Chapter 11

Chapter 11

Angular MomentumAngular Momentum

The Vector Product and Torque The torque vector lies in a The torque vector lies in a

direction perpendicular to direction perpendicular to the plane formed by the the plane formed by the position vector and the position vector and the force vectorforce vector

The torque is the vector The torque is the vector (or cross) product of the (or cross) product of the position vector and the position vector and the force vectorforce vector

F r

Torque Vector ExampleTorque Vector Example

Given the force and locationGiven the force and location

Find the torque producedFind the torque produced

ˆ ˆ(2.00 3.00 )

ˆ ˆ(4.00 5.00 )

N

m

F i j

r i j

ˆ ˆ ˆ ˆ [(4.00 5.00 )N] [(2.00 3.00 )m]

ˆ ˆ ˆ ˆ[(4.00)(2.00) (4.00)(3.00)

ˆ ˆ ˆ ˆ(5.00)(2.00) (5.00)(3.00)

ˆ2.0 N m

r F i j i j

i i i j

j i i j

k


Consider a particle of mass Consider a particle of mass mm located at the located at the vector position vector position and moving with linear and moving with linear momentum momentum

Find the net torqueFind the net torque

Add the term sinceit 0

( )

d

dtd

dtd

dt

pr F r

rp

r p

p r

Angular Momentum The instantaneous angular The instantaneous angular

momentum of a particle momentum of a particle relative to the origin relative to the origin OO is is defined as the cross product defined as the cross product of the particle’s instantaneous of the particle’s instantaneous position vector position vector and its and its instantaneous linear instantaneous linear

momentum momentum

L

r

p

L r p

Torque and Angular MomentumTorque and Angular Momentum

The torque is related to the angular momentumThe torque is related to the angular momentum Similar to the way force is related to linear momentumSimilar to the way force is related to linear momentum

The torque acting on a particle is equal to the The torque acting on a particle is equal to the time rate of change of the particle’s angular time rate of change of the particle’s angular momentummomentum

This is the rotational analog of Newton’s Second This is the rotational analog of Newton’s Second Law Law must be measured about the same originmust be measured about the same origin

d

dt L

and L


The The SISI units of angular momentum are units of angular momentum are ((kgkg..mm22)/ s)/ s

Both the magnitude and direction of the Both the magnitude and direction of the angular momentum depend on the choice of angular momentum depend on the choice of originorigin

The magnitude is The magnitude is LL = = mvrmvr sin sin is the angle between is the angle between and and

The direction of The direction of is perpendicular to the is perpendicular to the plane formed by and plane formed by and

p

r

L

p

r

Angular Momentum of a Particle, Angular Momentum of a Particle, ExampleExample

The vector is The vector is pointed out of the diagrampointed out of the diagram

The magnitude is The magnitude is

LL = = mvrmvr sin 90 sin 90oo = = mvrmvr sin 90sin 90oo is used since is used since vv is is

perpendicular to perpendicular to rr A particle in uniform circular A particle in uniform circular

motion has a constant motion has a constant angular momentum about angular momentum about an axis through the center an axis through the center of its pathof its path

= L r p

Angular Momentum of a System of Angular Momentum of a System of ParticlesParticles

The total angular momentum of a system of The total angular momentum of a system of particles is defined as the vector sum of the particles is defined as the vector sum of the angular momenta of the individual particlesangular momenta of the individual particles

Differentiating with respect to timeDifferentiating with respect to time

tot ii

i i

d d

dt dt L L

1 2tot n ii

L L L L L

Angular Momentum of a System of Particles

Any torques associated with the internal forces Any torques associated with the internal forces acting in a system of particles are zeroacting in a system of particles are zero

Therefore, Therefore,

The net external torque acting on a system about The net external torque acting on a system about some axis passing through an origin in an inertial some axis passing through an origin in an inertial frame equals the time rate of change of the total frame equals the time rate of change of the total angular momentum of the system about that originangular momentum of the system about that origin

totext

d

dt L

Angular Momentum of a System of Particles

The resultant torque acting on a system The resultant torque acting on a system about an axis through the center of mass about an axis through the center of mass equals the time rate of change of angular equals the time rate of change of angular momentum of the system regardless of the momentum of the system regardless of the motion of the center of massmotion of the center of mass This applies even if the center of mass is This applies even if the center of mass is

accelerating, provided accelerating, provided are are evaluated relative to the center of massevaluated relative to the center of mass

and L

System of Objects, Example

The masses are The masses are connected by a light connected by a light cord that passes over cord that passes over a pulley; find the a pulley; find the linear accelerationlinear acceleration

ConceptualizeConceptualize The sphere falls, the The sphere falls, the

pulley rotates and the pulley rotates and the block slidesblock slides

Use angular Use angular momentum approachmomentum approach


CategorizeCategorize The block, pulley and sphere are a nonisolated The block, pulley and sphere are a nonisolated

systemsystem Use an axis that corresponds to the axle of the Use an axis that corresponds to the axle of the

pulleypulley AnalyzeAnalyze

At any instant of time, the sphere and the block At any instant of time, the sphere and the block have a common velocity have a common velocity vv

Write expressions for the total angular momentum Write expressions for the total angular momentum and the net external torqueand the net external torque


Solve the expression for the linear Solve the expression for the linear accelerationacceleration The system as a whole was analyzed so that The system as a whole was analyzed so that

internal forces could be ignoredinternal forces could be ignored Only Only externalexternal forces are needed forces are needed

Angular Momentum of a Rotating Rigid Object

Each particle of the object Each particle of the object rotates in the rotates in the xyxy plane plane about the about the zz axis with an axis with an

angular speed of angular speed of The angular momentum of The angular momentum of

an individual particle is an individual particle is

LLii = = mmii r rii22

and and are directed are directed along the along the zz axis axis

L


To find the angular momentum of the entire To find the angular momentum of the entire object, add the angular momenta of all the object, add the angular momenta of all the individual particlesindividual particles

This also gives the rotational form of This also gives the rotational form of Newton’s Second LawNewton’s Second Law

2z i i i

i i

L L m r I

extzdL d

I Idt dt


The rotational form of Newton’s Second Law is The rotational form of Newton’s Second Law is also valid for a rigid object rotating about a also valid for a rigid object rotating about a moving axis provided the moving axis:moving axis provided the moving axis:(1) passes through the center of mass(1) passes through the center of mass(2) is a symmetry axis(2) is a symmetry axis

If a symmetrical object rotates about a fixed axis If a symmetrical object rotates about a fixed axis passing through its center of mass, the vector form passing through its center of mass, the vector form holds: holds:

where where is the total angular momentum measured is the total angular momentum measured with respect to the axis of rotationwith respect to the axis of rotation

IL

L

Angular Momentum of a Bowling Ball

The momentum of The momentum of inertia of the ball is inertia of the ball is 2/52/5MR MR 22

The angular The angular momentum of the ball momentum of the ball is is LLzz = = II

The direction of the The direction of the angular momentum is angular momentum is in the positivein the positive zz directiondirection

Conservation of Angular Conservation of Angular MomentumMomentum

The total angular momentum of a system is The total angular momentum of a system is constant in both magnitude and direction if the constant in both magnitude and direction if the resultant external torque acting on the system is resultant external torque acting on the system is zerozero Net torque = 0 -> means that the system is isolatedNet torque = 0 -> means that the system is isolated

For a system of particles, For a system of particles,

tot i f = constant or = L L L

tot n = = constantL L

Conservation of Angular Momentum

If the mass of an isolated system undergoes If the mass of an isolated system undergoes redistribution, the moment of inertia changesredistribution, the moment of inertia changes The conservation of angular momentum requires The conservation of angular momentum requires

a compensating change in the angular velocitya compensating change in the angular velocity IIii ii = I = Iff ff = constant = constant

This holds for rotation about a fixed axis and for This holds for rotation about a fixed axis and for rotation about an axis through the center of mass of rotation about an axis through the center of mass of a moving systema moving system

The net torque must be zero in any caseThe net torque must be zero in any case

Conservation Law Summary

For an isolated system -For an isolated system -

(1) Conservation of Energy: (1) Conservation of Energy: EEii = = EEff

(2) Conservation of Linear Momentum: (2) Conservation of Linear Momentum:

(3) Conservation of Angular Momentum:(3) Conservation of Angular Momentum:

i fp p

i fL L

A solid cylinder of radiusA solid cylinder of radius 15 cm15 cm and massand mass 3.0 kg3.0 kg rolls without slipping at a constant speed ofrolls without slipping at a constant speed of 1.6 m/s1.6 m/s. . (a) What is its angular momentum about its (a) What is its angular momentum about its symmetry axis? (b) What is its rotational kinetic symmetry axis? (b) What is its rotational kinetic energy? (c) What is its total kinetic energy? (c) What is its total kinetic

energy? (energy? ( II cylinder= )cylinder= )22

1MR

A light rigid rodA light rigid rod 1.00 m1.00 m in in length joins two particles, with length joins two particles, with massesmasses 4.00 kg4.00 kg and and 3.00 kg3.00 kg, , at its ends. The combination at its ends. The combination rotates in the rotates in the xyxy plane about a plane about a pivot through the center of the pivot through the center of the rod. Determine the angular rod. Determine the angular momentum of the system momentum of the system about the origin when the about the origin when the speed of each particle isspeed of each particle is 5.00 5.00 m/s.m/s.

A conical pendulum consists of a A conical pendulum consists of a bob of mass bob of mass mm in motion in a circular in motion in a circular path in a horizontal plane as shown. path in a horizontal plane as shown. During the motion, the supporting During the motion, the supporting wire of length maintains the wire of length maintains the constant angle with the vertical. constant angle with the vertical. Show that the magnitude of the Show that the magnitude of the angular momentum of the bob about angular momentum of the bob about the center of the circle isthe center of the circle is

2/1432

cos

sin

gm

L

The position vector of a particle of mass The position vector of a particle of mass 2.00 kg2.00 kg is is

given as a function of time by given as a function of time by

Determine the angular momentum of the particle Determine the angular momentum of the particle

about the origin, as a function of time.about the origin, as a function of time.

r 6.00̂ i 5.00t̂ j m

A particle of massA particle of mass mm is shot with an initial velocityis shot with an initial velocity vvii making an making an

angle with the horizontal as shown. The particle moves in the angle with the horizontal as shown. The particle moves in the gravitational field of the Earth. Find the angular momentum of gravitational field of the Earth. Find the angular momentum of the particle about the origin when the particle is (a) at the origin, the particle about the origin when the particle is (a) at the origin, (b) at the highest point of its trajectory, and (c) just before it hits (b) at the highest point of its trajectory, and (c) just before it hits the ground. (d) What torque causes its angular momentum to the ground. (d) What torque causes its angular momentum to change?change?

A wad of sticky clay with mass A wad of sticky clay with mass mm and velocity and velocity vvii is fired at a is fired at a

solid cylinder of mass solid cylinder of mass MM and radius and radius RR. The cylinder is initially at . The cylinder is initially at rest, and is mounted on a fixed horizontal axle that runs through rest, and is mounted on a fixed horizontal axle that runs through its center of mass. The line of motion of the projectile is its center of mass. The line of motion of the projectile is perpendicular to the axle and at a distance perpendicular to the axle and at a distance dd < < RR from the from the center. (a) Find the angular speed of the system just after the center. (a) Find the angular speed of the system just after the clay strikes and sticks to the surface of the cylinder. (b) Is clay strikes and sticks to the surface of the cylinder. (b) Is mechanical energy of the clay-cylinder system conserved in this mechanical energy of the clay-cylinder system conserved in this process? Explain your answer.process? Explain your answer.

Two astronauts, each having a mass of Two astronauts, each having a mass of 75.0 kg75.0 kg, are connected , are connected by a by a 10.0-m 10.0-m rope of negligible mass. They are isolated in rope of negligible mass. They are isolated in space, orbiting their center of mass at speeds of space, orbiting their center of mass at speeds of 5.00 m/s5.00 m/s. . Treating the astronauts as particles, calculate (a) the magnitude Treating the astronauts as particles, calculate (a) the magnitude of the angular momentum of the system and (b) the rotational of the angular momentum of the system and (b) the rotational energy of the system. By pulling on the rope, one of the energy of the system. By pulling on the rope, one of the astronauts shortens the distance between them to astronauts shortens the distance between them to 5.00 m5.00 m. (c) . (c) What is the new angular momentum of the system? (d) What What is the new angular momentum of the system? (d) What are the astronauts’ new speeds? (e) What is the new rotational are the astronauts’ new speeds? (e) What is the new rotational energy of the system? (f) How much work does the astronaut energy of the system? (f) How much work does the astronaut do in shortening the rope?do in shortening the rope?

A uniform solid disk is set into rotation with an angular speed A uniform solid disk is set into rotation with an angular speed ωωii about an axis through its center. While still rotating at this about an axis through its center. While still rotating at this

speed, the disk is placed into contact with a horizontal surface speed, the disk is placed into contact with a horizontal surface and released as in Figure. (a) What is the angular speed of and released as in Figure. (a) What is the angular speed of the disk once pure rolling takes place? (b) Find the fractional the disk once pure rolling takes place? (b) Find the fractional loss in kinetic energy from the time the disk is released until loss in kinetic energy from the time the disk is released until pure rolling occurs. (pure rolling occurs. (HintHint: Consider torques about the center of : Consider torques about the center of mass.)mass.)

Conservation of Angular Momentum:The Merry-Go-Round

The moment of inertia of the The moment of inertia of the system is the moment of inertia system is the moment of inertia of the platform plus the of the platform plus the moment of inertia of the personmoment of inertia of the person Assume the person can be Assume the person can be

treated as a particletreated as a particle As the person moves toward As the person moves toward

the center of the rotating the center of the rotating platform, the angular speed will platform, the angular speed will increaseincrease To keep the angular To keep the angular

momentum constantmomentum constant

Chapter 11

Documents

Transcript of Chapter 11