10/03/2013PHY 113 C Fall 2013-- Lecture 111 PHY 113 C General Physics I 11 AM - 12:15 PM MWF Olin...

PHY 113 C Fall 2013-- Lecture 11 110/03/2013

PHY 113 C General Physics I11 AM - 12:15 PM MWF Olin 101

Plan for Lecture 11:

Chapter 10 – rotational motion

1. Angular variables

2. Rotational energy

3. Moment of inertia

4. Torque


Comments about Exam 1• Solutions posted online• Please review the problems while the ideas are

“fresh” in your mind

iclicker question (serious)In future exams, would you like to have an additional “take” home component? Of course it would be assumed that all portions of the exam would be subject to strict honor code guidelines

A. YesB. No

http://users.wfu.edu/natalie/f13phy113/solutions


Review:iclicker questionIn which of the following cases does gravity do positive work?

yf

yi

yi

yf

b.a.


Review:iclicker questionIn which of the following cases does the work of gravity have a great magnitude?

rf

ri ri

rf

b.a.


2

2

:Define

:mass ofcenter for relation General

dt

dM

mMM

m

CMtotal

ii

ii

iii

CM

rFF

rr

Review of center of mass:

Example:

1 kg 3 kg

1 m

m 0.75m 4

13

:mass kg 1

torelative measured mass ofCenter

CM

ii

iii

CM m

mr

rr


Center of mass example from Webassign

Finding the center of massFor a solid object composed of constant density material, the center of mass is located at the center of the object.

cm ˆ3333.13ˆ6667.11

cm 6

ˆ25215153ˆ25115253

1

ji

ji

rrr

i

ii

ii

iii

CM m

m

m

m


Another Webassign question from Assignment #9

m/skgˆ6109.11ˆ8755.8

m/skgˆ99.3ˆ05.391.2

ji

jivp m


Another Webassign question from Assignment #9

?

final initial

Tf

TfTCfCTiTCiC

ii

ii

mmmm

v

vvvv

pp

Ti

T

CfCiCTf m

mv

vvv


Examples of two-dimensional collision; balls moving on a frictionless surface

smsmsm

v

smsmsm

vvv

smsm

vv

vmvm

vmvmvm

smv

smvkgmm

oof

o

fif

o

ff

ff

ffi

f

i

/11.188.17cos

/060.1

88.17sin

/342.0

88.71 060.1

342.0tan

/060.120cos/1/2

coscos

/342.020sin/1

sinsin

sinsin0

coscos

20 ,/1

,/2 ,06.0 :Suppose

1

o

211

21

2211

221111

o2

121


Angular motion

angular “displacement” q(t)

angular “velocity”

angular “acceleration” dtd

(t)ω

α dtd

(t)θ

ω

“natural” unit == 1 radian

Relation to linear variables: sq = r (qf-qi)

vq = r w

aq = r a

so180 radians 3.14159 radians


Rotation at constant angular velocity w

w

r

vq

vq = r w


Special case of constant angular acceleration: a = a0:

(w t) = wi + a0 t

(q t) = qi + wi t + ½ a0 t2

( (w t))2 = wi2 + 2 a0 ( (q t) - qi )

w

r1v1=r1w

v2=r2w

r2


A wheel is initially rotating at a rate of f=30 rev/sec.

smRv

R

πfω

/247.94)0.5m)(rad/s 495.188(

0.5m? radius aat wheel theof

rim on thedot a of speed theisWhat

rad/s 188.495

rad/s )30(22

elocity?angular v theisWhat

R


A wheel is initially rotating at a rate of f=30 rev/sec. Because of a constant angular deceleration, the wheel comes to rest in 3 seconds.

22

2

/42.31)0.5m)(rad/s 83.62(

0.5m? radius aat wheel theof

rim on thedot a ofon decelerati theisWhat

rad/s 83.26

3

rad/s )30(2

3

20

on?deceleratiangular theisWhat

smRa

R

ss

πf

R


Example: Compact disc motion

In a compact disk, each spot on the disk passes the laser-lens system at a constant linear speed of vq = 1.3 m/s.

w1=vq/r1=56.5 rad/s

w2=vq/r2=22.4 rad/s

What is the average angular acceleration of the CD over the time interval Dt=4473 s as the spot moves from the inner to outer radii?

a = (w2-w1)/Dt =-0.0076 rad/s2

w1

w2


Object rotating with constant angular velocity (a = 0)

w

v=0

v=Rw

Kinetic energy associated with rotation:

“moment of inertia”

iii

iii

iii

rmI

IrmvmK

2

22

1222

122

1

:where

;ωω

R


Moment of inertia: i

iirmI 2

iclicker exercise: Which case has the larger I? A. a B. b


Moment of inertia: i

iirmI 2

22MaI 22 22 mbMaI


Note that the moment of inertia depends on both

a) The position of the rotational axisb) The direction of rotation

m m

d d

I=2md2

m m

d d

I=m(2d)2=4md2


iclicker question:

Suppose each of the following objects each has the same total mass M and outer radius R and each is rotating counter-clockwise at an constant angular velocity of w=3 rad/s. Which object has the greater kinetic energy?

(a) (Solid disk) (b) (circular ring)


Various moments of inertia:

solid cylinder:

I=1/2 MR2

solid sphere:

I=2/5 MR2

solid rod:

I=1/3 MR2

R

RR


Calculation of moment of inertia:

Example -- moment of inertia of solid rod through an axis perpendicular rod and passing through center:

2222

31

22MRdrr

RM

rdrR

MrmI

R

R

R

Riii

R


iclicker exercise:Three round balls, each having a mass M and radius R, start from rest at the top of the incline. After they are released, they roll without slipping down the incline. Which ball will reach the bottom first?

AB C

2MRI A

22 5.02

1MRMRIB

22 4.05

2MRMRIC


22

2

1

2

1

:object rolling

ofenergy kinetic Total

CM

CMrollingtotal

MvI

KKK

CMvRdt

dR

dt

dsdt

d

: thatNote

2

2

22

2

2

1

2

1

2

1

CM

CM

CMrollingtotal

vMR

I

MvRR

I

KKK



AB C

2MRI A

22 5.02

1MRMRIB

22 4.05

2MRMRIC


Review of rotational energy associated with a rigid body

iii

iii

iii

iiirot

rmI

Irm

rmvmK

2

222

22

here w

2

1

2

1

2

1

2

1

:energy Rotational


Note that for a given center of rotation, any solid object has 3 moments of inertia; some times two or more can be equal

j

i

k

iclicker exercise:Which moment of inertia is the smallest? (A) i (B) j (C) k

d dm m

IB=2md2 IC=2md2IA=0


222222 24222223 mkgxmIi

iiyy

From Webassign:


22

2

1

2

1

:object rolling

ofenergy kinetic Total

CM

CMrottotal

MvI

KKK

CMvRdt

dR

dt

dsdt

d

: thatNote

2

2

22

2

2

1

2

1

2

1

CM

CM

CMrottotal

vMR

I

MvRR

I

KKK

CM CM



AB C

2MRI A

22 5.02

1MRMRIB

22 4.05

2MRMRIC


How to make objects rotate.

Define torque:

t = r x F

t = rF sin q

r

F

q

αarτFr

aF

Im

m

sinr

q

F sin q

Note: We will define and use the “vector cross product” next time. For now, we focus on the fact that the direction of the torque determines the direction of rotation.


Another example of torque:


clockwise)(counter

:T from Torque

)(clockwise

:T from Torque

222

2

111

1

TR

TR

clockwise)(counter 52

(1)(5)(0.5)(15)

: torqueTotal

15N 0.5m,

5N 1m, :Example

1122

22

11

Nm .

Nm

TRTR

TR

TR


Newton’s second law applied to rotational motion

dt

dM

dt

dm CM

totali

ii

ii

vF

vF

Newton’s second law applied to center-of-mass motion

mi Fi

ri

dt

dm

dt

dm i

iiiii

iiv

rFrv

F

axis) principalabout rotating(for αω

τ

rωrτ

rωv

Frτ

Idtd

I

dt

dm

total

iiii

ii

iii

i

iidmI 2

di


An example:

A horizontal 800 N merry-go-round is a solid disc of radius 1.50 m and is started from rest by a constant horizontal force of 50 N applied tangentially to the cylinder. Find the kinetic energy of solid cylinder after 3 s.

K = ½ I w2 = = t I a w wi + at = atIn this case I = ½ m R2 and t = FR

R F

Js

N

N

mg

tFg

RI

tFt

I

FRIIK

Rg

mgIt

I

FRtIFR

625.275)3(800

50m/s8.9

/2

1

2

1

2

1

2

1

22

222

2

2222

2


Re-examination of “Atwood’s” machine

T1

T2

T2T1

IT1-m1g = m1a

T2-m2g = -m2a

t =T2R – T1R = I a = I a/R

R

212

12

212

12

/RIg

τ

/g a

RImm

mm

RImm

mm


Another example: Two masses connect by a frictionless pulley having moment of inertia I and radius R, are initially separated by h=3m. What is the velocity v=v2= -v1 when the masses are at the same height? m1=2kg; m2=1kg; I=1kg m2 ; R=0.2m .

m1 m2v1 v2

hh/2

sm

RImm

mmv

hgmhgm

vvmvmghm

UKUK

RI

ffii

/19.02.0/112

12

/

0

:energy ofon Conservati

2

221

21

21

221

1

2212

2212

121

1 2



i

iirot rmIIK 2221

Distance to axis of rotation

Rolling: rotcomtot KKK

2

221 1

:slipping no is thereIf

comtot

com

vMR

IMK

Rv

Rolling motion reconsidered:


Newton’s law for torque:

αω

τ Idtd

Itotal

F

fs

FfMRI

IMRFf

I

RfaRIaIRf

MafF

s

s

sCMCMs

CMs

312

21

2

2

cylinder, solid aFor

/1

1

/

Note that rolling motion is caused by the torque of friction:


Bicycle or automobile wheel:

t

fs

τ/Rf

MRI

I/MR

τ/Rf

RIaIf

Maf

s

s

CMs

CMs

2

1

For

1

/αR-τ

2

2


iclicker exercise:What happens when the bicycle skids?

A. Too much torque is appliedB. Too little torque is appliedC. The coefficient of kinetic friction is too smallD. The coefficient of static friction is too smallE. More than one of these

10/03/2013PHY 113 C Fall 2013-- Lecture 111 PHY 113 C General Physics I 11 AM - 12:15 PM MWF Olin...

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Transcript of 10/03/2013PHY 113 C Fall 2013-- Lecture 111 PHY 113 C General Physics I 11 AM - 12:15 PM MWF Olin...