Chapter 11: Rotational Vectors and Angular Momentum Vector (cross) products Definition of vector product and its properties Axis of rotation unit vector

Chapter 11: Rotational Vectors and Angular Momentum

Vector (cross) products Definition of vector product and its properties

Axis of rotation

a

b

ba

sinˆ banba

n

unit vector normal to the planedefined by a

and b

abba

0

aaaa

Vector (cross) products (cont’d)

Properties of vector product

kji ˆˆˆ

x

y

z

i

jk

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

ikj ˆˆˆ jik ˆˆˆ Consider three vectors:

),,( zyx aaaa

),,( zyx bbbb

),,( zyx cccc

Then

kbaba

jbabaibaba

kbjbibkajaiabac

xyyx

zxxzyzzy

zyxzyx

ˆ)(

ˆ)(ˆ)(

)ˆˆˆ()ˆˆˆ(

Vector (cross) products (cont’d)

Properties of vector product (cont’d)

x

y

z

i

jk

yzzyx babac zxxzy babac

xyyxz babac

bac

zyx

zyx

bbb

aaa

kji ˆˆˆ

Torque

Case for 1 point-like object of mass m with 1 force

x

y

r

m

F

r

massless rigid rod

• r is constant only tangential componentof causes rotation F

FFt

mr

rrrm

amF

ˆ)ˆˆ(

ˆˆ

2

0

I

mrrFt

2

torque moment of inertiaunit Nm

tF

Torque is a quantitative measure of the tendency of a force to causeor change the rotational motion.

Torque

Case for 1 point-like object of mass m with 1 force

x

y

r

F

r

sinrrt lever arm

Fr

FrFrrF tt

sin

Define Fr

is aligned with the rotation axis

tF

Torque (cont’d)

Case for 2 point-like objects with 2 forces

21 1r 2r

tF2

tF1

2211 rFrF ttnet

0 0

increases

decreases

x

y

Example: A see-saw in balance

M mR r

Mg mg

02211

mgrMgR

rFrF ttnet

rRMm //

Work & energy (I)

A massive body on massless rigid rod

x

y

r

m

tF Work done by the force:

drdFdsFW tt

Also

ddtdIdIW )/(

ddtdddI )/)(/(

1

0

1

0

|)2/1( 2

IdI

KII 20

21 )2/1()2/1(

W=K

Work & energy (I) (cont’d)

Power in rotational motion

Work done by the force:

drdFdsFW tt

ddW

dtddtdW //

PPower :

Correspondence between linear & angular quantities

linear angular

displacement

velocity

acceleration

mass

force

Newton’s law

kinetic energy

work

x dtdxv / dtd / dtdva / dtd /

m 2iirmI

F

ImaF Fr

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

FdxW dW

Work & energy (II)

A massive body in rotational & translational motion

Kinetic energy: iii vvmK

)2/1(

Vvv ii

'

Now where 'iv

is the velocity with respect to

V

the center of mass (COM) and the velocity of COM. Then

22'

'22'

'22'

2'2'

''

)2/1()2/1(

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

/)2/1()2/1(

)2/1()2/1(

)()()2/1(

)2/1(

MVvm

dtrmdVMVvm

dtrdmVmVvm

VmvVmvm

VvVvm

vvmK

ii

iiii

iiiii

iiiii

iii

iii

0

com

'iv

V

Work & energy (II) cont’d

A massive body in rotational & translational motion (cont’d)

22' )2/1()2/1(

)2/1(

MVvm

vvmK

ii

iii

2''' ; iiii rmIvr

22 )2/1()2/1( IMVK

kinetic energy due totranslational motion

kinetic energy dueto rotational motionabout a rotation axisthrough COM



Example: A rolling cylinder (without sliding due to friction)

P P

O O

s

s

COMv

COMv

Rsx

RdtRd

dtdsdtdxvCOM

/

//

View point 1:

COMvv

COMvv

COMvv

COMvv

COMv

COMvv2

0 COMCOM vvv

+ =O O O

COMv

This view point was used in the last few slides

R

rota

tion

tra

nsla

tion



Example: A rolling cylinder (without sliding) (cont’d)

View point 2: Any point in the cylinder rotates around P

rotation axisP

2)2/1( PIK From the parallel axis theorem

2MRII COMP

22

222

)2/1()2/1(

)2/1()2/1(

COMCOM

COM

MvI

MRIK

rotational translational

RvCOM

Angular momentum & torque

x

y

r

p

r

sinrrt lever arm

tp tt rpprprprL

sinDefine angular momentum as:

prL

Since Fr

and dtpdF /

dtLdnet /

dtpdrvmv

dtpdrpdtrddtLd

/)(

///

0 F

angular momentum

net torque

Angular momentum of a particle


iiiiii LvrmprL

As before let’s break down vectors into two components

VvvRrr iiii

'' ;

Then

MVRpr

MVRdtrmdRpr

mVRdtrdmRVrmvrm

VRmvRmVrmvrm

VvRrmL

ii

iiii

iiiiiiii

iiiiiiii

iii

''

'''

''''

''''

''

/)(0

/)(

)()(

)()(

ang. mom. about COM + ang. mom. of COM0

position vector for com

velocity of com

Angular momentum of a multi-particle system

Conservation of angular momentum

iiiiii LvrmprL

netii dtLddtLd //

If the net torque is zero, the total angular momentumof the system is conserved.

.0/ constLdtLd

Conservation of angular momentum


x

z

plane A

plane B

plane A

i i iii

iiiiii

IrmLL

rmrrmL

)(

)(2

2

IL

plane B

1m2m

1r

2r

1L

2L

In general,

IL because of non-zero x/y component of

angular momentum unless the object is symmetric about the

axis of rotation.

If the object is symmetric about the axis of rotation, then

IdtLdIL /,

Angular momentum vector and axis of rotation

Gyroscope Gyroscopic motion

pivot

support

mg

remove

pivot

precession

Gyroscope Principle of gyroscopic

pivotL

wrdtLd

/

r

gmw

zy

x

LrLwr

//,

LLd

Gyroscope

L

LdL

Ld x

y

d

)/()(//)/(/ IwrLdtLLddtd

Precession angular speed:

Assumption: The angular momentum vector is associated only with the spin of the flywheel and is purely vertical.

The procession is much slower than the rotation,

Gyroscope

Gyroscope

Spinning Top

Example

A disk of mass M and radius R rotates around the axis with angular velocity i. A second identical disk, initially not rotating, is dropped on top of the first. There is friction between the disks, and eventually they rotate together with angular velocity f.

1) f = i 2) f = ½ i 3) f = ¼ i

i

z

f

z

First realize that there are no external torques acting on

the two-disk system. Angular momentum will be conserved!

ii MRL 211 2

1I

i

z2

1

f

z

ff MRL 22211 II

fi MRMR 22

21

fi 21

Example

You are sitting on a freely rotating bar-stool with your arms stretched out and a heavy glass mug in each hand. Your friend gives you a twist and you start rotating around a vertical axis though the center of the stool. You can assume that the bearing the stool turns on is frictionless, and that there is no net external torque present once you have started spinning. You now pull your arms and hands (and mugs) close to your body.

What happens to the angular momentum as you pull in your arms?

1. it increases 2. it decreases 3. it stays the same

L1 L2

CORRECT

1 2

I2 I1

L L

What happens to your angular velocity as you pull in your arms?


CORRECT

What happens to your kinetic energy as you pull in your arms?


CORRECT

1 2

I2 I1

L L

K 12

2I 12

2 2

II 1

22

IL (using L = I )

Example

A student sits on a barstool holding a bike wheel. The wheel is initially spinning CCW in the horizontal plane (as viewed from above). She now turns the bike wheel over. What happens?

1. She starts to spin CCW.2. She starts to spin CW.3. Nothing

CORRECT

Since there is no net external torque acting on the student

stool system, angular momentum is conserved.

Remember, L has a direction as well as a magnitude!

Initially: LLINI = LLW,I

Finally: LLFIN = LLW,F + LLS

LLW,F

LLS

LLW,I LLW,I = LLW,F + LLS

Example

A puck slides in a circular path on a horizontal frictionless table. It is held at a constant radius by a string threaded through a frictionless hole at the center of the table. If you pull on the string such that the radius decreases by a factor of 2, by what factor does the angular velocity of the puck increase?

(a) 2 (b) 4 (c) 8

Since the string is pulled through a hole at the center of rotation, there is no torque: Angular momentum is conserved.

L1 = I11 = mR21

mR

mR21 = m R2241

41

1 = 2 2 = 41

Example A uniform stick of mass M and length D is pivoted at the center. A

bullet of mass m is shot through the stick at a point halfway between the pivot and the end. The initial speed of the bullet is v1, and the final speed is v2.

What is the angular speed F of the stick after the collision? (Ignore gravity)

v1 v2

M

F

initial final

mD

D/4

Set Li = Lf using

v1 v2

M

F

initial finalm

DD/4

I 1

122MD

mvD

mvD

MD F1 22

4 4

1

12 F

m

MDv v

31 2

ExercisesProblem 1

Find the acceleration of an object of mass m.

Solution

m

R

M

h

x

y

T

Mg

n

T

mg

For the object, Newton’s 2nd law gives:

yy maTmgF

For the cylinder, the total torque is:

zzz MRIRT 2)2/1(

The tangential acceleration of the cylinderis equal to that of the object:

zy Raa tan

(1)

(2)

(3)

(2)+(3):

yMaT )2/1( (4)

ay>0 in –y direction

ExercisesProblem 1 (cont’d)

Solution (cont’d)

m

R

M

h

x

y

T

Mg

n

T

mg

(1)+(4):

yy maMamg )2/1(

)]2/(1/[ mMgay (5)

(4)+(5):

)/21/( MmmgmamgT y The final velocity of the object when itwas at rest initially:

hahavv yy 2220

2

)]2/(1/[22 mMghhav y

ExercisesProblem 2

Solution

m

R

M

h

x

y

T

Mg

n

T

mg

)]/21/()3[(

)]/21/([

MmmMg

MmmMgMgTn

(a) The normal force on the cylinder is:

(b) Compare n with (m+M)g:

As the suspended mass is accelerating downthe tension is less than mg. Thereforen is less than the total weight (m+M)g.

(c) If the cylinder is initially rotating clockwise and so that the object gets initial velocity upward, what effect does this have on T and n?

As long as the cable remains taut, T and n remain the same.

ExercisesProblem 3

Solution

T1

T2

RMm1

m2

A glider of mass m1 slides without friction ona horizontal air track. It is attached to anobject of mass m2 by a massless string.The pulley is a thin cylindrical shell withmass M and radius R, and the string turnsthe pulley without slipping or stretching.Find the acceleration of each body, theangular acceleration of the pulley, and thetension in each part of the string.

m1

n1 T1

m1g

m2

m2g

T2T1

T2

Mgn2

glider hanging obj. pulley

glider: xx amTF 111

object: yy amTgmF 2222

pulley:

zzz MRIRTRT )( 212

no stretching and slipping:

Zyx Raa 21

y

x (1)

(2)

(3)

(4)

glider

hangingobject

pulley

+

y xy

x

Exercises

Solution cont’d

Problem 3 (cont’d)

(1)-(4):

Mmm

gmMmT

Mmm

gmmT

21

212

21

211

)(;

Exercises

Solution

Problem 4

A primitive yo-yo is made by wrapping a stringseveral times around a solid cylinder with massM and radius R. You hold the end of the stringstationary while releasing the cylinder with noinitial motion. The string unwinds but does notslip or stretch as the cylinder drops and rotates.Find the speed vcm of the center of mass of thesolid cylinder after it has dropped a distance h.

h

1

22

2,cmv

0

0

1

1,

cmv

2211 UKUK

RvMRI

IMvKK

UMghU

cm

cm

/,)2/1(

)2/1()2/1(,0

0,

2,22

22

22,21

21

Energy conservation

0)4/3(0 22, cmMvMgh ghvcm )3/4(2,

Exercises

Solution

Problem 5

B

A

C

(Atwood’s machine)

Find the linear accelerations of blocks A and B,the angular acceleration of the wheel C, and the tension in each side of the cord if there isno slipping between the cord and the surface ofthe wheel.

I

mA

mB

R

The accelerations of blocks A and B will have the same magnitude .As the cord does not slip, the angular acceleration of the pulley will be . If we denote the tensions in cord as and , the equationsof motion are:

Ra /

a

AT BT

)(;;2

IaR

ITTamgmTamTgm BABBBAAA

RIRmRm

mmg

R

a

RImm

mmga

BA

BA

BA

BA

// 2

2

2

/

/2)(

RImm

RImmmgagmT

BA

ABAAA

2

2

/

/2)(

RImm

RImmmgagmT

BA

BABBB

Exercises

Solution

Problem 6

h=50.0 m

rough

smooth

A solid uniform spherical boulderstarts from rest and rolls down a50.0 m high hill. The top half of thehill is rough enough to cause theboulder to roll without slipping, butthe lower half is covered with ice andthere is no friction. What is thetranslational speed of the boulderwhen it reaches the bottom of the hill?

1st half (rough):

12

222

221

)7/10(

)/]()5/2)[(2/1()2/1(

)2/1()2/1(

ghv

RvmRmv

Imvmgh

2nd half (smooth):

smghghv

vghgh

KmvKmvmgh

B

B

rotBrot

/0.292)7/10(

)2/1(])7/10)[(2/1(

)2/1()2/1(

21

212

222

Exercises

Solution

Problem 7

Occasionally, a rotating neutron star undergoes a sudden andunexpected speedup called a glitch. One explanation is that aglitch occurs when the crust of the neutron star settles slightly,decreasing the moment of inertia about the rotation axis. Aneutron star with angular speed 0=70.4 rad/s underwent sucha glitch in October 1975 that increased its angular speed to, where =2.01x10-6. If the radius of the neutronstar before the glitch was 11 km, by now how much did itsradius decrease in the starquake? Assume that the neutronstar is a uniform sphere.

Conservation of angular momentum: 20

2000 RRII

20000

20

02

0020

2

)()(

RRRR

RRRcmRR 1.1)/)(2/( 00

Exercises

Solution

Problem 8

A small block with mass 0.250 kg is attached to a string passing through a hole in a frictionless, horizontal surface. The block isoriginally revolving in a circle with a radius of 0.800 m about thehole with a tangential speed of 4.00 m/s. The string is then pulled slowly from below, shortening the radiusof the circle in which the block revolves.The breaking strength of the string is 30.0N. What is the radius of the circle when thestring breaks. The tension on the string: )/()/()(/ 32322 mrLmrmvrrmvT The radius at which the string breaks is obtained from:

mr

Nkg

msmkg

mTrmvmTLr

440.0

)]0.30)(250.0[(

/)]800.0)(/00.4)(250.0[(

)/()()/(2

max2

00max23

ExercisesProblem 9

pivot

pivotcatcher

v

r

M,I

A ball catcher whose mass is M and momentof inertia is I is hung by a frictionless pivot.A ball with a speed v and mass m is caught by thecatcher. The distance between the pivot point andthe ball is r which is much greater that the radius ofthe ball.

(a)Find the angular speed of the catcher right after it catches the ball.(b) After the ball is caught, the catcher – ball system swings up as high as h. Find the angular speed at the maximum height h.

Exercises

Solution

Problem 10pivot

pivotcatcher

v

r

M,I

(a)The initial angular momentum with respect to the pivot is: The final total moment of inertia is

mvr

2mrI

)/( 2 Imrmvr

(b) The kinetic energy after the collision is:

ghmMImrK )()()2/1( 22

,IL Since

potential energy at height h

h

)(

)(22 Imr

ghMm

ExercisesProblem 11

h= 58.0 m

d= 42.0 cm

g/cmA 42.0 cm diameter wheel, consisting ofa rim and six spokes, is constructed froma thin rigid plastic material having a linearmass density of 25.0 g/cm. This wheel isreleased from rest at the top of a hill 58.0 mhigh.(a)How fast is it rolling when it reaches the bottom of the hill?(b) How would your answer change if the linear mass density and the diameter of the wheel were each doubled?

No sliding

Exercises

Solution

Problem 11 (cont’d)

h= 58.0 m

d= 42.0 cm

g/cm

(a) Conservation of energy:

2211 KUKU

)3/(6

)2/1()2/1(

6,0,0,

22

222

211

rmrmIII

IMvK

mmMUKMghU

spokerimspokesrim

spokesrim

)3(2,2 rMrmrm spokerim

322

2

)]3/(62)[2/1(

))(3)(2)(2/1()3(2

rrrr

rrghR

sradRgh /124)]2(/[])3[( 2

(b) Doubling the density would have no effect! As , doubling the diameter would reduce the angular velocity by half. But would be unchanged.

r/1rv

No sliding

Documents

Chapter 11: Rotational Vectors and Angular Momentum Vector (cross) products Definition of vector product and its properties Axis of rotation unit vector