angular momentum, torque, and...

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Frank L. H. Wolfs Department of Physics and Astronomy, University of Rochester

Physics 121, March 27, 2008.Angular Momentum, Torque, and Precession.


Physics 121.March 27, 2008.

• Course Information

• Quiz

• Topics to be discussed today:

• Review of Angular Momentum

• Conservation of Angular Momentum

• Precession


Physics 121.March 27, 2008.

• Homework set # 7 is now available and is due on SaturdayApril 5 at 8.30 am.

• Homework set # 7 has two components:• WeBWork (75%)• Video analysis (25%)

• Exam # 2 will be graded this weekend and the results will bedistributed via email on Monday March 29.

• Make sure you pick up the results of exam # 2 in workshopnext week.

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Homework Set # 7.

Rolling Motion.

Precession.

Moment of inertia.

Angular acceleration. Conservation

Of AngularMomentum.

ConservationOf AngularMomentum.

Pulley problem.


Video analysis.Is angular momentum conserved?

Ruler

Origin


Physics 121.Quiz lecture 18.

• The quiz today will have 4 questions!

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Angular momentum.A quick review.

• We have seen many similaritiesbetween the way in which wedescribe linear and rotationalmotion.

• Our treatment of these types ofmotion are similar if werecognize the followingequivalence: linear rotational mass m moment I force F torque τ = r x F

• What is the equivalent to linearmomentum? Answer: angularmomentum.



• The angular momentum isdefined as the vector productbetween the position vector andthe linear momentum.

• Note:• Compare this definition with the

definition of the torque.• Angular momentum is a vector.• The unit of angular momentum is

kg m2/s.• The angular momentum depends

on both the magnitude and thedirection of the position and linearmomentum vectors.

• Under certain circumstances theangular momentum of a system isconserved!



• Consider an object carrying outcircular motion.

• For this type of motion, theposition vector will beperpendicular to the momentumvector.

• The magnitude of the angularmomentum is equal to theproduct of the magnitude of theradius r and the linear momentump:

L = mvr = mr2(v/r) = Iω• Note: compare this with p = mv!

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Conservation of angular momentum.A quick review.

• Consider the change in the angular momentum of a particle:

• Consider what happens when the net torques is equal to0 Nm:

dL/dt = 0 Nm → L = constant (conservation of angular momentum)• When we take the sum of all torques, the torques due to the

internal forces cancel and the sum is equal to torque due toall external forces.

• Note: notice again the similarities between linear androtational motion.

d!L

dt=d

dt

!r !!p( ) = m

!r !

d!v

dt+d!r

dt!!v

"#$

%&'=

= m!r !!a +!v !!v( ) =!r !

!F( =

!)(



L = r mv sinθ = m (r v dt sinθ)/dt = 2m dA/dt = constant



• The connection between the angularmomentum L and the torque τ

is only true if L and τ are calculatedwith respect to the same referencepoint (which is at rest in an inertialreference frame).

• The relation is also true if L and τ arecalculated with respect to the centerof mass of the object (note: center ofmass can accelerate).

!!" =

d!L

dt

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The angular momentum of rotating rigidobjects.

• Consider a rigid object rotatingaround the z axis.

• The magnitude of the angularmomentum of a part of a smallsection of the object is equal to

li = ri (mi vi)

• Due to the symmetry of the objectwe expect that the angularmomentum of the object will bedirected on the z axis. Thus we onlyneed to consider the z component ofthis angular momentum.

Note the direction of li !!!!(perpendicular to ri and pi)


The angular momentum of rotating rigidobjects.

• The z component of the angularmomentum is

• The total angular momentum of therotating object is the sum of the angularmomenta of the individual components:

• The total angular momentum is thus equalto

Li,z= l

isin! = r

isin!( ) mi

vi( ) = Ri mi

vi( )

Lz= L

i,z! = Rim

ivi( )! =

= Rim

i"R

i( )! =" miRi

2

!

Lz= I!


Conservation of angular momentum.Sample problem.

• The rotational inertia of a collapsing spinning star changesto one-third of its initial value. What is the ratio of the newrotational kinetic energy to the initial rotational kineticenergy?

• When the star collapses, it is compresses, and its moment ofinertia decreases. In this particle case, the reduction is afactor of 3:

• The forces responsible for the collapse are internal forces,and angular momentum should thus be conserved.

I f =1

3Ii

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• The initial kinetic energy of the star can be expressed interms of its initial angular momentum:

• The final kinetic energy of the star ca also be expressed interms of its angular momentum:

• Note: the kinetic energy increased! Where does this energycome from?

Ki=1

2Ii!

i

2=1

2

Li

2

Ii

K f =1

2

Lf

2

I f=1

2

Li2

1

3Ii

!"#

$%&

= 3Ki



• A cockroach with mass m runscounterclockwise around the rim of alazy Susan (a circular dish mounted ona vertical axle) of radius R androtational inertia I with frictionlessbearings. The cockroach’s speed (withrespect to the earth) is v, whereas thelazy Susan turns clockwise withangular speed ω (ω < 0). Thecockroach finds a bread crumb on therim and, of course, stops. (a) What isthe angular speed of the lazy Susanafter the cockroach stops? (b) Ismechanical energy conserved?

ω

v

R

x-axis

y-axis



• The initial angular momentum of thecockroach is

• The initial angular momentum of thelazy Susan is

• The total initial angular momentumis thus equal to

ω

v

R

x-axis

y-axis

!Lc=!r !!p = Rmvk̂

!Ld= I! k̂

!L =!Lc+!Ld= Rmvk̂ + I! k̂ = Rmv + I!( ) k̂

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• When the cockroach stops, it willmove in the same way as the rim ofthe lazy Susan. The forces that bringthe cockroach to a halt are internalforces, and angular momentum isthus conserved.

• The moment of inertia of the lazySusan + cockroach is equal to

• The final angular velocity of thesystem is thus equal to

ω

v

R

x-axis

y-axis

I f = I + mR2

! f =Lf

I f=Rmv + I!

I + mR2



• The initial kinetic energy of the system is equal to

• The final kinetic energy of the system is equal to

• The change in the kinetic energy is thus equal to

Cockroach Lazy Susan

Loss ofKinetic Energy!!

Ki=1

2mv

2+1

2I!

2

K f =1

2I f! f

2=1

2I + mR

2( )Rmv + I!I + mR

2

"#$

%&'2

=1

2

Rmv + I!( )2

I + mR2

!K = K f " Ki = "1

2

mI

I + mR2v " R#( )

2


Precession.

• Consider a rotating rigid objectspinning around its symmetryaxis.

• The object carries a certainangular momentum L.

• Consider what will happen whenthe object is balanced on the tipof its axis (which makes an angleθ with the horizontal plane).

• The gravitational force, which isan external force, will generate atoques with respect of the tip ofthe axis.

mg

!

L

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Precession.

• The external torque is equal to

• The external torque causes achange in the angularmomentum:

• Thus:• The change in the angular

momentum points in the samedirection as the direction of thetorque.

• The torque will thus change thedirection of L but not itsmagnitude.

mg

!

L

! =!r "!F = rF sin

#2+$%

&'()*= Mgr cos$

d!L =!!dt


Precession.

• The effect of the torque can bevisualized by looking at themotion of the projection of theangular momentum in the xyplane.

• The angle of rotation of theprojection of the angularmomentum vector when theangular momentum changes bydL is equal to

d! =dL

L cos"=Mgr cos"dt

L cos"=Mgrdt

L


Precession.

• Since the projection of theangular momentum during thetime interval dt rotates by anangle dφ, we can calculate therate of precession:

• We conclude the following:• The rate of precessions does not

depend on the angle θ.• The rate of precession decreases

when the angular momentumincreases.

! =d"

dt=Mgr

L=Mgr

I#

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Torque and Angular Momentum.

• Let’s test our understanding of the basic aspects of torqueand angular momentum by working on the followingconcept problems:

• Q18.1

• Q18.2


Done for today! Next week we stop movingand focus on equilibrium.

Spirit Pan from Bonneville Crater's Edge Credit: Mars Exploration Rover Mission, JPL, NASA

angular momentum, torque, and...

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