PHYS 1441 – Section 001
Lecture # 15
Wednesday, June 25, 2008
Dr. Jaehoon Yu
•
•
•
Wednesday, June 25,
2008
Moment of Inertia
Work, Power and Energy in Rotation
Rotational Kinetic Energy
PHYS 1441-001, Summer 2008
Dr. Jaehoon Yu
1
Announcements
• Quiz Tomorrow, June 26
– Beginning of the class
– Covers CH8 – 9
• Final exam
– 8 – 10am, Monday, June 30, in SH103
– Comprehensive exam: Covers CH 1 – 9 + Appendices A – E
• There will be a review session by Dr. Satyanand in class tomorrow,
after the quiz
– Please take full advantage of this review
Wednesday, June 25,
2008
PHYS 1441-001, Summer 2008
Dr. Jaehoon Yu
2
Moment of Inertia
Rotational Inertia:
For a group
of objects
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!!!
Wednesday, June 25,
2008
PHYS 1441-001, Summer 2008
Dr. Jaehoon Yu
3
Ex. 9 The Moment of Inertia Depends on
Where the Axis Is.
Two particles each have mass and are fixed at the
ends of a thin rigid rod. The length of the rod is L.
Find the moment of inertia when this object rotates
relative to an axis that is perpendicular to the rod at
(a) one end and (b) the center.
2
2
2
(a) I    mr   m1r1  m2 r2
r1  0 r2  L
m1  m2  m
I  m  0   m  L   mL2
2
(b) I 
2
2
2
2
m
r

m
r
mr

  1 1 2 2
m1  m2  m
r1  L 2 r2  L 2
2
I  m  L 2   m  L 2   12 mL
2
Wednesday, June 25,
2008
2
PHYS 1441-001, Summer 2008
Dr. Jaehoon Yu
Which case is easier to spin?
Case (b)
Why? Because the moment
of inertia is smaller
4
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.
1 2 1
K R  Iw  2 Ml 2 w 2  Ml 2w 2
2
2
Why are some 0s?

Thus, the rotational kinetic energy is

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
Wednesday, June 25,
2008

1
1
K R  Iw 2  2 Ml 2  2mb2 w 2  Ml 2  mb2 w 2
2
2
PHYS 1441-001, Summer 2008
Dr. Jaehoon Yu
5
Check out Table
9.1 for moment
of inertia for
various shaped
objects
Wednesday, June 25,
2008
PHYS 1441-001, Summer 2008
Dr. Jaehoon Yu
6
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 the radial force Fr
Ft  mat  mr
The torque due to tangential force Ft is   Ft r  mat r  mr 2  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 d Ft  d mat  d mr
dFt
2
r
d m 
d
F
r


d

The
torque
due
to
tangential
force
F
is
t
t
dm
The total torque is d   r 2d m  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.
Wednesday, June 25,
PHYS 1441-001, Summer 2008
7
2008
Dr. Jaehoon Yu
Ex. 12 Hosting a Crate
The combined moment of inertia
of the dual pulley is 50.0 kg·m2.
The crate weighs 4420 N. A
tension of 2150 N is maintained in
the cable attached to the motor.
Find the angular acceleration of
the dual pulley.
  mg  ma
T
F

y
 y 2
T2'  mg  ma y
'


T

T
 11
2
2

T1 1   mg  ma y 
since
ay 
Solve for 

2
T  mg
 1 1
I  m 22
Wednesday, June 25,
2008
2

 I
T1 1  mg  m 2 
2
2
 I
 2150 N  0.600 m    451 kg   9.80 m s 2   0.200 m 
 6.3rad
2
2
46.0 kg  m   451 kg  0.200 m 
PHYS 1441-001, Summer 2008
Dr. Jaehoon Yu
8
s2
Work, Power, and Energy in Rotation
Let’s consider the motion of a rigid body with a
single external force F exerting tangentially, moving
the object by s.
The rotational work done by the force F as the
object rotates through the distance s=rq is
W  Fs  Frq
W  Frq  q
Since the magnitude of torque is rF,
What is the unit of the rotational work? J (Joules)
The rate of work, or power, of
the constant torque  becomes
P

What is the unit of the rotational power?
Wednesday, June 25,
2008
W
q
 w

t
t
How was the power
defined in linear motion?
J/s or W (watts)
PHYS 1441-001, Summer 2008
Dr. Jaehoon Yu
9
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 Ti
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 
KER   Ki   mi ri w    mi ri  w
2 i
2 i

i
Since moment of Inertia, I, is defined as
I   mi ri 2
i
The above expression is simplified as
Wednesday, June 25,
2008
1 
KER  I w
2
PHYS 1441-001, Summer 2008
Dr. Jaehoon Yu
Unit?
10
J
Ex. 13 Rolling Cylinders
A thin-walled hollow cylinder (mass = mh,
radius = rh) and a solid cylinder (mass =
ms, radius = rs) start from rest at the top of
an incline. Determine which cylinder has
the greatest translational speed upon
reaching the bottom.
Total Mechanical Energy = KE+ KER+ PE
2
2
E  12 mv  12 I w  mgh
From Energy Conservation
1
2
mv  Iw  mghi
2
f
Solve for vf
1
2
2
f
2mgho
vf 
m  I r2
The final speeds of
the cylinders are
v hf 
s
v

Wednesday, June 25, f
2008
1
2
mv2f  12 Iw 2f  mgh f  12 mvi2  122 I wi2  mgh0
since w f 
vf
r
What does
this tell you?
2mgho

m  Ih r 2
1
2
mv 
2
f
1
2
I
vf
r
2

mgh0
The cylinder with the smaller moment of inertia
will have a greater final translational speed.
2mgho

m  mrh2 rh2
2mgho

2m
2mgho
2mgho
2mgho



1441-001,
1 2 Summer
32008
m  I s r 2 PHYS
2
m  Dr.mr
rs
m
s
2 Jaehoon Yu 2
gho
4
4
gho  v hf  1.15vhf
3
3
11
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 the linear momentum to solve physical problems
with linear motions, the 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
u
r
r ur
The angular momentum L of this
L  r p sin 
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  mr 2  I
What do you learn from this?
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 a rim.
Wednesday, June 25,
2008
PHYS 1441-001, Summer 2008
Dr. Jaehoon Yu
12
Conservation of Angular Momentum
Remember under what condition the linear momentum is conserved?
ur
ur
p
F

0

Linear momentum is conserved when the net external force is 0. 
t
ur
By the same token, the angular momentum of a system
is constant in both magnitude and direction, if the
resultant external torque acting on the system is 0.
What does this mean?
p  const
u
r
r
L


ext
0

t
ur
L  const
Angular momentum of the system before and
after a certain change is the same.
r
r
Li  L f  constant
Three important conservation laws K i  U i  K f  U f
r
r
for isolated system that does not get p

p
i
f
affected by external forces
r
r
Li  L f
Wednesday, June 25,
2008
PHYS 1441-001, Summer 2008
Dr. Jaehoon Yu
Mechanical Energy
Linear Momentum
Angular Momentum
13
Example for Angular Momentum Conservation
A star rotates with a period of 30 days about an axis through its center. After the star
undergoes a supernova explosion, the stellar core, which had a radius of 1.0x104km, collapses
into a neutron star of radius 3.0km. Determine the period of rotation of the neutron star.
What is your guess about the answer?
Let’s make some assumptions:
The period will be significantly shorter,
because its radius got smaller.
1. There is no external torque acting on it
2. The shape remains spherical
3. Its mass remains constant
Li  L f
Using angular momentum
conservation
I iw  I f w f
The angular speed of the star with the period T is
Thus
w
I iw
mri 2 2


f
If
mrf2 Ti
Tf 
2
wf
Wednesday, June 25,
2008
 r f2
 2
r
 i
2
w
T
2

3
.
0


6
Ti  

2
.
7

10
days  0.23s

30
days

4

1
.
0

10



PHYS 1441-001, Summer 2008
Dr. Jaehoon Yu
14
Ex. 14 A Spinning Skater
An ice skater is spinning with both arms and
a leg outstretched. She pulls her arms and
leg inward and her spinning motion changes
dramatically. Use the principle of
conservation of angular momentum to
explain how and why her spinning motion
changes.
The system of the ice skater does not have any net external torque
applied to her. Therefore the angular momentum is conserved for her
system. By pulling her arm inward, she reduces the moment of inertia
(Smr2) and thus in order to keep the angular momentum the same, her
angular speed has to increase.
Wednesday, June 25,
2008
PHYS 1441-001, Summer 2008
Dr. Jaehoon Yu
15
Ex. 15 A Satellite in an Elliptical Orbit
A satellite is placed in an elliptical orbit about the
earth. Its point of closest approach is 8.37x106m
from the center of the earth, and its point of greatest
distance is 25.1x106m from the center of the earth.
The speed of the satellite at the perigee is 8450
m/s. Find the speed at the apogee.
Angular momentum is
L  Iw
From angular momentum conservation
since I  mr and
2
Solve for vA
w v r
rP vP

vA 
rA
Wednesday, June 25,
2008
I AwA  I PwP
vA
2 vP
mr
 mrP
rA
rP
2
A
6
8.37

10
m  8450 m s 

25.110 m
6
PHYS 1441-001, Summer 2008
Dr. Jaehoon Yu
rA v A  rP vP
 2820 m s
16
Similarity Between Linear and Rotational Motions
All physical quantities in linear and rotational motions show striking similarity.
Quantities
Mass
Length of motion
Speed
Acceleration
Force
Work
Power
Momentum
Kinetic Energy
Wednesday, June 25,
2008
Linear
Mass
M
Distance
r
t
v
a
t
L
v
ur r
P  F v
ur
r
p  mv
Kinetic
I  mr 2
Angle q (Radian)
q
t
w

t
w
ur
r
Force F  ma
r r
Work W  F  d
K
Rotational
Moment of Inertia
1
mv 2
2
PHYS 1441-001, Summer 2008
Dr. Jaehoon Yu
r ur
Torque   I 
Work W  q
P  w
ur
ur
L  Iw
Rotational
KR 
1
Iw 2
2
17
A thought problem
•
•
1.
Moment of Inertia
– Hollow cylinder: I  mr 2 2.
h
h
–
1 2 3.
Solid Cylinder: I s  mrs
2
Wednesday, June 25,
2008
Consider two cylinders – one
hollow (mass mh and radius rh) and
the other solid (mass ms and radius
rs) – on top of an inclined surface
of height h0 as shown in the figure.
Show mathematically how their
final speeds at the bottom of the
hill compare in the following cases:
Totally frictionless surface
With some friction but no energy
loss due to the friction
With energy loss due to kinetic
friction
PHYS 1441-001, Summer 2008
Dr. Jaehoon Yu
18
Congratulations!!!!
You all have done very well!!!
I certainly had a lot of fun with ya’ll!
Good luck with your exams!!!
Have a safe summer!!
Wednesday, June 25,
2008
PHYS 1441-001, Summer 2008
Dr. Jaehoon Yu
19