Rotational Motion and Angular Momentum.teacher.pas.rochester.edu/phy121/LectureSlides/Lecture17/Lecture17... · describe linear and rotational motion. ... the linear momentum. •Note:

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

Physics 121, March 25, 2008.Rotational Motion and Angular Momentum.


Physics 121.March 25, 2008.

• Course Information

• Topics to be discussed today:

• Review of Rotational Motion

• Rolling Motion

• Angular Momentum

• Conservation of Angular Momentum


Physics 121.March 25, 2008.

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

• There will be no workshops and office hours for the rest ofthe week. We will be busy grading exam # 2.

• The grades for exam # 2 will be distributed via email onMonday March 31.

• You should pick up your exam in workshop next week.Please check it carefully for any errors.

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Physics 121.Quiz lecture 17.

• The quiz today will have 4 questions!


Rotational variables.A quick review.

• The variables that are used todescribe rotational motion are:

• Angular position θ

• Angular velocity ω = dθ/dt

• Angular acceleration α = dω/dt

• The rotational variables arerelated to the linear variables:

• Linear position l = Rθ

• Linear velocity v = Rω

• Linear acceleration a = Rα


Rotational Kinetic Energy. A quick review.

• Since the components of a rotating object have a non-zero(linear) velocity we can associate a kinetic energy with therotational motion:

• The kinetic energy is proportional to square of the rotationalvelocity ω. Note: the equation is similar to the translationalkinetic energy (1/2 mv2) except that instead of beingproportional to the the mass m of the object, the rotationalkinetic energy is proportional to the moment of inertia I ofthe object:

Note: units of I: kg m2

K =1

2m

ivi

2

i

! =1

2m

i"r

i( )2

i

! =1

2m

iri

2

i

!#$%

&'(" 2

=1

2I" 2

I = miri

2

i

! or I = r2dm"

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Torque.• In general the torque associated with

a force F is equal to

• The arm of the force (also called themoment arm) is defined as rsinφ.The arm of the force is theperpendicular distance of the axis ofrotation from the line of action ofthe force.

• If the arm of the force is 0, thetorque is 0, and there will be norotation.

• The maximum torque is achievedwhen the angle φ is 90°.

!! = rF sin" =

!r #!F


Torque. A quick review.

• The torque τ of the force F is relatedto the angular acceleration α:

τ = Ια

• This equation looks similar toNewton’s second law for linearmotion:

F = ma

• Note: linear rotational mass m moment I force F torque τ

Ar

φ

F


Rolling motion. A quick review.

• Rolling motion is a combinationof translational and rotationalmotion.

• The kinetic energy of rollingmotion has thus twocontributions:

• Translational kinetic energy =(1/2) M vcm

2.

• Rotational kinetic energy =(1/2) Icm ω 2.

• Assuming the wheel does notslip: ω = v / R.

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How different is a world with rotationalmotion? Sample problem.

• Consider the loop-to-loop. Whatheight h is required to make it to thetop of the loop?

• First consider the case withoutrotation:• Initial mechanical energy = mgh.• Minimum velocity at the top of the

loop is determine by requiring thatmv2/R > mg

orv2 > gR

• The mechanical energy is thus equalto

(1/2)mv2 + 2mgR > (5/2)mgR• Conservation of energy requires

h > (5/2)R

hR



• What changes when the objectrotates?• The minimum velocity at the top

of the loop will not change.• The minimum translational kinetic

energy at the top of the loop willnot change.

• But in addition to translationalkinetic energy, there is now alsorotational kinetic energy.

• The minimum mechanical energyis at the top of the loop has thusincreased.

• The required minimum heightmust thus have increased.

• OK, let’s now calculate by howmuch the minimum height hasincreased.

hR



• The total kinetic energy at the topof the loop is equal to

• This expression can be rewrittenas

• We now know the minimummechanical energy required toreach this point and thus theminimum height:

hR

Note: without rotation h ≥ 25/10 R !!!

K f =1

2I! 2

+1

2Mv

2=1

2

I

r2+ M

"#$

%&'v2

K f =1

2

2

5M + M

!"#

$%&v2=7

10Mv

2

h !27

10R

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Torque and rotational motion.

• Let’s test our understanding of the basic aspects of torqueand rotational motion by working on the following conceptproblems:

• Q17.1

• Q17.2

• Q17.3

• Q17.4

• Q17.5


Torque.

• The torque associated with aforce is a vector. It has amagnitude and a direction.

• The direction of the torque can befound by using the right-handrule to evaluate r x F.

• For extended objects, the totaltorque is equal to the vector sumof the torque associated with each“component” of this object.


Angular momentum.

• 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.

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Angular momentum.

• 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!


Angular momentum.

• 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!


Angular momentum.

• An object does not need to carryout rotational motion to have anangular moment.

• Consider a particle P carrying outlinear motion in the xy plane.

• The angular momentum of P(with respect to the origin) isequal to

and will be constant (if the linearmomentum is constant).

P

θ

p

r

y-axis

x-axisr⊥

!L =!r !!p = mrvsin" k̂ =

= mvr#k̂ = pr

#k̂

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Conservation of angular momentum.

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

• Consider what happens when the net torque is equal to 0:

dL/dt = 0 → L = constant (conservation of angular momentum)

• When we take the sum of all torques, the torques due to theinternal forces cancel and the sum is equal to torque due to allexternal forces.

• Note: notice again the similarities between linear and rotationalmotion.

d!L

dt=d

dt

!r !!p( ) = m

!r !

d!v

dt+d!r

dt!!v

"#$

%&'=

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

!F( =

!)(



• 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 τare calculated with respect to thecenter of mass of the object (note:center of mass can accelerate).

!!" =

d!L

dt



• Ignoring the mass of the bicyclewheel, the external torque will beclose to zero if we use the centerof the disk as our reference point.

• Since the external torque is zero,angular momentum thus shouldbe conserved.

• I can change the orientation ofthe wheel by applying internalforces. In which direction will Ineed to spin to conserve angularmomentum?

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Done for today!More about rotational motion on Thursday.

The Orion Nebula from CFHT Credit & Copyright: Canada-France-Hawaii Telescope, J.-C. Cuillandre (CFHT), Coelum

Documents

Rotational Motion and Angular Momentum.teacher.pas.rochester.edu/phy121/LectureSlides/Lecture17/Lecture17... · describe linear and rotational motion. ... the linear momentum. •Note: