1. Chap. 11B - Rigid Body Rotation
A PowerPoint Presentation by
Paul E. Tippens, Professor of Physics
Southern Polytechnic State University
© 2007

2. Objectives: After completing this module, you should be able to:
Define and calculate the moment of inertia for simple systems.
Define and apply the concepts of Newton’s second law, rotational kinetic energy, rotational work, rotational power, and rotational momentum to the solution of physical problems.
Apply principles of conservation of energy and momentum to problems involving rotation of rigid bodies.

3. Force does for translation what torque does for rotation:
Inertia of Rotation
Consider Newton’s second law for the inertia of rotation to be patterned after the law for translation.
F = 20 N
a = 4 m/s2
Linear Inertia, m
m = = 5 kg
24 N 4 m/s2
F = 20 N
R = 0.5 m
a = 2 rad/s2
Rotational Inertia, I
I = = = 2.5 kg m2
(20 N)(0.5 m) 4 m/s2
t a
Force does for translation what torque does for rotation:

4. Rotational Kinetic Energy
Consider tiny mass m: m2 m3 m4 m m1
axis w
v = wR
Object rotating at constant w.
K = ½mv2
K = ½m(wR)2
K = ½(mR2)w2
Sum to find K total:
K = ½(SmR2)w2
Rotational Inertia Defined:
I = SmR2
(½w2 same for all m )

5. I = (3 kg)(1 m)2 + (2 kg)(3 m)2 + (1 kg)(2 m)2
Example 1: What is the rotational kinetic energy of the device shown if it rotates at a constant speed of 600 rpm?
First: I = SmR2
3 kg
2 kg
1 kg
1 m
2 m
3 m w
I = (3 kg)(1 m) (2 kg)(3 m)2 + (1 kg)(2 m)2
I = 25 kg m2
w = 600 rpm = 62.8 rad/s
K = ½Iw2 = ½(25 kg m2)(62.8 rad/s) 2
K = 49,300 J

6. Common Rotational Inertias
I = mR2
I = ½mR2
Hoop
Disk or cylinder
Solid sphere

7. Example 2: A circular hoop and a disk each have a mass of 3 kg and a radius of 30 cm. Compare their rotational inertias. R
I = mR2
Hoop
I = kg m2 R
I = ½mR2
Disk
I = kg m2

8. Important Analogies t m I x f
For many problems involving rotation, there is an analogy to be drawn from linear motion. R
4 kg w t
wo = 50 rad/s t = 40 N m m x f I
A resultant torque t produces angular acceleration a of disk with rotational inertia I.
A resultant force F produces negative acceleration a for a mass m.

9. Newton’s 2nd Law for Rotation
4 kg w F
wo = 50 rad/s
R = 0.20 m
F = 40 N
t = Ia
How many revolutions required to stop?
FR = (½mR2)a
2aq = wf2 - wo2
a = 100 rad/s2
q = 12.5 rad = 1.99 rev

10. Example 3: What is the linear accel- eration of the falling 2-kg mass?
R = 50 cm
6 kg
2 kg
a = ? M
Apply Newton’s 2nd law to rotating disk:
t = Ia
TR = (½MR2)a
a = aR; a =
but aR
T = ½MRa
R = 50 cm
6 kg
2 kg +a T mg
T = ½MR( ) ; aR
and
T = ½Ma T
Apply Newton’s 2nd law to falling mass:
mg - T = ma
mg = ma
½Ma
(2 kg)(9.8 m/s2) - ½(6 kg) a = (2 kg) a
19.6 N - (3 kg) a = (2 kg) a
a = 3.92 m/s2

11. Work and Power for Rotation
Work = Fs = 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

12. Example 4: The rotating disk has a radius of 40 cm and a mass of 6 kg
Example 4: 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

13. 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:

14. Applying the Work-Energy Theorem:
Work = DKr
What work is needed to stop wheel rotating: R
4 kg w F
wo = 60 rad/s
R = 0.30 m
F = 40 N
First find I for wheel: I = mR2 = (4 kg)(0.3 m)2 = 0.36 kg m2
Work = -½Iwo2
Work = -½(0.36 kg m2)(60 rad/s)2
Work = -648 J

15. Combined Rotation and Translation
vcm
First consider a disk sliding without friction. The velocity of any part is equal to velocity vcm of the center of mass.  v R P
Now consider a ball rolling without slipping. The angular velocity  about the point P is same as  for disk, so that we write: Or

16. 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:

17. Angular/Linear Conversions
In many applications, you must solve an equation with both angular and linear parameters. It is necessary to remember the bridges:
Displacement:
Velocity:
Acceleration:

18. 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:

19. Total energy: E = ½mv2 + ½Iw2
Example (a): Find velocity v of a disk if given its total kinetic energy E.
Total energy: E = ½mv2 + ½Iw2

20. Total energy: E = ½mv2 + ½Iw2
Example (b) Find angular velocity  of a disk given its total kinetic energy E.
Total energy: E = ½mv2 + ½Iw2

21. Strategy for Problems Draw and label a sketch of the problem.
List givens and state what is to be found.
Write formulas for finding the moments of inertia for each body that is in rotation.
Recall concepts involved (power, energy, work, conservation, etc.) and write an equation involving the unknown quantity.
Solve for the unknown quantity.

22. Total energy: E = ½mv2 + ½Iw2
Example 5: 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

23. 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?

24. = v = 8.85 m/s mgho mghf ½Iwo2 ½Iwf2 ½mvo2 ½mvf2
Example 6: 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

25. mgho = ½mv2 + ½mv2; mgho = mv2
Example 7: 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

26. 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

27. m = 4 kg
L = 2 m
Example 8: 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

28. 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:

29. Impulse = change in angular momentum
Example 9: 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

30. Conservation of Momentum
In the absence of external torque the rotational momentum of a system is conserved (constant).
Ifwf - Iowo = t Dt
Ifwf = Iowo
Io = 2 kg m2; wo = 600 rpm
If = 6 kg m2; wo = ?
wf = 200 rpm

31. 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)

32. 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

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

34. CONCLUSION: Chapter 11B Rigid Body Rotation