Physics 151: Lecture 23 Today’s Agenda
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Transcript of Physics 151: Lecture 23 Today’s Agenda
Physics 151: Lecture 23, Pg 1
Physics 151: Lecture 23Physics 151: Lecture 23
Today’s AgendaToday’s Agenda
TopicsMore on Rolling MotionCh. 11.1 Angular Momentum Ch. 11.3-5
Physics 151: Lecture 23, Pg 2
Example :Example : Rolling MotionRolling Motion
A cylinder is about to roll down an inclined plane. What is its speed at the bottom of the plane ?
M
h
Mv ?
Cylinder has radius R
M M
M
M
M
Physics 151: Lecture 23, Pg 3
Lecture 22, Lecture 22, ACT 4aACT 4aRolling MotionRolling 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
h
M?
M2
Physics 151: Lecture 23, Pg 4
Lecture 22, Lecture 22, ACT 4bACT 4bRolling MotionRolling 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
C) It will be a tie
R1
h
M?
R2
animation
Physics 151: Lecture 23, Pg 5
Lecture 22, Lecture 22, ACT 4cACT 4cRolling MotionRolling 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
h
M?
M2
animation
Physics 151: Lecture 23, Pg 6
Remember our roller coaster.
Perhaps now we can get the ball to go around the circle without anyone dying.
Note:Radius of loop = RRadius of ball = r
Physics 151: Lecture 23, Pg 7
How high do we have to start the ball ?
1
2
-> The rolling motion added an extra 2/10 R to the height)
h
h = 2.7 R = (2R + 1/2R) + 2/10 R
Physics 151: Lecture 23, Pg 8
Angular Momentum:Angular Momentum:Definitions & DerivationsDefinitions & Derivations
We have shown that for a system of particles
Momentum is conserved if
What is the rotational version of this ??
Fp
EXTddt
FEXT 0
See text: 11.3
L r p
r F The rotational analogue of force F F is torque
Define the rotational analogue of momentum pp to be
angular momentum
p=mv
Animation
Physics 151: Lecture 23, Pg 9
Definitions & Derivations...Definitions & Derivations...
First consider the rate of change of LL: ddt
ddt
Lr p
ddt
ddt
ddt
r pr
p rp
v vm
0
So ddt
ddt
Lr
p (so what...?)
See text: 11.3
Physics 151: Lecture 23, Pg 10
Definitions & Derivations...Definitions & Derivations...
ddt
ddt
Lr
p
Fp
EXTddt
EXTFdtd r
L Recall that
Which finally gives us: EXTddt
L
Analogue of !! Fp
EXTddt
EXT
See text: 11.3
Physics 151: Lecture 23, Pg 11
What does it mean?What does it mean?
where andEXTddt
L EXT EXT r FL r p
EXTddt
L0 In the absence of external torquesIn the absence of external torques
Total angular momentum is conservedTotal angular momentum is conserved
See text: 11.5
Physics 151: Lecture 23, Pg 12
i
j
Angular momentum of a rigid bodyAngular momentum of a rigid bodyabout a fixed axis:about a fixed axis:
k̂vrmmi
iiii
iiiii
i vrprL
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:
rr1
rr3
rr2
m2
m1
m3
vv2
vv1
vv3
We see that LL is in the z direction.
Using vi = ri , we get
LI
(since ri , vi , are
perpendicular)
Analogue of p = mv !!
k̂rmLi
2ii
See text: 11.4
Physics 151: Lecture 23, Pg 13
Lecture 23, Lecture 23, ACT 2ACT 2Angular momentumAngular 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. 40b. 10c. 30d. 20e. 50
Physics 151: Lecture 23, Pg 14
Angular momentum of a rigid bodyAngular momentum of a rigid bodyabout a fixed axis: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 theright hand rule (same as ).
We will omit the ”Z” subscript for simplicity,and write L = I LZ I
z
See text: 11.4
Physics 151: Lecture 23, Pg 15
Lecture 23, Lecture 23, ACT 2ACT 2Angular momentumAngular 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 (c) not enough info
I1 < I2
L I1 1
disk 2
L I2 2
disk 1
K1 > K2
Physics 151: Lecture 23, Pg 16
Example: Two DisksExample: 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 ?
0
z
F
z
Physics 151: Lecture 23, Pg 17
Example: Two DisksExample: 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:
0
z
L MRINI I1 12
01
2
2
1
Physics 151: Lecture 23, Pg 18
Example: Two DisksExample: 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 dueto both disks spinning:
F
z
L MRFIN F I I1 1 2 22 2
1
Physics 151: Lecture 23, Pg 19
Example: Two DisksExample: Two Disks
Since LINI = LFIN
0
z
F
z
LINI LFIN
1
22
02MR MR F
F 1
2 0
An inelastic collision,since E is not
conserved (friction) !
Physics 151: Lecture 23, Pg 20
Example: Two DisksExample: Two Disks
Let’s use conservation of energy principle:
0
z
F
z
EINI EFIN
1/2 I 02 = 1/2 (I + I) F
2
F2 = 1/2 0
2
F = 0 / 21/2
EINI = EFIN
Physics 151: Lecture 23, Pg 21
Using conservation of angular momentum:
LINI = LFIN we got a different answer !
1
22
02MR MR F F
1
2 0
Example: Two DisksExample: Two Disks
F’ = 0 / 21/2 Conservation of energy !
Conservation of momentum !
F’ > F Which one is correct ?
F = 0 / 2
Physics 151: Lecture 23, Pg 22
Example: Two DisksExample: Two Disks
• Is the system conservative ?
• Are there any non-conservative forces involved ?
• 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 !
F
z
W = E = 1/2 I02 - 1/2 (I+I) (0/2)2
= 1/2 I02 (1 - 2/4)
= 1/4 I 02
= 1/8 MR2 02
This is 1/2 of initial Energy !
Physics 151: Lecture 23, Pg 23
Lecture 23, Lecture 23, ACT 2ACT 2Angular momentumAngular momentum
Physics 151: Lecture 23, Pg 24
Recap of today’s lectureRecap of today’s lecture
Chapter 11.1-5, Rolling MotionAngular Momentum
For next time: Read Ch. 11.1-11.