inertia of a long thin rod of mass M and length L rotating about an axis perpendicular to the length and located at one end.

39. Solution 13 . . . Long, thin rod I = mr 2
However, r is a variable so we need to integrate. (ain’t that fun!)
A small mass m of length dr must = M/L dr
I = M/L  r2 dr
I = (M/L)(L3 / 3 )
I = 1/3 ML2

40. Problem 14 . . . In the middle ID = ICM + MD2
Suppose the rod spins about its C.M. One can use the Parallel Axis Theorem to calculate ICM
ID = ICM + MD2
D is the distance between the C.M. and the other axis of rotation

41. Solution 14 . . . In the middle ID = ICM + MD2 1/3 ML2 = ICM + M(L/2)2
ICM = 1/3 ML2 - 1/4 ML2
ICM = 1/12 ML2

42. Problem 1 The race of the century!
Will it be the hoop or the disc?

43. Solution 1 . . . Race of the Century Hoop Loses ! ! !
P.E. = K.E. (linear) + K.E. (angular)
mgh = 1/2 mv2 + 1/2 I2
mgh = 1/2 mv2 + 1/2 I (v/r)2
For the disc, I = 1/2 mr2
So mgh = 1/2 mv2 + 1/2 (1/2 mr2)(v/r)2
Disc v = (4/3 g h)1/2
Similarly Hoop v = (g h)1/2