PHYS 1443 – Section 003
Lecture #18
Monday, Nov. 10, 2002
Dr. Jaehoon Yu
1.
2.
3.
4.
5.
6.
Torque and Vector Products
Rotational Kinetic Energy
Work, Power and Energy in Rotation
Angular Momentum
Angular Momentum and Torque
Conservation of Angular Momentum
Monday, Nov. 10, 2003
PHYS 1443-003, Fall 2003
Dr. Jaehoon Yu
1
Announcements
• 2nd term exam changes
– Problem #7, the correct answer is c, 0.6.
•  Those who answered b and c get credit
– Problem #21, there is no correct answer
•  All of you received credit for this problem
• Class average: 52.1 (term 1: 53)
– The two exams are identical in average
Monday, Nov. 10, 2003
PHYS 1443-003, Fall 2002
Dr. Jaehoon Yu
2
Torque and Vector Product
z
O
r
rxF
Let’s consider a disk fixed onto the origin O and
the force F exerts on the point p. What happens?
p
The disk will start rotating counter clockwise about the Z axis
y
The magnitude of torque given to the disk by the force F is
q
  Fr sin 
F
x
But torque is a vector quantity, what is the direction?
How is torque expressed mathematically?
What is the direction?
The direction of the torque follows the right-hand rule!!
The above quantity is called
Vector product or Cross product
What is the result of a vector product?
Another vector
Monday, Nov. 10, 2003
  rF
C  A B
C  A  B  A B sin q
What is another vector operation we’ve learned?
Scalar product
PHYS 1443-003, Fall 2002
Dr. Jaehoon Yu
C  A  B  A B cosq
Result? A scalar
3
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
K i  mi vi2  mi ri 2 
Kinetic energy of a masslet, mi,
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     mi ri 
2 i
2 i

i
Since moment of Inertia, I, is defined as
I   mi ri2
i
The above expression is simplified as
Monday, Nov. 10, 2003
1 
K R  I
2
PHYS 1443-003, Fall 2002
Dr. Jaehoon Yu
4
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 P 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 P  I CM  MR   I CM   MR 
2
2
2
2
1
1
Since vCM=R, the above
2
2
K  I CM   MvCM
relationship can be rewritten as
2
2
What does this equation mean?
Rotational kinetic
energy about the CM
Total kinetic energy of a rolling motion is the sum
of the rotational kinetic energy about the CM
Monday, Nov. 10, 2003
PHYS 1443-003, Fall 2002
Dr. Jaehoon Yu
Translational Kinetic
energy of the CM
And the translational
kinetic of the CM
5
Kinetic Energy of a Rolling Sphere
Let’s consider a sphere with radius R
rolling down a 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
2
 I CM 
  MvCM
2
 R  2
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. 10, 2003
PHYS 1443-003, Fall 2002
Dr. Jaehoon Yu
2 gh
1  I CM / MR 2
6
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.
What must we know first?
R

h
I CM   r 2 dm 
q
vCM 
2
vCM

The linear acceleration
of the CM is
Monday, Nov. 10, 2003
2 gh
2 
1  I CM / MR
10
gx sin q
7
aCM
2
MR 2
5
Thus using the formula in the previous slide
vCM
Since h=xsinq,
one obtains
The moment of inertia the
sphere with respect to the CM!!
2 gh

1 2 / 5
Using kinematic
relationship
2
vCM
5

 g sin q
2x 7
PHYS 1443-003, Fall 2002
Dr. Jaehoon Yu
10
gh
7
2
vCM
 2aCM x
What do you see?
Linear acceleration of a sphere does
not depend on anything but g and q.
7
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. 10, 2003
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 1443-003, Fall 2002
Dr. Jaehoon Yu
aCM 
5
g sin q
7
8
Work, Power, and Energy in Rotation
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  d s  F sin  rdq
What is Fsin?
What is the work done by
radial component Fcos?
Since the magnitude of torque is rFsin,
The rate of work, or power becomes
The tangential component of force F.
Zero, because it is perpendicular to the
displacement.
dW  rF sin  dq  dq
P 
The rotational work done by an external force
equals the change in rotational energy.
dW dq
 

dt
dt
How was the power
defined in linear motion?
 d 


I

I




 d  dq   I  d 
 I




d
q


 dq  dt 
 dt 
dW   dq  Id
The work put in by the external force then
1
1
q

W  q  dq   Id  I 2f  I i2
2 9 2
Monday, Nov. 10, 2003
PHYS 1443-003, Fall 2002
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 instantaneous angular momentum
L of this particle relative to origin O is
L  r p
2
2
2 2
What is the unit and dimension of angular momentum? kg m / s [ ML T ]
Note that L depends on origin O. Why?
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. 10, 2003
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 a10rim.
PHYS 1443-003, Fall 2002
Dr. Jaehoon Yu
Angular Momentum and Torque
Can you remember how net force exerting on a particle
and the change of its linear momentum are related?
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.


Net torque acting on a particle is
F  r d p
  r
dt
z
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
dL
  dt
The net torque acting on a particle is the same as the time rate change of its angular momentum
Monday, Nov. 10, 2003
PHYS 1443-003, Fall 2002
Dr. Jaehoon Yu
11
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  L 2  ......  L n   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 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, Nov. 10, 2003
PHYS 1443-003, Fall 2002
Dr. Jaehoon Yu

ext
dL

dt
12
Example for Angular Momentum
A particle of mass m is moving in 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   mrv sin 90  mrv
So the angular momentum vector can be expressed as
L  mrvk
Find the angular momentum in terms of angular velocity .
Using the relationship between linear and angular speed
2
2
L  mrvk  mr  k  mr   I 
Monday, Nov. 10, 2003
PHYS 1443-003, Fall 2002
Dr. Jaehoon Yu
13
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 zaxis 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
Thus 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. 10, 2003
PHYS 1443-003, Fall 2002
Dr. Jaehoon Yu
14
Example for Rigid Body Angular Momentum
A rigid rod of mass M and length l pivoted without friction at its center. Two particles of mass
m1 and m2 are connected to its ends. The combination rotates in 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
1

  M  m1  m2  L  I 
 M  m1  m2 
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 net
torque is 0.
If q/p/, at equilibrium
so no angular momentum.
Monday, Nov. 10, 2003
First compute net
external torque
   m1 g
l
cosq
2
 ext     2

 2  m2 g
gl cos q m1  m2 
2
1
m1  m1 gl cos q
2



Thus 
ext


2
l
1


I
 M  m1  m2 
becomesPHYS 1443-003, Fall
2002 4  3

Dr. Jaehoon Yu
l
cosq
2
2m1  m1  cos q
g /l
1

 M  m1  m2 
3
15 
Conservation of Angular Momentum
Remember under what condition the linear momentum is conserved?
Linear momentum is conserved when the net external force is 0.  F  0 
dp
dt
p  const
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?
dL


ext
0

dt
L  const
Angular momentum of the system before and
after a certain change is the same.
Li  L f  constant
Three important conservation laws
for isolated system that does not get
affected by external forces
Monday, Nov. 10, 2003
 K i  U i  K f  U f Mechanical Energy


Linear Momentum
 pi  p f


Angular Momentum
Li  L f
PHYS 1443-003, Fall 2002
Dr. Jaehoon Yu
16
Example for Angular Momentum Conservation
A star rotates with a period of 30days 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 start 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:
Using angular momentum
conservation
The period will be significantly shorter,
because its radius got smaller.
1. There is no torque acting on it
2. The shape remains spherical
3. Its mass remains constant
Li  L f
I i  I f  f
The angular speed of the star with the period T is
Thus

Tf 
I i
mri 2 2p


f
If
mrf2 Ti
2p
f
Monday, Nov. 10, 2003
 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 1443-003, Fall 2002
Dr. Jaehoon Yu
17
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, Nov. 10, 2003
Linear
Mass
M
Distance
L
Rotational
Moment of Inertia
I   r 2 dm
Angle q (Radian)
v
dr
dt

dq
dt
a
dv
dt
 
d
dt
Force F  ma
Work W   Fdx
xf
xi
Kinetic
Torque
Work
  I
W 
P  F v
P  
p  mv
L  I
K 
1
mv 2
2
PHYS 1443-003, Fall 2002
Dr. Jaehoon Yu
Rotational
qf
q
d q
i
KR 
1
I 2
2
18