AP Physics C: Mechanics Chapter 11

Angular Momentum

The Vector ProductThere are instances where the product of two vectors is another vector.

Earlier we saw where the product of two vectors was a scalar.

This was called the dot product.

The vector product of two vectors is called the cross product.

Section 11.1

The Vector Product and TorqueThe torque vector lies in a direction perpendicular to the plane formed by the position vector and the force vector.The torque is the vector (or cross) product of the position vector and the force vector.

Section 11.1

The Vector Product Defined

Given two vectors, and The vector (cross) product of and is defined as a third vector, .

C is read as “A cross B”.The magnitude of vector C is AB sin

is the angle between and.

Section 11.1

More About the Vector ProductThe quantity AB sin is equal to the area of the parallelogram formed by .The direction of is perpendicular to the plane formed by . The best way to determine this direction is to use the right-hand rule.

Section 11.1

Properties of the Vector ProductThe vector product is not commutative. The order in which the vectors are multiplied is important.

To account for order, rememberIf is parallel to ( = 0o or 180o), then

Therefore If is perpendicular to , then The vector product obeys the distributive law.

Section 11.1

Final Properties of the Vector Product

The derivative of the cross product with respect to some variable such as t is

where it is important to preserve the multiplicative order of the vectors.

Section 11.1

Vector Products of Unit Vectors

Section 11.1

Signs in Cross Products Signs are interchangeable in cross products

and

Section 11.1

Using DeterminantsThe cross product can be expressed as

Expanding the determinants gives

Section 11.1

Vector Product ExampleGivenFindResult

Section 11.1

Torque Vector ExampleGiven the force and location

Find the torque produced

Section 11.1

Angular MomentumConsider a particle of mass m located at the vector position and moving with linear momentum . Find the net torque.

This looks very similar to the equation for the net force in terms of the linear momentum since the torque plays the same role in rotational motion that force plays in translational motion.

Section 11.2

April 20, 2023

Angular Momentum Same basic techniques that were used in

linear motion can be applied to rotational motion. F becomes m becomes I a becomes v becomes ω x becomes θ

Linear momentum defined as What if mass of center of object is not

moving, but it is rotating? Angular momentum

mvp

IL

April 20, 2023

Angular Momentum I 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: kg m2/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

April 20, 2023

Angular Momentum II Angular momentum of a particle

Angular momentum of a particle

r is the particle’s instantaneous position vector p is its instantaneous linear momentum Only tangential momentum component

contribute r and p tail to tail form a plane, L is

perpendicular to this plane

sinsin2 rpmvrrmvmrIL

)( vrmprL

April 20, 2023

Angular Momentum of a Particle in Uniform Circular Motion

The angular momentum vector points out of the diagram

The magnitude is L = rp sin = mvr sin (90o) = mvr

A particle in uniform circular motion has a constant angular momentum about an axis through the center of its path

O

Example: A particle moves in the xy plane in a circular path of radius r. Find the magnitude and direction of its angular momentum relative to an axis through O when its velocity is v.

April 20, 2023

Angular momentum III 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

April 20, 2023

Example: calculating angular momentum for particles

Two objects are moving as shown in the figure . What is their total angular momentum about point O?

m2

m1

221121 prprLLLnet

skgm

mvrmvr

mvrmvrLnet

/ 8.945.2125.31

2.25.65.16.31.38.2

sinsin

2

2211

222111

April 20, 2023

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

April 20, 2023

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

April 20, 2023

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

)(

April 20, 2023

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

April 20, 2023

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

April 20, 2023

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

I

April 20, 2023

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

April 20, 2023

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

April 20, 2023

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.

April 20, 2023

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

April 20, 2023

constant0 L τ axis about z - net

final

ffinitial

ii ωI ωI L

Moment of inertia changes

Isolated System

April 20, 2023

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

April 20, 2023

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.

April 20, 2023

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

April 20, 2023

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