# Lecture 22

```Physics 207: Lecture 22
Today’s Agenda
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Elastic collisions in two dimensions
&Iacute;Examples (billiards, nuclear scattering)
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Rotational Kinematics
&Iacute;Analogy with one-dimensional kinematics
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Kinetic energy of a rotating system
&Iacute;Moment of inertia
&Iacute;Discrete particles
&Iacute;Continuous solid objects
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Parallel axis theorem
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2-D Elastic Collision:
Nuclear Scattering
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A particle of unknown mass M is initially at rest. A particle of
known mass m is “shot” at it with initial momentum pi . After
the particles collide, the new momentum of the shot particle pf
is measured.
&Iacute; Figure out what M is in terms of pi and pf and m.
at rest
m
P
M
M
pi
m
pf
initial
final
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Page 1
2-D Elastic Collision:
Nuclear Scattering
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Using momentum conservation: pi = pf + P
&Iacute; So P2 = (pi -pf )2
pf
P
pi
Using kinetic energy conservation:
p i2
p2 P2
= f +
2m 2m 2M
and using
⎛ p2
p2 ⎞
P 2 = 2M ⎜ i − f ⎟
⎝ 2m 2m ⎠
2
P 2 = (p i − p f )
⎡ (p − p )2 ⎤
M = m ⎢ i2 f 2 ⎥
⎢⎣ p i − p f ⎥⎦
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2-D Elastic Collision:
Nuclear Scattering
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⎡ (pi − pf )2 ⎤
So we find that M = m ⎢ 2
2 ⎥
⎢⎣ p i − p f ⎥⎦
pf
P
pi
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If we measure pi and pf and we know m we can measure M.
&Iacute; We can learn about something we can’t see!
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This is the basic idea behind a large body of work done in
atomic, nuclear and particle physics.
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Page 2
Rutherford Backscattering
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Shoot a beam of α particles (helium nuclei) having known
energy Ei into a sample of unknown composition. Measure
the energy Ef of the α particles that bounce back out at
~180o with respect to the incoming beam.
Ei
unknown
stuff
Ei
particle detector
(measures energy)
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Rutherford Backscattering
pf
⎡ (p − p )2 ⎤
M = m ⎢ i2 f 2 ⎥
⎢⎣ p i − p f ⎥⎦
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P
pi
In the 180o case, this simplifies significantly:
⎡ (p + p )2 ⎤
(v + v f )(v i + v f )
M = m ⎢ i2 f 2 ⎥ = m i
−
(v i + v f )(v i − v f )
p f ⎥⎦
⎢⎣ p i
M =m
(v i
(v i
+ vf )
− vf )
vf = v i
2
2
1 mv f
⎛M − m⎞
Ef
⎟
⎜
= 2
=
1 mv i 2 ⎝ M + m ⎠
Ei
2
(M − m )
(M + m )
Solving for M:
⎛
⎞
m⎜1 + Ef ⎟
E
i
⎠
M= ⎝
⎛
⎞
E
f
⎜1 −
Ei ⎟⎠
⎝
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Page 3
Rutherford Backscattering
z
Shoot a beam of α particles (helium nuclei) having known
energy Ei into a sample of unknown composition. Measure
the energy Ef of the α particles that bounce back out at
~180o with respect to the incoming beam.
Ei
unknown
stuff
Ei
particle detector
(measures energy)
⎛
⎞
m⎜1 + Ef ⎟
E
i ⎠
M= ⎝
⎛
⎞
E
f
⎜1 −
Ei ⎟⎠
⎝
So we learn about the mass of the nuclei
in the unknown stuff. (We learn what the
stuff is).
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Rutherford Backscattering
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For example: Suppose we are shooting α particles that
have an initial energy of Ei = 2 MeV at a target made of an
unknown material. The α particles return with final energy
Ef = 1.1 MeV. What is the weight of the unknown material?
&Iacute; m(α) = 4
(2 protons, 2 neutrons)
So
⎛
⎞
m⎜1 + Ef ⎟
Ei ⎠
⎝
M=
⎛
Ef ⎞
⎜1 −
Ei ⎟⎠
⎝
=
(
(1 −
)
1.1 )
2
4 1 + 1.1
2
M = 27
Aluminum!! (13 protons,
14 neutrons)
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Page 4
Another example of 2-D elastic
collisions: Billiards.
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If all we know is the initial velocity of the cue ball, we don’t
have enough information to solve for the exact paths after
the collision. But we can learn some useful things...
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Billiards.
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Consider the case where one ball is initially at rest.
pf
pi
vcm
Pf
F
initial
The final direction of the
red ball will depend on
where the balls hit.
final
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Page 5
Billiards
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We know momentum is conserved: pi = pf + Pf
pi2 = (pf + Pf )2 = pf2 + Pf2 + 2 pf • Pf
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We also know that kinetic energy is conserved:
p 2i
p2 P 2
= f + f
2m 2m 2m
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p 2i = pf2 + Pf 2
Comparing these two equations tells us that:
pf • Pf = 0
Pf
pf
Therefore, pi and pf must be orthogonal!
pi
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Billiards.
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The final directions are separated by 90o.
pf
pi
vcm
Pf
F
initial
final
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Page 6
Billiards.
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So, we can sink the red ball without sinking the white ball.
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Billiards.
z
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So, we can sink the red ball without sinking the white ball.
However, we can also scratch. All we know is that the
angle between the balls is 90o.
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Page 7
Billiards.
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Tip: If you shoot a ball spotted on the “dot”, you will sink
both balls !
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Lecture 22, Act 1
Elastic Collisions in 2-D
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A moving ball initially traveling in the direction shown hits an
identical but stationary ball. The collision is elastic.
&Iacute;Describe one possible direction of both balls just after the
collision.
(a)
(b)
(c)
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Page 8
Lecture 22, Act 1
Solution
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In the first solution, the angle between the balls is not 90o.
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In the second solution, there are no downward y components to
balance out the upward y components.
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Lecture 22, Act 1
Solution
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The third choice both balances the y components and has 90o
between the final direction vectors of the two balls.
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As a result, the third choice is the only one of the three that fits all
necessary criteria.
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Page 9
Rotation
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Up until now we have gracefully avoided dealing with the
rotation of objects.
&Iacute;We have studied objects that slide, not roll.
&Iacute;We have assumed pulleys are without mass.
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Rotation is extremely important, however, and we need to
understand it!
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Most of the equations we will develop are simply rotational
analogues of ones we have already learned when studying
linear kinematics and dynamics.
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Lecture 22, Act 2
Rotations
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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.
&Iacute;Klyde’s angular velocity is:
(a) the same as Bonnie’s
(b) twice Bonnie’s
(c) half Bonnie’s
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Page 10
Lecture 22, Act 2
Rotations
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The angular velocity ω of any point on a solid object rotating
about a fixed axis is the same.
&Iacute;Both Bonnie &amp; Klyde go around once (2π radians) every
two seconds.
(Their “linear” speed v will be different since v = ωr).
ω
1
VKlyde = VBonnie
2
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Rotational Variables.
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z
an axis through its center:
First, recall what we learned about
Uniform Circular Motion:
ω=
θ
ω
dθ
dt
(Analogous to v =
dx
)
dt
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Page 11
Rotational Variables...
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z
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Now suppose ω can change as a function of time:
We define the
dω d 2θ
= 2
angular acceleration: α =
dt
dt
Consider the case when α
is constant.
&Iacute; We can integrate this to
find ω and θ as a function of time:
α = constant
ω = ω 0 + αt
θ
ω
α
1
2
θ = θ 0 + ω 0t + αt 2
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Rotational Variables...
α = constant
ω = ω 0 + αt
v
1
θ = θ 0 + ω0 t + αt 2
2
Recall also that for a point at a
distance R away from the axis of
rotation:
&Iacute;x = θR
&Iacute;v = ωR
And taking the derivative of this we find:
&Iacute;a = αR
θ
R
x
ω
z
α
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Page 12
Summary
(with comparison to 1-D kinematics)
Angular
Linear
a = constant
α = constant
ω = ω 0 + αt
v = v 0 + at
1
θ = θ0 + ω 0 t + αt 2
2
x = x0 + v 0t +
1 2
at
2
And for a point at a distance R from the rotation axis:
x = Rθ
v = ωR
a = αR
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Example: Wheel And Rope
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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?
a
R
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Page 13
Wheel And Rope...
Use a = αR to find α:
α = a / R = 4 m/s2 / 0.4 m = 10 rad/s2
z Now use the equations we derived above just as you would
use the kinematic equations from the beginning of the
semester.
1
1
θ = θ 0 + ω 0 t + α t 2 = 0 + 0(10) + (10)(10)2 = 500 rad
2
2
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1 rot
a
≈ 80 rev
α
R
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Rotation &amp; Kinetic Energy
z
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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
ω
r4
r3
r2
m2
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Page 14
Rotation &amp; Kinetic Energy...
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1
K = ∑ mi v i2
i 2
So:
K=
but vi = ωri
1
1 2
2
2
∑ m i (ωri ) = ω ∑ m i ri
2 i
2
i
v1
which we write as:
K=
m4
1
I ω2
2
v4
m3
I = ∑ mi ri 2
v2
r2
r3
i
Define the moment of inertia
r1 m1
ω
r4
m2
v3
I has units of kg m2.
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Rotation &amp; Kinetic Energy...
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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 ω2
2
ω is angular velocity
I is the moment of inertia
K=
I = ∑ mi ri 2
i
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Page 15
Moment of Inertia
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So K =
2
where I = ∑ mi ri
1
I ω2
2
i
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Notice that the moment of inertia I depends on the
distribution of mass in the system.
&Iacute;The further the mass is from the rotation axis, the bigger
the moment of inertia.
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For a given object, the moment of inertia will depend on
where we choose the rotation axis (unlike the center of
mass).
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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!
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Calculating Moment of Inertia
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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
Page 16
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Calculating Moment of Inertia...
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The squared distance from each point mass to the axis is:
2
L2
⎛L⎞
r 2 = 2⎜ ⎟ =
2
⎝2⎠
N
so I = ∑ mi ri 2 = m
i =1
Using the Pythagorean Theorem
L2
L2
L2
L2
L2
+ m + m + m = 4m
2
2
2
2
2
I = 2mL2
L/2
m
m
r
L
m
m
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Calculating Moment of Inertia...
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Now calculate I for the same object about an axis through
the center, parallel to the plane (as shown):
N
I = ∑ mi ri = m
i =1
2
L2
L2
L2
L2
L2
+ m + m + m = 4m
4
4
4
4
4
r
I = mL2
m
m
m
m
L
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Page 17
Calculating Moment of Inertia...
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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
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Calculating Moment of Inertia...
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For a single object, I clearly depends on the rotation axis!!
I = 2mL2
I = mL2
m
m
m
m
I = 2mL2
L
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Page 18
Lecture 22, Act 3
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.
&Iacute;Which of the following is correct:
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a
(a)
Ia &gt; Ib &gt; Ic
(b)
Ia &gt; Ic &gt; Ib
b
(c)
Ib &gt; Ia &gt; Ic
c
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Lecture 22, Act 3
Moment of Inertia
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Label masses and lengths:
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Calculate moments of inertia:
I a = 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 &gt; Ic &gt; Ib
L
c
m
m
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Page 19
Calculating Moment of Inertia...
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For a discrete collection of point
masses we found:
I=
N
∑ mi ri 2
i =1
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For a continuous solid object we have to add up the mr2
contribution for every infinitesimal mass element dm.
&Iacute;We have to do an
integral to find I :
dm
I = ∫ r 2 dm
r
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Moments of Inertia
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Some examples of I for solid objects:
I = MR 2
R
R
I=
Thin hoop (or cylinder) of mass M and
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.
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Page 20
Moments of Inertia...
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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=
1
MR 2
2
Solid disk or cylinder of mass M and
through its center.
R
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Lecture 22, Act 4
Moment of Inertia
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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.
&Iacute;Which one has the biggest moment of inertia about an axis
through its center?
(a) solid aluminum
(b) hollow gold
(c) same
hollow
solid
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Page 21
Lecture 22, Act 4
Moment of Inertia
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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.
&Iacute;The spherical shell (gold) will have a bigger moment of
inertia.
ISOLID &lt; ISHELL
hollow
solid
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Moments of Inertia...
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Some examples of I for solid objects (see textbook, Table 10-1):
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.
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Page 22
Parallel Axis Theorem
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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
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So if we know ICM , it is easy to calculate the moment of
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Parallel Axis Theorem: Example
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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
We know ICM =
So
1
ML2
12
IEND =
IEND
ICM
1
L 2 1
ML2 + M ⎛⎜ ⎞⎟ = ML2
⎝2⎠
12
3
which agrees with the result on a previous slide.
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