Chapter 9 Angular Momentumnewton.uor.edu/facultyfolder/eric_hill/Phys231/09...296 Chapter 9: Angular...

Chapter 9

Angular Momentum

9.1 Rotational kinetic energy and moment of inertia 296

9.2 Angular momentum 2979.2.1 Another way to calculate rotational kinetic energy 299

9.3 The angular momentum principle for a rotating object 299

9.4 Applications to systems with no torques 3019.4.1 An ice skater’s spin 3019.4.2 The dumbbell demonstration 3019.4.3 A high dive 302

9.5 Translational angular momentum 3039.5.1 Visualizing the magnitude of translational angular momentum 3039.5.2 Another view of rotational angular momentum 3049.5.3 Another example of conservation of angular momentum 3049.5.4 An alternative way to calculate translational angular momentum306

9.6 Direction of angular momentum 3069.6.1 The vector cross product 308

9.7 Angular momentum of multiparticle systems 3109.7.1 Rotational angular momentum as a vector 3119.7.2 More exercises on angular momentum 312

9.8 The angular momentum principle 3139.8.1 Multiparticle systems 3149.8.2 Summary: Angular momentum and torque 316

9.9 The three fundamental principles of mechanics 316

9.10 Conservation of angular momentum 317

9.11 Atomic and nuclear angular momentum 3189.11.1 The Bohr model of the hydrogen atom 3189.11.2 Particle spin 3229.11.3 Consequences of angular momentum quantization 322

9.12 Calculating torque 3239.12.1 Example: Falling object 3249.12.2 More exercises on the angular momentum principle 325

9.13 Applications to systems with nonzero torques 3259.13.1 A meter stick on the ice 3259.13.2 A puck with string wound around it 3279.13.3 Static equilibrium 327

9.14 *Gyroscopes 3289.14.1 *Magnetic Resonance Imaging (MRI) 3309.14.2 *Precession of spin axes in astronomy 331

9.15 *More complex rotational situations 333

9.16 *Vector rate of change of a rotating vector 333

9.17 Summary 334

9.18 Example problem: A satellite with solar panels 335

9.19 Review questions 336

9.20 Homework problems 337

9.21 Answers to exercises 345

Copyright 2003 John Wiley & Sons. Adopters of Matter & Interactions by Ruth Chabay and Bruce Sherwoodmay provide this revised version of Chapter 9 to their students.

296 Chapter 9: Angular Momentum

Chapter 9

Angular Momentum

Rotational motion is such an important type of motion that special tech-niques have been developed to describe and predict it. A particularly impor-tant concept that arises from the study of rotational motion is “angularmomentum,” the rotational analogue of the ordinary “linear” momentumwith which we are already familiar. Angular momentum is conserved, just asenergy and ordinary momentum are. In the atomic world angular momen-tum is quantized, which has far-reaching consequences for the structure ofatoms and nuclei.

9.1 Rotational kinetic energy and moment of inertia

An important special case of rotational motion is one where a system is ro-tating on an axis, with all the atoms in the system sharing the same “angularspeed” in radians per second but with different linear speeds in meters persecond, depending on their distances from the axis (Figure 9.1). This is thesituation for a rigid object.

As we saw in Chapter 2, angular speed, normally denoted by ω (lowercaseGreek omega), is a measure of how fast something is rotating. If an objectmakes one complete turn of 360 degrees (2π radians) in a time T, we saythat its angular speed is radians per second. The time T is calledthe period. We measure angular speed ω in the “natural” units of radiansper second rather than degrees per second, because the fundamental geo-metrical relationship between angle and arc length (arc length = rθ) is validonly if the angle θ is measured in radians. The speed of an object at a per-pendicular distance from the center is the circumference divided by theperiod:

The kinetic energy of a rotating rigid object like the one shown in Figure 9.1is . Since the speed of each mass is , wecan write the rotational kinetic energy like this:

=

The quantity in brackets is called the “moment of inertia” and is usually de-noted by the letter I:

MOMENT OF INERTIA

With this definition of moment of inertia we have the following convenientway to calculate rotational kinetic energy:

ROTATIONAL KINETIC ENERGY

v1 = ωr1 v2 = ωr2

r1

r2

r3

r4

v4 = ωr4 v3 = ωr3

m4

m2

m3

m1

ω 2πT------=

Figure 9.1 A case of rigid rotation about anaxis with angular speed ω. Note the differ-ent speeds, with vi = ωri.

ω 2π T⁄=

r⊥

v2πr⊥

T------------

2πT------

⎝ ⎠⎛ ⎞ r⊥ ωr⊥= = =

12---m1v1

2 12---m2v2

2 12---m3v3

2+ + + ... v ωr⊥=

Krot12--- m1 r⊥1ω( )2 m2 r⊥2ω( )2 m3 r⊥3ω( )2 + ...+ +[ ]=

12--- m1r⊥1

2m2r⊥2

2m3r⊥3

2+ + + ...[ ]ω2

I m1r⊥12 m2r⊥2

2 m3r⊥32 m4r⊥4

2+ + + + ...=

Krot12---Iω2=

9.2: Angular momentum 297

Review the derivation of this formula and notice that this is not some newkind of energy. It is ordinary kinetic energy of the atoms that are movingwith respect to the axle. The new formula is simply a compactand convenient way of expressing the total kinetic energy of all the atomsin the case of an object that is in “pure rotation” (stationary center of mass).

As you saw in Chapter 7, the total kinetic energy of a system can be ex-pressed in general as translational kinetic energy plus kinetic energy rela-tive to the center of mass:

, where

For the system shown in Figure 9.1, the center of mass isn’t moving, so, and since there is no vibrational kinetic energy. In

Chapter 7 we often calculated rotational kinetic energy by using the energyprinciple. Now we have an additional way of evaluating rotational kineticenergy, in terms of how fast the object is spinning.

Example: Find the moment of inertia of a “linear” triatomicmolecule such as CO2 (three atoms in a row) consisting of a centralatom of mass M flanked by two atoms each of mass m (Figure 9.2).The distance between the centers of adjacent atoms is d.Remember that almost all of the mass in an atom is concentratedin the tiny nucleus, whose diameter is much smaller than d.

Solution: Using the definition of moment of inertia, we have

Even if the nucleus of the central atom has a large mass, itdoesn’t contribute to the moment of inertia. This reflects thefact that if the molecule rotates, nucleons in the tiny central nu-cleus have almost no rotational kinetic energy.

Ex. 9.1 What is the moment of inertia of a diatomic molecule suchas N2 around its center of mass if the mass of each atom is M andthe distance between nuclei is d? It helps to draw a diagram.Remember that the electrons hardly contribute to the moment ofinertia because their mass is so small.

Ex. 9.2 A bicycle wheel has almost all its mass M located in theouter rim at radius R. If it rotates on a stationary axle with angularspeed ω, what is the rotational kinetic energy of the wheel? Hint:It’s helpful to think of dividing the wheel into the atoms it is madeof.

Ex. 9.3 The moment of inertia of a uniform-density disk rotatingabout an axle through its center can be shown to be . Thisresult is obtained by using integral calculus to add up thecontributions of all the atoms in the disk (see Problem 9.3). Thefactor of 1/2 reflects the fact that some of the atoms are near thecenter and some are far from the center; the factor of 1/2 is anaverage of the square distances. A uniform-density disk whose massis 10 kg and radius is 0.4 m makes one complete rotation every 0.2s. What is its kinetic energy?

9.2 Angular momentum

A single particle moving at a slow speed (v << c) has translational kinetic en-ergy (a scalar) and momentum (a vector). We have seen in the

Krot12---Iω2=

K Ktrans Krel+= Ktrans12---Mtot vcm

2=

Ktrans 0= Krel Krot=

d

m M m

d

Figure 9.2 A “linear” triatomic molecule,such as CO2, with the three atoms in a row.The masses are concentrated in the tinynuclei (drawn much larger than the nucleireally are!).

I M 02( ) m d2( ) m d2( )+ + 2md2= =

12---MR2

12---m v 2 mv


previous chapters that both energy and momentum are useful concepts,and the energy principle and the momentum principle tell how the energyand momentum change as the result of influences by the surroundings.

Note that the rotating device shown in Figure 9.1 has zero momentum,because the momentum of a multiparticle system is , and the centerof mass of the device is located at the axle, which is stationary. But does thedevice have some kind of “rotational” momentum?

The answer is yes, and this new quantity is called “angular momentum.”There are two kinds of angular momentum, and the kind associated with anobject that is rotating but not translating (stationary center of mass), is “ro-tational angular momentum,” with the symbol :

ROTATIONAL ANGULAR MOMENTUM

The relationship between rotational kinetic energy and rotational an-gular momentum is very similar to the relationship between translation-al kinetic energy and ordinary momentum . In fact, isreally the z component (into or out of the page) of an angular momentumvector, which makes the analogy to ordinary momentum even stronger. Thislets us account for the two different possible directions of rotation of the de-vice:

DIRECTION OF ROTATIONAL ANGULAR MOMENTUM

Counterclockwise: positive, angular momentum vector in direction

Clockwise: negative, angular momentum vector in direction

It may seem odd to say that angular momentum is a vector pointing in the zdirection when the rotation is in the x-y plane. It’s simply a convenient wayto describe the orientation of a plane of rotation by describing a vector per-pendicular to that plane. Once you’ve done that, you can also describewhether the rotation is clockwise or counterclockwise (viewed from some-where along the axis, looking back toward the x-y plane) by saying wheth-er your descriptive vector points in the or direction.

Because it is difficult to sketch a situation in three dimensions on paper,whenever possible we will work with two-dimensional projections onto thex-y plane. A special notation is used to indicate “out of the page” ( ) and“into the page” ( ); see Figure 9.3. The symbol can be thought of asshowing the tip of an arrow (a vector) pointing out at you, while the symbol

represents the feathers of an arrow (a vector) pointing away from you.It can be easy to confuse the similar names and concepts “momentum”

and “angular momentum.” We will sometimes call the older, familiar con-cept “linear momentum” to help distinguish between the two concepts.

Example: A bicycle wheel of mass M and radius R has almost all itsmass concentrated in the rim. It spins on its axis, making onerevolution in T seconds. What is the magnitude of its rotationalangular momentum?

Solution:

Ex. 9.4 What are the units of moment of inertia I?

Ex. 9.5 What are the units of angular speed ω?

Ex. 9.6 What are the units of rotational angular momentum ?

Mtotvcm

Lrot

Lrot Iω=

12---Iω2

Iω12---m v 2 mv Lrot Iω=

ω +z

ω z–

+z+z z–Rotational angular momentum

counterclockwise, out of page (+z)

Rotational angular momentumclockwise, into page (–z)

ω

ω

Figure 9.3 2-D projections: symbols forrotational angular momentum out of pageand into page.

Lrot Iω MR2( ) 2π T⁄( )= =

Lrot

9.3: The angular momentum principle for a rotating object 299

Ex. 9.7 A uniform-density disk of mass 3 kg and radius 0.4 m spinswith angular speed 300 radians/s. What is the magnitude of itsrotational angular momentum?

Ex. 9.8 The moment of inertia of a uniform-density sphere can beshown to be . Approximating the Earth as a sphere of mass

and radius , what is the magnitude of theEarth’s rotational angular momentum? Remember that the Earthtakes 24 hours to rotate once.

Ex. 9.9 The Earth rotates from west to east. That is, if you lookdown on the Earth from above the North Pole, the Earth rotatescounterclockwise. What is the direction of the rotational angularmomentum vector for the Earth?

9.2.1 Another way to calculate rotational kinetic energy

For v << c, the kinetic energy of a particle can be calculated in two differentways:

or , since

The second formula is particularly useful in situations where we happen toknow the momentum but not the speed. Similarly, we can calculate the ro-tational kinetic energy of a rotating object in two different ways:

ROTATIONAL KINETIC ENERGY: ALTERNATIVE FORMULA

or , since

Again, the second formula is particularly useful in situations where we hap-pen to know the rotational angular momentum but not the angular speed.

Ex. 9.10 If an object has a moment of inertia and themagnitude of its rotational angular momentum is ,what is its rotational kinetic energy?

9.3 The angular momentum principle for a rotating object

By looking at how work changes rotational kinetic energy, we can show whyit makes sense to define something called “the rotational angular momen-tum” of a rotating object as , and what it takes to change that angular mo-mentum. In Figure 9.4 we apply a force to one arm of the system, at adistance r from the axis, and turn the object through a small angle . Weshow the force split into two components and , parallel and perpen-dicular to the arm.

? Which of these force components does no work on the system, andwhy?

will do no work, because it is at right angles to the movement of the pointof application of the force; there is no displacement in the direction of thisforce component and therefore it does no work.

We will calculate the amount of work in terms of the angle through whichwe apply our force. We can express the distance through which the force

acts as , where is the angle through which the system rotates. It

25---MR2

6 24×10 kg 6.4 6×10 m

Ktrans12---m v 2=

p 2

2m-------- p mv=

Krot12---Iω2=

Lrot2

2I--------- Lrot Iω=

5 kg m2⋅20 kg m2/s⋅

Iω

v1 = ωr1 v2 = ωr2

v4 = ωr4 v3 = ωr3

m4

m2

m3

m1Fr

∆θ

Figure 9.4 Apply a force to change therotational kinetic energy.

F||

F⊥

∆θF|| F⊥

F ||

∆lF⊥ r∆θ ∆θ


is a fact of geometry that the length of an arc of a circle is when the an-gle is measured in radians. For example, when the angle is radians, thelength of the arc is , the circumference of the circle.

Given the geometrical fact that we apply a force through a distance ,here is the energy principle applied to the current situation:

The quantity is called the “torque,” which is Latin for “twist”. The unitsof torque are , which technically is the same as joules, but since torqueis physically not the same thing as energy, we say that a torque has units of

rather than joules.We’ll see why rotational angular momentum is defined to be for a ro-

tating object if we look at the rate of change of rotational kinetic energy. Di-vide by the time it takes to do the work:

In the limit of infinitesimal time interval we have derivatives:

? See if you can do these derivatives with respect to time yourself.Note in particular that is simply the angular speed , and that

is a function of time, since we’re making the object spin faster.

Here’s the result of taking the derivatives with respect to time:

Cancel the common factor , and since the moment of inertia I of our ro-tating object isn’t changing, write the result like this:

In words, we see that the rate of change of the angular momentum of a ro-tating object is equal to the torque applied to that object. This is very similarto the momentum principle, which says that the rate of change of ordinary“linear” momentum is equal to the net force acting on the system.

If more than one force acts on the rotating object, we need to add up allthe torques due to all the forces:

THE ANGULAR MOMENTUM PRINCIPLEFOR A ROTATING OBJECT

To keep the derivation simple, we assumed that the moment of inertiawasn’t changing. However, a more complete treatment shows that the angu-lar momentum principle as stated here does correctly describe the motionof a rotating object even if the moment of inertia changes. Our result ismore general than our simplified derivation.

In reality the angular momentum principle is a vector principle, and whatwe have dealt with in this particular case is the z component of the angularmomentum (and the z component of the torque, which is also a vector).

r∆θ2π

2πrr∆θ

∆Krot F⊥ r∆θ( ) rF⊥∆θ= =

rF⊥N m⋅

N m⋅Iω

∆Krot

∆t------------- rF⊥

∆θ∆t-------=

dKrot

dt------------ rF⊥

dθdt------=

ddt----- 1

2---Iω2( ) rF⊥

dθdt------=

dθ dt⁄ ωω

Iωdωdt------- rF⊥ω=

ω

ddt----- Iω( ) rF⊥=

ddt----- Iω( ) net torque=

9.4: Applications to systems with no torques 301

9.4 Applications to systems with no torques

We will apply the new principle to some situations where no torques are ap-plied. If the net torque is zero, the angular momentum cannot change.

9.4.1 An ice skater’s spin

You may have seen an ice skater spin vertically onthe tip of one skate, with her arms and one legoutstretched, then pull her leg in and bring herarms to a vertical position above her head. Shethen spins much faster.

What’s going on? There is some frictionalforce of the ice on the tip of the skate, but thisforce is applied so close to the axis of rotationthat the torque is small. That is, is very smallbecause r is small and the frictional force is small.

? During the short time when the skaterquickly changes her configuration, what canyou say about the rotational angularmomentum? Why?

Small torque implies small change in angularmomentum per unit time, so in a short time theangular momentum of the skater will hardlychange.

? But if the rotational angular momentum hardly changes, how canthe skater spin faster?

When she changes her configuration, the moment of inertia decreases, be-cause some parts of her body are now closer to the axis of rotation. For therotational angular momentum not to change, the angular velocity must in-crease to compensate for the decreased moment of inertia: ,and therefore .

Now you may quite legitimately be puzzled about how this actually works!Why does moving her arms and leg closer to the spin axis increase her an-gular speed? At one level of discussion, you can close your eyes (and maybehold your nose) and say, “Well, we did all that general analysis of the effectof torques on the angular momentum of a rotating object, so if that’s whatwe get when we apply these general principles, I guess that’s that.” But atanother level it would be very nice to get a better sense of the detailed mech-anisms involved in this odd phenomenon.

One approach is to analyze the changes in energy involved in this skatingmaneuver. The angular momentum of the skater doesn’t change, so as hermoment of inertia decreases, the kinetic energy of the skater ac-tually increases.

? Where does this increased energy come from?

Evidently the skater has to expend chemical energy in order to increase herkinetic energy. In fact, at high spin rates it takes a noticeable effort to pullher arms and leg toward the spin axis. This effect is even more dramaticwhen holding heavy weights, as in the next example we will discuss.

9.4.2 The dumbbell demonstration

A popular physics demonstration is to sit on a rotating stool holding adumbbell in each hand. Start spinning slowly with the arms held out, thenpull your hands in toward your chest. You find yourself spinning much fast-

rF⊥

I1ω1 I2ω2=ω2 I1ω1 I2⁄=

Lrot2 2I( )⁄


er, because your smaller moment of inertia I is compensated by a larger an-gular speed in order that the product stay constant in the absence ofsignificant torques. This works only if the frictional torques exerted on thestool by the axle are relatively small.

It requires considerable effort to pull the dumbbells in, costing you chem-ical energy. After all, the dumbbells would tend to move in straight lines,and you must exert a radial force just to keep turning them in a circle(though in that case you do no mechanical work, because the motion of thedumbbells has no radial component in the direction of the force you apply).To move the dumbbells into a smaller radius requires applying an even larg-er force which does work on the dumbbells, because there is now a radialcomponent of the motion, in the direction of the radial force you exert.

9.4.3 A high dive

You may have seen a skilled diver leap off a high board, tuck himself into atight ball (holding onto his ankles), and rotate quite fast for a few turns.Then he straightens out and enters the water like a knife, hardly ruffling thesurface (Figure 9.5).

? What can you say about the diver’s rotational angular momentumwhile he is in the air? Why?

We can probably neglect air resistance, because the diver is very far fromreaching terminal speed. Then the only significant force acting is theEarth’s gravitational force, which effectively acts through the center of mass.Therefore there is negligible torque about the center of mass, and the rota-tional angular momentum does not change.

? Early in the dive the diver is spinning rapidly. If rotational angularmomentum is constant during the dive, where did that rapid spincome from?

When the diver jumps off the board, he must thrust with his feet in such away that the force of the board on his feet has a sizable lever arm about hiscenter of mass, thereby exerting a torque around the center of mass. By thetime his feet lose contact with the board, the diver already has acquired asizable rotational angular momentum due to the torque.

? When the diver pulls out of the tucked position, why does he stopspinning rapidly?

There is a large increase in the moment of inertia, because many atoms arenow much farther from the center of mass than they were when the diverwas in the tucked position. Larger moment of inertia implies smaller angu-lar speed, since the rotational angular momentum Iω does not change.

? Can his body go straight from then on?

His body can’t really go completely straight, because he still has rotationalangular momentum. But his angular speed may be so small that you hardlynotice it, especially in comparison with the very rapid spin in the precedingtucked position. Moreover, his body could approximately follow the curvingpath of his center of mass, with the body rotating just enough to stay tangentto the trajectory. This enhances the illusion of straight motion. The most im-portant aspect for good form is to arrange that the body rotate into the ver-tical position at the time of entering the water, so as to make little splash.

See Problem 9.24, which asks you to analyze a diver’s motion.

ω Iω

Figure 9.5 A diver spins, then straightensout.

9.5: Translational angular momentum 303

9.5 Translational angular momentum

In Figure 9.6 a piece of clay starting from rest falls from rest a height h andsticks to a wheel that is free to rotate on a low-friction axle at location A. Sev-eral different possible locations for the collision are shown, each with thesame momentum of the clay just before it hits and sticks.

? In which case will the device rotate fastest?

As you probably expect, the device will rotate fastest if the clay has the big-gest impact parameter , where is the perpendicular distance from theaxle to the line of motion of the falling clay as shown In Figure 9.6. The fast-est rotation of the device will occur in case (d), with the largest impact pa-rameter.

? In which case will the device not rotate at all?

As you probably expect, if the impact parameter is zero (the line of motionof the clay passes through the axle) the device won’t rotate; case (b).

? What is the difference in the rotation in case (a) compared withcase (c)?

In case (a), the device will rotate counterclockwise (rotational angular mo-mentum vector out of the page, positive z component). In case (c) it will ro-tate clockwise (rotational angular momentum into the page, negative zcomponent).

? If we increase the mass of the clay, how will that affect the rotationof the device? If the clay is going faster just before it sticks to the ball,how will that affect the rotation of the device?

As you probably expect, the device will rotate faster if the clay has more massor hits with a higher speed. In other words, the magnitude of the mo-mentum of clay at the time of the collision is significant.

The effect of the falling clay in making the device rotate is related to boththe impact parameter and also to the magnitude of the momentum .A particle moving in a straight line has angular momentum because it canmake something else rotate. Both straight-line motion and rotation involveangular momentum.

We define the magnitude of the angular momentum of the clay rel-ative to location A (the location of the axle) to be the product of the impactparameter and the magnitude of the momentum, , and we callthis “translational” angular momentum. Although translational angularmomentum is a vector, at the moment we’re just interested in its magni-tude.

TRANSLATIONAL ANGULAR MOMENTUMRELATIVE TO LOCATION A

9.5.1 Visualizing the magnitude of translational angular momentum

There is a convenient way to visualize the magnitude of translational angu-lar momentum relative to some location A. On a diagram of the situation,draw these elements, as shown in Figure 9.7:

• Draw a vector representing the momentum of the particle.• Draw a line through location A, parallel to ,• Draw the impact parameter from this line to the particle.• Draw a rectangle bounded by and .• The area of the rectangle is the magnitude of the translational angular

momentum, .

h

(a)

h

(b)

h

(c)

r⊥

r⊥

h

(d)

r⊥

r⊥ = 0

A

A

A

A

Figure 9.6 A falling piece of clay could hitand stick to the wheel in different loca-tions, with different impact parameters( ).r⊥

r⊥ r⊥

p

r⊥ p

LA

LA r⊥ p=

LA r⊥ p=

A

Figure 9.7 Visualizing the magnitude ofangular momentum for the falling clay.

p

r⊥

LA r⊥ p=

pp

r⊥r⊥ p

LA r⊥ p=


Example: A dog whose mass is 10 kg runs at 20 m/s along the leftside of a highway that is 5 meters wide. What is the magnitude of itstranslational angular momentum relative to location A on the rightside of the highway (Figure 9.8)? Relative to location B on the rightside of the highway? Relative to location C on the left side of thehighway? Include appropriate units.

Solution: In Figure 9.8 we show the relevant rectangle forlocation A. The magnitude of the translational angularmomentum relative to location A is

The same rectangle is valid for location B, so is also equal to. Relative to location C, the impact parameter is

zero, and the rectangle has zero area: .

Ex. 9.11 In Figure 9.9 are seven particles each with the samemagnitude of momentum but with differentpositions relative to location A, and different directions of ordinarylinear momentum. Calculate the magnitude in eachcase. In each case, draw the appropriate rectangle, showing and

that bound the rectangle.

9.5.2 Another view of rotational angular momentum

Our new formula for calculating translational angular momentum can beused to rederive the formula for rotational angular momentum, thereby il-luminating the relationship between these two kinds of angular momen-tum. At a particular instant the translational angular momenta of one of theatoms in a rotating object is this:

Add up the contributions of all of the atoms:

From this you can see that rotational angular momentum is just the sum ofthe translational angular momenta of all the atoms in the rotating object.The formula for the translational angular momentum of a single particle,

, is really the fundamental definition of angular momentum.Nevertheless it is often very useful to express the angular momentum of anobject that rotates about a fixed axis as rotational angular momentum, eval-uated with the convenient formula .

9.5.3 Another example of conservation of angular momentum

Angular momentum, whether rotational or translational or a combinationof both kinds of angular momentum, is a conserved quantity, like linear mo-mentum and energy. That is, the change in angular momentum of a systemplus the change in angular momentum of the surroundings is equal to zero.We can use this to analyze the situation of the clay hitting the wheel.

Take the wheel and the clay as the system. The angular momentum of thecombined system (device plus clay) will stay constant from just before the

A

B

C

dog

Figure 9.8 The translational angularmomentum of a dog running along theleft side of a highway.

r⊥ 5 m=

p 10 kg( ) 20 m/s( )= LA r⊥ p=

LA 10 kg( ) 20 m/s( ) 5 m( ) 1000 kg m2/s⋅= =

LB

1000 kg m2/s⋅LC 0=

8 m

5 m 5 m

7 m

(1)

(7)

(6)(5)

(4)

(3)(2)

A

Figure 9.9 Draw the translational angularmomentum rectangles and calculate themagnitude of the translational angularmomentum in each case.

p 10 kg m/s⋅=

p 10 kg m/s⋅=

LA r⊥ p=r⊥

p

r⊥ p r⊥ mv( ) r⊥ mωr⊥( ) mr⊥2( )ω= = =

Lrot mr⊥2( )1ω mr⊥

2( )2ω mr⊥2( )3ω .....+ + +=

Lrot mr⊥2( )1 mr⊥

2( )2 mr⊥2( )3 .....+ + +[ ]ω=

Lrot Iω=

LA r⊥ p=

Lrot Iω=

9.5: Translational angular momentum 305

collision to just after, if the axle is nearly frictionless and therefore exertsnegligible torque on the system during the collision.

If the device is initially not rotating, the angular momentum just beforethe collision is due solely to the clay:

This initial angular momentum is equal to the angular momentum of thesystem just after the collision. The magnitude of the (rotational) angularmomentum of the wheel is , where I is the moment of inertia of thewheel and is the angular speed of the device just after the collision. If al-most all of the mass of the wheel is concentrated in the rim of radius R (low-mass spokes), we saw in Ex. 9.2 on page 297 that . If the wheel isa solid disk, .

In addition to the rotational angular momentum of the wheel, the lumpof clay has angular momentum; just after the collision (Figure 9.10). It istraveling with the wheel and has a speed , so it has translational an-gular momentum . It is true that its motion willnow be circular, but we treat it in terms of its instantaneous momentum asthough it were going to continue in a straight line. We now equate the ini-tial (translational) angular momentum of the clay just before impact to thefinal (rotational and translational) angular momentum of the wheel andclay just after impact (later the unbalanced device will oscillate):

, which gives

? If the impact parameter is zero, what is the final angular speed of the wheel? Does this make sense?

If the impact parameter is zero, our result indicates that . Thatmeans that the clay falls straight toward the axle, and the wheel won’t rotate,which makes sense.

The momentum principle and the energy principle are not enough

The analysis of the collision between the clay and the device is an exampleof the power of the angular momentum concept. It is important to see thatwe could not have calculated using only momentum conservationand/or energy conservation. The wheel itself has zero momentum both be-fore and after the collision, because its center of mass does not move, and

. The clay’s momentum however changes in both magnitudeand direction, consistent with the axle exerting a sizable force on the systemwhen the clay hits the wheel. As far as the energy principle is concerned, theforce exerted by the axle acts through no distance and does no work, butthe clay and the struck wheel experience a temperature rise (inelastic colli-sion).

Angular momentum is a new tool that lets us analyze some phenomenaquite easily that would be difficult or impossible to analyze solely in termsof linear momentum and energy.

Ex. 9.12 Consider a system of the kind justdiscussed. The radius is 0.3 m, and the wheel has amass of 1.6 kg, with all the mass concentrated in therim. The wheel is initially not rotating. A piece ofclay falls and sticks to the wheel as shown. The massof the clay is 0.04 kg and its speed just before thecollision is 5 m/s. Just after the collision, what is thespeed of the clay?

LA initial, r⊥ pclay=

Iωω

I MR2=I 1

2---MR2=

A

ω

m

R

Figure 9.10 After the collision.

pf

v ωR=r⊥ pf R mωR( ) mR2ω= =

r⊥ pclay Iω mR2ω+= ωr⊥ pclay

I mR2+-------------------=

r⊥ω

r⊥ ω 0=

ω

Ptot Mtotvcm=

vf


9.5.4 An alternative way to calculate translational angular momentum

There is another useful way to calculate the magnitude of the translationalangular momentum of a particle. In Figure 9.11 we draw a vector fromlocation A to the particle. The impact parameter can be read off the di-agram as having the value because is the side opposite theangle θ, the angle of turn from the direction of to the direction of .Therefore the magnitude of the angular momentum can be calculated likethis:

MAGNITUDE OF ANGULAR MOMENTUMFOR A PARTICLE RELATIVE TO LOCATION A

Moreover, there is an alternative visualization of the magnitude. The paral-lelogram bounded by the vectors and has the same area as the rectan-gle bounded by and . In Figure 9.12, the parallelogram has the same“base” and the same “altitude” as the rectangle, so the areas are thesame. In Figure 9.13 we show several parallelograms that all have the sameareas because they all have the same base and altitude.

Example: What is in Figure 9.14? ? ?

Solution: and are zero. The angle between and is 0, and . Similarly, the angle between and is

, and . Only is nonzero. The angle of turnfrom the direction of to the direction of is .

In Figure 9.15 is shown the corresponding parallelogram. Thebase of the parallelogram is , and the altitudeis . The area ofthe parallelogram is .

Ex. 9.13 What is the magnitude of the angular momentum

about location D?

Ex. 9.14 Draw the parallelogramwhose area is equal to .

9.6 Direction of angular momentum

If the clay falls on the left the wheel rotates counterclockwise, but if the clayfalls on the right it rotates clockwise. This can be expressed in terms of thedirection of angular momentum. For a rotating object we can use a “right-hand rule.” You curl the fingers of your right hand in the direction of rota-tion, and your thumb points along the axis in a direction that we use to in-dicate the direction of an angular momentum vector. If the fingers of yourright hand curl around in a counterclockwise rotation in the x-y plane (a “+”rotation), we say that there is an angular momentum vector pointing out-wards, in the +z direction. If the fingers of your right hand curl clockwise (a“–” rotation), we say that there is an angular momentum vector pointing in-wards, in the –z direction.

A

θ

θ

Figure 9.11 An alternative way to calculatethe magnitude of the angular momentumof a particle.

rA

LA r⊥ p rA p θsin= =

r⊥

prA

r⊥r⊥ rA θsin= r⊥

rA p

LA r⊥ p rA p θsin= =

A

θ

base

altitude

Figure 9.12 The parallelogram has thesame area as the rectangle because it hasthe same base and the same altitude

rA

p

r⊥rA p

r⊥ pp r⊥

base basebase

alti

tude

A CB

Figure 9.13 Different parallelograms allwith the same areas: .LA LB LC= =

rA rB rC

pp p

LA

AC

B

Figure 9.14 Find the magnitude of angularmomentum relative to these locations.

rA 5 m=p 10 kg m/s⋅=

θ 130°=

LB LC

LB LC rB p0sin 0= rC p

180° 180°sin 0= LA

rA p 130°

LA rA p θsin 5 m( ) 10 kg m/s⋅( ) 130°sin 38 kg m2/s⋅= = =

A

Figure 9.15 The par-allelogram whose area is the magnitude of

θ 130°=

50°

θ

r⊥ 5 m( ) 50°sin=

p 10 kg m/s⋅=r⊥ rA θsin 5 m( ) 130°sin 5 m( ) 50°sin= = =

LA

D

rD 6 m=

p 15 kg m/s⋅=

40°

LD

LD

9.6: Direction of angular momentum 307

This works for particles moving in straight lines, too. Consider the paral-lelogram in the x-z plane in Figure 9.16 whose area represents the magni-tude of the angular momentum. This parallelogram is bounded by thevectors and , with the vector pointing from location A to the parti-cle. Bend the fingers of your right hand around the perimeter of the paral-lelogram, in the direction of the turn from to . The angularmomentum vector is defined to point in the direction of your rightthumb, perpendicular to the parallelogram (in the y direction), and themagnitude of this vector is the area of the parallelogram,

.The other case is shown in Figure 9.17, where we go around the parallel-

ogram in the other direction. Hold your right hand in such a way that yourfingers bend in the direction of the turn from to . Now your thumbpoints in the opposite direction to what you found in Figure 9.16, and thedirection of the angular momentum vector is the direction of your rightthumb, perpendicular to the parallelogram (in the –y direction). Again, themagnitude of this vector is the area of the parallelogram,

.We will use this “right-hand rule” to determine the direction of angular

momentum vectors, which are defined to be perpendicular to the parallel-ogram bounded by and . The actual motion is in the plane of the par-allelogram, but it is convenient to describe the orientation of a plane inspace by referring to a vector perpendicular to that plane. We need a right-hand rule to establish the direction of the motion in the plane (essentially,whether the motion is clockwise or counterclockwise around the locationA) when viewed from above). Notice that in the case of a wheel rotating onan axle, the angular momentum vector lies along the axle.

Things to watch out for

The rotation of the wrist is an important part of the right-hand rule. Con-sider the situation in Figure 9.18. With your right hand in this position, youcan’t bend your fingers backwards from the vector toward the vector

—it is a physical impossibility. You need to rotate your wrist into a position from which it is possible to

bend the fingers, as shown in Figure 9.19. The direction of your thumb isdown, which is the correct direction in this situation. The right-hand ruleforces you to rotate your wrist into an orientation that permits bending yourfingers from toward , in which case your thumb will point in the cor-rect direction of the angular momentum.

? Here we show two differentparticles. What is the direction oftheir angular momentum vectorsrelative to location A?

You should find that the right-handrule gives a direction for thatis into the page, and has a direction out of the page.

Two-dimensional projections

As we mentioned earlier in the chapter, whenever possible we will work withtwo-dimensional projections onto the x-y plane. If and lie in the x-yplane, the angular momentum points in the +z direction (out of thepage) or in the –z direction (into the page). A special notation is used toindicate “out of the page” ( ) and “into the page” ( ); see Figure 9.20on the following page. The symbol can be thought of as showing the tipof an arrow (a vector) pointing out at you, while the symbol representsthe feathers of an arrow (a vector) pointing away from you.

AIn the xz plane

Figure 9.16 The fingers of your right handbend from toward and your thumbpoints in the direction of .

rA pLA

rA

p

LA

rA p rA

rA pLA

LA rA p θsin=

A

Figure 9.17 Going around a parallelogramin the other direction.

prA

LA

rA p

LA

LA rA p θsin=

rA p

A

Figure 9.18 You can’t bend your fingersbackward. You must rotate the wrist into aposition that lets you bend the fingers.

rAp

r⊥p

A

Figure 9.19 After rotating the wrist, it ispossible to bend the fingers.

rA p

rA p

A

1 2

rA 1,rA 2,

p2

LA 1,LA 2,

r⊥ pLA


A bar graph of angular momentum

A different visualization of the direction of is the following: Consider arotating wheel lying in the xz plane (Figure 9.21). If we represent the angu-lar momentum as a bar graph, it makes sense to have the angular momen-tum be positive if the wheel rotates counterclockwise and negative if thewheel rotates clockwise, and the bar is perpendicular to the plane of the mo-tion.

Ex. 9.15 In Figure 9.22 determine both the magnitude and direction ofthe angular momentum of the particle at location O relative to eachlocation: A, B, C, D, E, F, G, and H. In each case, draw with its tip at theparticle, and sketch the parallelogram formed by the vectors and .The area of the parallelogram (or of the equivalent rectangle formed by

and ) is the magnitude of the angular momentum, and the directionof the angular momentum vector is perpendicular to the parallelogram,in the direction given by the right-hand rule. For example, the angularmomentum relative to location D has magnitude , out of thepage (in the +z direction).

Ex. 9.16 In Figure 9.23, what are the magnitude and direction ofthe angular momentum about location K?

9.6.1 The vector cross product

The angular momentum vector is an example of something called the “vec-tor cross product.” The vector cross product of any two vectors and hasas its magnitude the area of the parallelogram bounded by the two vectors( ), and the direction is defined to be perpendicular to the paral-lelogram and given by the right-hand rule. It is written with the notation

. Angular momentum is an example of such a vector cross product:

ANGULAR MOMENTUM AS A VECTOR CROSS PRODUCT

In many cases the easiest way to determine the magnitude and direction ofa vector cross product such as angular momentum is to use the methodswe’ve used so far: Draw the parallelogram bounded by the two vectors, findthe area of this parallelogram, and use the right-hand rule to determine thedirection perpendicular to the parallelogram. However, this is difficult todo with arbitrary 3D vectors. For example, what is the vector cross productgiven by ? Here is the general formula for evaluatingany vector cross product of two vectors:

EVALUATING A VECTOR CROSS PRODUCT

It is easier to remember and use this formula than its complicated formmight imply, because the subscripts occur cyclically: xyz, yzx, zxy. That is, thefirst term in the x component is , the first term in the y component is

, and the first term in the z component is . Here is a way to picturethis cycle:

x y z x y zx y z x y zx y z x y z

A

A

Angular momentum out of page (+z)

Angular momentum into page (–z)

Figure 9.20 2-D projections: symbols forangular momentum out of page and intopage.

rA

rA

p

p

Figure 9.21 A bar graph of angularmomentum.

LA y, 0>

LA y, 0<

LA

C

E D

BA4 m

3 m

5 m

4 m

F

G

H

O

p 10 kg m/s⋅=

Figure 9.22 Ex. 9.15: Determine both themagnitude and direction of the angularmomentum relative to locations A, B, C, D,E, F, G, and H.

rr p

r⊥ p

50 kg m2/s⋅

K

130°

25 cm

p 2 kg m/s⋅=

Figure 9.23 Ex. 9.16: Find the magnitude

and direction of .LK

A B

A B θsin

A B×

LA rA p×=

3 5 2–,–,⟨ ⟩ 7 4 1, ,–⟨ ⟩×

A B× < AyBz AzBy–( ) AzBx AxBz–( ), AxBy AyBx–( )>,=

AyBzAzBx AxBy

9.6: Direction of angular momentum 309

The negative second terms for each component have their subscriptsswitched: , , . For example, the z component is

.We will sketch a proof that this is the correct formula for evaluating a vec-

tor cross product. Consider calculating the angular momentum of a particlemoving in the xy plane (Figure 9.24). Split the momentum vector into itscomponents and .

? If were zero, what would be the vector angular momentumexpressed in the form ?

If , the magnitude of the angular momentum is , and accordingto the right-hand rule it is out of the page, in the +z direction, so we have

.

? If were zero, what would be the vector angular momentumexpressed in the form ?

If , the magnitude of the angular momentum is , and accordingto the right-hand rule it is into the page, in the –z direction, so we have

.

? Therefore, what is ?

Adding together the contributions of the two components of , we have. We could do a similar analysis for the x and y com-

ponents of the angular momentum if had a z component. The final resultis this:

ALGEBRAIC EVALUATION OF ANGULAR MOMENTUM

Ex. 9.17 What is the angular momentum if and ?

Ex. 9.18 What is the angular momentum if and ?

yz zy→ zx xz→ xy yx→AxBy AyBx–( )

py

pxy

xA

Figure 9.24 Evaluating a vector cross prod-uct.

p

rA

ppx py

pxLx Ly Lz, ,⟨ ⟩

px 0= xpy

LA 0 0 xpy, ,⟨ ⟩=

pyLx Ly Lz, ,⟨ ⟩

py 0= ypx

LA 0 0 ypx–, ,⟨ ⟩=

LA rA p×=

pLA 0 0 xpy ypx–( ), ,⟨ ⟩=

p

LA rA p× < ypz zpy–( ) zpx xpz–( ) xpy ypx–( )>, ,= =

LA rA 5 3 0, ,⟨ ⟩ m=p 4 2 0, ,⟨ ⟩ kg m/s⋅=

LA rA 3 5 2–,–,⟨ ⟩ m=p 7 4 1, ,–⟨ ⟩ kg m/s⋅=


9.7 Angular momentum of multiparticle systems

As was shown in Chapter 7, the kinetic energy of a multiparticle system canbe split into overall translational kinetic energy plus rotational and vibra-tional kinetic energy relative to the center of mass:

where the translational kinetic energy is , and isthe velocity of the center of mass of the object.

In an analogous fashion, the angular momentum of a multiparticle sys-tem can be split into a term associated with the motion of the center of mass(translational angular momentum, sometimes called “orbital” angular mo-mentum) plus a term associated with rotation relative to the center of mass(which we call rotational angular momentum, sometimes called “spin” an-gular momentum):

Splitting the angular momentum into these two terms is very useful in ana-lyzing multiparticle systems such as the Earth, which orbits the Sun but alsorotates on its axis. You may also know that electrons orbiting the nucleus ofan atom have both translational and rotational angular momentum.

We will calculate the angular momentum relative to location A of a 3-masssystem (Figure 9.25). We specify the positions of each mass relative to loca-tion A by going from A to the center of mass (vector ) and then to theindividual masses. (The location A need not be in the same plane as themasses, nor are the momentum vectors necessarily in the plane of the mass-es.)

The position of mass m1 relative to location A is , the position ofmass m2 is , and the position of mass m3 is (Figure 9.26).Us-ing these vectors we calculate the total angular momentum relative to loca-tion A:

Collect common terms:

Since is the total momentum , we have the following:

“translational” “rotational”

The first term in the total angular momentum,

is the angular momentum the system would have if all the mass were con-centrated at the center of mass (this is the point particle system discussed inChapter 7). We call this term the “translational” angular momentum

, and it is also what we use in calculating the angular momentum ofa single particle. This term is also called the “orbital” angular momentum.Note that is equal to .

The second term in the total angular momentum,

is the angular momentum of the system relative to the center of mass of thesystem. We call this term the “rotational” angular momentum . An ex-ample of rotational angular momentum is the angular momentum due to

Ktot Ktrans Krel+ K trans Krot Kvib+ += =

Ktrans12---Mtotal vcm

2= vcm

LA Ltrans,A Lrot+=

rcm

Center of mass

A

r1

r3

r2rcm

p1

p2

p3

Figure 9.25 Position vectors relative to thecenter of mass.

rcm r1+rcm r2+ rcm r3+

A

rcm r1+rcm r3+

rcm r2+

Figure 9.26 Position vectors relative to loca-tion A.

LA rcm r1+( ) p1× rcm r2+( ) p2× rcm r3+( ) p3×+ +=

LA rcm p1 p2 p3+ +( )×[ ] r1 p1× r2 p2× r3 p3×+ +[ ]+=

p1 p2 p3+ +( ) Ptot

LA rcm Ptot×[ ] r1 p1× r2 p2× r3 p3×+ +[ ]+=

rcm Ptot×

Ltrans,A

Ptot Mtotalvcm

r1 p1× r2 p2× r3 p3×+ +

Lrot

9.7: Angular momentum of multiparticle systems 311

the rotation of the Earth on its axis. This term is also called the “spin” angu-lar momentum. In Figure 9.27 we show both the translational and rotation-al angular momentum vectors for the Earth relative to the Sun.

TRANSLATIONAL PLUS ROTATIONAL ANGULAR MOMENTUM

Translational: , where

Rotational:

The translational angular momentum differs for different choices of the lo-cation A, so we need the “A” subscript on the translational angular momen-tum and the total angular momentum to remind us of this dependence.The rotational angular momentum doesn’t need a subscript because it iscalculated relative to the center of mass, and this calculation is unaffectedby our choice of the point A, and by the motion of the center of mass, whichmay even be accelerating.

For a point particle, or the “collapsed” point-particle version of a multi-particle system discussed in Chapter 7, the translational angular momen-tum is the only kind of angular momentum there is, just as translationalkinetic energy is the only kind of kinetic energy a point particle can have.

Ex. 9.19 A rod rotates in the vertical plane around a horizontalaxle. A wheel is free to rotate on the rod, as shown in Figure 9.28.A vertical stripe is painted on the wheel. As the rod rotatesclockwise, the vertical stripe on the wheel remains vertical. Is thetranslational angular momentum of the wheel zero or nonzero? Ifnonzero, what is its direction? Is the rotational angular momentumof the wheel zero or nonzero? If nonzero, what is its direction?

Ex. 9.20 Consider a system similar to that in the previous exercise,but with the wheel welded to the rod (so it is forced to turn with therod). As the rod rotates clockwise, does the stripe on the wheelremain vertical? Is the translational angular momentum of thewheel zero or nonzero? If nonzero, what is its direction? Is therotational angular momentum of the wheel zero or nonzero? Ifnonzero, what is its direction?

Ex. 9.21 Pinocchio rides a horse on a merry-go-round turningcounterclockwise as viewed from above, with his long nose alwayspointing forward, in the direction of his velocity. Is Pinocchio’stranslational angular momentum relative to the center of themerry-go-round zero or nonzero? If nonzero, what is its direction?Is his rotational angular momentum zero or nonzero? If nonzero,what is its direction?

9.7.1 Rotational angular momentum as a vector

An important special case of rotational motion is one where a system is ro-tating on an axis, with all the atoms in the system sharing the same “angularspeed” in radians per second but with different linear speeds in meters persecond, depending on their distances from the axis (Figure 9.29). This isthe situation for a rigid object.

As we saw earlier in this chapter, the magnitude of the rotational angularmomentum is . We are now in a position to express rotational an-gular momentum as a vector, by defining angular velocity as a vector:

A

Figure 9.27 Translational angular momen-tum of the Earth relative to the Sun, withrotational angular momentum relative tothe Earth’s center of mass. At this instantthe linear momentum of the Earth is intothe page.

Ltrans,ALrot

LA Ltrans,A Lrot+=

Ltrans,A rcm,A Ptot×= Ptot Mtotalvcm=

Lrot r1 p1× r2 p2× r3 p3×+ +=

Figure 9.28 A rod rotating in the verticalplane with a wheel attached.

v1 = ωr1 v2 = ωr2

r1

r2

r3

r4

v4 = ωr4 v3 = ωr3

A

m4

m2

m3

m1ω 2π

T------=

Figure 9.29 A case of rigid rotation aboutan axis with angular speed ω. Note the dif-ferent speeds, with vi = ωri.

ω

Lrot Iω=


ANGULAR VELOCITY VECTOR

magnitude: radians/second

direction: see Figure 9.30

Start with your fingers headed radially outward, bend them in the directionof the rotation, and your thumb points in the direction of the angular veloc-ity, along the axis of rotation (an angular velocity vector is also shown in Fig-ure 9.29 on the preceding page).

Since the rotational angular momentum points in the direction of , wehave the following for the angular momentum relative to the center of mass,even if the center of mass is moving in some complicated way:

ROTATIONAL ANGULAR MOMENTUM

This way of evaluating rotational angular momentum is often easier to usethan the more basic expression in terms of cross products , especiallybecause the moments of inertia for common objects can be looked up in ref-erence books.

Complications

We will deal with situations where the axis of rotation of a rigid body passesthrough the center of mass, and the object is symmetric or nearly so, inwhich case we can calculate the rotational angular momentum and rotation-al kinetic energy using the formulas and . In Sec-tion 9.15 on page 334 we discuss some complications that apply in the moregeneral case. We will avoid these more complex rotational situations, whichare treated in more advanced courses.

9.7.2 More exercises on angular momentum

The following exercises constitute an important worksheet that gives youpractice in calculating angular momentum. Be sure to do all of these exer-cises and check your answers at the end of the chapter.

Ex. 9.22 A comet orbits the Sun (Figure 9.31). When it is atlocation 1 it is a distance d1 from the Sun, and has momentum .Location A is at the center of the Sun.

(a) When the comet is at location 1, what is the direction of ?

(b) When the comet is at location 1, what is the magnitude of ?

When the comet is at location 2, it is a distance d2 from the Sun,and has momentum .

(c) When the comet is at location 2, what is the direction of ?

(d) When the comet is at location 2, what is the magnitude of ?

Ex. 9.23 A barbell spins around a pivot at its center at A (Figure9.32). The barbell consists of two small balls, each with mass m, atthe ends of a very low mass rod of length d. The barbell spinsclockwise with angular speed ω0 radians/s.

(a) Calculate (both direction and magnitude).

(b) Calculate (both direction and magnitude).

(c) Calculate I for the barbell.

(d) What is the direction of ?

ω

Figure 9.30 Finding the direction of theangular velocity vector.

ω

ω

Lrot Iω=

r p×

Lrot Iω= Krot12---Iω2=

p1

LA

LA

d1

p1

p2

d2

α

Α

Figure 9.31 A comet orbits the Sun.Location A is at the center of the Sun.

p2

LA

LA

m

m

d

A

1

2

Figure 9.32 A rotating barbell.

Ltrans,1 A,

Ltot,A

ω0

9.8: The angular momentum principle 313

(e) Calculate (both direction and magnitude).

(f) How does compare to ?

(g) Calculate Krot.

Ex. 9.24 The barbell in the previous exercise is mounted on theend of a low mass rigid rod of length b (Figure 9.33). The apparatusis started in such a way that although the rod rotates clockwise withangular speed ω1 rad/s, the barbell maintains its verticalorientation.



(c) Calculate (both direction and magnitude).

Ex. 9.25 The apparatus in the previous exercise is restarted in sucha way that it again rotates clockwise with angular speed ω1 rad/s,but in addition, the barbell rotates clockwise about its center, withan angular speed ω2 rad/s (Figure 9.34).



(c) Calculate (both direction and magnitude).

9.8 The angular momentum principle

Earlier in this chapter we derived the angular momentum principle for a ro-tating object. Now we are in a position to do a more careful and more gen-eral derivation.

For a single particle, the momentum principle says that the time deriva-tive of the momentum is the net force: . Let’s see what thetime derivative of angular momentum for a single particle turns out to be.By applying the product rule for derivatives, we have

Since and , we have

since . Therefore we have proved that the followingis true for a single particle:

We have previously defined the quantity to be the (translation-al) angular momentum of a particle relative to location A. The quantity

is the “torque” exerted about location A (Figure 9.35).Torque is usually written as τ, Greek lowercase letter tau. Torque is a wordof Latin origin and simply means “twist.” Twisting a system changes its an-gular momentum. The magnitude of torque is or .

We can read the equation as saying “the rate of change of the angular mo-mentum of a particle relative to location A is equal to the torque applied tothe particle about location A.”

Lrot

Lrot Ltot,A

b

d

B

Figure 9.33 A barbell pivoted on a low-mass rotating rod. The barbell does notrotate.

Lrot

Ltrans,B

Ltot,B

b

d

B

Figure 9.34 A barbell pivoted on a low-mass rotating rod. The barbell rotates.

Lrot

Ltrans,B

Ltot,B

dp dt⁄ Fnet=

d rA p×( )dt

-----------------------drA

dt-------- p× rA

dpdt------×+=

drA

dt-------- v= p mv

1 v2 c2⁄( )–--------------------------------=

drA

dt-------- p× v m× v

1 v2 c2⁄( )–-------------------------------- 0= =

v v× v2 0°( )sin 0= =

d rA p×( )dt

----------------------- rA Fnet×=

A

Figure 9.35 Torque relative to location A is

defined as .τA rA Fnet×=

rA

Fnet

p

LA rA p×=

τA rA Fnet×=

rF⊥ r⊥F


THE ANGULAR MOMENTUM PRINCIPLE FOR A POINT PARTICLE

The angular momentum principle is similar to the momentum principle. In the following section we derive the angular momentum

principle for multiparticle systems.

9.8.1 Multiparticle systems

The angular momentum principle for multiparticle sys-tems makes it possible to understand the counterintuitivebehavior of spinning tops and gyroscopes. We will derivea multiparticle version of the angular momentum princi-ple, following closely the derivation in Chapter 2 of a mul-tiparticle version of the momentum principle.

To be as concrete as possible, we’ll consider a systemconsisting of just three particles. It is easy to see how thisgeneralizes to larger systems, including a block consistingof an astronomically huge number of atoms.

In Figure 9.36 we show all of the forces acting on eachparticle, where the lower case ’s are internal forces, andthe upper case ’s are external forces, such as the gravi-tational attraction of the Earth. We write the angular mo-mentum principle applied to each of the three particles.

We measure angular momenta and torques relative to location A, but toavoid clutter we don’t write the subscript A in any of these equations untilthe end. ( is the angular momentum of m1 relative to A.)

Nothing new so far. But now we add up these three equations. That is, wecreate a new equation by adding up all the terms on the left sides of thethree equations, and adding up all the terms on the right sides, and settingthem equal to each other. In doing so, we take into account that many ofthese terms cancel. For electric and gravitational forces, which obey the rec-iprocity principle (Newton’s third law), we have

Therefore the torques of the internal forces cancel. For example, considerthis piece of the sum:

The vector points from m2 toward m1, so the angle between it and is either 0° or 180°, and sinθ is zero. Consequently all of the torques as-

sociated with the internal forces cancel, and all that remains is this:

d LAdt

--------- rA Fnet× τA= =

dp dt⁄ Fnet=

F3 ext,

F2 ext,A

m1

m2

m3

Figure 9.36 The angular momentum prin-ciple in a multiparticle system.

F1 ext,

f1 3,f3 1,

f3 2,

f1 2,

f2 1,

f2 3,r1

r3

r2 fF

L1

dL1

dt--------- r1 Fext,1 f1,2 f1,3+ +( )×=

dL2

dt--------- r2 Fext,2 f2,1 f2,3+ +( )×=

dL3

dt--------- r3 Fext,3 f3,1 f3,2+ +( )×=

f1,2 f2,1–=

f1,3 f3,1–=

f2,3 f3,2–=

r1 f1,2× r2 f2,1×+ r1 f1,2× r2 f1,2×– r1 r2–( ) f1,2×= =

r1 r2–f1,2

d L1 L2 L3+ +( )dt

--------------------------------------- r1 Fext,1× r2 Fext,2× r3 Fext,3×+ +=

9.8: The angular momentum principle 315

The right side of this equation represents the net torque due to externalforces, . The great importance of this equation is that the reciproc-ity of electric and gravitational forces has allowed us to get rid of all thetorques due to internal forces, the torques that the particles in the systemexert on each other. We rewrite our result like this:

THE ANGULAR MOMENTUM PRINCIPLE FOR A MULTIPARTICLE SYSTEM

or

In words, the rate of change of the “total angular momentum” of a systemrelative to a location A, , is equal to the nettorque due to external forces exerted on that system, relative to location A.The internal forces and torques do not appear in this multiparticle versionof the angular momentum principle.

Often it pays to choose location A to be at the place where the center ofmass happens to be at that instant. In that case, the angular momentumprinciple simplifies:

, since

That is, the position of the center of mass, relative to the center of mass it-self, is of course a zero vector. Therefore we have the important and usefulspecial case for the motion relative to the center of mass

THE ANGULAR MOMENTUM PRINCIPLERELATIVE TO THE CENTER OF MASS

or

We will initially focus on situations where there are no external torques.This is the case for isolated systems, but the net external torque may be zeroeven if there are forces acting on the system. For example, an axle exerts aforce to hold up a wheel mounted on the axle, but the “lever arm” is verysmall if the axle is thin, so the torque associated with the axle force may benegligible. Another case where we can neglect small torques is during a col-lision.

Ex. 9.26 Consider a rotating star far from other objects. Its rate ofspin stays constant, and its axis of rotation keeps pointing in thesame direction. Why?

Ex. 9.27 Because the Earth is nearly perfectly spherical,gravitational forces act on it effectively through its center. Explainwhy the Earth’s axis points at the North Star all year long. Alsoexplain why the Earth’s rotation speed stays the same throughoutthe year (one rotation per 24 hours). In your analysis, does itmatter that the Earth is going around the Sun?

In actual fact, the Earth is not perfectly spherical. It bulges out a bitat the equator, and tides tend to pile up water at one side of theocean. As a result, there are small torques exerted on the Earth byother bodies, mainly the Sun and the Moon. Over many thousandsof years there are changes in what portion of sky the Earth’s axispoints toward (change of direction of rotational angular momen-tum), and changes in the length of a day (change of magnitude ofrotational angular momentum). See page 332.

τnet,ext,A

dLtot,A

dt------------- τnet,ext,A= ∆Ltot,A τnet,ext,A∆t=

Ltot,A L1,A L2,A L3,A .....+ + +=

dLcm

dt------------

ddt----- rcm,cm Ptot×( ) Lrot+[ ] dLrot

dt------------= = rcm,cm 0=

dLrot

dt------------ τnet,cm= ∆Lrot τnet,cm∆t=

r⊥


9.8.2 Summary: Angular momentum and torque

At last we have all the tools we need to investigate a variety of situations in-volving angular momentum. Let’s summarize these new tools and concepts:

• A force exerts a torque relative to a chosen location A: .• Torque causes changes in angular momentum; the rate of change of

the angular momentum is equal to the net torque:

• If the net torque relative to A is zero, the angular momentum relative toA does not change. (And if we know the angular momentum isn’tchanging about some location, as in static equilibrium, we know that thenet torque must be zero about that location.)

• The angular momentum can be divided into two pieces: translationalangular momentum of the system as a whole (as though all the masswere concentrated at the center-of-mass point), plus rotational angularmomentum relative to the center of mass: .

• The rate of change of the rotational angular momentum is equal to thetorque about the center of mass:

• Rotational angular momentum is given by moment of inertia times an-

gular velocity: , where

• Rotational kinetic energy about the center of mass can be calculated as

9.9 The three fundamental principles of mechanics

We now have three fundamental principles that relate a change in someproperty of a system to the interactions of the system with external objects.In the special case in which there are no external interactions, these princi-ples predict that the initial and final values of momentum, angular momen-tum, or energy are equal to each other.

Momentum Angular momentum Energy

If external forces:momentum changes.

If external torques:angular momentum changes.

If energy inputs:energy of system changes.

If no external forces:the momentum of a system is constant.

If no external torques:the angular momen-tum of a system is con-stant.

If no energy inputs:the energy of asystem is constant.

Location of object doesnot matter.

Location of object rela-tive to point A is impor-tant.

Location of objectdoes not matter.

τA rA F×=

dLA

dt--------- τnet,ext,A=

LA Ltrans,A Lrot+=

dLrot

dt------------ τnet,ext,cm=

Lrot Iω= I m1r⊥12 m2r⊥2

2 m3r⊥32+ + + ...=

Krot12---Iω2 1

2--- Iω( )2

I-------------

Lrot2

2I---------= = =

dPdt------ Fnet,ext= dLA

dt--------- τnet,ext,A= ∆E W Q+=

9.10: Conservation of angular momentum 317

9.10 Conservation of angular momentum

If the torque around a location A is zero, the angular momentum about thatlocation does not change. There may be forces acting, and causing changesin the linear momentum, but if these forces don’t exert any torque, the rateof change of angular momentum is zero. A major reason for definingtorque is to be able to identify situations in which the torque about a loca-tion A is zero, in which case angular momentum is constant.

Angular momentum is a conserved quantity:

CONSERVATION OF ANGULAR MOMENTUM

If a system gains angular momentum, the surroundings lose that amount.An important special case that merits separate attention is the case of notorque acting on the system, in which case the angular momentum does notchange. Comet orbits provide a nice example. Most comets have very longelliptical orbits around the Sun (Figure 9.37). We can treat the comet as apoint particle, because it is tiny compared to the distances involved.

Ex. 9.28 Relative to the center of the Sun, the torque exerted bythe Sun’s gravitational force on the comet is zero at every pointalong the orbit, because the force points toward the Sun, so

. What does that say about the angular momentum ofthe comet relative to the center of the Sun?

Ex. 9.29 In terms of the comet’s mass m and speed v1 when nearestthe Sun (distance r1 from the Sun), what is the comet’s angularmomentum relative to the center of the Sun (direction andmagnitude)? What is the comet’s angular momentum relative tothe center of the Sun (direction and magnitude) when it is farthestfrom the Sun (distance r2 from the Sun) and traveling at speed v2?What can you conclude about the relationship between v1 and v2?

Comments about comets

Some comets go far beyond Pluto and spend most of their time there, be-cause they’re traveling very slowly, as you have just shown. We see themwhen they come near the Sun, but only for a few months, because they’renow traveling fast.

The result at the closest and farthest points in the orbit is cor-rect no matter what kind of “central” force is involved. For any force thatacts along a line connecting two objects, there is no torque about a point atthe center of one of the objects, so angular momentum is constant aboutthat location. In the particular case of the gravitational force, we also havean energy equation relating v1 and v2:

Historical note: Kepler and elliptical orbits

Based on careful, accurate naked-eye measurements made by Tycho Brahebefore the invention of the telescope, Johannes Kepler in 1609 announcedhis discovery that the planets follow elliptical orbits around the Sun.

Kepler also stated that he had found that “a radius vector joining anyplanet to the Sun sweeps out equal areas in equal lengths of time.” This isequivalent to conservation of angular momentum, as can be seen with the

∆LA system, ∆LA surroundings,+ 0=

v2

r2

v1

r1

Figure 9.37 A long elliptical orbit of acomet around the Sun.

r Fext× 0=

r1v1 r2v2=

12---mv1

2 GMm r1⁄– 12---mv2

2 GMm r2⁄–=


aid of Figure 9.38. The area swept out in a time ∆t is the area of a trianglewhose base is v∆t and whose altitude is rsinθ. This area is , whichis proportional to the angular momentum .

In 1618 Kepler announced his discovery that the square of the time ittakes a planet to go around the Sun is proportional to the cube of the meandistance from the Sun (a result you derived in Chapter 2 for the simpler caseof circular orbits, using the momentum principle).

Later in the 1600’s Newton explained all three of these discoveries as de-rivable from his second law of motion (the momentum principle) plus hisuniversal law of gravitation. Kepler’s insights provided important tests forNewton’s theories.

9.11 Atomic and nuclear angular momentum

Many elementary particles have “spin,” rotational angular momentum (or,if you wish to be very precise, these particles behave as though they have ro-tational angular momentum, which comes down to essentially the samething). Electrons bound in an atom may have translational angular momen-tum relative to the nucleus as well as their own rotational angular momen-tum. Atoms as a whole may have angular momentum, as do many nuclei.The total angular momentum is as expected the sum of the translational an-gular momenta plus the rotational angular momenta.

The surprising thing about angular momentum at the atomic and sub-atomic level is that it is quantized. The basic quantum of angular momen-tum is this:

The angular momentum quantum =

where h is “Planck’s constant,” joule-second. Whenever you mea-sure a vector component of the angular momentum (that is, along the x ory or z axis), you get either a half-integer or integer multiple of this quantumof angular momentum.

Ex. 9.30 Show that and angular momentum have the same units.

9.11.1 The Bohr model of the hydrogen atom

As discussed in the previous chapter, in 1911 Rutherford and his group dis-covered the nucleus in atoms. Stimulated by this discovery, in 1913 the Dan-ish physicist Niels Bohr made a bold conjecture that a hydrogen atom couldbe modeled as an electron going around a proton in circular orbits, but onlyin those orbits whose translational angular momentum is an integer multi-ple of . Consequently, these orbits would have only certain radii, and onlycertain values of energy. The differences in energies between these quan-tized orbits match the observed energies of photons emitted by atomic hy-drogen (Figure 9.39).

Bohr’s basic hypothesis—that the angular momentum of electrons in aatom is quantized—has proven to be an important insight, and has been re-tained as a fundamental tenet of the quantum mechanical model of an at-om. As this model has been refined, a probabilistic view of the motion ofelectrons has been adopted. In more sophisticated models electrons do nothave precise trajectories, but only a “probability density”—a probability ofbeing found at a particular location. These more complex quantum me-chanical models explain a much wider range of atomic and molecular phe-nomena than does Bohr’s original model.

v∆t

r

r

v∆t

rsinθθ

Figure 9.38 “Equal area in equal time.”

12--- rv θsin( )∆ t

rmv θsin

h h2π------ 1.05 34–×10 joule second⋅= =

6.63 34–×10

h

hr1

r4

r3

r2

Figure 9.39 In the Bohr model of thehydrogen atom, electrons can be in onlythose circular orbits for which angularmomentum is a multiple of . In a transi-tion between allowed orbits, a photon isemitted.

h

9.11: Atomic and nuclear angular momentum 319

The simple Bohr model does, however, predict correctly the allowed elec-tronic energy levels for atomic hydrogen (as determined from the emissionand absorption spectra of hydrogen atoms). We will work with this model tosee how quantization of angular momentum leads to the prediction of elec-tronic energy levels in a hydrogen atom.

Allowed radii of electron orbits

Because the proton has much more mass than an electron (about 2000times as much), we’ll make the approximation that the proton is at rest,with the electron in a circular orbit around the proton.

? If the magnitude of the electron momentum is p, and the radius ofthe circular orbit is r, what is the translational angular momentum?

The translational angular momentum of the electron relative to the loca-tion of the proton is , where p is the magnitude of the momen-tum (Figure 9.40). Bohr proposed that the only possible states of thehydrogen atom are those where the electron is in a circular orbit whosetranslational angular momentum is an integer multiple of :

BOHR: ANGULAR MOMENTUM IS QUANTIZED

, where N is an integer (1, 2, 3,...)

The electric force and a circular orbit

? What is the magnitude of the electric force that the proton exertson the electron?

where +e is the electric charge on the proton (–e for the electron).

? Use the momentum principle to relate this force to the circularmotion of the electron.

In a circular orbit at constant speed we know that , whereω is the angular speed of the electron in orbit, so the momentum principlegives us this relationship:

Solving for r

We are looking for the allowed values of the orbit radius r, so we look forways to express ω and p in terms of r. Bohr’s angular momentum conditiongives us p in terms of r:

We can write ω in terms of r, in the approximation that v << c, so that, where m is the mass of the electron:

Putting these relations into , we have

Solving the latter relation for r, we obtain the following result (Figure 9.41):

r

p

C

LC = rp

Figure 9.40 The angular momentum ofthe electron in a circular Bohr orbit rela-tive to the proton. LC is out of the page.

Ltrans C, rp=

h

Ltrans,C rp Nh= =

Fel1

4πε0------------ e2

r2----=

dp dt⁄ mω2r–=

mω2r Fel1

4πε0------------e2

r2----= =

p Nhr

-------=

p mv=

ω vr-- mv

mr------- p

mr------ Nh

mr2---------= = = =

mω2r F=

mr Nhmr2---------

⎝ ⎠⎛ ⎞

2 14πε0------------ e2

r2----=


ALLOWED BOHR RADII FOR ELECTRON ORBITS IN HYDROGEN

Let’s evaluate this result numerically, so that later we can compare with mea-surements of the spectrum of light emitted by excited atomic hydrogen. Us-ing

J·s,

N·m2/coulomb2,

coulomb,

and the mass of the electron , we find this:

This is a striking result. The simple Bohr model predicts that the smallestpermissible electron radius (n = 1) is , and atoms are in factobserved to have radii of approximately this size. (Most atoms have a radiusof about .)

Energy for a circular orbit

We cannot observe directly the radius of the orbit of an electron in an atom.What we can observe are photons emitted by excited atoms, which tell us thedifferences in energies between various energy levels of the atom. Using ourmodel we can predict the energies of these energy levels, and see if the dif-ferences between these levels match the energies of photons observed in anatomic spectrum.

Given a set of possible values for r, we can calculate the possible values forenergy of the electron. Starting from our result for circular motion,

, where

we can obtain after a bit of rearranging the kinetic energy of the electron:

? Now write a formula for the kinetic energy plus electric potentialenergy of the hydrogen atom, using the Bohr model.

Kinetic energy:

(from previous equation)

Electric potential energy:

So the energy is this (omitting the rest energies):

This is negative, corresponding to these being bound states. According tothe Bohr model only certain radii will actually occur. Inserting our previousexpression for r, we get:

r N2 h2

14πε0------------

⎝ ⎠⎛ ⎞ e2m---------------------------=

1 2 3 4 5

Figure 9.41 Bohr radii for N=1 through 5.The nucleus is not shown.

h 1.05 34–×10=

1 4πε0⁄ 8.99 9×10=

e 1.60 19–×10=

m 9.11 31–×10=

r N2 0.53 10–×10 m( )=

0.53 10–×10 m

1 10–×10 m

mω2r Fel1

4πε0------------ e2

r2----= = ω v

r--=

12---mv2 1

2--- 1

4πε0------------ e2

r----

⎝ ⎠⎛ ⎞=

K 12---mv2 1

2--- 1

4πε0------------ e2

r----

⎝ ⎠⎛ ⎞= =

Uel 14πε0------------e2

r----–=

E K Uel+ – 12--- 1

4πε0------------ e2

r----

⎝ ⎠⎛ ⎞= =


BOHR MODEL ENERGY LEVELS

Evaluating this expression we get

Converting to electron-volts (1 eV = J) we have (Figure 9.42)

, N = 1, 2, 3,...

This prediction for the quantized energies of atomic hydrogen agrees wellwith the observed electronic spectrum of hydrogen. We quoted this resultfor the quantized energy levels in Chapter 6. is negative, correspond-ing to bound states.

Photons emitted by atomic hydrogen

Bohr proposed that electromagnetic radiation would be emitted whenthere was a sudden change from a higher energy level to a lower one. Wesaw in Chapter 6 that the Bohr energy formula does correctly predict theenergies of photons emitted by excited atomic hydrogen.

? In the Bohr model, does photon emission correspond to the radiusof the circular orbit getting larger or smaller? Does the quantumnumber N increase or decrease?

The kinetic plus potential energy is negative, corresponding to a boundstate, so a higher energy is one that is less negative and therefore corre-sponds to a larger radius (since the energy is proportional to –1/r). Also,higher energy corresponds to a larger value of N, which is also associatedwith a larger radius. Therefore photon emission is associated with a de-crease in r and a decrease in N (Figure 9.43). Photon absorption involves anincrease in r and N.

Refinements of the Bohr model

The Bohr model was an important predecessor to current quantum me-chanical models of the atom. It predicts the main aspects of the hydrogenenergy levels correctly (neglecting various small effects). The idea that pho-ton emission is associated with a drop from a higher energy level to a lowerenergy level, with the photon energy equal to the difference in the hydro-gen energy levels, is fundamental to the extended quantum-mechanicaltreatment.

The notion of deterministic circular orbits has been abandoned and re-placed with a picture of a probabilistic electron “cloud,” but even in moresophisticated models the average radius of this electron cloud is larger forhigher energies, just as in the Bohr model. From your study of chemistryyou are probably familiar with the shapes of the “probability density” func-tions or “orbitals” describing the electron’s location in the atom in differentenergy states, some of which are spherical, others dumbbell shaped.

The idea that angular momentum is quantized has far-reaching conse-quences and is a major element in modern quantum mechanics. However,it has become clear that the actual quantization rules in a hydrogen atomare more complex than those that are assumed in the Bohr model. For ex-ample, the translational angular momentum in the ground state (N = 1) isactually zero, not , and for the next higher state (N = 2), the z componentof translational angular momentum can be either 0 or .

E

14πε0------------

⎝ ⎠⎛ ⎞

2e4m

2N2h2-------------------------------–=

E – 2.17 18–×10 JN2

-------------------------------=

1.6 19–×10

r (distance from proton)

E1

E4E3

E2

E5

Figure 9.42 Bohr model prediction of elec-tronic energy levels for a hydrogen atom.The potential energy curve for the elec-tron–proton system is also shown.

E – 13.6 eVN2

-------------------=

K U+

1 2 3 4 5

absorptionemission

Figure 9.43 A photon is emitted in a transi-tion from a higher to a lower energy level.A photon is absorbed in a transition from alower to a higher energy level.

hh


9.11.2 Particle spin

An important example of angular momentum quantization is the observa-tion that many elementary particles have rotational angular momentum,and it is quantized. We know this in part because of observations of the in-teraction of these particles with other particles having nonzero angular mo-mentum; in these interactions total angular momentum is constant.

For example, the electron, muon, and neutrino are said to have spin, because measurement of a component of their rotational angular mo-

mentum always yields . Every quark has spin , and particlesbuilt out of three quarks, such as protons and neutrons, necessarily havehalf-integral spin. The spin of the proton and neutron is , as though twoof the quarks have their spins opposed (⇑⇓⇑), with no translational angularmomentum of the quarks. Some short-lived three-quark particles have spin

. One way to get this is for all the quarks to have their spins aligned(⇑⇑⇑), or there can be some translational angular momentum of thequarks. While rotational angular momentum can be half-integral, transla-tional angular momentum is always integral (0, 1, 2, etc.).

Particles made out of one quark and one antiquark, called mesons, haveintegral spin. For example, the pion spin is zero (quark and antiquark spinsopposed, ⇑⇓), and the spin of the “rho” meson is 1 (quark and antiquarkspins aligned, ⇑⇑).

Presumably the angular momentum of a macroscopic object such as abaseball is also quantized, but the angular momentum of a baseball is hugecompared to Planck’s constant, and you don’t notice the quantization,which is on an exceedingly fine scale (10–34 J·s!).

9.11.3 Consequences of angular momentum quantization

The quantization of angular momentum has far-reaching consequences.Angular momentum quantization plays a major role in determining thestructure of atoms, and the nature of the chemical periodic table. The twolowest-energy-state electrons in an atom always have zero translational angu-lar momentum and zero rotational angular momentum (because the tworotational angular momenta are always oppositely aligned and add up exact-ly to zero).

There is a very deep connection between angular momentum and the sta-tistical behavior of particles. Particles such as the electron and the protonthat have half-integral spin are called “fermions” and exhibit the peculiarbehavior that two fermions cannot be in the exact same quantum state. Thisis why only two electrons can be in the lowest energy state of an atom, withtheir spins opposed. Their spin directions are forbidden to be the same, be-cause then their energy and angular momentum would be exactly the same,which is forbidden for fermions. This prohibition is called the “Pauli exclu-sion principle.”

On the other hand, particles with integral spin, called “bosons,” are notsubject to the Pauli exclusion principle, and there is no limit to the numberof bosons that can be in the same energy state. In fact, there is a special stateof matter called a “Bose-Einstein condensation” in which very large num-bers of particles end up in exactly the same quantum state. This state of mat-ter was predicted long before it was actually created and observed in 1995.

In a diatomic gas molecule such as oxygen (O2), the kinetic energy asso-ciated with rotation of the molecule is of course . But the rotation-al angular momentum Lrot is quantized, so the rotational energies of oxygenare quantized. The phenomenon of angular momentum quantization af-fects the heat capacity of diatomic gases at low temperatures, as is discussedin Chapter 10. (You can of course refer to the angular momentum of a ox-

1 2⁄1 2⁄( )h± 1 2⁄

1 2⁄

3 2⁄

Lrot2 2I( )⁄


ygen molecule as translational angular momentum of the two atoms insteadof rotational angular momentum of the molecule.)

It is an important prediction of quantum mechanics that the x, y, or zcomponent of the angular momentum (Lx, Ly, or Lz) can only be an integeror half-integer multiple of , whereas the square magnitude of the angularmomentum has the quantized values , where l has integer or half-integer values, depending on whether a component has integer or half-in-teger values:

QUANTIZED VALUES OF L2

, where l is integer or half-integer

All nuclei that have an even number of protons and an even number ofneutrons (“even-even” nuclei) have a total angular momentum of zero. Ex-amples include carbon-12 (6 protons and 6 neutrons) and oxygen-16 (8protons and 8 neutrons). The spins of protons and neutrons in even-evennuclei are paired up in the lowest-energy state of a nucleus with each otherin such a way as to produce zero net angular momentum. This would be anexceedingly unlikely arrangement if angular momentum weren’t quan-tized.

The sections of Chapter 9 dealing with non-zero torque (pages 309-326 in the textbook)

have been omitted from this document.

hl l 1+( )h2

L2 l l 1+( )h2=


9.17 Summary

Fundamental Physical Principles

The angular momentum principle for a system:

or

New concepts

Angular momentum of a particle about location A (Figure 9.66):

, direction given by right-hand rule

Torque is a measure of the twist imparted by a force about location A:

Moment of inertia about an axis of rotation:

Angular momentum is quantized in microscopic systems:A component of angular momentum is an integer or half-integer mul-tiple of .The square magnitude of angular momentum has quantized values

, where l has integer or half-integer values.

Results

Angular momentum for a multiparticle system about location A can be di-vided into translational and rotational angular momentum (Figure 9.67):

,

Rotational angular momentum in terms of moment of inertia: The angular momentum principle for a rotating object:

or

Kinetic energy relative to center of mass in terms of moment of inertia androtational angular momentum:

Angular speed of precession for a gyroscope:

Bohr model of hydrogen: , N = 1, 2, 3,...

, for axis passing through center

Uniform solid cylinder of length L, radius R, about axis perpendicular to cyl-

inder, through center of cylinder:

dLtot,A

dt------------- τnet,ext,A= ∆Ltot,A τnet,ext,A∆t=

AIn the xz plane

Figure 9.66 The fingers of your right handbend from toward and your thumbpoints in the direction of .

rA pLA

rA

p

LA

LA rA p×=

LA rA p θsin=

LA < ypz zpy–( ) zpx xpz–( ) xpy ypx–( )>, ,=

τA rA F×=

I m1r⊥12 m2r⊥2

2 m3r⊥32+ + + ...=

h

L2 l l 1+( )h2=

Center of mass

A

r1

r3

r2rcm,A

p1

p2

p3

Figure 9.67 Calculating translational androtational angular momentum.

LA Ltrans,A Lrot+=

Ltrans,A rcm,A Ptot×= Lrot r1 p1× r2 p2× r3 p3×+ +=

Lrot Iω=

dLrot

dt------------ τnet,ext,cm= ∆Lrot τnet,cm∆t=

Krot12---Iω2 1

2--- Iω( )2

I-------------

Lrot2

2I---------= = =

Ω τLrot--------- RMg

Iω------------= =

E – 13.6 eVN2

-------------------=

Idisk Icylinder12---MR2= = Isphere

25---MR2=

I 112------ML2 1

4---MR2+=

9.18: Example problem: A satellite with solar panels 335

9.18 Example problem: A satellite with solar panels

A satellite has four low-mass solar panels stick-ing out as shown. The satellite can be consid-ered to be approximately a uniform solidsphere. Originally it is traveling to the right withspeed v and rotating clockwise with angular speedωi.

A tiny meteor traveling at high speed v1 ripsthrough one of the solar panels and continuesin the same direction but at reduced speed v2.Afterwards, calculate the vx and vy componentsof the center-of-mass velocity of the satellite,and its angular velocity ωf (magnitude and di-rection). Additional data are provided on thediagram.

Solution

What fundamental principles are useful starting points? The momentumprinciple and the angular momentum principle. The energy principlewould not be a useful starting point, because we don’t know how much ther-mal energy increase is associated with the meteor ripping through the pan-el. Take the satellite and meteor together as the system of interest, in whichcase the momentum and the angular momentum will remain constant,there being no external forces or torques.

Momentum principle, from just before to just after collision:

=

Angular momentum principle, about the original center of the satellite,component into the page, from just before to just after collision:

, where

Calculating just before and just after the collision makes it particularly easyto see how to calculate the translational angular momentum of the meteor,since it is directly above the center of the satellite at those instants. Note thatwe need , not , in calculating the translational angular momen-tum of the meteor.

Solving for the unknown final quantities:

, clockwise, into the page

θ

v1

v2

m

hRM

ωi

v

< Mv m+ v1 θcos( ) m1v1 θ 0>,sin,

< Mvx mv2 θcos+( ) Mvy mv2 θsin+( ) 0>, ,

Iωi hmv1 θcos+ Iωf hmv2 θcos+= I 25---MR2=

θcos θsin

vx v mM----- v1 v2–( ) θcos+=

vymM----- v1 v2–( ) θsin=

ωf ωihm

25---MR2( )

------------------- v1 v2–( ) θcos+=


9.19 Review questions

In addition to these review questions, note that the exercises on page 312and page 325 constitute important worksheets that gives you practice in cal-culating angular momentum and torque.

Angular momentum

RQ 9.1 Give an example of a situation where an object is traveling in astraight line, yet has nonzero angular momentum.

RQ 9.2 Under what circumstances is angular momentum constant? Give anexample of a situation where the x component of angular momentum isconstant, but the y component isn’t.

Translational and rotational angular momentum

RQ 9.3 Give examples of translational angular momentum and rotationalangular momentum in our Solar System.

RQ 9.4 Give an example of a physical situation in which the angular mo-mentum is zero yet the translational and rotational angular momenta areboth nonzero.

The Bohr model of hydrogen

RQ 9.5 What features of the Bohr model of hydrogen are consistent with thelater, full quantum mechanical analysis? What features of the Bohr modelhad to be abandoned?

Particle spin

RQ 9.6 Give some examples of atomic or nuclear phenomena associatedwith particle spin.

Torque

RQ 9.7 Under what conditions is the torque about some location equal tozero?

RQ 9.8 Make a sketch showing a situation where the torque due to a singleforce about some location is 20 N·m in the positive z direction, whereasabout another location the torque is 10 N·m in the negative z direction.

Static equilibrium

RQ 9.9 What must be true for a system to be in static equilibrium (that is,not moving)?

Gyroscopes

RQ 9.10 Two gyroscopes are made exactly alike except that the spinningdisk in one is made of low-density aluminum, whereas the disk in the otheris made of high-density lead. If they have the same spin angular speeds andthe same torque is applied to both, which gyroscope precesses faster?

9.20: Homework problems 337

9.20 Homework problems

Problem 9.1 Angular momentum in an elliptical orbitIn Chapter 2 you wrote a program to model the motion of a planet goingaround a fixed star. In this problem you will build on that program.

(a) Use initial conditions that produce an elliptical orbit. At each step cal-culate the translational angular momentum of the planet with re-spect to a location A chosen to be in the orbital plane but outside the orbit.Display this in two ways (i and ii below), and briefly describe in words whatyou observe:

(i) Display as an arrow with its tail at location A, throughoutthe orbit. Since the magnitude of is quite different fromthe magnitudes of the distances involved, you will need to scale thearrow by some factor to fit it on the screen.

(ii) Graph the component of perpendicular to the orbitalplane as a function of time.

(b) Repeat part (a), but this time choose a different location B at the cen-ter of the fixed star, and calculate and display relative to that loca-tion B. As in part (a), display as an arrow (scaled appropriately), andalso graph the component of perpendicular to the orbital plane, asa function of time, and briefly describe in words what you observe.

(c) Choose a location C which is not in the orbital plane, and calculateand display as an arrow throughout the orbit. (You do not need tomake a graph.) Briefly describe in words what you observe.

Problem 9.2 Playground rideA playground ride consists of a disk of mass M and radius Rmounted on a low-friction axle (Figure 9.68). A child of mass mruns at speed v on a line tangential to the disk and jumps ontothe outer edge of the disk.

(a) If the disk was initially at rest, now how fast is it rotating?(The moment of inertia of a uniform disk is .)

(b) If you were to do a lot of algebra to calculate the kineticenergies in part (a), you would find that Kf < Ki. Where has thisenergy gone?

(c) Calculate the change in linear momentum of the system consisting ofthe child plus the disk (but not including the axle), from just before to justafter impact. What caused this change in the linear momentum?

(d) The child moves to a location a distance R/2 from the axle. Now whatis the angular speed?

(e) If you were to do a lot of algebra to calculate the kinetic energies inpart (d), you would find that Kf > Ki. Where has this energy come from?

(f) In part (a), estimate numerical values for all of the quantities and de-termine a value for the spin rate ω. Is your result reasonable?

Problem 9.3 Moment of inertia of a diskShow that the moment of inertia of a disk of mass M and radius R is .Divide the disk into narrow rings each of radius r and width dr. The contri-bution to I by one of these rings is simply r2dm, where dm is the amount ofmass contained in that particular ring. The mass of any ring is the mass perunit area times the area of the ring. The area of this ring is approximately2πrdr. Use integral calculus to add up all the contributions.

Problem 9.4 Rotating diskA disk of radius 0.2 m and moment of inertia 1.5 kg·m2 is mounted on anearly frictionless axle (Figure 9.69). A string is wrapped tightly around thedisk, and you pull on the string with a constant force of 25 N. After a whilethe disk has reached an angular speed of 2 radians/s. What is its angularspeed 0.1 seconds later? Explain briefly.

Ltrans,A

Ltrans,A

Ltrans,A

Ltrans,A

Ltrans,B

Ltrans,B

Ltrans,B

Ltrans,C

M

R

v

m

Figure 9.68 A child runs and jumps on aplayground ride (Problem 9.2).

12---MR2

12---MR2

25 N

0.2 m

1.5 kg·m2

2 rad/s

Figure 9.69 A disk rotates on a nearly fric-tionless axle (Problem 9.4).


Problem 9.5 Momentum and angular momentumTwo small objects each of mass m are connected by a lightweight rod oflength d (Figure 9.70). At a particular instant they have velocities as shownand are subjected to external forces as shown. The system is moving in outerspace.

In the following questions involving vectors, give components along the axesshown, and state which axis you’re using for each component.

(a) What is the total (linear) momentum of this system?(b) What is the velocity of the center of mass?(c) What is the total angular momentum of the system relative to

point A?(d) What is the rotational angular momentum of the system?(e) What is the translational angular momentum of the system rel-

ative to point A?After a short time interval ∆t, (f) What is the total (linear) momentum of the system?(g) What is the rotational angular momentum of the system?

Problem 9.6 Ice skaterAn ice skater whirls with her arms and one legstuck out as shown, making one complete turn inone second (Figure 9.71). Then she quicklymoves her arms up above her head and pulls herleg in as shown.(a) Estimate how long it now takes for her tomake one complete turn. Explain your calcula-tions, and state clearly what approximations andestimates you make.(b) Estimate the minimum amount of chemicalenergy she must expend to change her configura-tion.

Problem 9.7 Rotating stool and barbellsYou sit on a rotating stool and hold barbells in both hands with your armsfully extended horizontally. You make one complete turn in 2 seconds. Youthen pull the barbells in close to your body.

(a) Estimate how long it now takes you to make one complete turn. Beclear and explicit about the principles you apply and about your assump-tions and approximations.

(b) About how much energy did you expend?

Problem 9.8 An asteroid collisionA spherical non-spinning asteroid of mass M and radius R moving withspeed v1 to the right collides with a similar non-spinning asteroid movingwith speed v2, to the left, and they stick together (Figure 9.72). The impactparameter is d. Note that .

After the collision, what is the velocity vcm of the center of mass and theangular velocity ω about the center of mass? (Note that each asteroid rotatesabout its own center with this same ω.)

Ptotal

F1

F2

v1

v2

m

m

d

w

h

A

x

y

z (out of page)

Figure 9.70 Two masses connected by alightweight rod (Problem 9.5).

vcmLA

Lrot

Ltrans

Ptotal

Lrot

Figure 9.71 A spinning skater pulls in oneleg (Problem 9.6).

Isphere25---MR2=

M, R v1

v2

d

M, R

Figure 9.72 Two asteroids collide and sticktogether (Problem 9.8).


Problem 9.9 A collision with a rod with masses on the endsTwo small objects each of mass m1 are con-nected by a lightweight rod of length L (Fig-ure 9.73). At a particular instant the center ofmass speed is v1 as shown, and the object is ro-tating counterclockwise with angular speedω1. A small object of mass m2 traveling withspeed v2 collides with the rod at an angle θ2 asshown, at a distance b from the center of therod. After being struck, the mass m2 is ob-served to move with speed v4, at angle θ4. Allthe quantities are positive magnitudes. This alltakes place in outer space.

For the object consisting of the rod with thetwo masses, write equations that, in principle,could be solved for the center of mass speedv3, direction θ3, and angular speed ω3 in terms of the given quantities. Stateclearly what physical principles you use to obtain your equations.

Don’t attempt to solve the equations; just set them up.

Problem 9.10 Space junkA tiny piece of space junk of mass m strikes a glancingblow to a spherical satellite (Figure 9.74). After the colli-sion the space junk is traveling in a new direction andmoving more slowly. The space junk had negligible rota-tion both before and after the collision. The velocities ofthe space junk before and after the collision are shown inthe diagram. The center of mass of the satellite is at itsgeometrical center. The satellite has mass M, radius R,and moment of inertia I about its center. Before the col-lision the satellite was moving and rotating as shown inthe diagram.

(a) Just after the collision, what are the components of the center-of-massvelocity of the satellite (vx and vy) and its rotational speed ω?

(b) Calculate the rise in the thermal energy of the satellite and space junkcombined. You do not need to substitute in the values for quantities you al-ready calculated in part (a).

Problem 9.11 Space stationA space station has the form of a hoop of radius R, with mass M (Figure9.75). Initially its center of mass is not moving, but it is spinning with angu-lar speed ω0. Then a small package of mass m is thrown by a spring-loadedgun toward a nearby spacecraft as shown; the package has a speed v afterlaunch. Calculate the center-of-mass velocity of the space station (vx and vy)and its rotational speed ω after the launch.

Problem 9.12 Rotational spectrum of a diatomic molecule(a) Calculate the energies of the quantized rotational energy levels for

O2. The most common oxygen nucleus contains 8 protons and 8 neutrons.Estimate any quantities you need. See discussion of diatomic molecules onpage 322; the parameter l has values 0, 1, 2, 3...

(b) Describe the emission spectrum for electromagnetic radiation emit-ted in transitions among the rotational O2 energy levels. Include a calcula-tion of the lowest-energy emission in electron volts ( J).

(c) It is transitions among “electronic” states of atoms that produce visiblelight, with photon energies on the order of a couple of electron-volts. Eachelectronic energy level has quantized rotational and vibrational (harmonic

v2

m1

m1

bm2

L

ω1

ω1

Just before collision

v1

Just after collision

ω3

ω3

m1

m1

m2

v3

θ3

v4

θ4

L

θ2

Figure 9.73 A spinning rod is struck by asmall object (Problem 9.9).

ω1 v1

M, R, I

v2 mθ

v3

x

y

z

Figure 9.74 Space junk strikes a satellite(Problem 9.10).

R

M

m

v

θ

ω0

y

x

Figure 9.75 Launch a package from aspace station (Problem 9.11).

1.6 19–×10


oscillator) energy sublevels. Explain why this leads to a visible spectrum thatcontains “bands” rather than individual energies.

Problem 9.13 The Bohr modelThe Bohr model correctly predicts the main energy levels not only for atom-ic hydrogen but also for other “one-electron” atoms where all but one of theatomic electrons has been removed, such as in He+ (one electron removed)or Li++ (two electrons removed).

(a) Predict the energy levels in eV for a system consisting of a nucleus con-taining Z protons and just one electron. You need not recapitulate the en-tire derivation for the Bohr model, but do explain the changes you have tomake to take into account the factor Z.

(b) The negative muon (µ–) behaves like a heavy electron, with the samecharge as the electron but with a mass 207 times as large as the electronmass. As a moving µ– comes to rest in matter, it tends to knock electrons outof atoms and settle down onto a nucleus to form a “one-muon” atom. For asystem consisting of a lead nucleus (Pb208 has 82 protons and 126 neutrons)and just one negative muon, predict the energy of a photon in eV emittedin a transition from the first excited state to the ground state. The high-en-ergy photons emitted by transitions between energy levels in such “muonicatoms” are easily observed in experiments with muons.

(c) Calculate the radius of the smallest Bohr orbit for a µ– bound to a leadnucleus (Pb208 has 82 protons and 126 neutrons). Compare with the ap-proximate radius of the lead nucleus (remember that the radius of a protonor neutron is about 10–15 m, and the nucleons are packed closely togetherin the nucleus).

Comments: This analysis in terms of the simple Bohr model hints at theresult of a full quantum-mechanical analysis, which shows that in the groundstate of the lead-muon system there is a rather high probability for findingthe muon inside the lead nucleus. Nothing in quantum mechanics forbidsthis penetration, especially since the muon does not participate in thestrong interaction.

The eventual fate of the µ– in a muonic atom is that it either decays intoan electron, neutrino, and antineutrino, or it reacts through the weak inter-action with a proton in the nucleus to produce a neutron and a neutrino.This “muon capture” reaction is more likely if the probability is high for themuon to be found inside the nucleus, as is the case with heavy nuclei suchas lead.

Problem 9.14 Nuclear gamma rayThe nucleus dysprosium-160 (containing 160 nucleons) acts like a spinningobject with quantized angular momentum, , and for this nu-cleus it turns out that l must be an even integer (0, 2, 4...). When a Dy-160nucleus drops from the l = 2 state to the l = 0 state, it emits an 87 keV photon( ).

(a) What is the moment of inertia of the Dy-160 nucleus?(b) Given your result from part (a), find the approximate radius of the

Dy-160 nucleus, assuming it is spherical. (In fact, these and similar experi-mental observations have shown that some nuclei are not quite spherical.)

(c) The radius of a (spherical) nucleus is given approximately by, where A is the total number of protons and neutrons.

Compare this prediction with your result in part (b).

L2 l l 1+( )h2=

87 3×10 eV

1.3 15–×10 m( )A1 3/


Problem 9.15 A cometA certain comet of mass m at its closest approach to the Sun isobserved to be at a distance r1 from the center of the Sun,moving with speed v1. At a later time the comet is observed tobe at a distance r2 from the center of the Sun, and the anglebetween and the velocity vector is measured to be θ. Whatis v2? Explain briefly.

Problem 9.16 Yo-yoA yo-yo is constructed of three disks (Figure 9.76): two outerdisks of mass M, radius R, and thickness d, and an inner disk(around which the string is wrapped) of mass m, radius r, andthickness d. The yo-yo is suspended from the ceiling and then released withthe string vertical.

Calculate the tension in the string as the yo-yo falls. Note that when thecenter of the yo-yo moves down a distance y, the yo-yo turns through an an-gle y/r, which in turn means that the angular speed ω is equal to vcm/r. Themoment of inertia of a uniform disk is .

Problem 9.17 A hovering yo-yoString is wrapped around an object of mass M and moment of inertia I. Youpull the string with your hand straight up with some constant force F suchthat the center of the object does not move up or down, but the object spinsfaster and faster (Figure 9.77). This is like a yo-yo; nothing but the verticalstring touches the object.

When your hand is a height y0 above the floor, the object has an angularspeed ω0. When your hand has risen to a height y above the floor, what isthe angular speed ω of the object?

Your result should not contain F nor the (unknown) radius of the object.Explain what physics principles you are using.

Problem 9.18 Bullet and stickA stick of length L and mass M hangs from a low-friction axle (Figure 9.78).A bullet of mass m traveling at a high speed v strikes near the bottom of thestick and quickly buries itself in the stick.

(a) During the brief impact, is the linear momentum of the stick+bulletsystem constant? Explain why or why not. Include in your explanation asketch of how the stick shifts on the axle during the impact.

(b) During the brief impact, around what point does the angular momen-tum of the stick+bullet system remain constant?

(c) Just after the impact, what is the angular speed ω of the stick (with thebullet embedded in it)? (Note that the center of mass of the stick has aspeed ωL/2. The moment of inertia of a uniform rod about its center ofmass is .)

(d) Calculate the change in kinetic energy from just before to just afterthe impact. Where has this energy gone?

(e) The stick (with the bullet embedded in it) swings through a maxi-mum angle θmax after the impact, then swings back. Calculate θmax.

Problem 9.19 An object rolls down a rampA solid object of uniform density with mass M, radius R, and moment of in-ertia I rolls without slipping down a ramp at an angle θ to the horizontal.The object could be a hoop, a disk, a sphere, etc.

(a) Carefully follow the complete analysis procedure explained in Chap-ter 3, but with the addition of the angular momentum principle about thecenter of mass. Note that in your force diagram you must include a smallfrictional force f that points up the ramp. Without that force the object willslip. Also note that the condition of nonslipping implies that the instanta-neous velocity of the atoms of the object that are momentarily in contact

r2

r1

v1

r2

v2

θm

M Mmr

R

Figure 9.76 A yo-yo is released and movesdownward (Problem 9.16).

12---MR2

F

M, I y0

ω0

F

y

ω

Floor Floor

Figure 9.77 The center of this yo-yodoesn’t move up or down (Problem 9.17).

mM

L

v

Closeup

θmax

Figure 9.78 A bullet strikes a stick that issuspended from an axle (Problem 9.18).

112------ML2


with the ramp is zero, so f < µFN (no slipping). This zero-velocity conditionalso implies that vcm = ωR, where ω is the angular speed of the object, sincethe instantaneous speed of the contact point is vcm–ωR.

(b) The moment of inertia about the center of mass of a uniform hoop isMR2, for a uniform disk it is (1/2)MR2, and for a uniform sphere it is(2/5)MR2. Calculate the acceleration dvcm/dt for each of these objects.

(c) If two hoops of different mass are started from rest at the same timeand the same height on a ramp, which will reach the bottom first? If a hoop,a disk, and a sphere of the same mass are started from rest at the same timeand the same height on a ramp, which will reach the bottom first?

(d) Write the energy equation for the object rolling down the ramp, andfor the point-particle system. Show that the time derivatives of these equa-tions are compatible with the force and torque analyses.

Problem 9.20 Pulling a rotating deviceA string is wrapped around a uniform disk of mass M and ra-dius R. Attached to the disk are four low-mass rods of radius b,each with a small mass m at the end (Figure 9.79).The apparatus is initially at rest on a nearly frictionless sur-face. Then you pull the string with a constant force F. At theinstant when the center of the disk has moved a distance d, alength w of string has unwound off the disk.(a) At this instant, what is the speed of the center of the appa-ratus? Explain your approach.(b) At this instant, what is the angular speed of the apparatus?Explain your approach.(c) You keep pulling with constant force F for an additional

time ∆t. By how much (∆ω) does the angular speed of the apparatus in-crease in this time interval ∆t?

Problem 9.21 A rotating disk with masses sliding on a rodA rod of length L and negligiblemass is attached to a uniform diskof mass M and radius R (Figure9.80). A string is wrapped aroundthe disk, and you pull on the stringwith a constant force F. Two smallballs each of mass m slide along therod with negligible friction. Theapparatus starts from rest, andwhen the center of the disk hasmoved a distance d, a length of string s has come off the disk, and the ballshave collided with the ends of the rod and stuck there. The apparatus slideson a nearly frictionless table. Here is a view from above:

(a) At this instant, what is the speed v of the center of the disk?(b) At this instant the angular speed of the disk is ω. How much thermal

energy has been produced?(c) In the next short amount of time ∆t, by how much will the angular

speed change (∆ω)?

Problem 9.22 Asteroid collisionSuppose an asteroid of mass is nearly at rest outside the solar sys-tem, far beyond Pluto. It falls toward the Sun and crashes into the Earth atthe equator, coming in at an angle of 30 degrees to the vertical as shown,against the direction of rotation of the Earth (Figure 9.81). It is so large thatits motion is barely affected by the atmosphere.

(a) Calculate the impact speed. (b) Calculate the change in the length of a day due to the impact.

M, R

F

dF

w + d

m

m

m

m

b

Figure 9.79 A rotating disk with fourmasses attached (Problem 9.20).

M, R

m

m

F FωL

d

d+s

Figure 9.80 The masses slide on the rod(Problem 9.21).

Asteroid

Earth

Sun

30o

Figure 9.81 An asteroid crashes into theEarth (Problem 9.22; not to scale).

2 21×10 kg


Problem 9.23 Stick on iceA stick of mass M and length L is lying on ice (Figure 9.82). A small mass

m traveling at high speed vi strikes the stick a distance d from the center andbounces off with speed vf as shown in the diagram, which is a top view of thesituation. The magnitudes of the initial and final angles to the x axis of thesmall mass’s velocity are θi and θf. All of the symbols in the diagram repre-sent positive numbers.

(a) Afterwards, what are the velocity components vx and vy of the centerof the stick? Explain briefly.

(b) Afterwards, what is the magnitude and direction of the angular veloc-ity of the stick?

(c) What is the increase in thermal energy of the objects? You can leaveyour expression in terms of the initial quantities and vx, vy, and ω.

Problem 9.24 DiverA diver dives from a high platform. When he leaves the platform, he tuckstightly (Figure 9.83) and performs three complete revolutions in the air,then straightens out with his body fully extended before entering the water(Figure 9.84). He is in the air for a total time of 1.4 seconds. What is his an-gular speed ω just as he enters the water? Give a numerical answer.

Be explicit about the details of your model, and include (brief) explana-tions. You will need to estimate some quantities.

Problem 9.25 Gyroscope experiment—qualitativeThis problem requires that you have a toy gyroscope available. The purposeof this problem is to make as concrete as possible the unusual motions of agyroscope and their analysis in terms of fundamental principles. In all of thefollowing studies, the effects are most dramatic if you give the gyroscope aslarge a spin angular speed as possible.

(a) Hold the spinning gyroscope firmly in your hand, and try to rotate thespin axis quickly to point in a new direction. Explain qualitatively why thisfeels “funny.” Also explain why you don’t feel anything odd when you movethe spinning gyroscope in any direction without changing the direction ofthe spin axis.

(b) Support one end of the spinning gyroscope (on a pedestal or in anopen loop of the string) so that the gyroscope precesses counterclockwise asseen from above. Explain this counterclockwise precession direction; in-clude sketches of top and side views of the gyroscope.

(c) Again support one end of the spinning gyroscope so that the gyro-scope precesses clockwise as seen from above. Explain this clockwise preces-sion direction; include sketches of top and side views of the gyroscope.

Problem 9.26 Gyroscope experiment—quantitativeThis problem requires that you have a toy gyroscope available. The purposeof this problem is to make as concrete as possible the unusual motions of agyroscope and their analysis in terms of fundamental principles. In all of thefollowing studies, the effects are most dramatic if you give the gyroscope aslarge a spin angular speed as possible.

(a) If you knew the spin angular speed of your gyroscope, you could pre-dict the precession rate. Invent an appropriate experimental technique anddetermine the spin angular speed approximately. Explain your experimen-tal method and your calculations. Then predict the corresponding preces-sion rate, and compare with your measurement of the precession rate. Youwill have to measure and estimate some properties of the gyroscope andhow it is constructed.

(b) Make a quick measurement of the precession rate with the spin axishorizontal, then make another quick measurement of the precession ratewith the spin axis nearly vertical. (It you make quick measurements, friction

θi

θf

vi

vf

m

M

Ld

x

y

z

View from above

Figure 9.82 A stick lying on ice is struck bya small mass (Problem 9.23).

ω

Figure 9.83 Tuckedposition (side view).

Figure 9.84 Enter-ing water (frontview).


on the spin axis doesn’t have much time to change the spin angular speed.)Repeat, this time with the spin axis initially nearly vertical, then horizontal.Making all four of these measurements gives you some indication of howmuch the spin unavoidably changes due to friction while you are quicklychanging the angle. What do you conclude about the dependence of theprecession rate on the angle, assuming the same spin rate at these differentangles? What is the theoretical prediction for the dependence of the preces-sion rate on angle (for the same spin rate)?

Problem 9.27 Wood or steel top(a) A solid wood top spins at high speed on the floor, with a spin direction

shown in Figure 9.85. Using appropriately labeled diagrams, explain the di-rection of motion of the top (you do not need to explain the magnitude).

(b) How would the motion change if the top had a higher spin rate? Ex-plain briefly.

(c) If the top were made of solid steel instead of wood, explain how thiswould affect the motion (for the same spin rate).

Problem 9.28 Bicycle wheelA bicycle wheel with a heavy rim is mounted on a lightweight axle, and oneend of the axle rests on top of a post. The wheel is observed to precess in thehorizontal plane (Figure 9.86). With the spin direction shown, does thewheel precess clockwise or counterclockwise? Explain in detail, includingappropriate diagrams.

Problem 9.29 GyroscopeThe axis of a gyroscope is tilted at an angle of 30° to the vertical (Figure9.87). The rotor has a radius of 15 cm, mass 3 kg, moment of inertia 0.06kg·m2, and spins on its axis at 30 radians/s. It is supported in a cage (notshown) in such a way that without an added weight it does not precess. Thena mass of 0.2 kg is hung from the axis at a distance of 18 cm from the centerof the rotor.

(a) Viewed from above, does the gyroscope precess in a 1) clockwise or a2) counterclockwise direction? That is, does the top end of the axis move 1)out of the page or 2) into the page in the next instant? Explain your reason-ing.

(b) How long does it take for the gyroscope to make one complete pre-cession?

Problem 9.30 Angular momentum of the Earth(a) Calculate the magnitude of the translational angular momentum of

the Earth relative to the center of the Sun. See data on inside back cover.(b) Calculate the magnitude of the rotational angular momentum of the

Earth. How does this compare to your result in part (a)?(The angular momentum of the Earth relative to the center of the Sun is

the sum of the translational and rotational angular momenta. The rotation-al axis of the Earth is tipped away from a perpendicular to the planeof its orbit.)

Problem 9.31 Clay falls on wheelA stationary bicycle wheel of radius R is mounted in the vertical plane on ahorizontal frictionless axle (Figure 9.88). The wheel has mass M, all concen-trated in the rim (the spokes have negligible mass). A lump of clay with massm falls and sticks to the outer edge of the wheel at the location shown. Justbefore the impact the clay has a speed v.

(a) Just after the impact, what are the magnitude and direction of the an-gular velocity of the wheel?

(b) Calculate the impulse ( ) applied to the system by the axle duringthe collision.

Spin

Figure 9.85 A wood or steel top (Problem9.27).

CWCCW

Spin

Figure 9.86 A bicycle wheel on a pivot(Problem 9.28).

30°

3 kg

0.06 kg·m2

15 cm

0.2 kg

18 c

mSpin 30 rad/s

Figure 9.87 A gyroscope (Problem 9.29).

23.5°

M

m

v

R

C

Figure 9.88 A lump of clay falls onto awheel mounted on a frictionless axle(Problem 9.31).

F∆t


9.21 Answers to exercises

9.1 (page 297)

9.2 (page 297)

9.3 (page 297) 395 J

9.4 (page 298)

9.5 (page 298) radian/s, or simply 1/s

9.6 (page 298)

9.7 (page 299)

9.8 (page 299)

9.9 (page 299) The vector points up at the North Pole.

9.10 (page 299) 40 J9.11 (page 304) (1) 80 kg·m2/s (see Figure 9.89 for rectangles)

(2) 80 kg·m2/s(3) 80 kg·m2/s(4) 50 kg·m2/s(5) 0(6) 0(7) 70 kg·m2/s

9.12 (page 305) 0.012 m/s

9.13 (page 306) 57.8 kg·m2/s

9.14 (page 306) See Figure 9.90 for parallelogram

9.15 (page 308) A: 30 kg·m2/s, into page ( )B: 30 kg·m2/s, into page ( )C: 0D: 50 kg·m2/s, out of page ( )E: 50 kg·m2/s, out of page ( )F: 50 kg·m2/s, out of page ( )G: 0H: 30 kg·m2/s, into page ( )

9.16 (page 308) 0.38 kg·m2/s, into page ( )

9.17 (page 309) kg·m2/s

9.18 (page 309) kg·m2/s

9.19 (page 311) Nonzero; into the page; zero; (no direction)9.20 (page 311) No; nonzero; into the page; nonzero; into the page9.21 (page 311) Nonzero; up (toward the sky); nonzero; up

9.22 (page 312) (a) up; (b) ; (c) up; (d)

9.23 (page 312) (a) into page

(b) into page ( )

(c)

(d) into page

(e) into page ( )

(f) they are equal

(g)

2M d2---

⎝ ⎠⎛ ⎞

2 12---Md2

=12---MR2ω2

kg m2⋅

kg m2/s⋅

72 kg m2/s⋅

7 33×10 kg m2/s⋅

8 m

5 m 5 m

(1) (3)(2)

A

5 m 5 m

7 m

(7)

(6)

(4) A

Figure 9.89 The area of each rectangle isequal to the magnitude of the correspond-ing angular momentum.

p 10 kg m/s⋅=

D

rD 6 m=

p 15 kg m/s⋅=

40°

Figure 9.90 The area of the parallelogramis equal to the magnitude of the angularmomentum.

0 0 2–, ,⟨ ⟩

3 11 23–, ,⟨ ⟩

d1p1 αsin d2p2

d 2⁄( )p d 2⁄( ) m d 2⁄( )ω0[ ]=

2m d 2⁄( )2ω0

2m d 2⁄( )2

2m d 2⁄( )2ω0

m d 2⁄( )2ω02


9.24 (page 313) (a) 0

(b) into page ( )

(c) into page ( )

9.25 (page 313) (a) into page ( )

(b) into page ( )

(c) into page ( )

9.26 (page 315) Negligible external forces means negligible externaltorques.Therefore magnitude and direction of angular momen-tum constant.

9.27 (page 315) No torques around center of mass means no change inrotational angular momentum, so rotational angular mo-mentum stays constant in magnitude (which determineslength of day) and direction (which determines what“North Star” the axis points at). Doesn’t matter thatEarth is going around Sun; rotational angular momen-tum affected solely by torque around center of mass.

9.28 (page 317) Force directed toward center of Sun, so zero torqueabout the location of the Sun.Angular momentum about center of Sun constant.

9.29 (page 317) , into page; , into page; 9.30 (page 318)9.31 (page 324) 6 N·m9.32 (page 324) 2 N9.33 (page 324) kg·m2/s9.34 (page 325) 0.8 N·m, into the page, or N·m9.35 (page 325) (a) 0.0256 kg·m2/s, into page ( ), or

kg·m2/s(b) 0.1856 kg·m2/s, into page ( ), or

kg·m2/s(c) 145 rad/s, into page ( ), or

rad/s9.36 (page 325) 8.33 N

9.37 (page 327) , since no torque.

Differentiating, we get .

9.38 (page 327)

(+z out of page)

9.39 (page 327)

Substitute and get

9.40 (page 330) Same as in horizontal case:

b 2m( ) bω1( )

b 2m( ) bω1( )

2m d 2⁄( )2ω2

b 2m( ) bω1( )

2m b2ω1 d 2⁄( )2ω2+[ ]

r1mv1 r2mv2 r1v1 r2v2=

J s⋅ kg m2⋅ /s2( ) s( ) kg m2⋅ /s= =

4 3.8 0, ,⟨ ⟩

0 0 0.8–, ,⟨ ⟩

0 0 0.0256–, ,⟨ ⟩

0 0 0.1856–, ,⟨ ⟩

0 0 145–, ,⟨ ⟩

dLstring z,

dt---------------------

ddt----- RMvcm

12---MR2ω– 0= =

dωdt-------

2R---

dvcm

dt-----------

2FT

MR---------= =

dPz

dt-------- FN M1g– M2g– 0= =

dLsupport z,

dt------------------------ d1M1g dM2g– 0= =

dLleft person z,

dt------------------------------ d1FN d1 d+ 2( )M2g– 0= =

FN M1 M2+( )g=

d1M1g d2M2g– 0=

Ω RMgIω

------------=

Chapter 9 Angular Momentumnewton.uor.edu/facultyfolder/eric_hill/Phys231/09...296 Chapter 9: Angular...

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