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.