Physics 106: Mechanics Lecture 06 Wenda Cao NJIT Physics Department.

Physics 106: Mechanics Lecture 06

Wenda Cao

NJIT Physics Department

February 24, 2011

Conservation of Angular Momentum

Cross Product Comparison: definitions

of single particle torque and angular momentum

Angular Momentum of a system of particles of a rigid object

Conservation of angular momentum

Examples http://www.youtube.com/watch?v=AQLtcEAG9v0

February 24, 2011

Cross Product

The cross product of two vectors says something about how perpendicular they are.

Magnitude:

is smaller angle between the vectors Cross product of any parallel vectors = zero Cross product is maximum for

perpendicular vectors Cross products of Cartesian unit vectors:

sinABBAC

BAC

A

B

sinA

sinB

0ˆˆ ;0ˆˆ ;0ˆˆ

ˆˆˆ ;ˆˆˆ ;ˆˆˆ

kkjjii

ikjjkikji

y

x

z

ij

k

i

kj

February 24, 2011

Cross Product Direction: C perpendicular

to both A and B (right-hand rule)

Place A and B tail to tail Right hand, not left hand Four fingers are pointed

along the first vector A “sweep” from first vector

A into second vector B through the smaller angle between them

Your outstretched thumb points the direction of C

First practice ? ABBA

? ABBA

- ABBA

February 24, 2011

More about Cross Product The quantity ABsin is the area of

the parallelogram formed by A and B

The direction of C is perpendicular to the plane formed by A and B

Cross product is not commutative

The distributive law

The derivative of cross product obeys the chain rule Calculate cross product

dt

BdAB

dt

AdBA

dt

d

CABACBA

)(

- ABBA

kBABAjBABAiBABABA xyyxzxxzyzzyˆ)(ˆ)(ˆ)(

February 24, 2011

Torque Angular Momentum prL

Fr

)sin(rFFr mvp )sin(rpprL

)sin(rFFr mvp )sin(rprpL

Torque and Angular Momentum for a Single Particle

February 24, 2011

Angular momentum of a system of particles

Angular momentum of a system of particles

angular momenta add as vectors be careful of sign of each angular momentum

pr L L...LLL i all

ii i all

i n 2 1 net

p r - p r |L| 2211net

for this case:

221121 prprLLLnet

February 24, 2011

Angular Momentum of a Rigid Body

Angular momentum of a rotating rigid object

L has the same direction as L is positive when object rotates in CCW L is negative when object rotates in CW

Angular momentum SI unit: kgm2/s

Calculate L of a 10 kg disc when = 320 rad/s, R = 9 cm = 0.09 m

L = I and I = MR2/2 for disc L = 1/2MR2 = ½(10)(0.09)2(320) = 12.96 kgm2/s

IL

L

February 24, 2011

A solid sphere and a hollow sphere have the same mass and radius. They are rotating with the same angular speed. Which one has the higher angular momentum?

A) the solid sphereB) the hollow sphereC) both have the same angular momentumD) impossible to determine

Finding angular momentum

IL

February 24, 2011

Linear Momentum and Force

Linear motion: apply force to a mass The force causes the linear momentum to

change The net force acting on a body is the time

rate of change of its linear momentum

dt

pd

dt

vdmamFFnet

Lt

vmp

ptFI net

February 24, 2011

Angular Momentum and Torque

Rotational motion: apply torque to a rigid body The torque causes the angular momentum to

change The net torque acting on a body is the time rate of

change of its angular momentum

and to be measured about the same origin The origin should not be accelerating, should be

an inertial frame

dt

pdFFnet

dt

Ldnet

L

February 24, 2011

Demonstration

Start from

Expand using derivative chain rule

)()( vrdt

dmpr

dt

d

dt

Ld

dt

Ldnet

dt

pdFFnet

arvvmdt

vdrv

dt

rdmvr

dt

dm

dt

Ld

)(

netnetFramrarmarvvmdt

Ld

)(

February 24, 2011

What about SYSTEMS of Rigid Bodies?

• i = net torque on particle “i”• internal torque pairs are included in sum

i LLsys

• individual angular momenta Li

• all about same origin

i

isys

dt

Ld

dt

Ld

i

BUT… internal torques in the sum cancel in Newton 3rd law pairs. Only External Torques contribute to Lsys

Nonisolated System: If a system interacts with its environment in the sense that there is an external torque on the system, the net external torque acting on a system is equal to the time rate of change of its angular momentum.

dt

Ld i i :body single a forlaw Rotational

nd

2

Total angular momentumof a system of bodies:

net external torque on the system

net,

i

extisys

dt

Ld

February 24, 2011

a

a

Example: A Non-isolated System

A sphere mass m1 and a block of mass m2 are connected by a light cord that passes over a pulley. The radius of the pulley is R, and the mass of the thin rim is M. The spokes of the pulley have negligible mass. The block slides on a frictionless, horizontal surface. Find an expression for the linear acceleration of the two objects.

gRmext 1

February 24, 2011

a

a

Masses are connected by a light cord Find the linear acceleration a.

• Use angular momentum approach• No friction between m2 and table• Treat block, pulley and sphere as a non- isolated system rotating about pulley axis. As sphere falls, pulley rotates, block slides• Constraints:

/pulleyfor

sphere andblock for sa' and s v'Equal

dv/dt αR a

dtd αωR v

• Ignore internal forces, consider external forces only• Net external torque on system:

• Angular momentum of system: (not constant) ωMRvRmvRmIωvRmvRmLsys

22121

gRmτMR)aRmR(mαMRaRmaR mdt

dLnet

sys121

221

wheelofcenter about 1 gRmnet

21

1 mmM

gma

same result followed from earlier

method using 3 FBD’s & 2nd law

I

February 24, 2011

Isolated System Isolated system: net external torque

acting on a system is ZERO no external forces net external force acting on a system is ZERO

fitot LLL

or constant

0dt

Ld totext

February 24, 2011

Angular Momentum Conservation

where i denotes initial state, f is final state

L is conserved separately for x, y, z direction

For an isolated system consisting of particles,

For an isolated system is deformable

fitot LLL

or constant

constant321 LLLLL ntot

constant ffii II

February 24, 2011

First Example A puck of mass m = 0.5 kg is

attached to a taut cord passing through a small hole in a frictionless, horizontal surface. The puck is initially orbiting with speed vi = 2 m/s in a circle of radius ri = 0.2 m. The cord is then slowly pulled from below, decreasing the radius of the circle to r = 0.1 m.

What is the puck’s speed at the smaller radius?

Find the tension in the cord at the smaller radius.

February 24, 2011

Angular Momentum Conservation

m = 0.5 kg, vi = 2 m/s, ri = 0.2 m, rf = 0.1 m, vf = ?

Isolated system? Tension force on m exert zero

torque about hole, why?

fi LL

m/s421.0

2.0 i

f

if v

r

rv

)( vmrprL

iiiii vmrvmrL 90sin fffff vmrvmrL 90sin

N 801.0

45.0

22

f

fc r

vmmaT

February 24, 2011

constant0 L τ axis about z - net

final

ffinitial

ii ωI ωI L

Moment of inertia changes

Isolated System

February 24, 2011

How fast should the student spin?

L is constant… while moment of inertia changes

The student on a platform is rotating (no friction) with angular speed 1.2 rad/s.

• His arms are outstretched and he holds a brick in each hand.

• The rotational inertia of the system consisting of the professor, the bricks, and the platform about the central axis is 6.0 kg·m2.

By moving the bricks the student decreases the rotational inertia of the system to 2.0 kg·m2.

(a) what is the resulting angular speed of the platform?

(b) what is the ratio of the system’s new kinetic energy to the original kinetic energy?

February 24, 2011

ffi II L

LL L

i axis fixed aabout ...

initialfinal torqueexternal Zero

KE has increased!!

L is constant… while moment of inertia changes,

Ii = 6 kg-m2

i = 1.2 rad/s

If = 2 kg-m2

f = ? rad/s

Solution (a):rad/s 3.6 1.2

2

6 i

f

if I

I

Solution (b):3 )( )( 22

22

1

22

1

f

i

f

i

i

f

f

f

i

f

ii

ff

i

f

I

I

I

I

I

I

I

I

I

I

K

K

February 24, 2011

Controlling spin () by changing I (moment of inertia)

In the air, net = 0

L is constant

ffii IIL

Change I by curling up or stretching out- spin rate must adjust

Moment of inertia changes

February 24, 2011

Example: A merry-go-round problem

A 40-kg child running at 4.0 m/s jumps tangentially onto a stationary circular merry-go-round platform whose radius is 2.0 m and whose moment of inertia is 20 kg-m2. There is no friction.

Find the angular velocity of the platform after the child has jumped on.

February 24, 2011

The moment of inertia of the system = the moment of inertia of the platform plus the moment of inertia of the person. Assume the person can

be treated as a particle As the person moves

toward the center of the rotating platform the moment of inertia decreases.

The angular speed must increase since the angular momentum is constant.

The Merry-Go-Round

February 24, 2011

Solution: A merry-go-round problem

ffiitot IωIL

I = 20 kg.m2

VT = 4.0 m/smc = 40 kgr = 2.0 m0 = 0

rv mrvmII L TcTciii 0

fcfff ωrmIωIL )( 2

rvmωrmI Tcfc )( 2

rad/s 78.124010

244022

rmI

rvmω

c

Tcf