Physics 111: Lecture 17
Today’s Agenda

Rotational Kinematics
Analogy with one-dimensional kinematics

Kinetic energy of a rotating system
Moment of inertia
Discrete particles
Continuous solid objects

Parallel axis theorem
Physics 111: Lecture 17, Pg 1
Rotation

Up until now we have gracefully avoided dealing with the
rotation of objects.
We have studied objects that slide, not roll.
We have assumed pulleys are without mass.

Rotation is extremely important, however, and we need to
understand it!

Most of the equations we will develop are simply rotational
analogues of ones we have already learned when studying
linear kinematics and dynamics.
Physics 111: Lecture 17, Pg 2
Lecture 17, Act 1
Rotations

Bonnie sits on the outer rim of a merry-go-round, and Klyde
sits midway between the center and the rim. The merry-goround 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
Physics 111: Lecture 17, Pg 3
Lecture 17, Act 1
Rotations

The angular velocity w of any point on a solid object rotating
about a fixed axis is the same.
Both Bonnie & Klyde go around once (2p radians) every
two seconds.
(Their “linear” speed v will be different since v = wr).
w
VKlyde
1
 VBonnie
2
Physics 111: Lecture 17, Pg 4
Rotational Variables.


Spin round
blackboard
Rotation about a fixed axis:
Consider a disk rotating about
an axis through its center:
First, recall what we learned about
Uniform Circular Motion:
w

w
d
dt
(Analogous to v 
dx
)
dt
Physics 111: Lecture 17, Pg 5
Rotational Variables...



Now suppose w can change as a function of time:
We define the
dw d 2
angular acceleration:  
 2
dt
dt
Consider the case when 
is constant.
 We can integrate this to
find w and  as a function of time:
  constant
w  w 0  t

w

1
2
   0  w 0t  t 2
Physics 111: Lecture 17, Pg 6
Rotational Variables...
  constant
w  w0  t
1
   0  w0 t  t 2
2
Recall also that for a point at a
distance R away from the axis of
rotation:
x = R
v = wR
And taking the derivative of this we find:
a = R
v

R
x
w


Physics 111: Lecture 17, Pg 7
Summary
(with comparison to 1-D kinematics)
Angular
  constant
w  w0  t
1
  0  w 0 t  t 2
2
Linear
a  constant
v  v 0  at
x  x 0  v 0t 
1 2
at
2
And for a point at a distance R from the rotation axis:
x = Rv = wR
a = R
Physics 111: Lecture 17, Pg 8
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 = 2p radians)
a
R
Physics 111: Lecture 17, Pg 9
Wheel And Rope...
Use a = R to find :
 = a / R = 4 m/s2 / 0.4 m = 10 rad/s2
 Now use the equations we derived above just as you would
use the kinematic equations from the beginning of the
semester.
1
1
   0  w0 t   t 2 = 0 + 0(10) + (10)(10)2 = 500 rad
2
2

1 rot
 500 rad x
2p rad
 80 rev
a

R
Physics 111: Lecture 17, Pg 10
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
w
r4
r3
r2
m2
Physics 111: Lecture 17, Pg 11
Rotation & Kinetic Energy...

1
K   mi v i2
i 2
So:
K
but vi = wri
1
1 2
2
m
w
r

w  mi ri 2
 i i
2 i
2
i
v1
which we write as:
1
K  I w2
2
I   mi ri
2
m4
v4
m3
i
Define the moment of inertia
about the rotation axis
r1 m1
w
r4
v2
r3
r2
m2
v3
I has units of kg m2.
Physics 111: Lecture 17, Pg 12
Rotation & Kinetic Energy...

The kinetic energy of a rotating system looks similar to that
of a point particle:
Point Particle
K 
1
mv 2
2
v is “linear” velocity
m is the mass.
Rotating System
1
I w2
2
w is angular velocity
I is the moment of inertia
about the rotation axis.
K
I   mi ri 2
i
Physics 111: Lecture 17, Pg 13
Moment of Inertia

So K 
1
I w2
2
Inertia Rods
2
where I   mi ri
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).

We will see that in rotational dynamics, the moment of
inertia I appears in the same way that mass m does when
we study linear dynamics!
Physics 111: Lecture 17, Pg 14
Calculating Moment of Inertia

We have shown that for N discrete point masses distributed
about a fixed axis, the moment of inertia is:
I
N
 mi ri 2
where r is the distance from the mass
to the axis of rotation.
i 1
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
Physics 111: Lecture 17, Pg 15
Calculating Moment of Inertia...

The squared distance from each point mass to the axis is:
2
L2
L
2
r  2  
2
2
Using the Pythagorean Theorem
2
2
2
2
2
L
L
L
L
L
so I   mi ri  m  m  m  m  4 m
i 1
2
2
2
2
2
N
2
I = 2mL2
L/2
m
r
m
L
m
m
Physics 111: Lecture 17, Pg 16
Calculating Moment of Inertia...

Now calculate I for the same object about an axis through
the center, parallel to the plane (as shown):
L2
L2
L2
L2
L2
I   mi ri  m  m  m  m  4 m
i 1
4
4
4
4
4
N
2
r
I = mL2
m
m
m
m
L
Physics 111: Lecture 17, Pg 17
Calculating Moment of Inertia...

Finally, calculate I for the same object about an axis along
one side (as shown):
N
I   mi ri  mL2  mL2  m0 2  m0 2
2
i 1
r
I = 2mL2
m
m
m
m
L
Physics 111: Lecture 17, Pg 18
Calculating Moment of Inertia...

For a single object, I clearly depends on the rotation axis!!
I = 2mL2
I = mL2
m
m
m
m
I = 2mL2
L
Physics 111: Lecture 17, Pg 19
Lecture 17, Act 2
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 > Ib > Ic
(b)
Ia > Ic > Ib
b
(c)
Ib > Ia > Ic
c
Physics 111: Lecture 17, Pg 20
Lecture 17, Act 2
Moment of Inertia

Label masses and lengths:

Calculate moments of inerta:
Ia  m 2 L  m 2 L  8 mL2
2
2
I b  mL2  mL2  mL2  3 mL2
I c  m 2 L  4 mL2
2
m
a
L
b
So (b) is correct: Ia > Ic > Ib
L
c
m
m
Physics 111: Lecture 17, Pg 21
Calculating Moment of Inertia...

For a discrete collection of point
masses we found:
I
N
 mi ri 2
i 1

For a continuous solid object we have to add up the mr2
contribution for every infinitesimal mass element dm.
We have to do an
integral to find I :
dm
I   r 2 dm
r
Physics 111: Lecture 17, Pg 22
Moments of Inertia

Hoop
Some examples of I for solid objects:
I  MR 2
R
R
I
Thin hoop (or cylinder) of mass M and
radius R, about an axis through its center,
perpendicular to the plane of the hoop.
1
MR 2
2
Thin hoop of mass M and radius R,
about an axis through a diameter.
Physics 111: Lecture 17, Pg 23
Sphere and disk
Moments of Inertia...

Some examples of I for solid objects:
2
I  MR 2
5
Solid sphere of mass M and radius R,
R
about an axis through its center.
I
R
1
MR 2
2
Solid disk or cylinder of mass M and
radius R, about a perpendicular axis
through its center.
Physics 111: Lecture 17, Pg 24
Lecture 17, Act 3
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
Physics 111: Lecture 17, Pg 25
Lecture 17, Act 3
Moment of Inertia

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
Physics 111: Lecture 17, Pg 26
Moments of Inertia...

Rod
Some examples of I for solid objects (see also Tipler, Table 9-1):
I
L
L
1
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.
Physics 111: Lecture 17, Pg 27
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.
Physics 111: Lecture 17, Pg 28
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
2
IPARALLEL = ICM + MD
x
L
We know ICM 
So
1
ML2
12
IEND
IEND
ICM
1
L2 1

2
 ML  M    ML2
2
12
3
which agrees with the result on a previous slide.
Physics 111: Lecture 17, Pg 29
Connection with CM motion

Recall what we found out about the kinetic energy of a
system of particles in Lecture 15:
1
1
2
K NET   mi ui2  MVCM
2
2

 

KREL
KCM

For a solid object rotating about its center of mass, we
now see that the first term becomes:
1
K REL   mi ui 2
2
1
K REL  w 2  mi ri 2
2
K REL 
Substituting ui  w ri
but
 mi ri
2
 ICM
1
ICM w 2
2
Physics 111: Lecture 17, Pg 30
Connection with CM motion...

So for a solid object which rotates about its center or mass
and whose CM is moving:
K NET 
1
1
2
ICM w 2  MVCM
2
2
VCM
w
We will use this formula more in coming lectures.
Physics 111: Lecture 17, Pg 31
Recap of today’s lecture

Rotational Kinematics
(Text: 9-1)
Analogy with one-dimensional kinematics

Kinetic energy of a rotating system
Moment of inertia
Discrete particles
Continuous solid objects
(Text: 9-2, 9-3, Table 9-1)
(Text: 9-3)
(Text: 9-3)

Parallel axis theorem
(Text: 9-3)

Look at textbook problems Chapter 9: # 7, 11, 27, 31, 33, 37
Physics 111: Lecture 17, Pg 32