Chap. 11B - Rigid Body Rotation
A PowerPoint Presentation by
Paul E. Tippens, Professor of Physics
Southern Polytechnic State University
©
2007
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.
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
F = 20 N
R = 0.5 m
a = 2 rad/s2
Linear Inertia, m
24 N
m = 4 m/s2 = 5 kg
Rotational Inertia, I
t
(20 N)(0.5 m)
2
I=a =
=
2.5
kg
m
4 m/s2
Force does for translation what torque does for rotation:
Rotational Kinetic Energy
Consider tiny mass m:
K = ½mv2
K = ½m(wR)2
K=
½(mR2)w2
Sum to find K total:
K = ½(SmR2)w2
(½w2 same for all m )
v = wR
m
w
m1
axis
m
4
m3
m2
Object rotating at constant w.
Rotational Inertia Defined:
I = SmR2
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
m)2
I = (3 kg)(1
+ (2 kg)(3 m)2
+ (1 kg)(2 m)2
I = 25 kg m2
2 kg
3m
3 kg
1m
2m
w
1 kg
w = 600 rpm = 62.8 rad/s
K = ½Iw2 = ½(25 kg m2)(62.8 rad/s) 2
K = 49,300 J
Common Rotational Inertias
L
L
I
I=
1
I
2
3
mL
R
R
mR2
½mR2
Hoop
I=
Disk or cylinder
1
12
2
mL
R
I
2
5
mR
Solid sphere
2
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.
I  mR  (3 kg)(0.2 m)
2
2
I = 0.120 kg m2
R
I = ½mR2
Disk
R
I = mR2
Hoop
I  12 mR2  12 (3 kg)(0.2 m)2
I = 0.0600 kg m2
Important Analogies
For many problems involving rotation, there is an
analogy to be drawn from linear motion.
x
m
f
t
I
R
4 kg
A resultant force F
produces negative
acceleration a for a
mass m.
F  ma
w w  50 rad/s
o
t = 40 N m
A resultant torque t
produces angular
acceleration a of disk
with rotational inertia I.
t  Ia
Newton’s 2nd Law for Rotation
How many revolutions
required to stop?
t = Ia
FR = (½mR2)a
2F
2(40N)
a

mR (4 kg)(0.2 m)
a = 100 rad/s2
F
R
4 kg
w
wo  50 rad/s
R = 0.20 m
F = 40 N
0
2aq  wf2 - wo2
w02 (50 rad/s)2
q

2a
2(100 rad/s2 )
q = 12.5 rad = 1.99 rev
Example 3: What is the linear acceleration of the falling 2-kg mass?
Apply Newton’s 2nd law to rotating disk:
TR =
t  Ia
T = ½MRa
M 6 kg
a=?
2 kg
a
a
=
aR;
a
=
but
R
a
T = ½MR( ) ;
R
(½MR2)a
R = 50 cm
and
T = ½Ma
R = 50 cm
6 kg
Apply Newton’s 2nd law to falling mass:
mg - T = ma
mg - ½Ma
T = ma
T
+a
(2 kg)(9.8 m/s2) - ½(6 kg) a = (2 kg) a
19.6 N - (3 kg) a = (2 kg) a
T
a = 3.92 m/s2
2 kg
mg
Work and Power for Rotation
Work = Fs = FRq
t  FR
q
Work = tq
tq
Work
Power =
= t
t
s
q
w=
t
F
F
s = Rq
Power = t w
Power = Torque x average angular velocity
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.
Work = tq = FR q
s
20 m
q=
=
= 50 rad
R
0.4 m
q
2 kg
6 kg
Power =
Work
= 392 J
t
4s
F
F=W
s = 20 m
F = mg = (2 kg)(9.8 m/s2); F = 19.6 N
Work = (19.6 N)(0.4 m)(50 rad)
s
Work = 392 J
Power = 98 W
The Work-Energy Theorem
Recall for linear motion that the work done is equal
to the change in linear kinetic energy:
Fx  ½mv  ½mv
2
f
2
0
Using angular analogies, we find the rotational work
is equal to the change in rotational kinetic energy:
tq  ½Iw  ½Iw
2
f
2
0
Applying the Work-Energy Theorem:
F
What work is needed
to stop wheel rotating:
R
Work = DKr
4 kg
w
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
0
tq  ½Iw  ½Iw
2
f
2
0
Work = -½Iwo2
Work = -½(0.36 kg m2)(60 rad/s)2
Work = -648 J
Combined Rotation and Translation
vcm
vcm
vcm
First consider a disk sliding
without friction. The velocity of
any part is equal to velocity vcm
of the center of mass.
w
Now consider a ball rolling without
slipping. The angular velocity w
about the point P is same as w for
disk, so that we write:
v
w
R
Or
v
R
P
v  wR
Two Kinds of Kinetic Energy
Kinetic Energy
of Translation:
Kinetic Energy
of Rotation:
K=
w
½mv2
v
R
P
K = ½Iw2
Total Kinetic Energy of a Rolling Object:
KT  mv  Iw
1
2
2
1
2
2
Angular/Linear Conversions
In many applications, you must solve an equation
with both angular and linear parameters. It is
necessary to remember the bridges:
s qR
s
q
R
Velocity:
v  wR
Acceleration:
v aR
v
w
R
a
a
R
Displacement:
Translation or Rotation?
If you are to solve for a linear parameter, you
must convert all angular terms to linear terms:
s
q
R
v
w
R
a
a
R
I  (?)mR
2
If you are to solve for an angular parameter, you
must convert all linear terms to angular terms:
s qR
v  wR
v aR
Example (a): Find velocity v of a disk if
given its total kinetic energy E.
Total energy: E = ½mv2 + ½Iw2
v
E  mv  I w ; I  mR ; w 
R
1
2
2
1
2
2
2
1
2
2


v
2
2
1
1 1
E  2 mv  2  2 mR   2  ;
R 
3mv 2
E
4
or
E  12 mv 2  14 mv 2
4E
v
3m
Example (b) Find angular velocity w of a disk
given its total kinetic energy E.
Total energy: E = ½mv2 + ½Iw2
E  12 mv2  12 Iw2 ; I  12 mR2 ; v  wR
E  12 m(w R) 2  12  12 mR 2  w 2 ; E  12 mR 2w 2  14 mR 2w 2
3mR 2w 2
E
4
or
4E
w
3mR 2
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.
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
w
Two kinds of energy:
KT = ½mv2
v
Kr = ½Iw2
Total energy: E =
½mv2
+
½Iw2
2


v
2
2
Disk: E  ½mv  ½ ½mR  2 
R 
2

v 
2
2
Hoop: E  ½mv  ½ mR  2 
R 




v
w=
R
E = ¾mv2
E = mv2
v
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?
mgho
Rotation?
½Iwo2
velocity?
½mvo2
=
mghf
Height?
½Iwf2
Rotation?
½mvf2
velocity?
Example 6: Find the velocity of the 2-kg mass
just before it strikes the floor.
R = 50 cm
mgho
mghf
=
½Iwo2
½Iwf2
½mvf2
½mvo2
mgh0  12 mv2  12 Iw 2
(2)(9.8)(10)  (2)v  (6)v
2
2 kg
h = 10 m
I  12 MR2
2


v
2
2
1
1 1
mgh0  2 mv  2 ( 2 MR )  2 
R 
1
2
6 kg
1
4
2
2.5v2 = 196 m2/s2
v = 8.85 m/s
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
2


v
2
2
mgh0  ½ mv  ½(mR )  2 
R 
20 m
mgho = ½mv2 + ½mv2; mgho = mv2
v  gh0  (9.8 m/s 2 )(20 m)
Hoop:
Disk: I = ½mR2; mgho = ½mv2 + ½Iw2
2

v 
2
2
mgh0  ½ mv  ½(½ mR )  2 
R 
v = 14 m/s
v
4
3
gh0
v = 16.2 m/s
Angular Momentum Defined
Consider a particle m
moving with velocity v
in a circle of radius r.
Define angular momentum L:
L = mvr
Substituting v= wr, gives:
L = m(wr) r = mr2w
For extended rotating body:
L = (Smr2) w
v = wr
m
w
m1
axis
m
4
m3
m2
Object rotating at constant w.
Since I = Smr2, we have:
L = Iw
Angular Momentum
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.
1
For rod: I = 12mL2 =
1
(4
12
kg)(2 m)2
L=2m
m = 4 kg
I = 1.33 kg m2
rev  2 rad  1 min 

w   300


  31.4 rad/s
min  1 rev  60 s 

L = Iw  (1.33 kg m2)(31.4 rad/s)2
L = 1315 kg m2/s
Impulse and Momentum
Recall for linear motion the linear impulse is equal to
the change in linear momentum:
F Dt  mv f  mv0
Using angular analogies, we find angular impulse to
be equal to the change in angular momentum:
t Dt  Iw f  Iw0
Example 9: A sharp force of 200 N is applied to
the edge of a wheel free to rotate. The force acts
for 0.002 s. What is the final angular velocity?
I = mR2 = (2 kg)(0.4 m)2
I = 0.32 kg m2
Applied torque t  FR
D t = 0.002 s w  0 rad/s
w o
R
R = 0.40 m
F
2 kg
F = 200 N
Impulse = change in angular momentum
0
t Dt = Iwf  Iwo
FR Dt = Iwf
wf = 0.5 rad/s
Conservation of Momentum
In the absence of external torque the rotational
momentum of a system is conserved (constant).
0
Ifwf  Iowo = t Dt
Ifwf  Iowo
Io = 2 kg m2; wo = 600 rpm
I0w0 (2 kg  m2 )(600 rpm)
wf 

If
6 kg  m2
If = 6 kg m2; wo = ?
wf = 200 rpm
Summary – Rotational Analogies
Quantity
Linear
Rotational
Displacement
Displacement x
Radians q
Inertia
Mass (kg)
I (kgm2)
Force
Newtons N
Torque N·m
Velocity
v
“ m/s ”
w
Rad/s
Acceleration
a
“ m/s2 ”
a
Rad/s2
Momentum
mv (kg m/s)
Iw (kgm2rad/s)
Analogous Formulas
Linear Motion
Rotational Motion
F = ma
K = ½mv2
Work = Fx
t = Ia
K = ½Iw2
Work = tq
Power = Fv
Power = Iw
Fx = ½mvf2 - ½mvo2
tq =
½Iwf2 - ½Iwo2
Summary of Formulas:
K  Iw
1
2
Work  tq
2
tq  ½Iw  ½Iw
2
f
Height?
mgho
Rotation?
½Iwo2
velocity?
½mvo2
2
0
=
I = SmR2
Iowo  I f w f
Power 
tq
t
 tw
mghf
Height?
½Iwf2
Rotation?
½mvf2
velocity?
CONCLUSION: Chapter 11B
Rigid Body Rotation