10/12/2012phy 113 a fall 2012 -- lecture 181 phy 113 a general physics i 9-9:50 am mwf olin 101 plan...

PHY 113 A Fall 2012 -- Lecture 18 110/12/2012

PHY 113 A General Physics I9-9:50 AM MWF Olin 101

Plan for Lecture 18:

Chapter 10 – rotational motion

1. Torque

2. Conservation of energy including both translational and rotational motion


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


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


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


Kinetic energy associated with rotation:

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

10/12/2012phy 113 a fall 2012 -- lecture 181 phy 113 a general physics i 9-9:50 am mwf olin 101 plan...

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