PHYS 1443 – Section 001
Lecture #14
Monday, June 26, 2006
Dr. Jaehoon Yu
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Monday, June 26, 2006
Moment of Inertia
Parallel Axis Theorem
Torque and Angular Acceleration
Rotational Kinetic Energy
Work, Power and Energy in Rotation
Angular Momentum & Its Conservation
Similarity of Linear and Angular Quantities
PHYS 1443-001, Summer 2006
Dr. Jaehoon Yu
1
Announcements
• Quiz Results
– Average: 60.2
– Top score: 100
• Last quiz this Wednesday
– Early in the class
– Covers Ch. 10 – what we cover tomorrow
• Final exam
–
–
–
–
Date and time: 8 – 10am, Friday, June 30
Location: SH103
Covers: Ch 9 – what we cover by Wednesday
No class this Thursday
Monday, June 26, 2006
PHYS 1443-001, Summer 2006
Dr. Jaehoon Yu
2
Moment of Inertia
Rotational Inertia:
For a group
of particles
Measure of resistance of an object to
changes in its rotational motion.
Equivalent to mass in linear motion.
I   mi ri
2
i
What are the dimension and
unit of Moment of Inertia?
For a rigid
body
I   r 2 dm
ML 
2
kg m
2
Determining Moment of Inertia is extremely important for
computing equilibrium of a rigid body, such as a building.
Dependent on the axis of rotation!!!
Monday, June 26, 2006
PHYS 1443-001, Summer 2006
Dr. Jaehoon Yu
3
Example for Moment of Inertia
In a system of four small spheres as shown in the figure, assuming the radii are negligible
and the rods connecting the particles are massless, compute the moment of inertia and
the rotational kinetic energy when the system rotates about the y-axis at angular speed w.
y
m
Since the rotation is about y axis, the moment of
inertia about y axis, Iy, is
b
l
M
O
l
M
x
b
m
I   mi ri2  Ml2 Ml 2 m  02  m  02  2Ml 2
i
This is because the rotation is done about y axis,
and the radii of the spheres are negligible.
Why are some 0s?
KR 
Thus, the rotational kinetic energy is
1 2 1
2
2
2 2
Iw  2 Ml w  Ml w
2
2
Find the moment of inertia and rotational kinetic energy when the system rotates on
the x-y plane about the z-axis that goes through the origin O.

2
2
2
I   mi ri 2  Ml 
Ml 2 mb2 mb 2  2 Ml  mb
i
Monday, June 26, 2006

1
1
K R  Iw 2  2 Ml 2  2mb2 w 2  Ml 2  mb2 w 2
2
2
PHYS 1443-001, Summer 2006
Dr. Jaehoon Yu
4
Calculation of Moments of Inertia
Moments of inertia for large objects can be computed, if we assume
the object consists of small volume elements with mass, Dmi.
2
2
I

lim
r
D
m

r

i
i
The moment of inertia for the large rigid object is
 dm
Dm 0
i
It is sometimes easier to compute moments of inertia in terms
of volume of the elements rather than their mass
dm
Using the volume density, r, replace
r
dm in the above equation with dV.
dV
i
How can we do this?
The moments of
dm  rdV inertia becomes
I   rr 2 dV
Example: Find the moment of inertia of a uniform hoop of mass M and radius R
about an axis perpendicular to the plane of the hoop and passing through its center.
y
O
The moment
of inertia is
dm
R
x
Monday, June 26, 2006
What do you notice
from this result?
I   r 2 dm  R 2  dm  MR 2
The moment of inertia for this
object is the same as that of a
point of mass M at the distance R.
PHYS 1443-001, Summer 2006
Dr. Jaehoon Yu
5
Example for Rigid Body Moment of Inertia
Calculate the moment of inertia of a uniform rigid rod of length L and mass M about an
axis perpendicular to the rod and passing through its center of mass.
M


The line density of the rod is
y
L
so the masslet is
dm  dx  M dx
L
dx
x
L
x
The moment
of inertia is
M  L   L 

     
3L  2   2 
3
What is the moment of inertia
when the rotational axis is at
one end of the rod.
Will this be the same as the
above? Why or why not?
Monday, June 26, 2006
L/2
2
M 1 3 
x
M
x 
I   r dm  
dx 

L / 2
L  3  L / 2
L
L/2
2
3
 M  L3  ML2
  

3
L
12

 4
L
M 1
I  r 2 dm   x M dx   x3 
0
L 3 0
L
M
M 3
ML2
3
L   0  L  

3L
3L
3

2
L


Nope! Since the moment of inertia is resistance to motion, it makes perfect sense
for it to be harder to move when it is rotating about the axis at one end.
PHYS 1443-001, Summer 2006
Dr. Jaehoon Yu
6
Parallel Axis Theorem
Moments of inertia for highly symmetric object is easy to compute if the
rotational axis is the same as the axis of symmetry. However if the axis of
rotation does not coincide with axis of symmetry, the calculation can still be
2
done in simple manner using parallel-axis theorem. I  I CM  MD
y


2
2
2

x

y
dm (1)
I

r
dm
Moment of inertia is defined


(x,y)
x  xCM  x' y  yCM  y'
Since x and y are
r
yCM
y
y’
One can substitute x and y in Eq. 1 to obtain
CM
(xCM,yCM)
D
xCM
x’
x
What does this
Monday, June 26, 2006
theorem tell you?


I   xCM  x'   yCM  y ' dm

2

2


2
2
2
2
 xCM
 yCM
dm

2
x
x
'
dm

2
y
y
'
dm

x
'

y
'
dm
CM
CM




Since the x’ and y’ are the
x ' dm  0  y ' dm  0
x distances from CM, by definition 
Therefore, the parallel-axis theorem
2
2
 dm   x'2  y'2 dm  MD 2  I CM
I  xCM
 yCM
Moment of inertia of any object about any arbitrary axis are the same as
the sum of moment
of inertia
for a2006
rotation about the CM and that7of
PHYS 1443-001,
Summer
Jaehoon axis.
Yu
the CM about theDr.
rotation
Example for Parallel Axis Theorem
Calculate the moment of inertia of a uniform rigid rod of length L and mass M about an
axis that goes through one end of the rod, using parallel-axis theorem.
The line density of the rod is  
y
so the masslet is
CM
dm  dx 
dx
x
L
x The moment of
inertia about
the CM
M
L
M
dx
L
L/2
2
M 1 3 
x
M
I CM   r dm  
x 
dx 

L / 2
L
3
L

 L / 2
3
3
M  L   L   M  L3  ML2
  

       
3L  2   2   3L  4  12
2
L/2
2
Using the parallel axis theorem
I  I CM
ML2  L 
ML2 ML2 ML2
2
D M 
  M 


12  2 
12
4
3
The result is the same as using the definition of moment of inertia.
Parallel-axis theorem is useful to compute moment of inertia of a rotation of a rigid
object with complicated shape about an arbitrary axis
Monday, June 26, 2006
PHYS 1443-001, Summer 2006
Dr. Jaehoon Yu
8
Torque & Angular Acceleration
Ft
r F
r
Let’s consider a point object with mass m rotating on a circle.
What forces do you see in this motion?
m
The tangential force Ft and radial force Fr
Ft  mat  mr
2
The torque due to tangential force Ft is   Ft r  mat r  mr   I
The tangential force Ft is
What do you see from the above relationship?
What does this mean?
  I
Torque acting on a particle is proportional to the angular acceleration.
What law do you see from this relationship?
Analogs to Newton’s 2nd law of motion in rotation.
How about a rigid object?
The external tangential force dFt is dFt  dmat  dmr
dFt
d dFt r  r 2 dm
The torque due to tangential force Ft is
dm
The total torque is     r 2 dm I
r
Contribution from radial force is 0, because its
What is the contribution due
line of action passes through the pivoting
O
to
radial
force
and
why?
point, making the moment arm 0.
Monday, June 26, 2006
PHYS 1443-001, Summer 2006
9
Dr. Jaehoon Yu
Example for Torque and Angular Acceleration
A uniform rod of length L and mass M is attached at one end to a frictionless pivot and is
free to rotate about the pivot in the vertical plane. The rod is released from rest in the
horizontal position. What are the initial angular acceleration of the rod and the initial linear
acceleration of its right end?
The only force generating torque is the gravitational force Mg
L/2
  Fd  F
Mg
L
L
 Mg  I
2
2
L
Since the moment of inertia of the rod I  r 2 dm 
0
0
when it rotates about one end
L
L
We obtain
  MgL 
2I
MgL 3 g

2ML2 2 L
3
Monday, June 26, 2006
3


M
x
ML2


2
x dx      
3
 L  3  0
Using the relationship between tangential and
angular acceleration
3g
at  L 
2
What does this mean?
The tip of the rod falls faster than
an object undergoing a free fall.
PHYS 1443-001, Summer 2006
Dr. Jaehoon Yu
10
Rotational Kinetic Energy
y
vi
mi
ri
q
O
x
What do you think the kinetic energy of a rigid object
that is undergoing a circular motion is?
1
1
2
2 

m
v
Kinetic energy of a masslet, mi,
Ki

m
r
i i
i i w
2
2
moving at a tangential speed, vi, is
Since a rigid body is a collection of masslets, the total kinetic energy of the
rigid object is
1
1
2 
2 
K R   Ki   mi ri w    mi ri w
2 i
2 i

i
Since moment of Inertia, I, is defined as
I   mi ri2
i
The above expression is simplified as
Monday, June 26, 2006
1 
K R  Iw
2
PHYS 1443-001, Summer 2006
Dr. Jaehoon Yu
11
Total Kinetic Energy of a Rolling Body
What do you think the total kinetic
energy of the rolling cylinder is?
P’
CM
Since it is a rotational motion about the point
P, we can write the total kinetic energy
1
K  I Pw 2
2
2vCM
vCM
Where, IP, is the moment of
inertia about the point P.
Using the parallel axis theorem, we can rewrite

P

1
1
1
1
2
2
2
2
2 2
K  I Pw  I CM  MR w  I CM w  MR w
2
2
2
2
1
1
Since vCM=Rw, the above
2
2
K  I CM w  MvCM
relationship can be rewritten as
2
2
What does this equation mean?
Rotational kinetic
energy about the CM
Translational Kinetic
energy of the CM
Total kinetic energy of a rolling motion is the sum
of the rotational kinetic energy about the CM
Monday, June 26, 2006
And the translational
PHYS 1443-001, Summer 2006
kinetic of the CM
Dr. Jaehoon Yu
12
Kinetic Energy of a Rolling Sphere
R
w
h
q
vCM
Since vCM=Rw
What is the speed of the
CM in terms of known
quantities and how do you
find this out?
Let’s consider a sphere with radius R
rolling down a hill without slipping.
1
1
2
2 2
K  I CM w  MR w
2
2
2
1
 vCM   1 Mv 2
 I CM 
CM

2
2
 R 
1I
 2
  CM2  M vCM
2 R

Since the kinetic energy at the bottom of the hill must
be equal to the potential energy at the top of the hill
1  I CM
 2


M
 2
vCM  Mgh
K
2 R

vCM 
Monday, June 26, 2006
PHYS 1443-001, Summer 2006
Dr. Jaehoon Yu
2 gh
1  I CM / MR 2
13
Example for Rolling Kinetic Energy
For solid sphere as shown in the figure, calculate the linear speed of the CM at the
bottom of the hill and the magnitude of linear acceleration of the CM. Solve this problem
using Newton’s second law, the dynamic method.
What are the forces involved in this motion?
Gravitational Force, Frictional Force, Normal Force
Newton’s second law applied to the CM gives
n
f
M
h
Mg
F
F
q
x
 Mg sin q  f  MaCM
y
 n  Mg cosq  0
Since the forces Mg and n go through the CM, their moment arm is 0

and do not contribute to torque, while the static friction f causes torque CM
We know that
I CM 
2
MR 2
5
aCM  R
Monday, June 26, 2006
We
obtain
 fR  I CM 
2
MR 2
I CM 
2
f 
 aCM   MaCM
5
R  R  R  5


Substituting f in
dynamic equations
7
Mg sin q  MaCM
5
PHYS 1443-001, Summer 2006
Dr. Jaehoon Yu
aCM 
5
g sin q
7
14
Work, Power, and Energy in Rotation
F
f
dq r
O
ds
Let’s consider a motion of a rigid body with a single external
force F exerting on the point P, moving the object by ds.
The work done by the force F as the object rotates
through the infinitesimal distance ds=rdq is
dW  F  ds   F cos(  f )  rdq  F sin f rdq
What is Fsinf?
What is the work done by
radial component Fcosf?
Since the magnitude of torque is rFsinf,
The rate of work, or power becomes
The tangential component of force F.
Zero, because it is perpendicular to the
displacement.
dW  rF sin f dq  dq
P 
The rotational work done by an external force
equals the change in rotational energy.
dW dq
 w

dt
dt
How was the power
defined in linear motion?
 dw 


I

I




 dw  dq   Iw  dw 
 I




d
q


 dq  dt 
 dt 
dW   dq  Iwdw
The work put in by the external force then
1
1
q
w
W  q  dq  w Iwdw  Iw 2f  Iw i2
2 15 2
Monday, June 26, 2006
PHYS 1443-001, Summer 2006
f

Dr. Jaehoon Yu
f

Angular Momentum of a Particle
If you grab onto a pole while running, your body will rotate about the pole, gaining
angular momentum. We’ve used linear momentum to solve physical problems
with linear motions, angular momentum will do the same for rotational motions.
Let’s consider a point-like object ( particle) with mass m located
at the vector location r and moving with linear velocity v
The angular momentum L of this
Lrp
particle relative to the origin O is
What is the unit and dimension of angular momentum?
Note that L depends on origin O. Why?
kg  m2 / s [ ML2T 1 ]
Because r changes
What else do you learn? The direction of L is +z
Since p is mv, the magnitude of L becomes L  mvr sin f
What do you learn from this?
Monday, June 26, 2006
If the direction of linear velocity points to the origin of
rotation, the particle does not have any angular momentum.
If the linear velocity is perpendicular to position vector, the
particle
moves exactly the same way as a point on a16rim.
PHYS 1443-001, Summer 2006
Dr. Jaehoon Yu
Angular Momentum and Torque
dp
 F  dt
Can you remember how net force exerting on a particle
and the change of its linear momentum are related?
Total external forces exerting on a particle is the same as the change of its linear momentum.
The same analogy works in rotational motion between torque and angular momentum.


  r  F  r  dt
Net torque acting on a particle is
z
dr
dp
dp
d r  p
dL

 pr
 0r 

dt
dt
dt
L=rxp
dt
dt
O
x
r
y
m
f
Why does this work?
p
Thus the torque-angular
momentum relationship
dp
Because v is parallel to
the linear momentum
 
dL
dt
The net torque acting on a particle is the same as the time rate change of its angular momentum
Monday, June 26, 2006
PHYS 1443-001, Summer 2006
Dr. Jaehoon Yu
17
Angular Momentum of a System of Particles
The total angular momentum of a system of particles about some point
is the vector sum of the angular momenta of the individual particles
L  L1  L2  ......  Ln 
L
i
Since the individual angular momentum can change, the total
angular momentum of the system can change.
Both internal and external forces can provide torque to individual particles. However,
the internal forces do not generate net torque due to Newton’s third law.
Let’s consider a two particle
system where the two exert
forces on each other.
Since these forces are action and reaction forces with
directions lie on the line connecting the two particles, the
vector sum of the torque from these two becomes 0.
Thus the time rate change of the angular momentum of a
system of particles is equal to the net external torque
acting on the system
Monday, June 26, 2006
PHYS 1443-001, Summer 2006
Dr. Jaehoon Yu

ext
dL

dt
18
Example for Angular Momentum
A particle of mass m is moving on the xy plane in a circular path of radius r and linear
velocity v about the origin O. Find the magnitude and direction of angular momentum with
respect to O.
Using the definition of angular momentum
y
v
L  r  p  r  mv  mr  v
r
O
x
Since both the vectors, r and v, are on x-y plane and
using right-hand rule, the direction of the angular
momentum vector is +z (coming out of the screen)

The magnitude of the angular momentum is L  mr  v  mrv sin f  mrv sin 90  mrv
So the angular momentum vector can be expressed as
L  mrvk
Find the angular momentum in terms of angular velocity w.
Using the relationship between linear and angular speed
2
L  mrvk  mr 2wk  mr w  Iw
Monday, June 26, 2006
PHYS 1443-001, Summer 2006
Dr. Jaehoon Yu
19