rotation, ang momentum

7/29/2019 Rotation, Ang Momentum

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LECTURE 25

Rotational Dynamics and Angular Momentum

Introduction

To introduce the subject of this lecture,consider this spinning wheel suspended by a ropetied to one end of its axle

RotatingBicycle wheel

Weight ofbicycle wheeel

Suspension

cord

The curious thing about this wheel is that it

doesnt fall so that its axis becomes verticalas, it would if it werent spinning. This despitethe fact that the weight of the wheel, which acts atthe center of the wheel, gives a torque whichshould do just that.


Suspension

cord

But rather, when the wheel is spinning theaxle turns about the suspension point so that theplane of rotation of the wheel remains vertical.

Weight of

bicycle wheeel

Suspension

cord

Some of you may recognize this to be anexample of the gyroscope, a central piece ofequipment for navigating in space and often a toysold in science shops.

But did you actually ever figure out how itworks? Why it does what it does? In this lectureI hope to lead to some understanding of that andto show the importance of this understanding inbiomechanics.

The Equations of Circular Motion

The basic kinematic equations of circularmotion were introduced earlier in the course.These are the angular displacement expressed inradians and the angular velocity expressed inradians per second. These have their directanalogs in linear displacement, s, and linearvelocity, v.


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Physics 101A - Physics for the life sciences 2

x

y

r

LinearDisplacement

s AngularDisplacement

Linear Velocity v Angular Velocity

If the velocities are constant

s vt t = =

Although I have no yet emphasized it in thiscourse, angular acceleration is defined as therate of change of angular velocity, just as linearacceleration is defined as the rate of change oflinear velocity.

r

LinearDisplacement



LinearAcceleration

a AngularAcceleration

If the accelerations are constant

s v t at t t o o= + = +

1

2

1

2

2 2

Thus the equations relating lineardisplacement, velocity and accelerations have theirdirect analogs in angular motion.

Continuing from a study of kinematics tomechanics, the concept of a force in linearmechanics has its analog, torque, in rotational

mechanics.

LinearDisplacement



LinearAcceleration


Force F Torque

That torque is the direct analog of force isshown by the work involved in pushing an object

in a circle.

r Distancetravelled

= r

F

The distance the force moves the object isr . The work the force does is then F r .Rearranging this gives

W Fr T = =

LinearDisplacement



LinearAcceleration


Force F Torque

Work sF Work

Power vF Power T


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Lecture 25 - Rotational Dynamics and Angular Momentum 3

Continuing on to dynamics, the ubiquitous

F ma=

has a direct analog in rotational dynamics. Thisis shown by again considering the case of anobject being forced to move in a circle.

F

r

a

As an example, suppose the object of mass6 kg was moved in a circle of radius 0.1 m by aforce of 600N. The tangential acceleration of theobject, that is the acceleration along thecircumference of the circle, is calculated from theequation

F maa

t

t

=

= =600 6 100 2/ m/s

This tangential acceleration is related to theangular acceleration is the same way that thelinear displacement and velocity are related to theangular displacement and velocity:

s r

v r

a rt

=

=

=

This results inF mr=

Multiplying both sides of this equation by theradius r results in

Fr mr=2

The left hand side of this equation is now thetorque. The equation can therefore be made into

a form analogous to F= ma by representing theproduct mr2 by the symbolI.

T I=

This symbol can also be used to simplify theequation for the kinetic energy of rotationalmotion.

Kinetic Energy = = ( ) =1

2

1

2

1

2

2 2 2 2mv m r mr

Kinetic Energy =1

2

2I

So our table of analogies has become evenmore extensive:

LinearDisplacement


LinearVelocityv

AngularVelocity

LinearAcceleration


Force F Torque

Work sF Work

Power vF Power T

Dynamics F=ma Dynamics =

KineticEnergy

1

2

2mvKineticEnergy

1

2

2I

The product mr2 is so important in dynamicsthat it is given its own name; The moment ofinertia. Again the word moment here meansimportance. It implies that the: moment ofinertia is the importance of the inertia.Essentially, the importance of the mass of anobject is proportional to the square of its distancefrom the center of rotation.

The only item left in linear dynamics forwhich there should be an analog in rotationaldynamics is then momentum. Is there a rotationalconcept analogous to linear momentum?

Indeed there is, and for reasons I will try tomake clear later, it is the most important conceptthere is regarding rotational motion.

Following the sequence above it would seemthat if linear momentum is the product mv, thenangular momentum should beI. But is thisreasonable?

The momentum of an object is changed by aforce action on it. In fact, the rate of change of


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the momentum of an object is equal to theunbalanced force on the object.

Fdp

dt=

For the analogy with rotational motion to be

complete then the rate of change of angularmomentum would have to be equal to theunbalanced torque on the object. In equationform:

Rate of change of angular momentum

= Rate of change ofI

IfIis a fixed quantity, such as when a massm moves at a constant radius from a center ofrotation, then the only thing that can change is .Since the rate of change ofis just , then

Rate of change of angular momentum = I

This is, of course, just the torque on the objectand the table of equivalences is complete.

LinearDisplacement


LinearVelocity

v AngularVelocity

LinearAcceleration


Force F Torque

Work sF

Power vF T

Dynamics F=ma =

KineticEnergy

1

2

2mv

1

2

2I

Linearmomentum

mv Angularmomentum

I

Angular momentum is such an importantquantity that by International agreement it isdesignated asL.

An Application of the Equations of CircularMotion

At first it might seem that the equations ofangular motion are just another way ofexpressing motion that could be just as easilyexpressed, perhaps even more so, by the moreeasily understood linear equations. To show howthey can be used to actually simplify the analysisof a motion that can indeed be described by linear

equations consider again the pendulum. You mayrecall, with a feeling that is certainly not nostalgia,how the analysis of this device was carried out inthe linear equations. (If you like self-flagellationyou might go back to the notes for Lecture 21.)Now proceed with the analysis using rotationaldynamics.

m

W= mg

Wsin

l

The torque of the weight of the object is thecomponent of the weight that is perpendicular tothe string times the string length:

T mgl= sin

Here the sign is negative because the torque isclockwise. In the small angle approximation thesine is equal to the angle in radians and so

T mgl= sin

But the torque is just the moment of inertiamultiplied by the angular acceleration, and themoment of inertia is just the mass times thelength of the pendulum squared

T I ml= = 2

so that

=mgl ml 2

Flipping the equation left to right, doing alittle manipulation and noting that is the secondtime derivative ofgives


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d

dt

g

l

2

2

=

Since the equation for simple harmonicmotion is

ddt

2

22

=

the equation for the angular frequency of thependulum is

2

= =g

l

g

l;

This is considerably more straightforwardthan the treatment by linear dynamics. Of course,one of the reasons this is so is that the motion ofthe mass on the string is in a circle.

Power and the Equations of Circular MotionAs an example of the use of the rotational

equations of motion in power calculations,consider a set of specifications for a NorthAmerican automobile engine; peak torque of 220ft-lbs at 3600 rpm and 200 HP at 5000 rpm.Why is the torque peak at 3600 while the peakHorsepower (it must the peak horsepower,otherwise why would they advertise it) at 5000rpm?

First it is necessary to wade into the thicket ofunits that gives engineering in North America acertain mystic. Converting the information intothe units of the World (outside the US), a ft-lb(that is a foot-pound in real English) becomes0.305m times 0.454 kg or 0.1386 kg-m. This isconverted to the scientific N-m by multiplying by9.8 to get 1.36 N-m. Thus 220 ft-lb is 298.9 or,to the accuracy of these car specifications,300 Nm. The power of this torque at 3600 rpmis T. This is 300 2 3600/60 = 113 kWatt.Since 1 HP is 746 Watt, then the power of theengine at 3600 rpm is 152 HP.

How come the peak engine power was not atthis peak torque? Because while the torque is

less at higher engine speeds, the higher speedgives more horsepower even at this reducedtorque. This can be checked by calculating thetorque that comes with the peak HP at 5000 rpm.Here the calculation is the reverse of that we justdid. 200 HP is 200 746 = 149.2 kW.5000 rpm is 5000/60 2 = 523.6 rad/s and so149.2 kW is achieved with only 149200/523.6 =285 N-m, or 285/1.36 = 209 ft-lb of torque.

There are many such examples whereknowledge of the rotational equations of motionis necessary to understand the motion. However,the most important aspect of rotational motionthat I want you to concentrate on here is that ofangular momentum. Is it, like its linearcounterpart, conserved?

The Conservation of Angular Momentum

As a demonstration that introduces theconservation of angular momentum, consider thecase of me spinning in a chair with a very goodbearing that allows me to spin for severalminutes.

I now take two 4-kg masses, one in eachhand, and, holding them as far from my body as Ican, I use my foot against the support structure togive myself a gentle spin, of about 2 seconds perrevolution.

2 s perrevolution


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I now pull the masses in as close as I can tomy body.

Less than a secondper revolution

It is easy to see that my rotation has sped up.You can estimate it to be now less than a secondper revolution.

But how could this have happened? I appliedno torque to my body, I just pulled the masses intoward the center. Any torque to my system (me,the chair and the masses) would have to beapplied from outside, such as the torque I had toapply to get myself going at the beginning, whichcame from my foot being pressed against thesupport structure.

The solution to this conundrum comes inconsidering the moment of inertia of the completesystem before and after pulling in the weights.Pulling in the weights, and my arms broughtmore of the mass toward the center, therebydecreasing the overall moment of inertia of thesystem. Since angular momentum is the productof the moment of inertia times the angularvelocity and the moment of inertia is reduced thenfor the angular momentum to be preserved theangular velocity has to increase:

L L

I I

I I

before after

before after

before before after after

=

( ) = ( )

=

But do the numbers add up? To get anestimate of the actual moment of inertia of mysystem we have to consider how moments ofinertia are calculated.

How to Calculate Moments of Inertia

As an introduction to the calculation ofmoments of inertia, consider a body made up of anumber of masses, each small enough to beconsidered a point but distributed as in thediagram below.

0.5 kg

2 kg

1 kg1

3kg

1.5 m 0.5 m

2m

0.5m

CM

The total moment of inertia of the structureabout its center of mass is the sum of theproducts of each mass multiplied by the square ofits distance from the center. Going clockwisefrom the upper mass this is

Itotal = + + +

= + + +

=

0 5 2 1 0 0 5 2 0 0 51

31 5

2 0 0 25 0 5 0 75

3 5

2 2 2 2

2

. . . . . .

. . . .

. kg m

Note that the 0.5 kg mass has 4 times themoment of inertia of the much larger 2 kg massbecause of its greater distance from the masscenter. This shows the meaning of moment inmoment of inertia. Masses that are far fromthe center are the ones whose inertia has the

greatest moment, i.e. the greatest importance.This sort of calculation can be carried out

with any structure of masses. If the structure is ashape that can be described by a mathematicalfunction then the moment of inertia can becalculated using integral calculus (whichessentially sums up to contributions of an infinitenumber of elements of the system). This is not asubject for this course, and is really only a subject


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of importance in mechanical engineering, andeven there modern computer techniques havemade calculus of less and less importance even inthat subject. However, whatever the origin it isuseful for what follows to know a few of thesimple formulas.

First consider a ring:

I=MR

R

Mass =M

2

RING

All of the mass of a ring is at the same radiusfrom its center. The contribution of each part ofthis mass to the overall moment of inertia of the

ring is therefore the mass of that part times thesquare of the ring radius. The total moment ofinertia of the hoop is therefore its total mass,M,multiplied by this square.

I MRring ring=2

Now consider a disk spinning about itscenter.

I= MR

R

Mass =M

2

DISK

1

2

This disk can be considered to be made up ofmany rings, all very thin and all contributing tothe total moment of inertia in proportion to thesquare of their radii. Since the inner rings have asmaller radius a ring of the same mass willcontribute less to the moment of inertia. Theactual contribution can be calculated usingintegral calculus but again that is not a subject for

this course. What is important here is theformula that results:

I MRdisk disk =1

2

2

Of course this formula applies to a long rodrotating about its axis, even though it does notlook like a disk. (It can be thought of as a wholestack of disks.)

R

Mass =M

I= MR21

2

Formulae for the moment of inertia for almostany object that has a regular geometry can befound in handbooks of Physics, Chemistry andEngineering. For example the moment of inertiaof a sphere rotating about its center is 2/5MR2.For the discussion that follows, two particular

formulae are of importance. First, the moment ofinertia of a long thin rod rotating about its center.

l

Mass =M

I= Ml21

12

Its moment of inertia is

I Mlrod =1

12

2


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And finally, the moment of inertia of a rodrotating about one of its ends:

l

Mass =M

I= Ml21

3

Its moment of inertia is

I Mlrod =1

3

2

Note that the moment of inertia of an objectdepends on the choice of axis about which itrotates. In general, the more of the mass that is

distributed away from the center of rotation, thegreater is the moment of inertia.In summary, the total moment of inertia of an

object is the sum of the moments of inertia of itscomponent parts.

In most of the examples above the moment ofinertia has been calculated for rotation about thecenter of mass. However, there is a very usefulrule, that can be derived from integral calculus,that the moment of inertia for rotation of an objectabout any center is the moment of inertia forrotation about the center of mass plus the mass ofthe object multiplied by the square of the distance

of the center of mass from the center of rotation,

CM

R

Center ofrotation

Mass =M

Object

I I MRCM= +2

You can check this formula by comparing theequations for the rotation of a rod about its centerof mass and about one of its ends. For themoment of inertia about its end we merely have toadd the mass of the rod times the square of halfits length:

I Ml Ml

Ml Mlend = +

= +

=1

12 2

1

124

1

3

2

2

2 2

The Moment of Inertia of Humans

You can use the rules just given forcalculating moments of inertia to estimate themoment of inertia of human beings in variousactivities, such as figure skating. First considerthe range of possible values for an averagehuman. Consider a person of mass 75 kg andheight 1.8 m. Modeling that person as basically acylinder of diameter 0.25 m, the moment ofinertia about a vertical axis through the center ofmass gives a moment of inertia

I Mr= = =1

2

1

275 0 125 0 62 2. . kg - m2

Now consider that same person stretched outhorizontal and rotating about a vertical axisthrough the center of mass. The moment ofinertia is now

I Ml= = =112

112

75 1 8 202 2. kg - m2

These are the two extremes of the moment ofinertia of the person rotating about the personscenter of mass. They cover a range of about 30. In between will be the situation where theperson is rolled into a ball. Using the equationfor the moment of inertia for a sphere and notingthat the radius of a 75 kg sphere of body masswould be about 0.25 m gives

I Mr= = =2

5

2

5

75 0 25 22 2. kg - m2

Now consider a typical figure skater. Againassume the skater weighs 75 kg and is 1.8 m tall.A simplified mathematical model of such a skaterpreparing for a spin is shown below.

Mechanical model of figure skater(starting position for a spin)

Body 50 kg

Upright leg15 kg

Arms10 kg

1.8 m

0.3 m


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The moment of inertia of the skater in thisposition is approximated by a vertical cylinder forthe upright leg with the skate that the skater isspinning on, a horizontal cylinder representingthe upper body and the other leg, and anothertransverse cylinder representing the outstretched

arms. The radius of the upright leg cylinder willbe about 0.1 m. The moments of inertia of theupright leg and the horizontal body are:

I Mr Ml= + = +

= + =

1

2

1

2

1

215 0 1

1

1250 1 8

0 075 13 5 13 6

2 2 2 2. .

. . . kg - m2

It is seen that the moment of inertia of theupright leg is insignificant.

To the above figure we must add the momentof inertia of the outstretched arms:

Mechanical model of figure skater(view from above)

Arms 10 kg

1.6 m

This is the moment of inertia of the armsabout the center of mass of the arms plus themass of the arms times the square of the distanceof the center of mass of the arms from the centerof rotation.

I Ml Mr= +

= +

= + =

1

12

1

1210 1 6 10 0 3

2 1 0 9 3 0

2 2

2 2. .

. . . kg - m2

The total moment of inertia of the skater isthen about 17 kg-m2. Comparing this to the

moment of inertia calculated earlier for a personrotating about an axis along the body length, youcan see that it is about 30 times greater.

This implies that if the skater executes thismaneuver his or her angular velocity mustincrease by 30 times if angular momentum is tobe preserved. Starting at a slow turn of about 3

seconds per revolution will therefore result in anincredible 10 revolutions per second at the finish.Of course, this would imply no loss of angularmomentum due to friction of the skate on the icebut a good figure skater can indeed achieveincredibly high angular velocities by simplychanging their moment of inertia during a spin.(You might notice that due to the tremendouscentrifugal forces in a typical spin the skater hasgreat difficulty in bringing the elbows and thefree skate inward toward the center of the body.The ratio of final angular velocity to initial

angular velocity is therefore probably more like10:1Return now to my very tame rotation in a

comfortable chair on a rotary bearing. A crudemathematical model of my system is shownbelow.

Mechanical model of me on chair

(chair mass is negligible)

0.2 m

0.2 m

0.6 m

Upper body30 kg(R = 0.1 m)

Arms10 kg

Thighs 25 kg

Lower legs10 kg(R= 0.05 m)

Rotatingplatform

5 kg1.0 m

Here my upper body in a sitting position isrepresented as an upright cylinder of radius 0.1 mand mass 30 kg at a distance of 0.2 m from thecenter of rotation, my thighs as a 0.6 m long


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horizontal cylinder of mass 25 kg with its centerof mass at the center of rotation and my lowerlegs are represented by a vertical cylinder ofradius 0.05 m and of mass 10 kg at a distance of0.2 m from the center of rotation. In addition, mytwo arms are represented by a cylinder 1.6 mlong and of mass 10 kg with its center of mass at

0.2m from the center of rotation. In additionthere are, of course, the two masses of 4 kg eachat a distance of 0.8 m from the center of rotation.

Arms 10 kg

1.6 m

4 kg 4 kg

Mechanical model of me on chair(view from above)

The moment of inertia of my upper body isthat of a cylinder rotating about an axis that is0.2 m from its own axis. That is

1

230 0 1 30 0 2

0 15 1 2 1 35

2 2 +

= + =

. .

. . . kg - m2

The moment of inertia of my thighs is that ofa cylinder rotating transverse to its axis. That is

1

1225 0 3 0 122 =. . kg - m2

The moment of inertia of my lower legs isthat of a cylinder rotating about an axis that is0.2 m from its own axis. That is

12

10 0 05 10 0 2

0 0125 0 4 0 41

2 2 +

= + =

. .

. . . kg - m2

The moment of inertia of the platform is thatof a disk rotating about its center of mass. It is

1

25 0 5 0 6252 =. . kg - m2

The total moment of my system, not countingmy arms and the 4 kg masses, is then 2.5 kg-m2.For my initial moment of inertia I must add themoments of inertia of my arms and the masses.For my arms, as for the figure skater, it is themoment of inertia for a rotation about their centerof mass plus their mass times the square of the

distance of their center of mass from the center ofrotation.

1

12

1

1210 1 6 10 0 2

2 1 0 4 2 5

2 2

2 2

Ml Mr+

= +

= + =

. .

. . . kg - m2

For the masses the moment of inertia issimply the sum of the two masses times thesquare of their distance from the center of

rotation.Mr2 28 0 8 5 12= =. . kg - m2

The chair is of plastic and its mass andmoment of inertia are negligible compared to theremainder of the system.

The total moment of inertia of my systemwith my arms stretched out is then

Iinitial = + +

=

=

2 5 2 5 5 12

10 12

10 2

. . .

.

kg m (to the accuracy of the model)

The vertical view of when I brought theweights in toward my body is shown below.

Arms 10 kg

1.0 m

4 kg

Mechanical model of me on chair(arms pulled in)

4 kg

r = 0.2 m


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My arms now formed a cylinder of onlyabout 1 meter length. Their moment of inertia isthen

1

12

1

12 10 1 0 10 0 2

0 83 0 4 1 2

2 2

2 2

Ml Mr+

= +

= + =

. .

. . . kg - m2

The moment of inertia of the masses is now

Mr2 28 0 2 0 32= =. . kg - m2

The total moment of inertia of my system isnow 2.5 + 1.2 + 0.32 = 4.02 = 4 kg-m2 . Thismoment of inertia is only 40% of what it wasbefore I brought the weights inward.Conservation of angular momentum thereforeimplies that my rotational velocity after I brought

in the weights must have been 2.5 times what itwas before. This indeed is compatible with whatwas observed.

Relationship of The Law of Conservation ofAngular Momentum to the OtherConservation Laws.

We now have three conservation laws:

The Conservation of EnergyThe Conservation of Linear Momentum

The Conservation of Angular Momentum

Do we really need all three? Cant at least thisnew intruder be shown to be just a result of othertwo?

No it cannot. And to show this consider acase of two masses of, say, 2 kg each rotating atan angular velocity of 5 rad per second at adistance of 1 meter from the center of rotation.Now suppose these masses moved inward to0.5 m from the center of rotation. By the law ofconservation of angular momentum the momentof inertia has decreased by a factor of 4 from

4 kg-m2 to 1 kg-m2 and the angular velocitywould then have increased by a factor of 4 to 20radians per second.

But this means that the velocity of each masshas actually increased, from initially 5 m/s tofinally 10 m/s. Regarding momentum as a scalarwould mean that the momentum has actuallydoubled!

We only save the Law of Conservation ofMomentum by noting that momentum is a vectorand since the two masses are moving in oppositedirections the total momentum both before andafter was zero. So the Law of Conservation ofAngular Momentum is completely independent ofthe Law of Conservation of Linear Momentum.

But what about the Law of Conservation ofEnergy? Since the velocities of the masses hasactually doubled, their kinetic energy hasquadrupled. Where did this energy come from?

Go back to the case of me pulling in theweights to get them closer to my body. To do thisI did have to pull against them. This was so as tocreate the radial acceleration that comes withcircular motion. This means that I had to dowork on the weights in order to pull them in. Itwill not be part of this course but if you evaluatedthe area under the force displacement graph for

me pulling in the weight you would find that thiswork is just equal to the kinetic energy theweights gained as they were being pulled in.

So indeed, all three laws are independent andall three are applicable.

The General Motion of Bodies in Space.

I will not pretend to derive the generaltheorems about the motion of bodies in space, butrather state them, hoping that by now you will seethem to be reasonable. They are:

1. The center of mass of a body will beaccelerated according to the equation

F ma=

where Fis the vector sum of all the forcesacting on the body.

2. The angular acceleration of the bodyabout its center of mass will beaccording to the equation

T I=

where Tis the sum of all the torques actingacting on the body.

An example of the consequences of theselaws is that of a diver. After leaving the board theonly force of any importance on the divers bodyis his or her weight, which acts on the center ofmass. If you follow this center of mass throughall the contortions that the diver may make you


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will see that it follows the smooth parabola of asimple mass thrown into the air to fall into thepool.

Also, once the diver has left the board there isno net torque on his or her body. Therefore, nomatter what contortions the diver may undergo,his or her angular momentum remains constant.In the picture to the left you can see that when thediver is curled into the tuck position, where hermoment of inertia could be modeled to be that of

a disk of about radius 0.2 m, she has a moment ofinertia, if her mass is 60 kg, of about 1.2 kg-m2.When stretched out as she enters the water hermoment of inertia, if she is 1.6 m tall, would beabout 13 kg-m2, or about 10 times greater. Herangular velocity will therefore be reduced by 10times from her tuck position to her dive position.This is very important in that it allows her tocomplete her somersaults in a very short timewhile enabling her to almost stop her rotation justbefore she enters the water. The sequence ofpictures to the left are approximately 0.1 second

apart and so you can get a rough estimate of herrotational velocity at various points in hertrajectory.

The conservation of angular momentum takesa particularly bizarre twist (pun intended I mustadmit) in the well-known case of dropping a catfrom a height of about a meter when it is held byits paws upside down. (I assure you that there hasnever been a cat that has been injured by thisexercise.) While it is in this short fall the catwill execute a maneuver in which it lands on itsfeet. If angular momentum is preserved, and ifyou allowed the cat to have no angular

momentum when you released it, how could ithave achieved the angular rotation required for itto get its feet under itself?

In fact, time-lapse photographs of thismaneuver show that the cat achieves it by rotatingpart of its body one way and the rest the otherway, thereby keeping its total angular momentumto be zero. As a result of this twisting it can get itslegs, which have very little moment of inertiacompared to the rest of its body, to rotate therequired 180 degrees while the remainder of itsbody rotates only a few degrees the other way.

And the cat does this without knowing

anything about the Law of Conservation ofAngular Momentum. It is just that the Law ofConservation of Angular Momentum requiresthat it do it that way. The same with humans.Very few Olympic divers know anything aboutthe Law of Conservation of Angular Momentum.However, if you are to understand anything aboutthe dynamics of such maneuvers and the strains


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they can put on body parts, then you shouldknow something about that Law.

Angular Momentum as a Vector!

Up to now you may have been taking solacein the appearance that while angular momentumwas somewhat abstract it was at least not a vector.

Sorry but not true. Consider anotherdemonstration. While sitting on the magic chair Ihave this heavy bicycle wheel in my hand. I thenput it into a spin in a horizontal plane, i.e. with itsaxis vertical.

I then lift my feet from the floor and, while Ihave no connection to the floor, I invert the axisof the wheel. You can see that the result is thatme and the chair are now spinning.

Where did I get this spin? It representsangular momentum; angular momentum that I didnot have before I turned over the wheel.

The only way I can recover the Law ofConservation of Angular Momentum is to saythat by turning over the wheel I changed the signof its angular momentum. Before I turned the

wheel it had an angular momentum that, say, waspositive. By turning it over I made this angularmomentum negative. Therefore I changed itsangular momentum by an amount that was twiceits initial value. If, for example, it was 2 kg-m2-rad/sec at the start then it would have been2 kgm2-rad/sec at the end. In order for theangular momentum of the system to be preserved,I would need to have 4 kg-m2-rad/sec of angularmomentum to make the new total the same asbefore, i.e. 2 kg-m2-rad/sec.

But what if I only turned the wheel half-way

so that it was spinning in a vertical plane? Youcan see my spinning is now only about half whatit was when I turned the wheel completely. Andif I only turned the wheel a little bit, I only turneda little bit.

By now you are probably convinced that thereis a vector lurking in the spin of this wheel. Andindeed there is. Angular momentum has amagnitude. All it needs to be a vector is adirection. That direction is along the axis ofrotation. By convention, the angular momentumis said to be positive in the direction of so-calledright-handed rotation. That is a rotation which, if

you look along the axis results in acounterclockwise motion. Another way ofthinking about it is to consider the thumb andforefingers of your right hand. By curling yourfingers your thumb will point in the direction ofthe angular momentum vector for a circularmotion in the direction of your forefingers.

So if the bicycle wheel was initially rotatingcounterclockwise, looking down, then its angularmomentum vector was pointing up. When Iturned it over it was spinning clockwise lookingdown and its angular momentum vector wasdown. This negative change in its angular

momentum vector required a positive change inmy angular momentum vector to counteract it.That means that I turned clockwise looking down.

Finally you may be in a position tounderstand the hanging wheel at the beginning ofthe lecture. This wheel was spinning in a verticalplane so that its angular momentum vector washorizontal.


14/14



Suspension

cord

T

L

Change inLproduced by T

New L

Now if torque is the rate of change of angularmomentum then the torque on this system will bechanging this angular momentum vector. Thistorque is the weight of the wheel times thedistance of its center from the point of support.

Suspension

cord

s

W

T=Ws

But what is the direction of this torque?Like angular momentum, torque is also a

vector and points in the direction of the axis ofrotation. In the diagram for the rotating wheelthis is perpendicular to the axis of the wheel andinto the plane of the above drawing, as shownabove.

Therefore the torque tries to change theangular momentum vector by adding to it in thedirection of the torque. This means that theangular momentum vector must be forced to pivotabout its origin in the manner shown in the

diagram. The action of the weight of the wheel istherefore to turn it in a horizontal circle.

The Origin of Conservation of AngularMomentum

Like the Conservation of Energy and theConservation of Linear Momentum, the

Conservation of Angular Momentum can bederived from a basic symmetry principle. And inthis case it is deceptively simple; the laws of theUniverse do not change when you look at theUniverse from a different angle.

This may seem to be such an obviousassumption that it should not have a consequenceas great as a fundamental Law. However, it does.And indeed the consequences of the assumptiongo even beyond the conservation of angularmomentum. This is because of the simple factthat when you turn a complete circle the universeis exactly back to what it was before you started.

The logic of this is very abstract but is central tomodern quantum mechanics. Whatever, the endresult is that angular momentum cannot becontinuous but must come in discrete units, justas matter is made of atoms. The unit of angularmomentum is now referred to as PlanksConstant, after Plank who discovered the discretenature of atomic interactions, and is the mustfundamental quantized unit (more fundamentalthen the quantized unit of matter; the atom). ByInternational agreement it is given the symbol hand has the value

h = 6 626176 10 24. Joule - seconds

(It may seem strange to give the units ofh as J-s,instead of the kg-m2-rad/s that comes fromI.However, note that kg-m/s2 is ma, of which theunit of that of force, i.e.the N. Since kg-m2-rad/scan be expressed as (kg-m/s2) times (m-s-rad)and since radians are unitless, the units becomeN-m-s. Since N-m are Joules the units ofangular momentum become simply J-s.)

Relevant material in Hecht

Chap 8 - Section 8.11. Also figure 8.28 andassociated text. Also recommended are theDiscussion Questions.

rotation, ang momentum

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