PHYS 1441 – Section 002
Lecture #20
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•
•
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Monday, Apr. 27, 2009
Dr. Jaehoon Yu
Torque and Angular Acceleration
Work, Power and Energy in Rotation
Rotational Kinetic Energy
Angular Momentum & Its Conservation
Conditions for Equilibrium & Mechanical
Equilibrium
A Few Examples of Mechanical Equilibrium
Today’s homework is HW #11, due 9pm, Wednesday, May 6!!
Monday, Apr. 27, 2009
PHYS 1441-002, Spring 2009 Dr.
Jaehoon Yu
1
Announcements
• Term exam resolution
– There will be another exam for those of you who wants to take in
the class 1 – 2:20pm this Wednesday, Apr. 25
• You are welcome to take it again but
– If you take this exam despite the fact you took last Wednesday, the grade from this exam
will replace the one from last Wednesday’s
– It will cover the same chapters and will be at the same level of
difficulties
– The policy of one better of the two term exams will be used for final
grading will still be valid
• The final exam
– Date and time: 11am – 12:30pm, Monday, May 11
– Comprehensive exam
– Covers: Ch 1.1 – What we finish next Monday, May 4
Monday,
Apr. 27, 2009
1441-002,
Spring 2009 Dr.
2
– There
will be a help PHYS
session
Wednesday,
May 6, during the class
Jaehoon Yu
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.
Monday, Apr. 27, 2009
PHYS 1441-002, Spring 2009 Dr.
3
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
F

T
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
Monday, Apr. 27, 2009
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 1441-002, Spring 2009 Dr.
Jaehoon Yu
4
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?
Monday, Apr. 27, 2009
W
q
 

t
t
How was the power
defined in linear motion?
J/s or W (watts)
PHYS 1441-002, Spring 2009 Dr.
Jaehoon Yu
5
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 
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     mi ri  
2 i
2 i

i
Since moment of Inertia, I, is defined as
I   mi ri 2
i
The above expression is simplified as
Monday, Apr. 27, 2009
1 
KER  I 
2
PHYS 1441-002, Spring 2009 Dr.
Jaehoon Yu
Unit?
6
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   mgh
From Energy Conservation
1
2
mv  I  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

Monday, Apr. 27, 2009 f
1
2
mv2f  12 I 2f  mgh f  12 mvi2  122 I i2  mgh0
since  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



1441-002,
2009
1 2 Spring
2 Dr.
m  I h r 2 PHYS
2
m  Jaehoon
mrh rh Yu
m
2
3
gho
3
3
gho  v hf  1.15vhf
2
2
7
Kinetic Energy of a Rolling Sphere
Let’s consider a sphere with radius R
rolling down the hill without slipping.
R

h
q
vCM
Since vCM=R
What is the speed of the
CM in terms of known
quantities and how do you
find this out?
1
1
2
2 2
K  I CM   MR 
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, Apr. 27, 2009
PHYS 1441-002, Spring 2009 Dr.
Jaehoon Yu
2 gh
1  I CM / MR 2
8
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, Apr. 27, 2009
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 1441-002, Spring 2009 Dr.
Jaehoon Yu
aCM 
5
g sin q
7
9
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.
Monday, Apr. 27, 2009
PHYS 1441-002, Spring 2009 Dr.
Jaehoon Yu
10
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
Monday, Apr. 27, 2009
PHYS 1441-002, Spring 2009 Dr.
Jaehoon Yu
Mechanical Energy
Linear Momentum
Angular Momentum
11
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 i  I f  f
The angular speed of the star with the period T is
Thus

I i
mri 2 2


f
If
mrf2 Ti
Tf 
2
f
Monday, Apr. 27, 2009
 r f2
 2
r
 i
2

T
2

3
.
0


6
Ti  

2
.
7

10
days  0.23s

30
days

4

1
.
0

10



PHYS 1441-002, Spring 2009 Dr.
Jaehoon Yu
12
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.
Monday, Apr. 27, 2009
PHYS 1441-002, Spring 2009 Dr.
Jaehoon Yu
13
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  I
From angular momentum conservation
since I  mr and
2
Solve for vA
 v r
rP vP

vA 
rA
Monday, Apr. 27, 2009
I AA  I PP
vA
2 vP
mr
 mrP
rA
rP
2
A
6
8.37

10
m  8450 m s 

25.110 m
6
PHYS 1441-002, Spring 2009 Dr.
Jaehoon Yu
rA v A  rP vP
 2820 m s
14
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
Monday, Apr. 27, 2009
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


t

ur
r
Force F  ma
r r
Work W  F  d
K
Rotational
Moment of Inertia
1
mv 2
2
PHYS 1441-002, Spring 2009 Dr.
Jaehoon Yu
r ur
Torque   I 
Work W  q
P  
ur
ur
L  I
Rotational
KR 
1
I 2
2
15
A thought problem
•
•
1.
Moment of Inertia
– Hollow cylinder: I  mr 2 2.
h
h
–
1 2 3.
Solid Cylinder: I s  mrs
2
Monday, Apr. 27, 2009
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 1441-002, Spring 2009 Dr.
Jaehoon Yu
16