Download Physics 207: Lecture 2 Notes

Physics 207, Lecture 16, Oct. 29
Agenda: Chapter 13
 Center of Mass
 Torque
 Moment of Inertia
 Rotational Energy
 Rotational Momentum
Assignment:
 Wednesday is an exam review session, Exam will be
held in rooms B102 & B130 in Van Vleck at 7:15 PM
 MP Homework 7, Ch. 11, 5 problems,
NOTE: Due Wednesday at 4 PM
 MP Homework 7A, Ch. 13, 5 problems, available soon
Physics 207: Lecture 16, Pg 1
Chap. 13: Rotational Dynamics
 Up until now rotation has been only in terms of circular
motion with ac = v2 / R and | aT | = d| v | / dt
 Rotation is common in the world around us.
 Many ideas developed for translational motion are
transferable.
Physics 207: Lecture 16, Pg 2
Conservation of angular momentum has consequences
How does one describe rotation (magnitude and direction)?
Physics 207: Lecture 16, Pg 3
Rotational Dynamics: A child’s toy, a physics
playground or a student’s nightmare
 A merry-go-round is spinning and we run and
jump on it. What does it do?
 We are standing on the rim and our “friends”
spin it faster. What happens to us?
 We are standing on the rim a walk towards
the center. Does anything change?
Physics 207: Lecture 16, Pg 4

Summary
(with comparison to 1-D kinematics)
Angular
  constant
Linear
a  constant
  0  t
v  v0  at
1 2
  0   0 t  t
2
1 2
x  x0  v 0t  at
2
And for a point at a distance R from the rotation axis:
x = R v =  R
a=R
Physics 207: Lecture 16, Pg 7
System of Particles (Distributed Mass):
 Until now, we have considered the behavior of very simple
systems (one or two masses).
 But real objects have distributed mass !
 For example, consider a simple rotating disk and 2 equal
mass m plugs at distances r and 2r.

1
2
 Compare the velocities and kinetic energies at these two
points.
Physics 207: Lecture 16, Pg 12
System of Particles (Distributed Mass):
 An extended solid object (like a disk) can be thought of as
a collection of parts.
 The motion of each little part depends on where it is in the
object!
1 K= ½ m v2 = ½ m (r)2

 The rotation axis matters too!
2 K= ½ m (2v)2 = ½ m (2r)2
Physics 207: Lecture 16, Pg 13
System of Particles: Center of Mass
 If an object is not held then it rotates about the
center of mass.
 Center of mass: Where the system is balanced !
 Building a mobile is an exercise in finding
centers of mass.
m1
+
m2
m1
+
m2
mobile
Physics 207: Lecture 16, Pg 14
System of Particles: Center of Mass
 How do we describe the “position” of a system made up of
many parts ?
 Define the Center of Mass (average position):
For a collection of N individual pointlike particles whose
masses and positions we know:

 mi ri
N

RCM  i 1
M
RCM
m1
r1
m2
r2
y
x
(In this case, N = 2)
Physics 207: Lecture 16, Pg 15
Sample calculation:
 Consider the following mass distribution:

 mi ri
N

RCM  i 1
M
 XCM î  YCM ĵ  ZCM k̂
XCM = (m x 0 + 2m x 12 + m x 24 )/4m meters
RCM = (12,6)
YCM = (m x 0 + 2m x 12 + m x 0 )/4m meters
(12,12)
2m
XCM = 12 meters
YCM = 6 meters
m
(0,0)
m
(24,0)
Physics 207: Lecture 16, Pg 16
System of Particles: Center of Mass
 For a continuous solid, convert sums to an integral.
dm
y



 r dm  r dm
RCM 

M
 dm
r
x
where dm is an infinitesimal
mass element.
Physics 207: Lecture 16, Pg 17
Rotational Dynamics: What makes it spin?
a
A force applied at a distance
FTangential
from the rotation axis
F
TOT = |r| |FTang| ≡ |r| |F| sin f

Frandial
r
Torque is the rotational equivalent of force
Torque has units of kg m2/s2 = (kg m/s2) m = N m

A constant torque gives constant angular
acceleration iff the mass distribution and the axis of
rotation remain constant.
Physics 207: Lecture 16, Pg 18
Lecture 16, Exercise 1
Torque
 In which of the cases shown below is the torque provided
by the applied force about the rotation axis biggest? In
both cases the magnitude and direction of the applied
force is the same.
 Remember torque requires F, r and sin 
or the tangential force component times perpendicular distance
L
F
F
A. Case 1
B. Case 2
L
axis
C. Same
case 1
case 2
Physics 207: Lecture 16, Pg 19
Lecture 16, Exercise 1
Torque
 In which of the cases shown below is the torque
provided by the applied force about the rotation axis
biggest? In both cases the magnitude and direction of
the applied force is the same.
 Remember torque requires F, r and sin f
or the tangential force component times perpendicular distance
(A) case 1
(B) case 2
(C) same
L
F
F
L
axis
case 1
case 2
Physics 207: Lecture 16, Pg 20
Rotational Dynamics: What makes it spin?
a
A force applied at a distance
FTangential
from the rotation axis
F
TOT = |r| |FTang| ≡ |r| |F| sin f

Frandial
r
Torque is the rotational equivalent of force
Torque has units of kg m2/s2 = (kg m/s2) m = N m
TOT =
r FTang = r m a
=rmr
= m r2 
For every little part of the wheel
Physics 207: Lecture 16, Pg 21
TOT = m r2 and inertia
a
The further a mass is away
from this axis the greater the
inertia (resistance) to rotation

FTangential
F
Frandial
r
This is the rotational version of FTOT = ma
Moment of inertia, I ≡ m r2 , (here I is just a point
on the wheel) is the rotational equivalent of mass.
 If I is big, more torque is required to achieve a
given angular acceleration.

Physics 207: Lecture 16, Pg 22
Calculating Moment of Inertia
N
I   mi ri
2
where r is the distance from
the mass to the axis of rotation.
i 1
Example: Calculate the moment of inertia of four point masses
(m) on the corners of a square whose sides have length L,
about a perpendicular axis through the center of the square:
m
m
m
m
L
Physics 207: Lecture 16, Pg 23
Calculating Moment of Inertia...
 For a single object, I depends on the rotation axis!
 Example: I1 = 4 m R2 = 4 m (21/2 L / 2)2
I1 = 2mL2
m
m
m
m
I2 = mL2
I = 2mL2
L
Physics 207: Lecture 16, Pg 24
Lecture 16, Exercise 2
Moment of Inertia
 A triangular shape is made from identical balls and
identical rigid, massless rods as shown. The moment
of inertia about the a, b, and c axes is Ia, Ib, and Ic
respectively.
 Which of the following is correct:
I   mi ri
2
i
(A)
Ia > Ib > I c
a
(B)
Ia > Ic > I b
b
(C)
Ib > Ia > Ic
c
Physics 207: Lecture 16, Pg 25
Calculating Moment of Inertia...
 For a discrete collection of point
masses we found:
N
I   mi ri
2
i 1
 For a continuous solid object we have to add up the mr2
contribution for every infinitesimal mass element dm.
dm
 An integral is required to find I :
I   r dm
2
r
Physics 207: Lecture 16, Pg 27
Moments of Inertia...
 Some examples of I for solid objects:
R
Solid sphere of mass M and radius R,
about an axis through its center.
I = 2/5 M R2
R
Thin spherical shell of mass M and
radius R, about an axis through its
center.
Use the table…
See Table 13.3, Moments of Inertia
Physics 207: Lecture 16, Pg 29
Rotation & Kinetic Energy
 Consider the simple rotating system shown below.
(Assume the masses are attached to the rotation axis by
massless rigid rods).
 The kinetic energy of this system will be the sum of the
kinetic energy of each piece:
4
K   mi v
1
2
i 1
2
i
 K = ½ m1v12+ ½ m2v22+ ½ m3v32+ ½ m4v42
m4

r4
m3
r3
r1
m1
r2
m2
Physics 207: Lecture 16, Pg 31
Rotation & Kinetic Energy
 Notice that v1 =  r1 , v2 =  r2 , v3 =  r3 , v4 =  r4
 So we can rewrite the summation:
4
4
K   mi v   mi r 
1
2
i 1
2
i
1
2
2 2
i
i 1
1
2
4
[  m r ]
i 1
2
i i
2
 We recognize the quantity, moment of inertia or I, and
write:
m4
K  I
1
2
2

r4
m3
r3
r1
r2
m1
m2
Physics 207: Lecture 16, Pg 32
Lecture 16, Exercise 2
Rotational Kinetic Energy
 We have two balls of the same mass. Ball 1 is attached
to a 0.1 m long rope. It spins around at 2 revolutions per
second. Ball 2 is on a 0.2 m long rope. It spins around at
2 revolutions per second.
2
K  12 I 
 What is the ratio of the kinetic energy
of Ball 2 to that of Ball 1 ?
A. ¼
B. ½
C. 1
D. 2
Ball 1
E. 4
I   mi ri
2
i
Ball 2
Physics 207: Lecture 16, Pg 33
Lecture 16, Exercise 2
Rotational Kinetic Energy
 K2/K1 = ½ m r22 / ½ m r12 = 0.22 / 0.12 = 4
 What is the ratio of the kinetic energy of Ball 2 to
that of Ball 1 ?
(A) 1/4 (B) 1/2
Ball 1
(C) 1
(D) 2
(E) 4
Ball 2
Physics 207: Lecture 16, Pg 34
Rotation & Kinetic Energy...
 The kinetic energy of a rotating system looks similar
to that of a point particle:
Point Particle
K  mv
1
2
Rotating System
2
v is “linear” velocity
m is the mass.
K  I
1
2
2
 is angular velocity
I is the moment of inertia
about the rotation axis.
I   mi ri
2
i
Physics 207: Lecture 16, Pg 35
Moment of Inertia and Rotational Energy
 So
K  I
1
2
2
where
I   mi ri
2
i
 Notice that the moment of inertia I depends on the
distribution of mass in the system.
 The further the mass is from the rotation axis, the
bigger the moment of inertia.
 For a given object, the moment of inertia depends on
where we choose the rotation axis (unlike the center of
mass).
 In rotational dynamics, the moment of inertia I appears
in the same way that mass m does in linear dynamics !
Physics 207: Lecture 16, Pg 36
Work & Kinetic Energy:
 Recall the Work Kinetic-Energy Theorem: K = WNET
 This is true in general, and hence applies to rotational
motion as well as linear motion.
 So for an object that rotates about a fixed axis:
K  I      WNET
1
2
2
f
2
i
Physics 207: Lecture 16, Pg 38
Connection with CM motion
 If an object of mass M is moving linearly at velocity VCM
without rotating then its kinetic energy is
2
K T  12 MVCM

If an object of moment of inertia ICM is rotating in place
about its center of mass at angular velocity  then its
kinetic energy is
K R  I CM
1
2

2
What if the object is both moving linearly and rotating?
K  I CM  MV
1
2
2
1
2
2
CM
Physics 207: Lecture 16, Pg 45
Angular Momentum:
 We have shown that for a system of particles,
momentum


p  mv
 dp
F
0
dt
is conserved if
 What is the rotational equivalent of this?
angular momentum
is conserved if



L  I
 

dL
dt
0
Physics 207: Lecture 16, Pg 59
Example: Two Disks
 A disk of mass M and radius R rotates around the z axis
with angular velocity 0. A second identical disk, initially
not rotating, is dropped on top of the first. There is
friction between the disks, and eventually they rotate
together with angular velocity F.
z
z
0
F
Physics 207: Lecture 16, Pg 60
Example: Two Disks
 A disk of mass M and radius R rotates around the z axis
with initial angular velocity 0. A second identical disk,
at rest, is dropped on top of the first. There is friction
between the disks, and eventually they rotate together
with angular velocity F.
No External Torque so Lz is constant
Li = Lf  I i i = I f  ½ mR2 0 = ½ 2mR2 f
z
z
0
F
Physics 207: Lecture 16, Pg 61
Lecture 16, Oct. 29
Assignment:
 Wednesday is an exam review session, Exam will be held
in rooms B102 & B130 in Van Vleck at 7:15 PM
 MP Homework 7, Ch. 11, 5 problems,
NOTE: Due Wednesday at 4 PM
 MP Homework 7A, Ch. 13, 5 problems, available soon
Physics 207: Lecture 16, Pg 62
p = mv
Angular Momentum:
Definitions & Derivations
 We have shown that for a system of particles

FEXT

dp

dt
Momentum is conserved if
FEXT  0
 What is the rotational equivalent of this?
  
  r F

The rotational analog of force F is torque

Define the rotational analog of momentum p to be
angular momentum, L or
  
L r p
Physics 207: Lecture 16, Pg 63
Recall from Chapter 9: Linear Momentum

Definition: For a single particle, the momentum p is
defined as:
p ≡ mv
(p is a vector since v is a vector)
So px = mvx etc.
 Newton’s 2nd Law:
F = ma

dv d

m
 (mv)
dt dt

 dp
F
dt
 Units of linear momentum are kg m/s.
Physics 207: Lecture 16, Pg 64
Linear Momentum and Angular Momentum

So from:


p  mv
 Newton’s 2nd Law:
F = ma

dv d

m
 (mv)
dt dt

 dp
F 
dt


L  I

 dL
 
dt

 dL  
 
 r F
dt
 Units of angular momentum are kg m2/s.
Physics 207: Lecture 16, Pg 65
Putting it all together

 dL  
 r F
  
dt
  
dL  r  F dt

 
Lrp
 EXT

dL

dt
 



L  r   F dt  r  p
L r p
 EXT  r  FEXT
In the absence of external torques
Total angular momentum is conserved
 EXT
dL

0
dt
Physics 207: Lecture 16, Pg 66
Angular momentum of a rigid body
about a fixed axis:
 Consider a rigid distribution of point particles rotating in
the x-y plane around the z axis, as shown below. The
total angular momentum around the origin is the sum of
the angular momentum of each particle:
(since ri , vi , are
perpendicular)
Lz   ri  pi   mi ri  v i   mi ri vi k̂
i
i
We see that L is in the z direction.
m2
Using vi =  ri , we get
L z   mi ri  k̂
2
i


L  I
v1
i
v2
r2
m3
j

r3
m1
i r1
v3
Physics 207: Lecture 16, Pg 67
Example: Two Disks
 A disk of mass M and radius R rotates around the z axis
with angular velocity 0. A second identical disk, initially
not rotating, is dropped on top of the first. There is
friction between the disks, and eventually they rotate
together with angular velocity F.
z
z
0
F
Physics 207: Lecture 16, Pg 68
Example: Two Disks
 A disk of mass M and radius R rotates around the z axis
with initial angular velocity 0. A second identical disk,
at rest, is dropped on top of the first. There is friction
between the disks, and eventually they rotate together
with angular velocity F.
No External Torque so Lz is constant
Li = Lf  I i i = I f  ½ mR2 0 = ½ 2mR2 f
z
z
0
F
Physics 207: Lecture 16, Pg 69
Demonstration:
Conservation of Angular Momentum
 Figure Skating :
z
z
Arm
A
LA =
LB
IA
IB
A
B
Arm
B
No External Torque so Lz is constant even if internal work done.
Physics 207: Lecture 16, Pg 70
Demonstration:
Conservation of Angular Momentum
 Figure Skating :
z
z
Arm
A
IAA= LA = LB = IBB
Arm
B
IA < I B
A > B
½ IAA2 > ½ IB B2 (work needs to be done)
No External Torque so Lz is constant even if internal work done.
Physics 207: Lecture 16, Pg 71
Angular Momentum Conservation
 A freely moving particle has a well defined angular
momentum about any given axis.
 If no torques are acting on the particle, its angular
momentum remains constant (i.e., will be conserved).
 In the example below, the direction of L is along the z
axis, and its magnitude is given by LZ = pd = mvd.
y
x
d
m v
Physics 207: Lecture 16, Pg 72
Example: Bullet hitting stick
 A uniform stick of mass M and length D is pivoted at the
center. A bullet of mass m is shot through the stick at a
point halfway between the pivot and the end. The initial
speed of the bullet is v1, and the final speed is v2.
 What is the angular speed F of the stick after the
collision? (Ignore gravity)
M
m
F
D
D/4
v1
v2
before
after
Physics 207: Lecture 16, Pg 73
Example: Bullet hitting stick
 What is the angular speed F of the stick after the
collision? (Ignore gravity).
 Process: (1) Define system (2) Identify Conditions
(1) System: bullet and stick (No Ext. torque, L is constant)
(2) Momentum is conserved (Istick = I = MD2/12 )
Linit = Lbullet + Lstick = m v1 D/4 + 0 = Lfinal = m v2 D/4 + I f
M
m
F
D
D/4
v1
v2
before
after
Physics 207: Lecture 16, Pg 74
Example: Throwing ball from stool
 A student sits on a stool, initially at rest, but which is
free to rotate. The moment of inertia of the student plus
the stool is I. They throw a heavy ball of mass M with
speed v such that its velocity vector moves a distance d
from the axis of rotation.
 What is the angular speed F of the student-stool
system after they throw the ball ?
M
v
F
I
Top view: before
d
I
after
Physics 207: Lecture 16, Pg 75
Example: Throwing ball from stool
 What is the angular speed F of the student-stool
system after they throw the ball ?
 Process: (1) Define system (2) Identify Conditions
(1) System: student, stool and ball (No Ext. torque, L is
constant)
(2) Momentum is conserved
Linit = 0 = Lfinal = m v d + I f
M
v
F
I
Top view: before
d
I
after
Physics 207: Lecture 16, Pg 76
An example: Neutron Star rotation
Neutron star with a mass of 1.5 solar masses has a
diameter of ~11 km.
Our sun rotates about once every 37 days
f / i = Ii / If = ri2 / rf2 = (7x105 km)2/(11 km)2 = 4 x 109
gives millisecond periods!
period of pulsar is 1.187911164 s
Physics 207: Lecture 16, Pg 77
Angular Momentum as a Fundamental Quantity
 The concept of angular momentum is also valid on a
submicroscopic scale
 Angular momentum has been used in the
development of modern theories of atomic, molecular
and nuclear physics
 In these systems, the angular momentum has been
found to be a fundamental quantity
 Fundamental here means that it is an intrinsic
property of these objects
Physics 207: Lecture 16, Pg 78
Fundamental Angular Momentum
 Angular momentum has discrete values
 These discrete values are multiples of a fundamental
unit of angular momentum
 The fundamental unit of angular momentum is h-bar
 Where h is called Planck’s constant
h
34 kg  m

 1.054 10
2
s
L  n (n  1,2,3,...)
Physics 207: Lecture 16, Pg 79
Intrinsic Angular Momentum
Physics 207: Lecture 16, Pg 80
Angular Momentum of a Molecule
 Consider the molecule as a rigid rotor, with
the two atoms separated by a fixed distance
 The rotation occurs about the center of mass
in the plane of the page with a speed of
Physics 207: Lecture 16, Pg 81
Angular Momentum of a Molecule
(It heats the water in a microwave over)
L  
   / I CM
E = h2/(82I) [ J (J+1) ]
J = 0, 1, 2, ….
Physics 207: Lecture 16, Pg 82
Center of Mass Example: Astronauts & Rope
 Two astronauts are initially at rest in outer space and 20
meters apart. The one on the right has 1.5 times the
mass of the other (as shown). The 1.5 m astronaut
wants to get back to the ship but his jet pack is broken.
There happens to be a rope connected between the
two. The heavier astronaut starts pulling in the rope.
(1) Does he/she get back to the ship ?
(2) Does he/she meet the other astronaut ?
m
M = 1.5m
Physics 207: Lecture 16, Pg 83
Example: Astronauts & Rope
(1) There is no external force so if the larger astronaut
pulls on the rope he will create an impulse that
accelerates him/her to the left and the small
astronaut to the right. The larger one’s velocity will
be less than the smaller one’s so he/she doesn’t let
go of the rope they will either collide (elastically or
inelastically) and thus never make it.
m
M = 1.5m
Physics 207: Lecture 16, Pg 84
Example: Astronauts & Rope
(2) However if the larger astronaut lets go of the rope
he will get to the ship. (Too bad for the smaller
astronaut!)
In all cases the center of mass will remain fixed
because they are initially at rest and there is no
external force.
To find the position where they meet all we need do is
find the Center of Mass
m
M = 1.5m
Physics 207: Lecture 16, Pg 85