PHY126 Summer Session I, 2008 Most of information is available at: chiaki/PHY126-08 including the syllabus and lecture.

PHY126 Summer Session I, 2008

• Most of information is available at: http://nngroup.physics.sunysb.edu/~chiaki/PHY126-08 including the syllabus and lecture slides. Read syllabus and watch for important announcements.

• Homework assignment for each chapter due nominally a week later. But at least for the first two homework assignments, you will have more time. All the assignments will be done through MasteringPhysics, so you need to purchase the permit to use it. Some numerical values in some problems will be randomized.

• In addition to homework problems and quizzes, there is a reading requirement of each chapter, which is very important.

Chapter 10: Rotational Motion

Movement of points in a rigid body

All points in a rigid body move in circles about the axis of rotation

z

y

x

P

orbit of point P

rigid body

Axis of rotation

In this specific example onthe left, the axis of rotationis the z-axis.

A rigid body has a perfectlydefinite and unchanging shapeand size. Relative position ofpoints in the body do not changerelative to one another.

Movement of points in a rigid body (cont’d)

At any given time, the 2-d projection of any point in the object is described by two coordinates ( r , )

y

x

P

r

s = rwhere s,r in m, and in rad(ian)

length of the arc from the x-axis s:

A complete circle: s = r

360o = 2 rad

57.3o= 1 rad

In our example, 2-d projection onto the x-y plane isthe right one.

1 rev/s = 2 rad/s1 rev/min = 1 rpm

Angular displacement, velocity and acceleration

At any given time, the 2-d projection of any point in the object is described by two coordinates ( r , )

y

x

P at t2

r

P at t1

Angular displacement:

in a time interval t = t2 – t1

Average angular velocity:

12

12

tttavg

rad/s

Instantaneous angular velocity:

dt

d

tt

0

lim rad/s

counter) clockwise rotation

Angular displacement, velocity and acceleration

(cont’d)

At any given time, the 2-d projection of any point in the object is described by two coordinates ( r , )

y

x

P at t2

r

P at t1

Average angular acceleration:

Instantaneous angular acceleration:

12

12

tttavg

rad/s2

dt

d

tt

0

lim rad/s2

Correspondence between linear & angular quantities

Linear Angular

Displacement

Velocity

Acceleration

x

dtdxv / dtd /

dtdva / dtd /

Case for constant acceleration (2-d)

0/ dtd

Consider an object rotating with constant angular acceleration

tt

dtdtdtd0 00

)/(

00td

t00

t00

t

t

t

t

dtt

dtdtddtdtd

00

000)/()/(

2000 )2/1( tt

2000 )2/1( tt

Eq.(1)

Eq.(2)

Case for constant acceleration (2-d) (cont’d)

Eliminating t from Eqs.(1) & (2):

Eq.(1) 00 /)( t Eq.(1’)

Eqs.(1’)&(2)

20

200

0000

/)()2/1(

/)(

)(2 0020

2

Vectors

Vectors in 3-dimension

x

y

z

i

jk

kji ˆ,ˆ,ˆ : unit vector in x,y,z direction

Consider a vector:

kajaiaaaaa zyxzyxˆˆˆ),,(

)1,0,0(ˆ),0,1,0(ˆ ),0,0,1(ˆ kji

Also

• Inner (dot) product

zzyyxx bababababa cos

baba

if 0

a

b

ba

Small change of radial vector Rotation by a small rotation angle

irr ˆ1

1for sin and 1cos ˆ

ˆ)ˆsinˆcos(12

jr

irjrirrrr

jrirr ˆsinˆcos2

12 rrr

ˆˆlim

ˆ ,ˆ define weIf

0

r

d

rdr/rr

ˆˆ0ˆˆ rr

Note:

Relation between angular & linear variables

x

y

r

r

i

jA point withFixed radius

unit vector in x direction

uni

t ve

cto

r in

y d

irect

ion

)ˆsinˆ(cosˆ jirrrr

ˆˆcosˆsin/ˆ jidrd

rjidd ˆˆsinˆcos/ˆ

)/ˆ(/)ˆ(/ dtrdrdtrrddtrdv

ˆ)/)(/ˆ( rdtddrdr

rvv

ˆˆ0ˆˆ rr

unittangentialvector

unitradialvector

)1,0(ˆ ),0,1(ˆ ji

)cos,sin(ˆ

r is const.

Relation between angular & linear variables (cont’d)

x

y

r

r

i

jA point withFixed radius

unit vector in x direction

uni

t ve

cto

r in

y d

irect

ion

rrvr

rrr

dtdddrr

dtdrdtdr

dtrddtvda

ˆ)/(ˆ

ˆˆ

)/)(/ˆ(ˆ

)/ˆ(ˆ)/(

/)ˆ(/

2

2

ta ratangentialcomponent

radialcomponent

Relation between angular & linear variables (cont’d)

Example v

How are the angular speeds of thetwo bicycle sprockets in Fig. relatedto the number of teeth on each sprocket?

The chain does not slip or stretch, so it moves at the sametangential speed v on both sprockets:

2

1

1

22211 r

rrrv

The angular speed is inversely proportional to the radius. Let N1 andN2 be the numbers of teeth. The condition that tooth spacing is thesame on both sprockets leads to:

2

1

2

1

2

2

1

1 22

N

N

r

r

N

r

N

r

Combining the above two equations:2

1

1

2

N

N

Description of general rotation Why it is not always right to define a rotation by a vector

x x x

x x x

y yy

y y y

z z z

rotation about x axis

rotation about y axis

rotation about y axis

rotation about x axis

original

original

The result depends on the order of operations

Description of general rotation

Up to this point all the rotations have been about the z-axis or in x-y plane. In this case the rotations are about a unit vector where is normal to the x-y plane. But in general, rotations are about a general direction.

n

nn

• Define a rotation about by as:

n n

RH rule

• In general1221

but if 21 &, infinitesimally small,

1221

ddddd

ndt

dn

dt

dˆ ; ˆ

we can define a vector by

d

• If the axes of rotation are the same,

1221

Kinetic energy & rotational inertia

x

y

rv

axis of rotation

2

1

2

2

1

1

2

)2/1(

)()2/1(

)2/1(

ii

N

i

i

N

i i

N

i ii

rm

rm

vmK

I

rotational inertia/moment of inertia

A point in aRigid body

Kinetic energy & rotational inertia (cont’d)

x

y

rv

A point in aRigid body

rotation axis

2)2/1( IK

Compare with:2)2/1( mvK

dVrrdmr

rmI ii

N

iN

)(

lim

22

2

1

density volumeelement

m

And remember conservation ofenergy:

otherWUKUK 2211

More precise definition of I :

Kinetic energy & rotational inertia (cont’d)

x

y

R

rotation axis (z-axis)

Moment of inertia of a thin ring (mass M, radius R) (I)

dsdm

dsRdmrI 22

2

2

0

32

2)(

MR

RRdR

Rdds ;

ds

linear mass density

Kinetic energy & rotational inertia (cont’d)

x

y

R

rotation axis (y-axis)

Moment of inertia of a thin ring (mass M, radius R) (II)

ds

dsdm

r

cosRr Rdds ;

2

0

2)cos( RdRI

2

0

23 cos dR

2

0

3 )2cos1)(2/1( dR 2

03 |]2sin)2/1([)2/1( R

23 )2/1()2()2/1( MRR 1cos2sincos2cos 222

Kinetic energy & rotational inertia (cont’d)

Table of moment of inertia

Kinetic energy & rotational inertia (cont’d)Tables of moment of inertia

Parallel axis theorem

x

yrotation axisthrough P

P

O

COM

dm

y-br

x-ad

a

b

x

y

rotation axisthrough COM

dmbyaxdmrI ])()[( 222

xdmadmyx 2)( 22

dmbaydmb )(2 22

MdbMyaMxI comcomcom222

2MdIcom

COM: center of mass

dmr

MM

rm

m

rmr i ii

i i

i iicom

1

The axis of rotation is parallel to the z-axis

totalmass

x, y measuredw.r.t. COM

Parallel axis theorem (cont’d)

x

yrotation axisthrough P

P

O

COM

dm

y-br

x-ad

a

b

x

y

rotation axisthrough COM

2MdII com

Knowing the moment of inertia about an axis through COM (center of mass) of a body, the rotational inertia for rotation about any parallel axis is :

Correspondence between linear & angular quantities

linear angular

displacement

velocity

acceleration

mass

kinetic energy

x

dtdxv / dtd /

dtdva / dtd / m I

2)2/1( mvK 2)2/1( IK

Problem 1

A meter stick with a mass of 0.160 kg is pivoted about one end so that it canrotate without friction about a horizontal axis. The meter stick is held in a horizontal position and released. As it swings through the vertical, calculate(a) the change in gravitational potential energy that has occurred; (b) the angular speed of the stick;(c) the linear speed of the end of the stick opposite the axis.(d) Compare the answer in (c) to the speed of a particle that has fallen 1.00m, starting from rest.

Solution y

x

cm

cm

01 v

2v

1.00 m

0.500 m

(a)

J

mmsmkg

yyMgU

MgyU

cmcm

cm

784.0

)00.150.0)(/80.9)(160.0(

)(2

12

Exercises

Problem 1 (cont’d)

(c) smsradmrv /42.5)/42.5)(00.1(

(d) 2000

20

2 /80.9;00.1,0);(2 smamyyvyyavv yyyy

smyyav y /43.4)(2 0

2211 UKUK 22

22121 )3/1(;)2/1(0 MLIIUUUKK

sradmkgJIU /42.5])00.1)(160.0/[()784.06(/)(2 22

(b)

Problem 2

The pulley in the figure has radius R and a moment of inertia I. The ropedoes not slip over the pulley, and the pulley spins on a frictionless axle.The coefficient of kinetic friction between block A and the tabletop is .kThe system is released from rest, and block B descends. Block A has massmA and block B has mass mB. Use energy methods to calculate the speed ofBlock B as a function of distance d that it has descended.

Solution A

B

I

d

y

x

00 21 00 21 vv

WUKUKE 2211

Energy conservation:

gdmUK B 11 ,0

RvgdmWU

IvmvmK

Ak

BA

/,,0

)2/1()2/1()2/1(

222

22

22

222

gdmvRImmgdm AkBAB 22

2 )/)(2/1(

22 /

)(2

RImm

mmgdv

BA

AkB

Problem 3

You hang a thin hoop with radius R over a nail at the rim of the hoop. Youdisplace it to the side through an angle from its equilibrium position and letit go. What is its angular speed when it returns to its equilibrium position?

Solution

x

y

the origin =the original location ofthe center of the hoop

pivot pointR

1,cmy

R

0),cos1(

2

)2/1(,0

;

21,1

2222

2221

2211

UMgRMgyU

MRMRMRMdII

IKK

UKUK

cm

cm

22

2)cos1( MRMgR

Rg /)cos1(2