CHAPTER 4: ROTATIONAL MOTION
4.1 Rotational kinematics
4.2 Moment of inertia
4.3 Parallel axis theorem
4.4 Angular momentum and
rotational energy
Part 1
Rotational kinematics
Riview of rotations
• Bonnie sits on the outer rim of a merry-goround, and Klyde sits midway between the
center and the rim. The merry-go-round
makes one complete revolution every two
seconds.
• Klyde’s angular velocity is:
(a) the same as Bonnie’s
(b) twice Bonnie’s
(c) half Bonnie’s
REVIEW of ANGLE VELOCITY
• The angular velocity  of any point on a solid
object rotating about a fixed axis is the same.
– Both Bonnie & Klyde go around once
(2pi radians) every two seconds.
How about their “linear” speed v ?
The same or different? ?

1
VKlyde  VBonnie
2
Their “linear” speed v will be different since v = r.
Review: Rotational Variables.
• Rotation about a fixed axis:
– Consider a disk rotating about
an axis through its center:
• First, recall what we learned about
Uniform Circular Motion:
d

dt
dx
(Analogous to v 
dt )


Rotational Variables...
• Now suppose  can change as a function
of time:
• We define the angular acceleration:

d d 
  2
dt dt
2

Consider the case when 
is constant.
 We can integrate
this to find  and 
as a function of time:



constant
  0  t
1 2
  0  0 t  t
2
Rotational Variables...

v
constant
  0  t
1 2
  0  0 t  t
2
• Recall also that for a point
at a distance R away from
the axis of rotation:
– x = R (distance in the circle)
– v = R
And taking the derivative of
this we find:
– a = R

R


x
Example: Wheel And Rope
• A wheel with radius R = 0.4 m rotates freely about a
fixed axle. There is a rope wound around the wheel.
Starting from rest at t = 0, the rope is pulled such that
it has a constant acceleration a = 4 m/s2. How many
revolutions has the wheel made after 10 seconds?
(One revolution = 2 radians)
a
R
Solution
• Use a=R to find  :
 = a / R = (4 m/s2 )/ 0.4 m = 10 rad/s2
• Now use this equations just as you
would use the kinematic equations
from the beginning of the semester.
1 2
  0  0 t   t
2
= 0 + 0(10) +
1 rot
rev  500 rad x
2 rad
 80 times
1(10)(10)2 = 500 rad
a
2

R
Part 2
Moment of Inertia
Rotation & Kinetic Energy
• Consider the simple rotating system
shown below. (Assume the masses are
attached to the rotation axis by massless
rigid rods).
• The kinetic energy of this system will be
the sum of the kinetic energy of each
piece:
m4
m3
r1 m1

r4
r3
r2
m2
Compute: Kinetic Energy of Rotation
system
•
1
So: K   mi v i2
i 2
but vi = ri
1
2 1 2
K   mi ri     mi ri 2
2i
2 i
v1
which we write as:
1
K  I 2
2
m4
v4
I   mi ri 2
m3
i
Define the moment of inertia
about the rotation axis
r1 m1

r4
v2
r3
r2
m2
v3
I has units of kg m2.
Rotation & Kinetic Energy...
• The kinetic energy of a rotating system
looks similar to that of a point particle:
Point Particle
1
K  mv 2
2
v is “linear” velocity
m is the mass.
Rotating System
1
I 2
2
 is angular velocity
I is the moment of inertia
about the rotation axis.
K
I   mi ri 2
i
Moment of Inertia

So
1 2
K  I
2
where
I   mi ri 2
i
• Notice that the moment of inertia I depends on the
distribution of mass in the system.
• The further the mass is from the rotation axis,
the bigger the moment of inertia.
* For a given object, the moment of inertia will
depend on where we choose the rotation axis
(unlike the center of mass).
* In rotational dynamics, the moment of inertia I
appears in the same way that mass m does
when we study linear dynamics!
Calculating Moment of Inertia
• We have shown that for N discrete point masses
distributed about a fixed axis, the moment of inertia
is:
N
I
 mi ri 2
i 1
where ri is the distance from the mass i
to the axis of rotation.
Example: Calculate the moment of inertia of four point
masses (m) on the corners of a square whose sides
have length L, about a perpendicular axis through the
center of the square:
m
m
m
m
L
Calculating Moment of Inertia...
• The squared distance from each point mass to
the axis is:
2
L
L2
 
2
r  2  
Using the Pythagorean Theorem
2
2
so
2
2
2
2
2
L
L
L
L
L
2
I   mi ri  m  m  m  m  4 m
i 1
2
2
2
2
2
N
I = 2mL2
L/2
m
r
m
L
m
m
Learning check
• Now calculate I for the same object about an axis
through the center, parallel to the plane (as
shown):
N
L2
L2
L2
L2
L2
2
I   mi ri  m  m  m  m  4 m
i 1
4
4
4
4
4
I=
r
mL2
m
m
m
m
L
Calculating Moment of Inertia...
• Finally, calculate I for the same
object about an axis along one side
(as shown):
N
2
I   mi ri  mL2  mL2  m0 2  m0 2
i 1
r
I = 2mL2
m
m
m
m
L
Calculating Moment of Inertia...
• For a single object, I clearly depends on
the rotation axis!!
I = 2mL2
I = mL2
m
m
m
m
L
I = 2mL2
Check: Moment of Inertia
• A triangular shape is made from identical
balls and identical rigid, massless rods
as shown. The moment of inertia about
the a, b, and c axes is Ia, Ib, and Ic
respectively.
– Which of the following is correct:
a
(a)
Ia > I b > Ic
(b)
Ia > I c > Ib
b
(c)
Ib > I a > Ic
c
Solution

Masses m and lengths L as show:

Calculate moments of inerta:
Ia  m2 L  m2 L  8 mL2
2
2
m
a
I b  mL  mL  mL  3 mL
2
2
2
2
L
b
Ic  m2 L  4 mL
2
2
L
c
m
So (b) is correct: Ia > Ic > Ib
m
Calculating Moment of Inertia...
For a continuous solid object
• For a discrete collection of point
masses we found:
N
I   mi ri 2
i 1
• For a continuous solid object we have to add up
the mr2 contribution for every infinitesimal mass
element dm.
dm
• We have to do an
integral to find I :
I   r dm
2
r
Learn by heart
Moments of Inertia
• Some examples of I for solid objects:
I  MR 2
R
Thin hoop (or cylinder) of mass M and
radius R, about an axis through its center,
perpendicular to the plane of the hoop.
I   r 2dm   R 2dm  R 2  dm  MR 2
R
1
I  MR 2
2
Thin hoop of mass M and radius R,
about an axis through a diameter.
Moments of Inertia...
• Some examples of I for solid objects:
I
R
I
R
2
MR 2
5
Solid sphere of mass M and radius R,
about an axis through its center.
1
MR 2
2
Solid disk or cylinder of mass M and
radius R, about a perpendicular axis
through its center.
Moment of Inertia
• Two spheres have the same radius and equal
masses. One is made of solid aluminum, and
the other is made from a hollow shell of gold.
– Which one has the biggest moment of inertia
about an axis through its center?
(a) solid aluminum
(b) hollow gold
(c) same
hollow
solid
same mass & radius
Hint
• Moment of inertia depends on mass (same for
both) and distance from axis squared, which is
bigger for the shell since its mass is located
farther from the center.
– The spherical shell (gold) will have a bigger
moment of inertia.
ISOLID < ISHELL
hollow
solid
same mass & radius
Moments of Inertia...
• Some examples of I for solid objects (see
also Tipler, Table 9-1):
L
L
1
I  ML2
12
Thin rod of mass M and length L, about
a perpendicular axis through its center.
1
I  ML2
3
Thin rod of mass M and length L, about
a perpendicular axis through its end.
Part 3
Parallel Axis Theorem
Parallel Axis Theorem
• Suppose the moment of inertia of a solid
object of mass M about an axis through
the center of mass, ICM, is known.
• The moment of inertia about an axis
parallel to this axis but a distance D away
is given by:
IPARALLEL = ICM + MD2
So if we know ICM , it is easy
to calculate the moment of inertia
about a parallel axis.
Parallel Axis Theorem: Example
• Consider a thin uniform rod of mass M and length
D. Figure out the moment of inertia about an axis
through the end of the rod.
D=L/2
M
CM
IPARALLEL = ICM + MD2
x
L
IEND
1
2
We know ICM  ML
12
1
L2 1

2
So IEND  ML  M    ML2
2
12
3
ICM
which agrees with the result on a previous slide.
Part 4
Angular momentum and
rotational energy
Angular momentum
Defination:
  



L  r xP  L   L i   ( ri x[m i v i ])
i
i
v
r
  m i ri  I CM
2
i
Example: compute L for the system m=100g,
a=10cm, =(2/5) rad/s

m
L=2ma2. =2x0.1x 10-4 x0.1 
a
=2  10-6.
L
Complete Motion
by linear + rotation
• The total kinetic energy of a system of particles
include 2 parts
1
1
2
2
K NET   m i u i  MVCM
2
2 

KR

KCM
For a solid object rotating about its center of mass, we
now see that the first term becomes:
1
2
K R   mi u i
2
1 2
2
K R    m i ri
2
K TOT
Substituting
ui   ri
 mi ri  ICM
1
1
2
2
 I CM   MVCM
2
2
but
2
Connection with CM motion...
• So for a solid object which rotates about its
center or mass and whose CM is moving:
1
1
2
2
K NET  I CM   MVCM
2
2
VCM

We will use this formula more in coming lectures.
Similarity Between Linear and Rotational Motions
All physical quantities in linear and rotational motions show striking similarity.
Quantities
Mass
Mass
Linear
Rotational
M
Moment of Inertia
Length of motion Distance
Speed
Acceleration
Force
Work
Power
Momentum
Kinetic Energy
v
a
t
Force F
I  mr 2
L
Angle
v
 ma
 (Radian)
r

t  
t
Torque


t
  I
W  
Work W  Fd cos Work
P  F v
P  
p  mv
L  I
Kinetic K  1 mv 2 Rotational K  1 I 2
R
2
2
Review of today’s lecture
• Rotational Kinematics
– Analogy with one-dimensional kinematics
• Kinetic energy of a rotating system
– Moment of inertia
– Discrete particles Continuous solid objects
• Parallel axis theorem
Problem
M=2kg,
r=4cm
Use the law of energy conservation to compute
the velocity of the ball when it hits to the ground.
V0=0, 0=0
Hint : There are 2 motions
(frictionless case)
Change the ball by
Cylinder have the
same M and r
H=3m
A=45
Deg