In the two cases shown below identical ladders are leaning against frictionless walls. In which case...

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In the two cases shown below identical ladders are leaning against frictionless walls. In which case is the force of friction between the ladder and the ground bigger? A) Case 1 B) Case 2 C) Same Catching up from Yesterday Case 1 Case 2 Mechanics Lecture 18, Slide 1

Transcript of In the two cases shown below identical ladders are leaning against frictionless walls. In which case...

Page 1: In the two cases shown below identical ladders are leaning against frictionless walls. In which case is the force of friction between the ladder and the.

In the two cases shown below identical ladders are leaning against frictionless walls. In which case is the force of friction between the ladder and the ground bigger?

A) Case 1 B) Case 2 C) Same

Catching up from Yesterday

Case 1 Case 2Mechanics Lecture 18, Slide 1

Page 2: In the two cases shown below identical ladders are leaning against frictionless walls. In which case is the force of friction between the ladder and the.

In the two cases shown below identical ladders are leaning against frictionless walls. In which case is the force of friction between the ladder and the ground bigger?

A) Case 1 B) Case 2 C) Same

Case 1 Case 2A) Because the bottom of the ladder is further away from the wall.

B) The angle is steeper, which means there is more normal force and thus more friction.

C) Both have same mass.

Mechanics Lecture 18, Slide 2

CheckPoint

Page 3: In the two cases shown below identical ladders are leaning against frictionless walls. In which case is the force of friction between the ladder and the.

CheckPoint

Suppose you hang one end of a beam from the ceiling by a rope and the bottom of the beam rests on a frictionless sheet of ice. The center of mass of the beam is marked with an black spot. Which of the following configurations best represents the equilibrium condition of this setup?

A) B) C)

Mechanics Lecture 18, Slide 3

Page 4: In the two cases shown below identical ladders are leaning against frictionless walls. In which case is the force of friction between the ladder and the.

CheckPointWhich of the following configurations best represents the equilibrium condition of this setup?

A) B) C)

If the tension has any horizontal component, the beam will accelerate in the horizontal direction.

Objects tend to attempt to minimize their potential energy as much as possible. In Case C, the center of mass is lowest.

Mechanics Lecture 18, Slide 4

DemoDemo

Page 5: In the two cases shown below identical ladders are leaning against frictionless walls. In which case is the force of friction between the ladder and the.

footprint

footprint

Stability & Potential Energy

Mechanics Lecture 18, Slide 5

CM DemosCM Demos

Page 6: In the two cases shown below identical ladders are leaning against frictionless walls. In which case is the force of friction between the ladder and the.

Physics 211Lecture 19

Today’s Concepts:Angular Momentum

Mechanics Lecture 19, Slide 6

Page 7: In the two cases shown below identical ladders are leaning against frictionless walls. In which case is the force of friction between the ladder and the.

Linear and Rotation

Mechanics Lecture 19, Slide 7

Page 8: In the two cases shown below identical ladders are leaning against frictionless walls. In which case is the force of friction between the ladder and the.

New Cast of Characters with Old Theme

Remember

Mechanics Lecture 19, Slide 8

Page 9: In the two cases shown below identical ladders are leaning against frictionless walls. In which case is the force of friction between the ladder and the.

Angular Momentum

We have shown that for a system of particles

What is the rotational version of this?

Momentum is conserved ifEXT

dpF

dt

0EXTF

The rotational analogue of force F is torque r F

Define the rotational analogue of momentum p to be Angular Momentum: L r p

For a symmetric solid object L I

Mechanics Lecture 19, Slide 9

Page 10: In the two cases shown below identical ladders are leaning against frictionless walls. In which case is the force of friction between the ladder and the.

Torque & Angular Momentum

where and

Total angular momentum is conserved

EXT

dL

dt

L r p

L I ext EXTr F

In the absence of external torques 0EXT

dL

dt

Mechanics Lecture 19, Slide 10

Page 11: In the two cases shown below identical ladders are leaning against frictionless walls. In which case is the force of friction between the ladder and the.

Example – Disk dropped on Disk

Mechanics Lecture 19, Slide 11

0, initialbottom

oinitialtop ,

ffinalbottomfinaltop ,,

Friction is an “internal force” because it is “internal” to the “system”. Friction is between two parts of the “system”.Friction may change the Kinetic Energy.But friction, being an internal force, does not change the Angular Momentum (nor the linear momentum in linear situations).

DemoDemo

Page 12: In the two cases shown below identical ladders are leaning against frictionless walls. In which case is the force of friction between the ladder and the.

Example – Disk dropped on Disk

02

1 2 oinitial MRL fffinal MRMRL 22

2

1

2

1

fo MRMR 22

2

1 02

1 f

Mechanics Lecture 19, Slide 12

0, initialbottom

oinitialtop ,

ffinalbottomfinaltop ,,

Page 13: In the two cases shown below identical ladders are leaning against frictionless walls. In which case is the force of friction between the ladder and the.

2

2afterafter

LK

I

2

2beforebefore

LK

I same

x2

What about Kinetic Energy?

221

2 2

LK I

I

Mechanics Lecture 19, Slide 13

0, initialbottom

oinitialtop ,

ffinalbottomfinaltop ,,

Page 14: In the two cases shown below identical ladders are leaning against frictionless walls. In which case is the force of friction between the ladder and the.

Case 1 Case 2

A) Case 1B) Case 2C) Same

In both cases shown below a solid disk of mass M, radius R, and initial angular velocity o is dropped onto an initially stationary second disk having the same radius. In Case 2 the mass of the bottom disk is twice as big as in Case 1. If there are no external torques acting on either system, in which case is the final kinetic energy of the system bigger?

Same initial L

CheckPoint

Mechanics Lecture 19, Slide 14

0, initialbottom

oinitialtop ,

M

M

0, initialbottom

oinitialtop ,

2M

M

Page 15: In the two cases shown below identical ladders are leaning against frictionless walls. In which case is the force of friction between the ladder and the.

Case 1 Case 2

A) Case 1B) Case 2C) Same

In which case is the final kinetic energy of the system bigger?

A) L is equal but bigger mass in case 2

B) Case 2 has greater mass

C) Conservation of angular momentum

CheckPoint

2

2

LK

I

Mechanics Lecture 19, Slide 15

M 2MM M

Demo: wheel rim dropDemo: wheel rim drop

Page 16: In the two cases shown below identical ladders are leaning against frictionless walls. In which case is the force of friction between the ladder and the.

Your Comments

Mechanics Lecture 19, Slide 16

This doesn’t mean anything.This doesn’t mean anything.

Sort of . . .Sort of . . .Work is done by friction so KE is Work is done by friction so KE is notnot conserved. conserved.

Okay; now Okay; now whatwhat does that mean? does that mean?

Find the Find the finalfinal KE in both cases. KE in both cases.

True but what about the True but what about the finalfinal KE? KE?

That’s for omega but what about the That’s for omega but what about the finalfinal KE? KE?True but what about the True but what about the finalfinal KE? KE?

True but what about the True but what about the finalfinal KE? KE?? ? ? ? ? ? ? ?

? ? ? ? ? ? ? ?

Page 17: In the two cases shown below identical ladders are leaning against frictionless walls. In which case is the force of friction between the ladder and the.

axis

Point Particle Moving in a Straight Line

L R p

pRL

11

1

)sin( 11 pRpRL

22

2

)sin( 22 pRpD pD pD

3

)sin( 33 pRpRL

33

Mechanics Lecture 19, Slide 17

2R

3R

1R

p

p

p

D

Page 18: In the two cases shown below identical ladders are leaning against frictionless walls. In which case is the force of friction between the ladder and the.

Direction given by right hand rule(out of the page in this case)

L R p

L mvD pD

Mechanics Lecture 19, Slide 18

axis

p

D

Demo: train on bike wheelDemo: train on bike wheel

Page 19: In the two cases shown below identical ladders are leaning against frictionless walls. In which case is the force of friction between the ladder and the.

R

M

mv

Before

Top View

Disk Kid

Playground Example

mvRLBefore )2

1( 22 mRMRILAfter

Before AfterL L mMR

v

21

1

Mechanics Lecture 19, Slide 19

After

Page 20: In the two cases shown below identical ladders are leaning against frictionless walls. In which case is the force of friction between the ladder and the.

Disk Kid

Algebra Details

mvRLBefore

Before AfterL L mMR

v

21

1

Mechanics Lecture 19, Slide 20

)2

1( 22 mRMRILAfter

Page 21: In the two cases shown below identical ladders are leaning against frictionless walls. In which case is the force of friction between the ladder and the.

The magnitude of the angular momentum of a freely rotating disk around its center is L. You toss a heavy block onto the disk along the direction shown. Friction acts between the disk and the block so that eventually the block is at rest on the disk and rotates with it. What is the magnitude of the final angular momentum of the disk-block system:

A) LB) LC) L

Instead of a block, imagine it’s a kid hopping onto a merry go round

CheckPoint

Mechanics Lecture 19, Slide 21

Top View

The block provides no angular momentum since its perpendicular distance from the axis of rotation is 0. No external forces act on this systems, so angular momentum is conserved.

Page 22: In the two cases shown below identical ladders are leaning against frictionless walls. In which case is the force of friction between the ladder and the.

What is the magnitude of the final angular momentum of the disk-block system:

A) LB) LC) L

CheckPoint

Mechanics Lecture 19, Slide 22

Top View

The block provides no angular momentum since its perpendicular distance from the axis of rotation is 0. No external forces act on this systems, so angular momentum is conserved.

Page 23: In the two cases shown below identical ladders are leaning against frictionless walls. In which case is the force of friction between the ladder and the.

Your Comments

Mechanics Lecture 19, Slide 23

Zero!>!>! Zero!>!>! HowHow can that be? can that be?

Will it be conserved in the Will it be conserved in the nextnext CheckPoint? CheckPoint?Good! Altho’ I wouldn’t call it an external force.Good! Altho’ I wouldn’t call it an external force.

Good! The block’s Good! The block’s perpendicularperpendicular distance from the axis of rotation is zero. . distance from the axis of rotation is zero. .

But the block is But the block is NOTNOT at rest. at rest.

What is the final angular momentum?What is the final angular momentum?What does that mean for the final angular momentum?What does that mean for the final angular momentum?

What about the What about the nextnext CheckPoint?CheckPoint?

! ! ! ! !! ! ! ! !

okayokay

Tell me more.Tell me more.

What about the block?What about the block?Tell me more.Tell me more.

Page 24: In the two cases shown below identical ladders are leaning against frictionless walls. In which case is the force of friction between the ladder and the.

Top View

ACT

Mechanics Lecture 19, Slide 24

The magnitude of the angular momentum of a freely rotating disk around its center is L. You toss a heavy block onto the disk along the direction shown. Friction acts between the disk and the block so that eventually the block is at rest on the disk and rotates with it. Is the total angular momentum of the disk-block system conserved during this?

A) Yes

B) No

Friction decreases the angular momentum of the disk

But the TOTAL angular momentum of the block+disk is CONSTANT!

Page 25: In the two cases shown below identical ladders are leaning against frictionless walls. In which case is the force of friction between the ladder and the.

Top View

CheckPoint

Mechanics Lecture 19, Slide 25

The magnitude of the angular momentum of a freely rotating disk around its center is L. You toss a heavy block onto the disk along the direction shown. Friction acts between the disk and the block so that eventually the block is at rest on the disk and rotates with it. What is the magnitude of the final angular momentum of the disk-block system:

Instead of a block, imagine it’s a kid hopping onto a merry go round

A) LB) LC) L

Page 26: In the two cases shown below identical ladders are leaning against frictionless walls. In which case is the force of friction between the ladder and the.

CheckPoint

What is the magnitude of the final angular momentum of the disk-block system:

B) L is conserved so Linitial=Lfinal

A) Since the block has angular momentum to begin with, that is added into L.

C) The block adds momentum going in the opposite direction, so the total momentum is smaller than the momentum of the disk.

A) LB) LC) L

Top View

Mechanics Lecture 19, Slide 26

Page 27: In the two cases shown below identical ladders are leaning against frictionless walls. In which case is the force of friction between the ladder and the.

Your Comments

Mechanics Lecture 19, Slide 27

Tell me more.Tell me more.

Smaller than what? In the previous CheckPoint, the lever arm was Smaller than what? In the previous CheckPoint, the lever arm was zerozero..

Okay but tell me more.Okay but tell me more.Good.Good.

What about the angular momentum of the What about the angular momentum of the blockblock??

No, not this time.No, not this time.

Any better by now?Any better by now?

Why bigger? Angular momentum has a Why bigger? Angular momentum has a directiondirection connected to it. connected to it.

What about the block?What about the block?

What about the block?What about the block?Bravo!Bravo!

Bravo!Bravo!

Page 28: In the two cases shown below identical ladders are leaning against frictionless walls. In which case is the force of friction between the ladder and the.

ACT

A student holding a heavy ball sits on the outer edge a merry go round which is initially rotating counterclockwise. Which way should she throw the ball so that she stops the rotation?

A) To her leftB) To her rightC) Radially outward

top view: initial final

A B

C

Mechanics Lecture 19, Slide 28

Page 29: In the two cases shown below identical ladders are leaning against frictionless walls. In which case is the force of friction between the ladder and the.

ACT

A student is riding on the outside edge of a merry-go-round rotating about a frictionless pivot. She holds a heavy ball at rest in her hand. If she releases the ball, the angular velocity of the merry-go-round will:

A) Increase B) Decrease C) Stay the same

Mechanics Lecture 19, Slide 29

Page 30: In the two cases shown below identical ladders are leaning against frictionless walls. In which case is the force of friction between the ladder and the.

Mechanics Lecture 19, Slide 30

Page 31: In the two cases shown below identical ladders are leaning against frictionless walls. In which case is the force of friction between the ladder and the.

m2

m10

21

1

2diskI m R

Mechanics Lecture 19, Slide 31

odiskinitial IL

Page 32: In the two cases shown below identical ladders are leaning against frictionless walls. In which case is the force of friction between the ladder and the.

m2

m1o

21

1

2diskI m R

Mechanics Lecture 19, Slide 32

2

2

1odiskinitial IK

Page 33: In the two cases shown below identical ladders are leaning against frictionless walls. In which case is the force of friction between the ladder and the.

( )final disk rod fL I I

f

Solve for f

m2

m10

0initial diskL I

22

1

12rodI m Lfinal initialL L

Mechanics Lecture 19, Slide 33

Page 34: In the two cases shown below identical ladders are leaning against frictionless walls. In which case is the force of friction between the ladder and the.

final initialL L

Mechanics Lecture 19, Slide 34

Page 35: In the two cases shown below identical ladders are leaning against frictionless walls. In which case is the force of friction between the ladder and the.

f

M

21( )2final disk rod fK I I

Mechanics Lecture 19, Slide 35

Page 36: In the two cases shown below identical ladders are leaning against frictionless walls. In which case is the force of friction between the ladder and the.

f

M

Just like

for linear momentum

average

PF

t

for angular momentumaverage

L

t

0( )rod rod fL I 0( )rod faverage

I

t

Mechanics Lecture 19, Slide 36

Page 37: In the two cases shown below identical ladders are leaning against frictionless walls. In which case is the force of friction between the ladder and the.

Mechanics Lecture 19, Slide 37

Conservation IdeasConservation Ideas

Internal forces do not change the momentum.Internal forces do not change the momentum.Internal torques do not change the angular Internal torques do not change the angular momentum.momentum.

However, internal forces or internal torques However, internal forces or internal torques can still do work and, thus, can still do work and, thus, change the Kinetic Energy.change the Kinetic Energy.

Page 38: In the two cases shown below identical ladders are leaning against frictionless walls. In which case is the force of friction between the ladder and the.

Heads Up! for next preLectureHeads Up! for next preLecture

Mechanics Lecture 19, Slide 38