1. Physics 203 College Physics I Fall 2012
S. A. Yost
Chapter 8 Part 2
Rotational Motion

2. Announcements
Problem set HW08 is due Thursday. It covers Chapter 8 except angular momentum, rolling, and some torque problems. Today: Rotational Motion, Ch. 10 , mostly sec7 – 8. Thursday: chapter 9, sec. 1, 2, and 4 . Please read these sections before class. A problem set HW09 on them and the remainder of Ch. 8 will be posted today and due next Tuesday. Deadline for making up Exam 2: Wed. 5 PM.

3. Rotational Analogs of Linear Motion
Linear Motion Rotational Motion
Kinematics: x , v, a q, w, a
Dynamics: m, F, Kt I, t, Kr
F = ma t = Ia
Kt = ½ mv Kr = ½ Iw 2
W = Fx W = t q
P = F v P = tw

4. Torque t = R F^ = R F sin q . The torque is defined to
be the perpendicular
component of the force
times the distance from
the pivot to where it acts: F F^ R q
t = R F^
where F^ = F sin q.
Counter-clockwise torque
is considered to be
positive, as for angles.
Then
= R F sin q .

5. Torque t = R ^ F t = R F^ = R F sin q The torque can also be
expressed in terms of the
magnitude of the force and
the distance from the axis
to the line of the force.
The distance R ^ is called
the lever arm of the torque. F R q R^
t = R ^ F
t = R F^
= R F sin q

6. Tightening a Nut with a Wrench
Which use of the wrench is most effective for tightening the nut? Which is least effective? Which of A and D is more effective? Choose E if they are the same.
A B
C D
The lever arms are the same.

7. Question
A force is applied to the rim of two wheels. Assuming the only significant mass is in the rim, what force F2 will give the wheels identical angular accelerations? (A) 0.25 N (B) 0.5 N (C) 1.0 N (D) 2.0 N (E) 4.0 N
I = mR2

8. Question t = RF = I a = mR2 a. = = → = F2 = 2 F1 = 2 N.
t R2F mR22 a F R2
t R1F mR12 a F R1
I = mR2

9. Example a Mass falling on rope wrapped around a massive pulley. R
Assume the pulley is a uniform disk as shown.
What is the acceleration of the hanging mass? R M a

10. Example FT a mg Isolate the hanging mass: Newton’s Law:
M a = Fnet = Mg – FT
where FT is the tension in the rope. FT M a mg

11. Example m
Isolate the pulley: t = I a with I = ½ m R2, t = RFT , a = a/R.
Then RFT = (½ mR2)(a/R).
Therefore, FT = ½ ma. R a FT

12. Example FT M a Combine results: R M a = Fnet = Mg – FT = Mg – ½ ma.
Then (M + m/2) a = Mg.
Result: a = R FT M
1 + m/2M g a

13. Example F2
Note that the tension does not need to be the same on two sides of a massive pulley.
Net torque =
R(F1 – F2) = Ia. a m R R F1

14. Rigid Body Motion
The relation t = I a holds for rigid body rotation in any inertial frame. This always holds in the CM frame of the rigid body, even if it is accelerating. The energy of a rigid body can be expressed as a sum K = K cm + K rot with K cm = ½ mvcm2, K rot = ½ Iw 2. “Newton’s Law” F = m acm (ch. 7), t = I a.

15. Dumbbell m → → F
A force F is applied for time t to a dumbbell in one of two ways shown.
Which gives the greater speed to the center of mass?
(a) A (b) B
(c) the same A m → F
→ →
Dp = Ft B

16. Dumbbell m → → F
A force F is applied for time t to a dumbbell in one of two ways shown.
Which gives the greater energy to the dumbbell?
(a) A (b) B
(c) the same A m → F B

17. Dumbbell F F The total kinetic energy is Case A:→ F
The total kinetic energy is
Case A:
K = Ktrans + Krot = ½ mvcm2 + ½ Iw2
Case B: no rotation:
K = ½ mvcm2
There is more energy in case A. A m → F B

18. Rolling
When an object rolls, its circumference moves a distance 2pr every period, so w and v are related:
v = 2pr/T = rw
2pr
2pr
2pr
2pr

19. Rolling
A solid wheel and a hollow wheel roll down a ramp, starting from rest at the same point.
Which gets to the bottom faster?

20. Rolling
If an object with mass m and moment of inertia I rolls down an inclined plane of height h and length L, how fast is it rolling when it gets to the bottom?
m,I h L

21. √ Rolling v = Energy conservation: Ui = Ktrans + Krot
mgh = ½ mv2 + ½ Iw2.
Rolling: w = v/R.
mgh = ½ (m + I/R2) v2. √ h L
m,I
v =
2gh
1 + I/(mR2)

22. Rolling
The solid wheel gets to the bottom first, because the object with the smaller moment of inertia relative to its mass and size moves faster.

23. Rotational Analog of Momentum
Linear Motion:
(one dimension)
Momentum:
p = mv
Impulse: Dp = Ft
Rotational Motion:
(fixed axis)
Angular momentum:
L = Iw
DL = t t

24. Angular Momentum Units of angular momentum:
L = I w = [kg m2][s-1] = kg . m2/s
DL = t t = [mN][s] = Nms = J.s
When there is no external torque on a system, angular momentum is conserved.
In particular, this applies to collisions between rigid bodies.

25. Figure Skater
A figure skater increases her rotational rate from 1.0 rev/s to 2.5 rev/s in 1.5 s. Her initial moment of inertia was 4.6 kg ∙ m2. (a) What was her final moment of inertia?

26. Figure Skater
A figure skater increases her rotational rate from 1.0 rev/s to 2.5 rev/s in 1.5 s. Her initial moment of inertia was 4.6 kg ∙ m2. I1w1 = I2 w2 w2 = 2.5 w1 I2 = I1 / 2.5 = 1.84 kg ∙ m2
≈ 1.8 kg ∙ m2

27. Figure Skater
A figure skater increases her rotational rate from 1.0 rev/s to 2.5 rev/s in 1.5 s. Her initial moment of inertia was 4.6 kg ∙ m2. (b) What average power did she apply to pull in her arms?

28. Figure Skater
P = W/t, W = DK = ½ Iw22 – ½ Iw12. w1 = 1.0 rev/s (2p rad/rev) = 2.0 p rad/s w2 = 2.5 rev/s (2p rad/rev) = 5.0 p rad/s I1 = 4.6 kg ∙ m2, I 2 = 1.84 kg ∙ m2 W = 277 J – 90.8 J ≈ 186 J, t = 1.5 s. P = 124 W.

29. Analog of Inelastic Collision
A bar of length 2R is dropped onto a rotating disk of radius R.
Suppose M = 2m. If the disk initially rotates at 120 rpm, how fast does it rotate if the stick drops onto it and rotates together with the disk? 2R m M R w0

30. Analog of Inelastic Collision
Angular momentum is conserved: I1w1 = I2 w2 .
I1 =½ MR2 = mR2 and I2 = Ibar + Idisk with
Ibar = mL2/12 = mR2/3, Idisk = ½ MR2 = mR2
I2 = 4mR2/3
w2= ( I1 / I2) w1 = ¾ w1
= ¾ (120 rpm) = 90 rpm. m R wf M

31. Angular Quantities as Vectors
If the axis is not fixed, we have to specify a direction for angular displacements and velocities. The convention use a vector pointing along the axis. For fixed axis rotations, the vectors q and w are parallel, but they won’t be if the axis direction changes. → q
→ →
w = d q /dt w → q
→ → w
Right-Hand Rule