Transcript Physics 131: Lecture 14 Notes

Physics 151: Lecture 23
Today’s Agenda

Topics
More on Rolling Motion
Ch. 11.1 Angular Momentum
Ch. 11.3-5
Physics 151: Lecture 23, Pg 1
Example : Rolling Motion

A cylinder is about to roll down an inclined plane.
What is its speed at the bottom of the plane ?
Cylinder has radius R
M
h
q
M
v?
Physics 151: Lecture 23, Pg 2
Lecture 22, ACT 4a
Rolling Motion

A race !!
Two cylinders are rolled down a ramp. They have
the same radius but different masses, M1 > M2.
Which wins the race to the bottom ?
A) Cylinder 1
B) Cylinder 2
C) It will be a tie
M1
M2
h
q
M?
Physics 151: Lecture 23, Pg 3
Lecture 22, ACT 4b
Rolling Motion

A race !!
Two cylinders are rolled down a ramp. They have
the same moment of inertia but different radius,
R1 > R2. Which wins the race to the bottom ?
A) Cylinder 1
B) Cylinder 2
R1
C) It will be a tie
R2
animation
h
q
M?
Physics 151: Lecture 23, Pg 4
Lecture 22, ACT 4c
Rolling Motion

A race !!
A cylinder and a hoop are rolled down a ramp. They
have the same mass and the same radius. Which
wins the race to the bottom ?
A) Cylinder
B) Hoop
C) It will be a tie
M1
M2
animation
h
q
M?
Physics 151: Lecture 23, Pg 5
Remember our roller coaster.
Perhaps now we can get the ball to go
around the circle without anyone dying.
Note:
Radius of loop = R
Radius of ball = r
Physics 151: Lecture 23, Pg 6
How high do we have to start the ball ?
1
h
2
h = 2.7 R = (2R + 1/2R) + 2/10 R
-> The rolling motion added an extra 2/10 R to the height)
Physics 151: Lecture 23, Pg 7
See text: 11.3
p=mv

Angular Momentum:
Definitions & Derivations
We have shown that for a system of particles
FEXT 
dp
Momentum is conserved if
dt
FEXT  0

What is the rotational version of this ??

The rotational analogue of force F is torque   r  F

Define the rotational analogue of momentum p to be
angular momentum L  r  p
Animation
Physics 151: Lecture 23, Pg 8
See text: 11.3
Definitions & Derivations...

First consider the rate of change of L:
dL
dt

d
dt
r  p 
dp 
 dr
 
r

p


p

r

 


 



dt
dt
dt 
d
 v  mv 
0
dL
So
dt
r
dp
dt
(so what...?)
Physics 151: Lecture 23, Pg 9
See text: 11.3
Definitions & Derivations...
dL
dt

r
Recall that
dp
dt
FEXT 
dp
dL
dt
dt
 r  FEXT

EXT


 EXT 
Which finally gives us:
Analogue of
FEXT 
dp
dL
dt
!!
dt
Physics 151: Lecture 23, Pg 10
See text: 11.5
What does it mean?

 EXT 
dL
dt
where
Lr p
and
 In the absence of external torques  EXT 
 EXT  r  FEXT
dL
dt
0
Total angular momentum is conserved
Physics 151: Lecture 23, Pg 11
See text: 11.4
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
momenta of each particle:
L   ri  pi   mi ri  v i   mi ri v i kˆ
i
i
i
We see that L is in the z direction.
m2
Using vi =  ri , we get
v2
L   mi ri  kˆ
2
j
r2

m3
i
L I
(since ri , vi , are
perpendicular)
v1
i r1 m1
r3
v3
Analogue of p = mv !!
Physics 151: Lecture 23, Pg 12
Lecture 23, ACT 2
Angular momentum
In the figure, a 1.6-kg weight swings in a vertical circle at the
end of a string having negligible weight. The string is 2 m
long. If the weight is released with zero initial velocity from a
horizontal position, its angular momentum (in kg · m2/s) at
the lowest point of its path relative to the center of the circle
is approximately
a.
b.
c.
d.
e.
40
10
30
20
50
Physics 151: Lecture 23, Pg 13
See text: 11.4
Angular momentum of a rigid body
about a fixed axis:

In general, for an object rotating about a fixed (z) axis we
can write LZ = I 

The direction of LZ is given by the
right hand rule (same as ).

z
We will omit the ”Z” subscript for simplicity,
and write L = I 
LZ  I 

Physics 151: Lecture 23, Pg 14
Lecture 23, ACT 2
Angular momentum

Two different spinning disks have the same angular momentum,
but disk 1 has more kinetic energy than disk 2.
Which one has the biggest moment of inertia ?
(a) disk 1
(b) disk 2
K1
L  I1 1
(c) not enough info
>
K2
1
2
I1
disk 1
L  I2  2
<
I2
disk 2
Physics 151: Lecture 23, Pg 15
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. What is F ?
z
z
0
F
Physics 151: Lecture 23, Pg 16
Example: Two Disks

First realize that there are no external torques acting on the
two-disk system.
Angular momentum will be conserved !

Initially, the total angular momentum
is due only to the disk on the bottom:
z
2
LINI  I1 1 
1
2
1
2
MR  0
0
Physics 151: Lecture 23, Pg 17
Example: Two Disks

First realize that there are no external torques acting on the
two-disk system.
Angular momentum will be conserved !

Finally, the total angular momentum is due
to both disks spinning:
z
2
1
2
LFIN  I1 1  I2  2  MR  F
F
Physics 151: Lecture 23, Pg 18
Example: Two Disks

1
Since LINI = LFIN
2
F 
z
2
2
MR  0  MR  F
1
2
0
An inelastic collision,
since E is not
conserved (friction) !
z
LINI
0
LFIN
F
Physics 151: Lecture 23, Pg 19
Example: Two Disks

Let’s use conservation of energy principle:
EINI = EFIN
1/2 I 02 = 1/2 (I + I) F2
F2 = 1/2 02
F = 0 / 21/2
z
z
EINI
0
EFIN
F
Physics 151: Lecture 23, Pg 20
Example: Two Disks
Using conservation of angular momentum:
LINI = LFIN
we got a different answer !

1
2
2
2
MR  0  MR  F
F’ = 0 / 21/2
F = 0 / 2
F’ > F
F 
1
2
0
Conservation of energy !
Conservation of momentum !
Which one is correct ?
Physics 151: Lecture 23, Pg 21
Example: Two Disks
z
• Is the system conservative ?
• Are there any non-conservative forces involved ?
F
• In order for top disc to turn when in contact with the bottom one
there has to be friction ! (non-conservative force !)
• So, we can not use the conservation of energy here.
• correct answer:
F = 0/2
• We can calculate work being done due to this friction !
W = DE = 1/2 I02 - 1/2 (I+I) (0/2)2
= 1/2 I02 (1 - 2/4)
= 1/4 I 02
This is 1/2 of initial Energy !
= 1/8 MR2 02
Physics 151: Lecture 23, Pg 22
Lecture 23, ACT 2
Angular momentum
Physics 151: Lecture 23, Pg 23
Recap of today’s lecture

Chapter 11.1-5,
Rolling Motion
Angular Momentum

For next time: Read Ch. 11.1-11.
Physics 151: Lecture 23, Pg 24