announcements capa #11 due this friday at 10 pm reading: finish chapter 8, start chapter 9.1-9.4...

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Announcements • CAPA #11 due this Friday at 10 pm • Reading: Finish Chapter 8, Start Chapter 9.1-9.4 • Section – this week Lab #4: Rotations Midterm Exam #3 on Tuesday November 8 th , 2011 details given in class on Wednesday practice exam and solutions on CULearn

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Page 1: Announcements CAPA #11 due this Friday at 10 pm Reading: Finish Chapter 8, Start Chapter 9.1-9.4 Section – this week Lab #4: Rotations Midterm Exam #3

Announcements• CAPA #11 due this Friday at 10 pm

• Reading: Finish Chapter 8, Start Chapter 9.1-9.4

• Section – this week Lab #4: Rotations

• Midterm Exam #3 on Tuesday November 8th, 2011 details given in class on Wednesday practice exam and solutions on CULearn formula sheet to be posted on web page

• Fraction of all clicker questions answered posted on CULearn. Email me with your clicker ID, name, student ID if you believe it is incorrect.

Page 2: Announcements CAPA #11 due this Friday at 10 pm Reading: Finish Chapter 8, Start Chapter 9.1-9.4 Section – this week Lab #4: Rotations Midterm Exam #3

I miri2

i

Which has a larger moment of inertia?

A) IA > IB B) IA < IB

C) IA = IB D) Impossible to tell.

IB 2mL2 IA 2(2m)L

2

2

mL2

Clicker Question Room Frequency BA

Consider two masses each of size 2m at the ends of a light rod of length L with an axis of rotation through the center of the

rod. The rod is doubled in length and the masses are halved.

Page 3: Announcements CAPA #11 due this Friday at 10 pm Reading: Finish Chapter 8, Start Chapter 9.1-9.4 Section – this week Lab #4: Rotations Midterm Exam #3

A bar has four forces, all of the same magnitude, exerted on it, as shown.

What is the sign of the net torque about the axis of rotation? Use the sign convention shown.

Clicker Question Room Frequency BA

A) torque is zero B) positive (+) C) negative (–)

tnet = + (F)(L) + (F)(L/2) + (F)(L/2) – (F)(L) = +FL

Page 4: Announcements CAPA #11 due this Friday at 10 pm Reading: Finish Chapter 8, Start Chapter 9.1-9.4 Section – this week Lab #4: Rotations Midterm Exam #3

Rotational Kinetic EnergyDoes this object have

translational kinetic energy?

No, zero net translational velocity of the object.

However, there is motion of each piece of the object and thus there must be kinetic energy.

Each piece of the donut has a velocity v = w r.

KE = ½ mv2 = ½ m (w r)2

KE = ½ I w2 Rotational KE

Page 5: Announcements CAPA #11 due this Friday at 10 pm Reading: Finish Chapter 8, Start Chapter 9.1-9.4 Section – this week Lab #4: Rotations Midterm Exam #3

Rolling Kinetic Energy

Translation Rotation

KE (total) = KE (translation) + KE (rotation)

KEtotal = ½ mv2 + ½ I w2

Both pieces in units of Joules.

* Rolling without slipping means v = w r.One revolution Dq=2p leads to displacement of 2pr

Page 6: Announcements CAPA #11 due this Friday at 10 pm Reading: Finish Chapter 8, Start Chapter 9.1-9.4 Section – this week Lab #4: Rotations Midterm Exam #3

KEtot 1

2Mv2

1

2I 2

Isphere 2

5MR2

Ihoop MR2

Idisk 1

2MR2

Which object has the largest total

kinetic energy at the bottom of the ramp?

A) Sphere B) Disk C) Hoop D) All the same.

Clicker Question Room Frequency BA

ffii PEKEPEKE 00 fKEMgH

MgHKE f

M

All have the same total KE.

Page 7: Announcements CAPA #11 due this Friday at 10 pm Reading: Finish Chapter 8, Start Chapter 9.1-9.4 Section – this week Lab #4: Rotations Midterm Exam #3

KEtot 1

2Mv2

1

2I 2

Isphere 2

5MR2

Ihoop MR2

Idisk 1

2MR2

M

ffii PEKEPEKE

02

1

2

10 22

IMvMgH

2

22

2

1

2

1

R

vIMvMgH

22

2

1

R

IMvMgH 2MRI

MgHRv

Page 8: Announcements CAPA #11 due this Friday at 10 pm Reading: Finish Chapter 8, Start Chapter 9.1-9.4 Section – this week Lab #4: Rotations Midterm Exam #3

KEtot 1

2Mv2

1

2I 2

Isphere 2

5MR2

Ihoop MR2

Idisk 1

2MR2

M

2MRI

MgHRv

gHMRMR

MgHRv

7

10

5

2 22

Sphere

Page 9: Announcements CAPA #11 due this Friday at 10 pm Reading: Finish Chapter 8, Start Chapter 9.1-9.4 Section – this week Lab #4: Rotations Midterm Exam #3

KEtot 1

2Mv2

1

2I 2

Isphere 2

5MR2

Ihoop MR2

Idisk 1

2MR2

vsphere 10

7gH

Which has the greater speed at the bottom of the ramp, the sphere that rolls down the ramp or a block of the same mass

that slides down the ramp?

(Assume sliding friction is negligible)

A) Block B) Sphere C) Both the same

Clicker Question Room Frequency BA

ffii PEKEPEKE

02

10 2 MvMgH

Block gHv 2

Page 10: Announcements CAPA #11 due this Friday at 10 pm Reading: Finish Chapter 8, Start Chapter 9.1-9.4 Section – this week Lab #4: Rotations Midterm Exam #3

KEtot 1

2Mv2

1

2I 2

I sphere 2

5MR2

Ihoop MR2

Idisk 1

2MR2

Sphere:

vhoop gH

Hoop:

vdisk 4

3gH

Disk:

Who wins the race to the bottom…… sphere, disk, hoop?

2MRI

MgHRv

gHv7

10

Smallest moment of inertia Iwill have the largest translational

velocity at the bottom.

Page 11: Announcements CAPA #11 due this Friday at 10 pm Reading: Finish Chapter 8, Start Chapter 9.1-9.4 Section – this week Lab #4: Rotations Midterm Exam #3

Demonstration

Page 12: Announcements CAPA #11 due this Friday at 10 pm Reading: Finish Chapter 8, Start Chapter 9.1-9.4 Section – this week Lab #4: Rotations Midterm Exam #3

Which object will go furthest up the incline?

A) Puck B) Disk C) Hoop D) Same height.

Ihoop MR2 , Idisk 1

2MR2

The hoop has the largest moment of inertia, and

therefore thehighest total kinetic

energy.

H

Clicker Question Room Frequency BA

ffii PEKEPEKE

MgHr

vIMv

22

2

1

2

1

Page 13: Announcements CAPA #11 due this Friday at 10 pm Reading: Finish Chapter 8, Start Chapter 9.1-9.4 Section – this week Lab #4: Rotations Midterm Exam #3

Recall:

Momentum p = mv L = Iω Angular momentum

Relation to force F = Δp/Δt τ = ΔL/Δt Relation to torque

No external force Δp = 0 ΔL = 0 No external torque (momentum is conserved) (angular momentum is conserved)

Ii ωi = If ωf

Conservation of Angular Momentum

Li = Lf

Page 14: Announcements CAPA #11 due this Friday at 10 pm Reading: Finish Chapter 8, Start Chapter 9.1-9.4 Section – this week Lab #4: Rotations Midterm Exam #3

L I

Lt

Ii i = I f f

(if Fext 0)

Ii largeωi small

Ii smallωi large

By changing the distribution of mass, the moment of inertia is changed.

By conservation of angular momentum, the angular velocity is therefore modified.

Conservation of L:

finalinitial LL

ffii II

if

if I

I

Page 15: Announcements CAPA #11 due this Friday at 10 pm Reading: Finish Chapter 8, Start Chapter 9.1-9.4 Section – this week Lab #4: Rotations Midterm Exam #3

I1 largeω1 small

By changing the distribution of mass, the moment of inertia is changed.

By conservation of angular momentum, the angular velocity is therefore modified.

Conservation of L:

I2 smallω2 large

I3 largeω3 small

L I

Lt

Ii i = I f f

(if Fext 0)

finalinitial LL

ffii II

if

if I

I

Page 16: Announcements CAPA #11 due this Friday at 10 pm Reading: Finish Chapter 8, Start Chapter 9.1-9.4 Section – this week Lab #4: Rotations Midterm Exam #3

Hoberman Sphere