Transcript Wednesday, Mar. 27, 2002

1443-501 Spring 2002
Lecture #16
Dr. Jaehoon Yu
1.
2.
3.
4.
Torque and Vector Product
Angular Momentum of a Particle
Angular Momentum of a Rotating Rigid Object
Conservation of Angular Momentum
Today’s Homework Assignment is the Homework #7!!!
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?
Mar. 27, 2002
Let’s consider a sphere with radius R
rolling down a hill without slipping.
1
1
2
K  I CM w  MR 2w 2
2
2
2
1
1
v

2
K  I CM  CM   MvCM
2
2
 R 
1  I CM
 2
  2  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
K   2  M vCM
 Mgh
2 R

2 gh
2
1443-501 Spring 20021  I CM / MR
vCM 
Dr. J. Yu, Lecture #15
2
Example 11.1
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
w
h
I CM   r 2 dm 
q
vCM
vCM 
Since h=xsinq,
one obtains
2
vCM

The linear acceleration
of the CM is
Mar. 27, 2002
2
MR 2
5
Thus using the formula in the previous slide
2 gh

2
1  I CM / MR
10
gx sin q
7
aCM
The moment of inertia the
sphere with respect to the CM!!
Using kinematic
relationship
2
vCM
5

 g sin q
2x
7
1443-501 Spring 2002
Dr. J. Yu, Lecture #15
2 gh

1 2 / 5
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.
3
Example 11.2
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 cos q  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
We know that
2
MR 2
5
 R
I CM 
aCM
Mar. 27, 2002
We
obtain
2
MR 2
I 
 aCM  2
f  CM  5

  MaCM
R
R
R

 5
Substituting f in
dynamic equations
Mg sin q 
1443-501 Spring 2002
Dr. J. Yu, Lecture #15
 CM  fR
 I CM 
7
5
MaCM ; aCM  g sin q
5
7
4
Torque and Vector Product
z
O
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
r
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
Mar. 27, 2002
  rF
C  A B
C  A  B  A B sin q
What is another vector operation we’ve learned?
Scalar product
1443-501 Spring 2002
Dr. J. Yu, Lecture #15
C  A  B  A B cosq
Result? A scalar
5
Properties of Vector Product
Vector Product is Non-commutative
What does this mean?
If the order of operation changes the result changes
Following the right-hand rule, the direction changes
A B  B  A
A  B  B  A
Vector Product of a two parallel vectors is 0.
C  A  B  A B sin q  A B sin 0  0
Thus,
A A  0
If two vectors are perpendicular to each other
A  B  A B sin q  A B sin 90  A B  AB
Vector product follows distribution law


A B  C  A B  A C
The derivative of a Vector product with respect to a scalar variable is


d A B
dA
dB

 B  A
dt
dt
dt
Mar. 27, 2002
1443-501 Spring 2002
Dr. J. Yu, Lecture #15
6
More Properties of Vector Product
The relationship between unit
vectors, i, j, and k
ii  j  j  k  k  0
i j   j i  k
j  k  k  j  i
k  i  i  k  j
Vector product of two vectors can be expressed in the following determinant form
i
j
A  B  Ax
Ay
Bx
By
k
Ay
Az  i
By
Bz
Az
Ax
 j
Bz
Bx
Ax
k
Bx
Bz
Ay
By
Az
 Ay Bz  Az B y i   Ax Bz  Az Bx  j  Ax B y  Ay Bx k
Mar. 27, 2002
1443-501 Spring 2002
Dr. J. Yu, Lecture #15
7
Example 11.3
Two vectors lying in the xy plane are given by the equations A=2i+3j and
B=-i+2j, verify that AxB= -BxA
(2,3)
(-1,2)
B
A



A  B  2i  3 j   i  2 j  4i  j  3 j  i
Since i  j  k
 
B  A   i  2 j  2i  3 j   3i  j  4 j  i
A  B  4k  3 j   i  4k  3k  7k
Using the same unit vector relationship as above
B  A  3k  4k  7k
Therefore, AxB= -BxA
Now prove this using determinant method
Mar. 27, 2002
1443-501 Spring 2002
Dr. J. Yu, Lecture #15
8
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.
z
Let’s consider a point-like object ( particle) with mass m located
at the vector location r and moving with linear velocity v
L=rxp
The instantaneous angular momentum
L  r p
L of this particle relative to origin O is
O
y
m
r
What is the unit and dimension of angular momentum? kg m2 / s 2
 p
Note that L depends on origin O. Why? Because r changes
x
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?
The point O has
to be inertial.
Mar. 27, 2002
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
the same way as a point on a 9rim.
1443-501
Springexactly
2002
Dr. J. Yu, Lecture #15
Angular Momentum and Torque
Can you remember how net force exerting on a particle
and the change of its linear momentum are related?
F 
dp
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   r   F  r  d p
dt
z
dL
d r p
dr
dp
dp


 pr
 0r
L=rxp
dt
dt
dt
dt
dt

O
r
x
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
Mar. 27, 2002
1443-501 Spring 2002
Dr. J. Yu, Lecture #15
10
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
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
Mar. 27, 2002
1443-501 Spring 2002
Dr. J. Yu, Lecture #15

ext
dL

dt
11
Example 11.4
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 w.
Using the relationship between linear and angular speed
L  mrvk  mr 2w k  mr 2 w  I w
Mar. 27, 2002
1443-501 Spring 2002
Dr. J. Yu, Lecture #15
12
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, w
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 w
p
Summing over all particle’s angular momentum about z axis
Lz   Li   mi ri 2w  What do
i
Lz 
you see?
i
 m r w  Iw
2
i i
i
dLz
dw
I
 I
dt
dt
Since I is constant for a rigid body
Thus the torque-angular momentum
relationship becomes

ext
 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.
Mar. 27, 2002
1443-501 Spring 2002
Dr. J. Yu, Lecture #15
13
Example 11.6
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 w. Find an expression for the magnitude of the angular momentum.
y
m2
l
q
O
m1
m1 g
m2 g
x
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/2, at equilibrium
so no angular momentum.
Mar. 27, 2002
The moment of inertia of this system is
1
1
1
2
2
I  I rod  I m1  I m2  Ml  m1l  m2l 2
12
4
4
2
l2  1

  M  m1  m2  L  Iw  wl  1 M  m1  m2 
4 3

4 3

First compute net
external torque
   m1 g
l
l
cos q ;  2  m2 g cos q
2
2
 ext      2 
gl cosq m1  m2 
2
1
m1  m1 gl cos q
 ext
2m1  m1  cos q

Thus 
2

 2

g /l
l 1
I
1


becomes 1443-501 Spring 2002  M  m1  m2   M  m1  m
142 
4 3
 3
Dr. J. Yu, Lecture #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?

ext
dL

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
Mar. 27, 2002
 K i  U i  K f  U f Mechanical Energy


Linear Momentum
 pi  p f


Angular Momentum
Li  L f
1443-501 Spring 2002
Dr. J. Yu, Lecture #15
15
Example 11.8
A start 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.
The period will be significantly shorter,
because its radius got smaller.
1. There is no torque acting on it
Let’s make some assumptions:
2. The shape remains spherical
3. Its mass remains constant
Li  L f
Using angular momentum
conservation
I iw  I f w f
What is your guess about the answer?
The angular speed of the star with the period T is
Thus
2p
w
T
I iw
mri 2 2p
wf 

If
mrf2 Ti
Tf 
Mar. 27, 2002
2p
wf
 r f2
 2
r
 i
2

3
.
0


6
Ti  

30
days

2
.
7

10
days  0.23s

4

 1.0  10 

1443-501 Spring 2002
Dr. J. Yu, Lecture #15
16