1. Work and Power for Rotation
Work = Fd = FRq
t = FR q F s
s = Rq
Work = tq
Power = =
Workt
tq t
w =
q t
Power = t w
Power = Torque x average angular velocity

2. Example 1: The rotating disk has a radius of 40 cm and a mass of 6 kg
Example 1: The rotating disk has a radius of 40 cm and a mass of 6 kg. Find the work and power if the 2-kg mass is lifted 20 m in 4 s. q F
F=W s
s = 20 m
2 kg
6 kg
Work = tq = FR q sR
q = = = 50 rad
20 m 0.4 m
F = mg = (2 kg)(9.8 m/s2); F = 19.6 N
Work = 392 J
Work = (19.6 N)(0.4 m)(50 rad)
Power = =
Workt
392 J 4s
Power = 98 W

3. The Work-Energy Theorem
Recall for linear motion that the work done is equal to the change in linear kinetic energy:
Using angular analogies, we find the rotational work is equal to the change in rotational kinetic energy:

4. Two Kinds of Kinetic Energy v R P
Kinetic Energy of Translation:
K = ½mv2
Kinetic Energy of Rotation:
K = ½I2
Total Kinetic Energy of a Rolling Object:

5. Translation or Rotation?
If you are to solve for a linear parameter, you must convert all angular terms to linear terms:
If you are to solve for an angular parameter, you must convert all linear terms to angular terms:

6. Total energy: E = ½mv2 + ½Iw2
Example 2: A circular hoop and a circular disk, each of the same mass and radius, roll at a linear speed v. Compare the kinetic energies. w v
Two kinds of energy:
KT = ½mv2
Kr = ½Iw2
w = vR
Total energy: E = ½mv2 + ½Iw2
Disk:
E = ¾mv2
Hoop:
E = mv2

7. Conservation of Energy
The total energy is still conserved for systems in rotation and translation.
However, rotation must now be considered.
Begin: (U + Kt + KR)o = End: (U + Kt + KR)f
Height?
Rotation?
velocity?
mgho
½Iwo2
½mvo2 =
mghf
½Iwf2
½mvf2
Height?
Rotation?
velocity?

8. = v = 8.85 m/s mgho mghf ½Iwo2 ½Iwf2 ½mvo2 ½mvf2
Example 3: Find the velocity of the 2-kg mass just before it strikes the floor.
h = 10 m
6 kg
2 kg
R = 50 cm
mgho
½Iwo2
½mvo2 =
mghf
½Iwf2
½mvf2
2.5v2 = 196 m2/s2
v = 8.85 m/s

9. mgho = ½mv2 + ½mv2; mgho = mv2
Example 4: A hoop and a disk roll from the top of an incline. What are their speeds at the bottom if the initial height is 20 m?
mgho = ½mv2 + ½Iw2
Hoop: I = mR2
20 m
mgho = ½mv2 + ½mv2; mgho = mv2
Hoop:
v = 14 m/s
Disk: I = ½mR2;
mgho = ½mv2 + ½Iw2
v = 16.2 m/s

10. Angular Momentum Defined
Consider a particle m moving with velocity v in a circle of radius r. m2 m3 m4 m m1
axis w
v = wr
Object rotating at constant w.
Define angular momentum L:
L = mvr
Substituting v= wr, gives:
L = m(wr) r = mr2w
Since I = Smr2, we have:
L = Iw
For extended rotating body:
L = (Smr2) w
Angular Momentum

11. Example
Rank the following from largest to smallest angular momentum.

12. answer

13. m = 4 kg
L = 2 m
Example 5: Find the angular momentum of a thin 4-kg rod of length 2 m if it rotates about its midpoint at a speed of 300 rpm.
For rod: I = mL2 = (4 kg)(2 m)2
1 12
I = 1.33 kg m2
L = Iw = (1.33 kg m2)(31.4 rad/s)2
L = 1315 kg m2/s

14. Impulse and Momentum
Recall for linear motion the linear impulse is equal to the change in linear momentum:
Using angular analogies, we find angular impulse to be equal to the change in angular momentum:

15. Impulse = change in angular momentum
Example 6: A sharp force of 200 N is applied to the edge of a wheel free to rotate. The force acts for s. What is the final angular velocity? R
2 kg w F
wo = 0 rad/s
R = 0.40 m
F = 200 N
D t = s
I = mR2 = (2 kg)(0.4 m)2
I = 0.32 kg m2
Applied torque t = FR
Impulse = change in angular momentum
t Dt = Iwf - Iwo
FR Dt = Iwf
wf = 0.5 rad/s

16. Conservation of Momentum
If no net external torques act on a system then the system’s angular momentum, L, remains constant. Speed and radius can change just as long as angular momentum is constant.
Ifwf = Iowo
Ifwf - Iowo = t Dt
Io = 2 kg m2; wo = 600 rpm
If = 6 kg m2; wo = ?
wf = 200 rpm

17. Example
The figure below shows two masses held together by a thread on a rod that is rotating about its center with angular velocity, ω. If the thread breaks, what happens to the system's (a) angular momentum and (b) angular speed. (Increase, decrease or remains the same)

18. Examples of the Conservation of Angular Momentum
Natalia Kanounnikova World Record Spin RPM

19. Diving 18 meter high dive Amazing Fouette Turns on Pointe
Playground Physics

20. Summary – Rotational Analogies
Quantity
Linear
Rotational
Displacement
Displacement x
Radians 
Inertia
Mass (kg)
I (kgm2)
Force
Newtons N
Torque N·m
Velocity
v “ m/s ”
 Rad/s
Acceleration
a “ m/s2 ”
 Rad/s2
Momentum
mv (kg m/s)
I (kgm2rad/s)

21. Analogous Formulas F = ma  = I K = ½mv2 K = ½I2 Work = Fx Work = tq
Linear Motion
Rotational Motion
F = ma
 = I
K = ½mv2
K = ½I2
Work = Fx
Work = tq
Power = Fv
Power = I
Fx = ½mvf2 - ½mvo2
 = ½If2 - ½Io2

22. = Summary of Formulas: I = SmR2 mgho mghf Height? ½Iwo2 ½Iwf2
½mvo2 =
mghf
½Iwf2
½mvf2
Height?
Rotation?
velocity?

23. CONCLUSION: Angular Motion