PHYS 1443 – Section 002
Lecture #19
Monday, Nov. 24, 2008
Dr. Jae Yu
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Rolling Motion of a Rigid Body
Angular Momentum
Conservation of Angular Momentum
Relationship between angular and linear quantities
Similarity between Linear and Angular Quantities
Today’s homework is HW #11, due 9pm, Monday, Dec. 1!!
Monday, Nov. 24, 2008
PHYS 1443-002, Fall 2008
Dr. Jaehoon Yu
1
Announcements
• 3rd term exam grading not completed
– Will be done by Monday, Dec. 1
• Thanksgiving is this Thursday, Nov. 27
– Vote for the class this Wednesday, Nov. 26
Monday, Nov. 24, 2008
PHYS 1443-002, Fall 2008 Dr. Jaehoon
Yu
2
Rolling Motion of a Rigid Body
What is a rolling motion?
A more generalized case of a motion where the
rotational axis moves together with an object
A rotational motion about a moving axis
To simplify the discussion, let’s
make a few assumptions
1.
2.
Limit our discussion on very symmetric
objects, such as cylinders, spheres, etc
The object rolls on a flat surface
Let’s consider a cylinder rolling on a flat surface, without slipping.
Under what condition does this “Pure Rolling” happen?
The total linear distance the CM of the cylinder moved is
R q s
s=Rq
Monday, Nov. 24, 2008
Thus the linear
speed of the CM is
vCM
s  Rq
ds
dq
 R
 R
dt
dt
The condition for a “Pure Rolling motion”
PHYS 1443-002, Fall 2008 Dr. Jaehoon
Yu
3
More Rolling Motion of a Rigid Body
The magnitude of the linear acceleration of the CM is
P’
CM
aCM
dvCM
d

R
 R
dt
dt
As we learned in rotational motion, all points in a rigid body
2vCM moves at the same angular speed but at different linear speeds.
vCM
CM is moving at the same speed at all times.
At any given time, the point that comes to P has 0 linear
speed while the point at P’ has twice the speed of CM
P
Why??
A rolling motion can be interpreted as the sum of Translation and Rotation
P’
CM
P
vCM
P’
vCM
CM
v=0
vCM
Monday, Nov. 24, 2008
+
v=R
v=R
2vCM
P’
=
P
PHYS 1443-002, Fall 2008 Dr. Jaehoon
Yu
CM
vCM
P
4
Kinetic Energy of a Rolling Sphere
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?
Let’s consider a sphere with radius R
rolling down the hill without slipping.
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, Nov. 24, 2008
PHYS 1443-002, Fall 2008 Dr. Jaehoon
Yu
2 gh
1  I CM / MR 2
5
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, Nov. 24, 2008
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 1443-002, Fall 2008 Dr. Jaehoon
Yu
aCM 
5
g sin q
7
6
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.
z
y
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
particle relative to the origin O is
What is the unit and dimension of angular momentum?
x
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 
What do you learn from this?
Monday, Nov. 24, 2008
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 7rim.
PHYS 1443-002, Fall 2008 Dr. Jaehoon
Yu
Angular Momentum and Torqueur
Can you remember how net force exerting on a particle
and the change of its linear momentum are related?
ur
dp
 F  dt
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
r
ur
  r   F
ur
r dp
 r
dt
Net torque acting on the particle is
z
r ur
ur
r
ur
ur
u
r
r
r
d
r

p
dr
dp
dp
dL

 pr
 0r

L=rxp
dt
dt
dt
dt
dt

O
x
r
y
m


Why does this work?
p
Thus the torque-angular
momentum relationship

Because v is parallel to
the linear momentum
u
r
r
dL
  dt
The net torque acting on a particle is the same as the time rate change of its angular momentum
Monday, Nov. 24, 2008
PHYS 1443-002, Fall 2008 Dr. Jaehoon
Yu
8
r

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
u
r ur u
r
u
r
L  L1  L 2 ......  L n 
u
r
 Li
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 the 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 only the net external torque
acting on the system
Monday, Nov. 24, 2008
PHYS 1443-002, Fall 2008 Dr. Jaehoon
Yu
r

ext
u
r
dL

dt
9
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 the direction of the angular
momentum with respect to O.
Using the definition of angular momentum
y
v
r ur
r
r
r r
ur
r
L  r  p  r  mv  mr  v
Since both the vectors, r and v, are on x-y plane and
O
using right-hand rule, the direction of the angular
x
momentum vector is +z (coming out of the screen)
ur
r r

The magnitude of the angular momentum is L  mr  v  mrv sin   mrv sin 90  mrv
So the angular momentum vector can be expressed as
ur
r
L  mrvk
Find the angular momentum in terms of angular velocity .
Using the relationship between linear and angular speed
ur
r
r
ur
ur
2
2
L  mrvk  mr  k  mr   I 
Monday, Nov. 24, 2008
PHYS 1443-002, Fall 2008 Dr. Jaehoon
Yu
10
Angular Momentum of a Rotating Rigid Body
z
Let’s consider a rigid body rotating about a fixed axis
Each particle of the object rotates in the xy plane about the z-axis
at the same angular speed, 
L=rxp
O
y
m
r

x
Magnitude of the angular momentum of a particle of mass mi
about origin O is miviri
2
Li  mi ri vi  mi ri 
p
Summing over all particle’s angular momentum about z axis
Lz   Li   mi ri 2 
i
i
Since I is constant for a rigid body
Thus the torque-angular momentum
relationship becomes
What do
you see?
Lz 
ext
2
i i
i
dL z
d
 I
dt
dt

 m r   I
 I
 is angular
acceleration
dLz

 I
dt
The net external torque acting on a rigid body rotating about a fixed axis is equal to the moment of
inertia about that axis multiplied by the object’s angular acceleration with respect to that axis.
Monday, Nov. 24, 2008
PHYS 1443-002, Fall 2008 Dr. Jaehoon
Yu
11
Example for Rigid Body Angular Momentum
A rigid rod of mass M and length l is pivoted without friction at its center. Two particles of mass
m1 and m2 are attached to either end of the rod. The combination rotates on a vertical plane with
an angular speed of . Find an expression for the magnitude of the angular momentum.
y
m2
l
q
m2 g
O
m1
m1 g
x
The moment of inertia of this system is
1
1
1
2
2

I

I

I

Ml

m
l

m2l 2
I
rod
m1
m2
1
12
4
4
2
l2  1
 L  I  l  1 M  m  m 
  M  m1  m2 

1
2
4
3


4 3

Find an expression for the magnitude of the angular acceleration of the
system when the rod makes an angle q with the horizon.
If m1 = m2, no angular
momentum because the net
torque is 0.
If q/p/2, at equilibrium
so no angular momentum.
Monday, Nov. 24, 2008
l
First compute the    m1 g cosq
2
net external torque
 ext     2

 2  m2 g
l
cosq
2
gl cos q m1  m2 
2
1
 m1  m2  gl cosq
2  m1  m2  cos q g
2



Thus 
ext
 2

1
 l
l 1

I
M

m

m
 M  m1  m2 

1
2 
becomes
3
4 3
PHYS 1443-002, Fall 2008 Dr. Jaehoon
12 

Yu
Conservation of Angular Momentum
Remember under what condition the linear momentum is conserved?
ur
ur
dp
F

0

Linear momentum is conserved when the net external force is 0. 
dt
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
dL


ext
0

dt
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, Nov. 24, 2008
PHYS 1443-002, Fall 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 i  I f  f
The angular speed of the star with the period T is
Thus

I i
mri 2 2p


f
If
mrf2 Ti
Tf 
2p
f
Monday, Nov. 24, 2008
 r f2
 2
r
 i
2p

T
2

3
.
0


6
Ti  

2
.
7

10
days  0.23s

30
days

4

1
.
0

10



PHYS 1443-002, Fall 2008 Dr. Jaehoon
Yu
14
Kepler’s Second Law and Angular Momentum Conservation
Consider a planet of mass Mp moving around the Sun in an elliptical orbit.
D
S
C
r
Since the gravitational force acting on the planet is
A always toward radial direction, it is a central force
dr
B Therefore the torque acting on the planet by this
force is always 0.
r
r ur r
  r  F  r  Frˆ  0
ur
ur
dL
  dt  0 L  const
r
r r
r ur r
 r  p  r  M p v  M p r  v  const
Since torque is the time rate change of angular
momentum L, the angular momentum is constant.
Because the gravitational force exerted on a
planet by the Sun results in no torque, the
angular momentum L of the planet is constant.
ur
L
r
r
Since the area swept by the
1 r
1 r r  L dt
dA  r  d r  r  vdt
motion of the planet is
2M p
2
2
dA  L
 const
2M p
dt
This is Keper’s second law which states that the radius vector from
the Sun to a planet sweeps out equal areas in equal time intervals.
Monday, Nov. 24, 2008
PHYS 1443-002, Fall 2008 Dr. Jaehoon
Yu
15
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
Linear
Mass
M
Distance
r
t
v
a
t
Monday, Nov. 24, 2008
I  mr 2
Angle q (Radian)
L
q
t


t
v

ur
r
Force F  ma
r r
Work W  F  d
ur r
P  F v
ur
r
p  mv
Momentum
Kinetic Energy
Rotational
Moment of Inertia
Kinetic
K
1
mv 2
2
r ur
Torque   I 
Work W  q
P  
ur
ur
L  I
Rotational
PHYS 1443-002, Fall 2008 Dr. Jaehoon
Yu
KR 
1
I 2
2
16