PHYS 1441 – Section 002
Lecture #22
Wednesday, Dec. 1, 2010
Dr. Jaehoon Yu
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•
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Rotational Kinetic Energy
Angular Momentum
Angular Momentum Conservation
Similarities Between Linear and Rotational
Quantities
Conditions for Equilibrium
Wednesday, Dec. 1, 2010
PHYS 1441-002, Fall 2010
Dr. Jaehoon Yu
1
Announcements
• The Final Exam
– Date and time: 11am, Monday Dec. 13
– Place: SH103
– Comprehensive exam
• Covers from CH1.1 – what we finish Wednesday, Dec. 8
• Plus appendices A.1 – A.8
• Combination of multiple choice and free response
problems
• Bring your Planetarium extra credit sheet to the class next
Wednesday, Dec. 8, with your name clearly marked on
the sheet!
• Colloquium this today
Wednesday, Dec. 1, 2010
PHYS 1441-002, Fall 2010 Dr. Jaehoon
Yu
2
Wednesday, Dec. 1, 2010
PHYS 1441-002, Fall 2010 Dr. Jaehoon
Yu
3
Rotational Kinetic Energy
y
vi
What do you think the kinetic energy of a rigid object
that is undergoing a circular motion is?
1
1
Kinetic energy of a masslet, mi, K i  mi vi2  mi ri 2 
2
2
moving at a tangential speed, vi, is
mi
ri

O
x
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
Wednesday, Dec. 1, 2010
1 
K R  I
2
PHYS 1441-002, Fall 2010 Dr. Jaehoon
Yu
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 .
y
m
Since the rotation is about y axis, the moment of
inertia about y axis, Iy, is
b
l
M
O
l
b
m
M
2
2
2
2
2
2
I
m
0
m
0

2Ml

m
r

Ml

Ml
 ii
x
i
Why are some 0s?
Thus, the rotational kinetic energy is
This is because the rotation is done about y axis,
and the radii of the spheres are negligible.
1
1
2
2Ml 2  2  Ml 2 2
K R  I 
2
2


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
I   mi ri2  Ml
Ml 2 mb2mb2  2 Ml 2  mb 2
i
Wednesday, Dec. 1, 2010

1
1
K R  I 2 2Ml 2  2mb 2  2 Ml 2  mb 2  2
2
2
PHYS 1441-002, Fall 2010 Dr. Jaehoon
Yu
5
Kinetic Energy of a Rolling Sphere
R

h
q
vCM
Since vCM=Rω
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 
Wednesday, Dec. 1, 2010
PHYS 1441-002, Fall 2010 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. 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
θ
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
We
obtain
2
2
I CM  5 MR  aCM  2

f 

  MaCM
R
R  R  5
Substituting f in
dynamic equations
Wednesday, Dec. 1, 2010
 fR  I CM 
7
Mg sin q  MaCM
5
PHYS 1441-002, Fall 2010 Dr. Jaehoon
Yu
aCM 
5
g sin q
7
7
Work, Power, and Energy in Rotation

Δs
Δθ
r
O
Let’s consider the motion of a rigid body with a single external
force F exerting on the point P, moving the object by Δs.
The work done by the force F as the object rotates through
the infinitesimalurdistance
r Δs=rΔθ is
W  F  s  F sin  rq
What is Fsinϕ?
What is the work done by
radial component Fcosϕ?
Since the magnitude of torque is rFsinϕ,
The rate of work, or power, becomes
The rotational work done by an external force
equals the change in rotational Kinetic energy.
The work put in by the external force then
Wednesday, Dec. 1, 2010
The tangential component of the force F.
Zero, because it is perpendicular to the
displacement.
W  rF sin  q  q
P
How was the power
W  q
  defined in linear motion?


t
t
  


I

I




 t 
1
2
q  I
1
2
W  I 2f  I i2
PHYS 1441-002, Fall 2010 Dr. Jaehoon
Yu
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 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
ur
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  mr 2  I
What do you learn from this?
Wednesday, Dec. 1, 2010
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 9rim.
PHYS 1441-002, Fall 2010 Dr. Jaehoon
Yu
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
ur u
r
ur
ur
L  L1  L2 ......  L n 
ur
 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 Since these forces are the action and reaction forces with
system where the two exert directions lie on the line connecting the two particles, the
vector sum of the torque from these two becomes 0.
forces on each other.
ur
ur
u
r
Thus the time rate change of the angular
r
p
 L Just
momentum of a system of particles is equal to
  ext  t like  F  t
only the net external torque acting on the system
Lz  I  I
 ext 


 I

PHYS 1441-002, Fall 2010 Dr. Jaehoon t
t
t10
For a rigid body, the external torque is written
Wednesday, Dec. 1, 2010
Yu
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

O
m1
m1 g
m2 g
x
The moment of inertia of this system is
1
1
1
2
2
I  I rod  I m1  I m2 
m
l

m2l 2
Ml 
1
4
4
12
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 θ with the horizon.
If m1 = m2, no angular
momentum because the net
torque is 0.
If θ//at equilibrium
so no angular momentum.
Wednesday, Dec. 1, 2010
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 α
ext
 2

1
 l
l 1

I
M

m

m
 M  m1  m2 

1
2 
becomes
3
4 3
PHYS 1441-002, Fall 2010 Dr. Jaehoon
11 

Yu
Conservation of Angular Momentum
Remember under what condition the linear momentum is conserved?
ur
ur
p
Linear momentum is conserved when the net external force is 0.  F  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
ur
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, Dec. 1, 2010
PHYS 1441-002, Fall 2010 Dr. Jaehoon
Yu
Mechanical Energy
Linear Momentum
Angular Momentum
12
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
 r f2
 2
r
 i
Wednesday, Dec. 1, 2010
2

T
2

3
.
0


6
Ti  

2
.
7

10
days  0.23s

30
days

4

1
.
0

10



PHYS 1441-002, Fall 2010 Dr. Jaehoon
Yu
13
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, Dec. 1, 2010
Linear
Mass
Rotational
Moment of Inertia
M
Distance
r
t
v
a
urt
I  mr 2
Angle q (Radian)
L
q
t


t
v

r
Force F  ma
r r
Work W  F  d
ur r
P  F v
ur
r
p  mv
Kinetic
K
1
mv 2
2
r ur
Torque   I 
Work W  q
P  
ur
ur
L  I
Rotational
PHYS 1441-002, Fall 2010 Dr. Jaehoon
Yu
KR 
1
I 2
2
14