Co

pyri

gh

t R

. J

an

ow

–S

pri

ng

2012

Physics 111 Lecture 10

Torque, Energy, Rolling

SJ 8th Ed.: Chap 10.6 –10.9

•Recap and Overview

•Torque

•Newton’s Second Law for Rotation

•Energy Considerations in Rotational Motion

•Rolling

•Energy Methods

•Second Law Applications

10

.6T

orq

ue

10

.7T

he

Rig

id B

od

y U

nd

er

a N

et

To

rqu

e

(Ro

tati

on

al

Sec

on

d L

aw

)

10

.8E

ne

rgy C

on

sid

era

tio

ns

in

Ro

tati

on

al

Mo

tio

n

(Wo

rk,

Po

we

r, a

nd

En

erg

y)

10

.9R

oll

ing

Mo

tio

n o

f a

Rig

id O

bje

ct

(co

ve

rs

ba

sic

s, en

erg

y m

eth

od

s,

2n

d l

aw

pro

ble

ms

)

Add

exa

mp

le 1

0.8

, 10

.11

, 10

.12

Co

pyri

gh

t R

. J

an

ow

–S

pri

ng

2012

Mom

ent

of

Ine

rtia

I

measu

res B

OT

H m

ass a

nd

its

dis

trib

uti

on

dep

en

ds o

n t

he c

ho

ice o

f ro

tati

on

axis

r pis

perp

en

dic

ula

r d

ista

nce f

rom

mass p

oin

t to

axis

masses o

naxis

have r

p=

0, n

o c

on

trib

uti

on

to

I

m

as

se

s

po

int

fo

r2

i,i

P

rm

I ∑ ∑∑∑

⊥ ⊥⊥⊥= ===

sam

e m

ass

small I

larg

er

I

“M

om

en

t o

f in

ert

ia”

pla

ys t

he r

ole

of

mass

“T

orq

ue p

lays t

he r

ole

of

forc

e

Rot

ati

onal 2

ndLaw:

I

ne

tα ααα

= ===τ τττ

ma

F net= ===

Ro

tati

on

al w

ork

:

Po

wer

used

wh

en

d

oin

g r

ota

tio

nal w

ork

:

θ θθθ∆ ∆∆∆τ τττ

= ===∆ ∆∆∆

rot

Wx

FW

∆ ∆∆∆= ===

∆ ∆∆∆

Pt

W

rot

rot

τω τωτω

τω= ===

≡ ≡≡≡∆ ∆∆∆

∆ ∆∆∆v

FP

o

tW= ===

≡ ≡≡≡∆ ∆∆∆∆ ∆∆∆

Tor

que τ τττ

TW

IST

-m

easu

res a

pp

lied

fo

rce A

ND

wh

ere

it

acts

MO

ME

NT

AR

M(l

ever

arm

) r p

= p

erp

en

dic

ula

r d

ista

nce

fro

m a

xis

to

lin

e o

f acti

on

of

forc

e

Fr

)sin

(rF

arm

le

ver

x

forc

e

× ×××= ===

θ θθθ= ===

= ===τ τττ

F

r p

r

θF

r pr

θ

Key Angular Dynamics Concepts

bo

die

s

co

nti

nu

ou

s

for

P

dm

xd

mr

I

∫ ∫∫∫∫ ∫∫∫

= ===≡ ≡≡≡

⊥ ⊥⊥⊥

22

the

ore

m

ax

is

pa

rall

el

c

mP

M

h

I

I2

+ +++= ===

Co

pyri

gh

t R

. J

an

ow

–S

pri

ng

2012

Torque and Moment arm (lever arm)

•T

he lever

arm

d =

r s

in(φ φφφ

), is

the p

erp

en

dic

ula

r d

ista

nce

fro

m t

he a

xis

of

rota

tio

n t

o a

lin

e d

raw

n a

lon

g t

he

dir

ecti

on

of

the f

orc

e

•T

orq

ue τ τττ

measu

res t

he e

ffect

of

a f

orc

e a

pp

lied

at

a p

oin

t

as it

cau

ses r

ota

tio

n a

bo

ut

so

me a

xis

�T

orq

ue i

s a

ve

cto

r, m

ag

nit

ud

e i

s:

τ τττ=

r F

sin

φ φφφ=

F d

�F

is t

he f

orc

e

�φ φφφ

is t

he a

ng

le t

he f

orc

e m

ak

es w

ith

th

e d

isp

lacem

en

t fr

om

th

e

rota

tio

n a

xis

�d

is t

he m

om

en

t arm

(or

lever

arm

) o

f th

e f

orc

e

•D

irecti

on

: c

ou

nte

rclo

ckw

ise � ���

torq

ue is p

osit

ive

clo

ckw

ise � ���

torq

ue is n

eg

ati

ve

Co

pyri

gh

t R

. J

an

ow

–S

pri

ng

2012

Torque in 3D –Force applied at a point � ���a vector

x

y

z

φ φφφ

⊥ ⊥⊥⊥r

⊥ ⊥⊥⊥F

rad

F

r

φ φφφ

line

of

acti

on

of F

F

P

φ φφφ

90

o

90

o

Fr

)sin

(F r

arm

mo

men

t

x fo

rce

× ×××

→ →→→φ φφφ

= ==== ===

τ τττr

Axis of rotation along z, through P

Moment arm = r sin(

φ φφφ)

Torque Frsin(

φ φφφ) (CCW about z)

Transverse component of the

force = F sin(

φ φφφ)

Moment arm is simply r

Torque is still Frsin(

φ φφφ)

The torque is the same if F is applied

anywhere along its line of action

Vector picture: torque vector is

perpendicular to plane of r and F,

(along z-axis in sketch, CCW sense)

⊥ ⊥⊥⊥r

⊥ ⊥⊥⊥F

Tor

que is

a v

ect

or p

erp

end

icul

ar

to

the p

lane

con

tain

ing

rand

F

Tor

ques

abou

t th

e s

am

e a

xis

add (

supe

rpos

itio

n)

Co

pyri

gh

t R

. J

an

ow

–S

pri

ng

2012

10.1

A c

on

sta

nt

no

n-z

ero

net

torq

ue i

s e

xert

ed

on

a w

heel.

W

hic

h o

f

the f

ollo

win

g q

uan

titi

es m

ust

be c

han

gin

g?

Effect of a Constant Net Torque

1.

an

gu

lar

po

sit

ion

2.

an

gu

lar

velo

cit

y

3.

an

gu

lar

accele

rati

on

4.

mo

men

t o

f in

ert

ia

5.

kin

eti

c e

nerg

y

6.

the m

ass c

en

ter

locati

on

A.

1, 2, 3

B.

4, 5, 6

C.

1,2, 5

D.

1, 2, 3, 4

E.

2, 3, 5

Co

pyri

gh

t R

. J

an

ow

–S

pri

ng

2012

Exam

ples:

Eva

luati

ng t

orqu

es

•M

assle

ss,

ho

rizo

nta

l b

ar:

φ φφφ

= 9

0o

F =

mg

l

φ φφφ•

Mo

men

t arm

= l

sin

(90

o)

= l

•T

orq

ue τ τττ

= -

mgl

clo

ckw

ise

CCW

� ���+

CW

� ���-

•B

ar

as a

bo

ve b

ut

NO

T h

ori

zo

nta

l

•M

om

en

t arm

= l

sin

(φ φφφ)

•T

orq

ue τ τττ

= +

m g

lsin

(φ φφφ)

C

CW

F =

mg

l

φ φφφ

φ φφφ

mo

men

t arm

Fre

e B

ody D

iagr

am

To

rqu

es c

an

ad

d (

they m

ust

refe

r to

th

e s

am

e r

ota

tio

n a

xis

(S

up

erp

osit

ion

)

m1

m2

W1=

m1g

W2=

m2g

NL

2L

1

•H

ori

zo

nta

l see s

aw

, u

neq

ual arm

s

•C

alc

ula

te t

orq

ues a

rou

nd

piv

ot

po

int

Nn

et

τ τττ+ +++

τ τττ+ +++

τ τττ= ===

τ τττ2

1

•M

om

en

t arm

= 0

fo

r fo

rce

N a

bo

ut

piv

ot

L g

m

L g

m

2

2 1

1n

et

− −−−+ +++

= ===τ τττ

∴ ∴∴∴

Co

pyri

gh

t R

. J

an

ow

–S

pri

ng

2012

Rotational Second Law

F rp

r

θ

FT

axis

m

rF

T

net

to

rqu

e= ===

τ τττ

m

a

F

TT

= ===

n)

(defi

nit

io

T

ra

α ααα≡ ≡≡≡

α ααα= ===

= ===τ τττ

∴ ∴∴∴2

mr

rm

a

Tn

et

inert

ia)

l(r

ota

tio

na

m

rI

2≡ ≡≡≡

Inf

er

2nd

law f

or a

for

ce a

ctin

g on

a s

ingl

e p

oint

mass

:

2nd

Law E

xam

ple:

Fin

d α

α

α

α

& a

T.

For

ce F

act

s at

the r

im o

f a r

igid

dis

k.

F rp

r

θFT

axis

inert

ia)

l(r

ota

tio

na

m

rI

2

21≡ ≡≡≡ α ααα

= ===α ααα

= ==== ===

τ τττ2

21m

rI

rF

T

net

Law

S

eco

nd

p

ly

A

p

mr

FT

ion

Ac

ce

lera

t A

ng

ula

rR

es

ult

ing

2= ===

α ααα

mFr

aT

T

:ri

m

of

on

accele

rati

(l

inear)

T

an

gen

tial

2= ===

α ααα= ===

“2”

com

es

from

I

Mom

ent

of

inert

ia

α ααα= ===

τ τττ∴ ∴∴∴

I

ne

tα ααα

= ===τ τττ

rr

tot

net

IA

PPLIES A

LSO

TO

RIG

ID B

ODIES

AXIS F

IXED

Co

pyri

gh

t R

. J

an

ow

–S

pri

ng

2012

Exam

ple:

Ang

ular

acc

ele

rati

on a

for

an

unbala

nced b

ar

Bar

is m

assle

ss

an

d o

rig

ina

lly h

ori

zo

nta

l

Ro

tati

on

ax

is t

hro

ug

h f

ulc

rum

po

int

� ���N

has

zero

to

rqu

eF

ind

an

gu

lar

ac

cele

rati

on

of

ba

r an

d t

he l

inear

accele

rati

on

of

m1 ju

st

aft

er

yo

u le

t g

o

N

m1g

m2g

L2

L1

fulc

rum

+y

a1

= -

α αααL

1

a2

= +

α αααL

2

Co

nstr

ain

ts:

tot

net

tot

net

I

I

τ τττ= ===

α ααα⇒ ⇒⇒⇒

α ααα= ===

τ τττU

se:

2 22

2 11

21

tot

Lm

Lm

I

I

I

+ +++= ===

+ +++= ===

21

gL

m g

Lm

2

1i,

on

et

− −−−+ +++

= ===τ τττ

= ===τ τττ

∑ ∑∑∑

wh

ere

:

Where

did

sin

(θ θθθ)

fact

ors

in m

oment

arm

go?

net

torq

ue

tota

l I

abou

t

pivo

t

L

m

L

m

gL

m g

Lm

2 22

2 11

21

+ +++− −−−= ===

α ααα2

1

What

if b

ar

is n

ot h

oriz

onta

l?

Clo

ckwis

e

Acc

ele

rate

s U

P

Eva

luati

on:

Let

m1

= m

2=

m

L1=

20 c

m,

L2

= 8

0 c

m

2

22

2 22 1

rad

/s

8.6

5

0.8

0.2

0.8

)-

g(0

.2

L

L

gL

g

L

− −−−= ===

+ +++= ===

+ +++− −−−= ===

α ααα2

1

21

m/s

1

.7

L

a

+ +++= ===

α ααα= ===

1

Doe

s ans

wer

chang

e if

axis

chang

es?

Co

pyri

gh

t R

. J

an

ow

–S

pri

ng

2012

�10.2

. S

up

po

se

every

thin

g i

s a

s i

t w

as in

th

e p

reced

ing

exam

ple

,b

ut

the b

ar

is

NO

T h

ori

zo

nta

l.

As

su

me b

oth

masses a

re e

qu

al.

Wh

ich

of

the f

oll

ow

ing

is t

he

co

rrect

eq

ua

tio

n f

or

the

an

gu

lar

accele

rati

on

?

See Saw

N

mg

L2

L1

fulc

rum

mg

θ θθθ α ααα= ===

τ τττto

tn

et

I

g L

L

)L

L(

A)

2 22 1

+ +++− −−−= ===

α ααα2

1

)sin

(g

L

L

)L

L(

C)

2 22 1

θ θθθ+ +++− −−−

= ===α ααα

21

)co

s(

g L

L

)L

L(

D)

2 2

2 1

θ θθθ+ +++− −−−

= ===α ααα

21

)sin

(g

L

L

)L

L(

E)

2 2

2 1θ θθθ

+ +++− −−−= ===

α ααα2

1

)co

s(

g

L

L

)L

L(

B)

2 2

2 1θ θθθ

+ +++− −−−= ===

α ααα2

1

Co

pyri

gh

t R

. J

an

ow

–S

pri

ng

2012

�10.2

. S

up

po

se

every

thin

g i

s a

s i

t w

as in

th

e p

reced

ing

exam

ple

,b

ut

the b

ar

is

NO

T h

ori

zo

nta

l.

As

su

me b

oth

masses a

re e

qu

al.

Wh

ich

of

the f

oll

ow

ing

is t

he

co

rrect

eq

ua

tio

n f

or

the

an

gu

lar

accele

rati

on

?

See Saw -Solution

N

m2g

L2

L1

fulc

rum

m1g

θ θθθ

tot

ne

tI/

τ τττ= ===

α ααα

2 22

2 11

21

tot

Lm

Lm

I

I

I

+ +++= ===

+ +++= ===

)co

s(

gL

m )

co

s(

gL

m

21

i,o

net

θ θθθ− −−−

θ θθθ+ +++

= ===τ τττ

= ===τ τττ

∑ ∑∑∑2

1

)co

s(

g

Lm

Lm

]L

m L

m[

2 2

22 1

1

2θ θθθ

+ +++− −−−= ===

α ααα2

11

For

equ

al m

ass

es

m1

= m

2= m )

gc

os

(

L

L

)L

L(

2 2

2 1

θ θθθ+ +++− −−−

= ===α ααα

∴ ∴∴∴2

1If

bar

is b

ala

nced w

hen

it’s

hor

izon

tal, d

oes

that

chang

e

when

it is

not

hor

izon

tal?

Co

pyri

gh

t R

. J

an

ow

–S

pri

ng

2012

Method for solving Second Law problems

Syst

em

s wit

h m

any

com

pone

nts

have

many

unk

nowns

….

…ne

ed a

n equ

al nu

mber

of ind

epe

ndent

equ

ati

ons

Meth

od:

•Dra

w o

r sk

etc

h s

yst

em

. A

dop

t co

ordin

ate

s, n

am

e t

he v

ari

able

s, ind

icate

rota

tion

axes,

list

the k

nown

and

unk

nown

quant

itie

s, …

•Dra

w f

ree b

ody d

iagr

am

s of

key p

art

s.

Show

for

ces

at

their

poi

nts

ofapp

lica

tion

. W

eig

hts

of

bod

ies

act

as

if t

hey a

re a

t th

e m

ass

cent

er

•Calc

ulate

tor

ques

abou

t a (

com

mon

) axis

.•

May n

eed t

o app

ly b

oth f

orm

s of

Seco

nd L

aw t

o each

part

�Tra

nsla

tion

:

�Rot

ati

on:

•Gene

rate

equ

ati

ons

usin

g Seco

nd L

aw.

•There

may b

e c

onst

rain

t equ

ati

ons

(extr

a c

ondit

ions

con

nect

ing

unkno

wns

).•

Make s

ure t

here

are

eno

ugh (

N)

equ

ati

ons.

•

Sim

plif

y a

nd s

olve

the s

et

of (

sim

ulta

neou

s) e

quati

ons.

•Int

erp

ret

the f

inal fo

rmul

as.

Do

they m

ake int

uiti

ve s

ens

e? R

efe

r back

to

the s

ketc

hes

and

ori

gina

l pr

oble

m

•Calc

ulate

num

eri

cal re

sult

s, a

nd s

ani

ty c

heck

anw

ers

(e.g

., r

ight

order

of

magn

itud

e?)

am

FF

in

et

rr

= ==== ===∑ ∑∑∑

Ii

ne

tα ααα

= ===τ τττ

= ===τ τττ

∑ ∑∑∑r

rr

No

te:

can

ha

ve

Fn

et.e

q.

0

bu

t τ τττ

ne

t.n

e. 0

Co

pyri

gh

t R

. J

an

ow

–S

pri

ng

2012

Example: Angular Acceleration of a W

heel

Standard approach:

�B

reak in

to t

wo

su

b-s

ys

tem

s�

wh

eel is

accele

rate

d a

ng

ula

rly b

y t

en

sio

n T

�b

lock is a

ccele

rate

d lin

earl

y b

y w

eig

ht

mg

, w

ith

ten

sio

n

op

po

sin

g

�D

raw

fre

e b

od

y d

iag

ram

s

�T

he w

heel is

ro

tati

ng

an

d s

o w

e a

pp

ly

Σ ΣΣΣτ τττ

=

=

=

= Ι

α ΙαΙαΙα�

Th

e t

en

sio

n (

tan

gen

tial)

su

pp

lies t

he t

orq

ue

�T

he m

ass m

m

oves

in

a s

tra

igh

t li

ne,

so

ap

ply

N

ew

ton

’s S

eco

nd

Law

Σ ΣΣΣF

y=

ma

y=

mg

-T

�H

ow

to

co

nn

ec

t th

e t

wo

pro

ble

ms a

bo

ve?

�A

co

nstr

ain

t lin

ks lin

ear

accele

rati

on

to

α ααα

a =

α αααr

r

a

mg

Co

pyri

gh

t R

. J

an

ow

–S

pri

ng

2012

FBD f

or m

ass

m:

T mg

y

Tm

gm

aF

y− −−−

= ==== ===

∑ ∑∑∑ )

ag(

m

T

− −−−= ===

Un

kn

ow

ns:

T, a

FBD f

or d

isk,

wit

h a

xis

at

“o”:

N Mg

T

2

21M

rI

= ===α ααα

= ===+ +++

= ===τ τττ

∑ ∑∑∑I

T

r

0

M

r

r)a

g(m

ITr

2

21

− −−−= ===

= ===α ααα

Elim

ina

ted

T

Un

kn

ow

ns:

a,

α ααα

So

far:

2 E

quati

ons,

3 u

nkno

wns

� ���N

eed a

con

stra

int:

r

a

α ααα+ +++

= ===fr

om

“n

o

slip

pin

g”

assu

mp

tio

nSub

stit

ute a

nd s

olve

:

α αααpr

opor

tion

al

to g

M

r

2m

g)

Mm(

= ===

+ +++α ααα

21

r

M

r2m

- M

r

2m

gr

2

2

2α ααα

= ===α ααα

)(

M

/2)r

(m

mg

r

2ra

d/s

2

42

= ===+ +++

= ===α ααα

)(

M

/2)

(m

mg

a

2m

/s

4.8

= ===

+ +++= ===

•C

ord

doe

s no

t sl

ip o

r st

retc

h (

cons

trai

nt)

•D

isk’s

rot

atio

nal

inert

ia s

low

s ac

cele

rati

ons

•Let

m =

1.2

kg,

M =

2.5

kg,

r =

0.2

m•

Fin

d a

ccele

rati

on o

f m

ass

m,

find

α αααfo

r dis

k

r

a

mg

supp

ort

forc

e N

at

axis

“O

”has

zero

tor

que

N

Meth

od u

sing

the S

eco

nd L

aw

Co

pyri

gh

t R

. J

an

ow

–S

pri

ng

2012

Sam

e p

roble

m –

usin

g sy

stem

poin

t of

vie

w (

caut

ion)

Defi

ne t

he s

yst

em

(dash

ed lin

e)

•ig

nore

int

ern

al to

rque

s & f

orce

s (e

.g.,

Tens

ions

)•

dete

rmin

e t

otal sy

stem

rota

tion

al in

ert

ia (

abou

t “O

”)

y

fo

rces

n

ox

gm

Mg

N

F

F

+ +++= ===

⇒ ⇒⇒⇒= ==== ===

∑ ∑∑∑∑ ∑∑∑00

App

ly lin

ear

Seco

nd L

aw (

no n

ew inf

o):

App

ly r

otati

onal Seco

nd L

aw t

o sy

stem

:

22

21 m

as

sp

ul

sys

rm

Mr

II

I+ +++

= ===+ +++

= ===

I sys

ex

t,

ne

tα ααα

= ===τ τττ

m

gr

mass

ext

,n

et

+ +++= ===

τ τττ= ===

τ τττ

m

rM

r

gr

m

212

2+ +++

= ===α ααα

m

M

gm

r

a

21T

+ +++= ===

α ααα= ===

r

a

mg

Mg

N

M

r

21I

2

pu

l= ===

Co

pyri

gh

t R

. J

an

ow

–S

pri

ng

2012

Modifications for Rotating Bodies…

Add R

otati

onal Kin

eti

c Ene

rgy

K

KK

K

KK

rot

cm

...s

o...

rot

cm

∆ ∆∆∆+ +++

∆ ∆∆∆= ===

∆ ∆∆∆+ +++

= ===

Wor

k-KE T

heor

em

is

the s

am

e a

s befo

re b

ut K

E a

nd W

ork inc

lude r

otati

on

2 021

2

21ω ωωω

− −−−ω ωωω

= ===∆ ∆∆∆

∑ ∑∑∑∑ ∑∑∑

II

Kf

rot

Tor

ques

do

rota

tion

al wor

k e

ven

when

zero

wor

k is

don

e o

n th

e m

ass

cent

er:

∫ ∫∫∫θ θθθ

τ τττ= ===

θ θθθτ τττ

= ===∆ ∆∆∆

lim

its

rot

dis

tan

ce

)

an

gu

lar

x

(t

orq

ue

ro

td

W

d

dW

hm

gU

gra

vit

y

e.g

.,

;d

iffe

ren

ce

en

erg

y

Po

ten

tial

U

∆ ∆∆∆

= ===∆ ∆∆∆

≡ ≡≡≡∆ ∆∆∆

E

nerg

y

Mech

an

ical

U

K

E

me

ch

+ +++

≡ ≡≡≡

vF

dts

dF

dt

dW

Po

wer

ro

rr

or

= ==== ===

≡ ≡≡≡

Lin

ear

Vers

ion

of M

ech

ani

cal Ene

rgy C

onse

rvati

on

on

ly

ve

co

nserv

ati

-n

on

mech

W

U

KE

∆ ∆∆∆= ===

∆ ∆∆∆+ +++

∆ ∆∆∆≡ ≡≡≡

∆ ∆∆∆

Work & Energy for Linear and Rotational Motion…

rota

tio

n

no

o

nly

,

mass

po

int

mv

K

:

en

erg

y

Kin

eti

c2

21= ===

∫ ∫∫∫= ===

≡ ≡≡≡∆ ∆∆∆

path

te...

...in

teg

ra

sd

FW

sd

Fd

W

:W

ork

ro

rr

or

Lin

ear

Vers

ion

of W

ork-

Kin

eti

c Ene

rgy T

heor

em

)p

ote

nti

als

(i

nc

lud

ing

fo

rce

s

ex

tern

al

b

y

do

ne

...w

ork

WK

∆ ∆∆∆= ===

∆ ∆∆∆

Co

pyri

gh

t R

. J

an

ow

–S

pri

ng

2012

Work Done by Pure Rotation

θ θθθφ φφφ

= ===

φ φφφ= ===≡ ≡≡≡

)drs

in(

F

)ds

-co

s(9

0

F

s

dF

W

d

n

tran

sla

tio

fo

r

as

ro

r

App

ly f

orce

F t

o m

ass

at

poin

t r,

caus

ing

rota

tion

-on

lyabou

t axis

.

Dis

place

ment

is

only

alo

ng θ θθθ

(tra

nsve

rse)

dir

ect

ion

�O

nly

th

e t

ran

sve

rse c

om

po

nen

t o

f

the f

orc

e (

alo

ng

th

e d

isp

lacem

en

t)

do

es w

ork

–th

e s

am

e c

om

po

nen

t th

at

co

ntr

ibu

tes t

o t

he t

orq

ue:

θ θθθτ τττ

= ===∴ ∴∴∴

d

W

d

)rs

in(

F

φ φφφ= ===

τ τττ

�T

he r

ad

ial

co

mp

on

en

t o

f th

e f

orc

e

do

es n

o w

ork

becau

se i

t is

perp

en

dic

ula

r to

th

e d

isp

lacem

en

t

Fr

θ θθθ= ===

dr

sd

�H

ow

mu

ch

wo

rk d

oes f

orc

e

do

as i

t ro

tate

s

thro

ug

h a

n in

fin

itesim

al

dis

tan

ce

Co

pyri

gh

t R

. J

an

ow

–S

pri

ng

2012

Instantaneous Power:

τω τωτωτω= ===

θ θθθτ τττ

= ===≡ ≡≡≡

dt

d

dt

dW

P

Div

ide b

oth s

ides

of d

W= τ τττ

dθ θθθ

by d

t

Watt

s746

hp

1≡ ≡≡≡

W

att

s10

x

1.4

9

hp

1 Watt

s746

h

p

200

hp

200

P5

= ===

= ===

= ===

Co

nve

rt t

he a

ng

ula

r velo

cit

y ω

ω

ω

ω

to

rad

/s:

rad

/s

628

s 60m

in 1

re

v 1

rad

2

re

v/m

in

6000

rev/m

in

6000

= ===

π πππ

= ==== ===

ω ωωω

τω τωτωτω= ===

P

Ap

ply

an

d s

olv

e f

or

the t

orq

ue:

N.m

237

rad

/s

628

N.m

/s

10

x

1.4

9

P

5

= ==== ===

ω ωωω= ===

τ τττ

Exam

ple:

The p

ower

outp

ut o

f a c

ert

ain

car

is 2

00 h

p at

6000 r

pm.

What

is t

he c

orre

spon

din

g to

rque

?

Co

nve

rt t

he p

ow

er

to W

att

s:

Co

pyri

gh

t R

. J

an

ow

–S

pri

ng

2012

Exam

ple:

An

ele

ctri

c m

otor

att

ach

ed t

o a g

rind

ston

e e

xert

s a c

onst

ant

to

rque

of

τ τττ= 1

0 N

.m.

The s

yst

em

sta

rts

from

rest

. The m

oment

of

inert

ia o

f th

e g

rind

ston

e is

I =

2.0

kg.

m2.

a)

Fin

d t

he k

ineti

c ene

rgy a

fter

8 s

eco

nds.

b)

Fin

d t

he w

ork d

one b

y t

he m

otor

dur

ing

this

tim

e.

c) F

ind t

he a

vera

ge p

ower

delive

red b

y t

he m

otor

.

a)

α ααα= ===

τ τττI

20

50

20

10

s/ra

d

.

.

/.

I/

= ===

= ===τ τττ

= ===α ααα

s/ra

d

.x

.

t

f40

08

05

0= ===

= ===α ααα

+ +++ω ωωω

= ===ω ωωω

J.

6

00

).

( x

.

x

I

K

K21

f21

f1

040

02

22

= ==== ===

ω ωωω= ===

= ===∆ ∆∆∆

Sh

ou

ld e

qu

al w

ork

∆ ∆∆∆W

do

ne b

y m

oto

r

b)

Fir

st,

an

gle

tu

rned

th

rou

gh

:

rad

(8.0

)

x 5.0

x

t

t

2

16

021

2

21

0= ===

= ===α ααα

+ +++ω ωωω

= ===θ θθθ

∆ ∆∆∆

∫ ∫∫∫θ θθθ θ θθθ

= ==== ===

θ θθθ∆ ∆∆∆τ τττ

= ===θ θθθ

τ τττ= ===

∆ ∆∆∆f i

J.

1600

160

x

10.0

d

W

Defi

nit

ion

of

wo

rk, co

nsta

nt

torq

ue

c)

Ave

rag

e p

ow

er:

.W

att

s

s.

8

J.

1600

tW

P

av

200

= ==== ===

∆ ∆∆∆∆ ∆∆∆= ===

bu

t in

sta

nta

neo

us

po

wer

is n

ot

co

nsta

nt:

τω τωτωτω≡ ≡≡≡

P

0 t

at

W

.0

= ===

= ===

s.

8

t at

W

.400

= ==== ===

s.

4

t

at

W

.200

= ===

= ==={

Co

pyri

gh

t R

. J

an

ow

–S

pri

ng

2012

Summary:

Pure Translation (Fixed Direction)

Pure Rotation (Fixed Axis)

Position coordinate x

Angular position

θ θθθ

Velocity v= dx/dy

Angular velocity

ω

ω

ω

ω = d

θ θθθ/dt

Acceleration a = dv/dt

Angular acceleration α ααα= d

ω ωωω/dt

Mass m

Rotational inertia I

Newton's second law Fnet= ma

Newton's second law τ τττnet= I

α ααα

Work dW

= Fdx

Work dW

= τ

τ τ τ d

θ θθθ

Kinetic energy Kcm= (m/2)v2

Kinetic energy Krot= (I/2)

ω ωωω2

Power (constant force) P= F.v

Power (constant torque) P= τ

ωτωτω

τω

Work–KE theorem

∆ ∆∆∆W = ∆ ∆∆∆K

Same, include K

rot

∆ ∆∆∆W

= ∆ ∆∆∆K

Ene

rgy C

onse

rvati

on:

for

trans

lati

on +

rot

ati

on a

bou

t m

ass

cent

er

(cm

)

W

KK

Kc

mro

tto

t∆ ∆∆∆

= ===∆ ∆∆∆

+ +++∆ ∆∆∆

≡ ≡≡≡∆ ∆∆∆

∆ ∆∆∆W includes both conservative and

non-conservative forces, treated as

external to system

OR

W

UK

En

cto

tm

ec

h∆ ∆∆∆

= ===∆ ∆∆∆

+ +++∆ ∆∆∆

≡ ≡≡≡∆ ∆∆∆

∆ ∆∆∆W

ncincludes only non-conservative

forces,

∆ ∆∆∆U contains the conservative

forces

Co

pyri

gh

t R

. J

an

ow

–S

pri

ng

2012

Rolling:A wheel rolls without slipping on a table

•T

he g

reen

lin

e a

bo

ve i

s t

he p

ath

of

the m

ass

cen

ter

of

a w

heel.

•T

he r

ed

cu

rve

sh

ow

s t

he p

ath

(called

a c

yclo

id)

sw

ep

t o

ut

by a

po

int

on

th

e r

im o

f th

e w

heel.

•W

hen

th

ere

is n

o s

lip

pin

g,

the r

ela

tio

nsh

ips

betw

een

th

e t

ran

sla

tio

nal

(mass c

en

ter)

an

d

rota

tio

nal

mo

tio

n a

re s

imp

le:

Ra

Rv

R s

cm

cm

α ααα= ===

ω ωωω= ===

θ θθθ= ===

ω ωωω

vcm

Co

pyri

gh

t R

. J

an

ow

–S

pri

ng

2012

Rolling = pure rotation around CM + pure translation of CM

•T

he b

ott

om

po

int

“P

”o

f th

e w

heel is

sta

tio

nary

(n

o s

lip

pin

g)

•T

he t

op

po

int

“T

”is

mo

vin

g a

t sp

eed

2v

co

m,

(fa

ste

st)

a)

Pu

re r

ota

tio

n

Som

eon

e m

ovin

g w

ith

th

e C

M o

f th

e w

heel

sees

tang

ent

ial

speed

v t

ang

= ω ωωω

r= v

com

for

no s

lipp

ing

(rim

)

As

seen

from

th

e “

lab

”w

ith

ro

tati

on t

urne

d o

ff,

all

poin

ts o

n th

e w

heel

wou

ld

mov

e a

t s

peed

vco

m

b)

Pu

re t

ran

sla

tio

n+

Tot

al v

elo

city

for

a p

oint

on t

he r

im

co

mg

tan

lab

vv

vr

rr

+ +++= ===

c)

Ro

llin

g m

oti

on

=

co

mla

b:

ce

nte

r

At

vv

+ +++

= ===0

0

vv

v c

om

co

mla

b :

bo

tto

m

At

= ===+ +++

− −−−= ===

co

mc

om

co

mla

b:

top

A

tv

vv

v

2= ===

+ +++= ===

Co

pyri

gh

t R

. J

an

ow

–S

pri

ng

2012

Stationary observer sees ROLLING as pure rotation about

contact point P, which is constantly changing

•Sna

pshot

in

tim

e:

Com

plem

ent

ary

vie

w t

o pr

evi

ous

•P

is t

he “

inst

ant

ane

ous

cent

er

of r

otati

on”

R

Rv

Pc

mω ωωω

= ===

cm

Pω ωωω

= ===ω ωωω

cm

pα ααα

= ===α ααα

P

ω ωωωP

Rv

Pg

tan

ω ωωω= ===

2

φ φφφ

φ φφφ

)co

s(

Rv

PA

ϕ ϕϕϕω ωωω

= ===2

AR

vP

cm

ω ωωω= ===

0= ===

gta

nv

Ang

ular

velo

city

and

acc

ele

rati

on a

re t

he s

am

e a

bou

t co

ntact

po

int

“P”

or a

bou

t CM

.

Co

pyri

gh

t R

. J

an

ow

–S

pri

ng

2012

Exam

ple:

Fin

d t

he s

peed o

f th

e b

owling

ball a

s it

rol

ls w

/o s

lipp

ing

to

the b

otto

m o

f th

e r

am

p. U

se e

nerg

y c

onse

rvati

on

�Rot

ati

on a

ccele

rate

s due

to

fric

tion

betw

een

the s

phere

and

the r

am

p�

Fri

ctio

n fo

rce s

uppl

ies

the n

et

torq

ue

that

caus

es

ang

ular

acc

ele

rati

on.

�Con

tact

poi

nt is

alw

ays

at

rest

rela

tive

to

the s

urfa

ce,

so n

o wor

k is

don

e

aga

inst

fri

ctio

n�

The m

ech

ani

cal ene

rgy d

oes

not

chang

e�

There

is

now a

rot

ati

on t

erm

in

the K

E

�L

et

Uf=

0 a

t th

e b

ott

om

of

the p

lan

e

�K

i= 0

at

the t

op

(sta

rt f

rom

rest)

�U

i=

Mg

hat

the t

op

, U

f=

0

�R

ollin

g c

on

dit

ion

:

ff

ii

co

nsta

nt

mech

UK

UK

E

+ +++= ===

+ +++= ===

= ===

ff

vR

= ===ω ωωω

�E

qu

ate

kin

eti

c &

po

ten

tia

l

en

erg

y c

han

ges

2

2

22

fcm

21f

21f

cm

21f

vM

RIM

v

I

K

+ +++

= ===+ +++

ω ωωω= ===

Mg

hv

MRI

K

fcm

21f

= ===

+ +++

= ===2

2

�S

up

po

se b

all

is

a s

oli

d s

ph

ere

2

52M

RI c

m= ===

Mg

hM

vf

21= ===

+ +++2

55

52

21

710

/

f

gh

v

= ===

∴ ∴∴∴

Note: Need Second Law approach

to find internal forces, like friction

Co

pyri

gh

t R

. J

an

ow

–S

pri

ng

2012

Example: Awheel accelerating while rolling without slipping

•Pulled b

y c

onsta

nt

horizonta

l fo

rce F

applied a

t th

e C

M t

o a

ccele

rate

wheel

•Friction forc

e f

sm

akes it

roll:

applies t

orq

ue t

o w

heel, c

auses a

ngula

r accele

ration

cm P

F

f s

CC

W =

+

r

Lin

ear

2n

dL

aw

mo

tio

n o

f cm

:m

gN

F

ma

fF

Fi

y,

cm

si,

x− −−−

= ==== ===

= ===− −−−

= ===∑ ∑∑∑

∑ ∑∑∑0

Ro

tati

on

al 2

nd

Law

axis

th

rou

gh

cm

:cm

cm

si,

cm

I

rf

α ααα= ===

− −−−= ===

τ τττ∑ ∑∑∑

No

slip

pin

g

co

nstr

ain

t:)

als

oP

cm

cm

( r

r

aα ααα

− −−−= ===

α ααα− −−−

= ===

2ra

If

cm

cm

s= ===

∴ ∴∴∴

subst

itut

e:

r

IF

raI

F

fF

mr

-

ma

cm

cm

cm

cm

scm

cm

α ααα+ +++

= ===− −−−

= ===− −−−

= ===α ααα

= ===2

solv

e f

or

α αααcm

:

m

rI

Fr

P

cm

cm

α ααα= ===

+ +++− −−−

= ===α ααα

2

Sam

e r

esu

lt b

y p

laci

ng

rota

tion

axis

at

P: I

nsta

ntane

ous

Rot

ati

on A

xis

find

acm

:

+ +++= ===

+ +++= ===

22

2

1

1

mr

/I

mF

m

rI

Fr

a

cm

cm

cm

•F/m

is

the n

o-fr

icti

on a

ccele

rati

on•

acm

= F

/m a

lso

if I

cm=0

•a

cm= F

/2m

for

a h

oop

(Icm

= m

r2)

•a

cm= 5

F/7

m f

or s

olid

sph

ere

,

mr

I

IF

r

I

f

cm

cm

cm

cm

s2

+ +++= ===

α ααα− −−−

= ===fi

nd f

s:•

f s� ���

0 if

Icm

=0

•If

F=0,

α ααα= 0

, ω ωωω

is c

onst

ant

, f s

=0

•f s

is t

o th

e left

Min

imum

µ µµµs fo

r no

slipp

ing:

m

g

f

mg

f

sm

in,

ss

s= ===

µ µµµ⇒ ⇒⇒⇒

µ µµµ≤ ≤≤≤

Co

pyri

gh

t R

. J

an

ow

–S

pri

ng

2012

α ααα,

I cm

Fg=

Mg

Example: Rolling on a ramp w/o slipping again, via Second Law

•Pla

ce r

ota

tion a

xis

at

mass c

ente

r -

ram

p a

pplies t

orq

ue t

o t

he e

dge o

f ro

llin

g o

bje

ct

•Friction forc

e f

sis

up t

he r

am

p, o/w

wheel slides. t

orq

ue is C

CW

aro

und a

xis

Ap

ply

lin

ear

2n

d

Law

to

cm

m

oti

on

:0

= ===θ θθθ

− −−−= ===

= ===

θ θθθ= ===

= ===

∑ ∑∑∑∑ ∑∑∑c

os

Mg

NM

aF

Mg

sin

-f

Ma

F

cm

,y

y

scm

,x

x

Ap

ply

ro

tati

on

al 2

nd

Law

fo

r axis

at

cm

:R

fI

scm

cm

cm

= ===α ααα

= ===τ τττ

∑ ∑∑∑•

all o

ther

forc

es

have

zero

mom

ent

arm

� ���ze

ro t

orqu

e

No

slip

pin

g

co

nstr

ain

t:)

(als

oP

cm

cm

,x

R

R

a

α ααα− −−−

= ===α ααα

− −−−= ===

•a

cmis

in

–x d

irect

ion

for

CCW

(+)

α αααcm

Sol

ve f

or f

rict

ion

forc

e in

term

s of

acm

:

R

I

f

cm

cm

s

α ααα= ===

RaI

f

cm

,x

cm

s2

− −−−= ===

•a

cm� ���

g si

nθ θθθif

we let

Icm

� ���ze

ro•

acm

is n

ega

tive

, so

fs

is p

osit

ive (

up)

Sam

e r

esu

lt u

sing

ins

tant

ane

ous

rota

tion

axis

“P”

I

MR

I

Mg

Rsin

Pn

et

,P

cm

cm

cm

τ τττ= ===

α ααα= ===

+ +++

θ θθθ+ +++

= ===α ααα

2

)g

sin

(aM

f

x,c

ms

θ θθθ+ +++

= ===

)M

R

I(

sin

ga

cm

x,c

m

21

+ +++

θ θθθ− −−−

= ===

Mg

sin

)

RI (M

a

cm

x,c

mθ θθθ

= ===+ +++

− −−−2

Sol

ve f

or m

ass

cent

er

acc

ele

rati

on: