Ch7 angular momentum

Chapter 7

Angular Momentum, Part I(Constant L)

The angular momentum of a point mass, relative to a given origin, is

L = r× p. (7.1)

For a collection of particles, the total L is simply the sum of the L’s of each particle.The quantity r×p is a useful thing to study because it has many nice properties.

One of these was presented in Theorem 6.1, which allowed us to introduce the“effective potential” in eq. 6.10. And later in this chapter we’ll introduce the conceptof torque, τ , which appears in the bread-and-butter statement, τ = dL/dt (which isanalogous to Newton’s F = dp/dt law). This equation is the basic ingredient, alongwith F = ma, in solving any rotation problem.

There are two basic types of angular momentum problems in the world. Sincethe solution to any rotational problem invariably comes down to using τ = dL/dt,we must determine how L changes in time. And since L is a vector, it can changebecause (1) its length changes, or (2) its direction changes (or through some com-bination of these effects). In other words, if we write L = LL, where L is theunit vector in the L direction, then L can change because L changes, or because Lchanges, or both.

The first of these cases, that of constant L, is the easily understood one. If youhave a spinning record (in which case L =

∑r × p is perpendicular to the record,

assuming that you’ve chosen the center as the origin), and if you give the record atangential force in the proper direction, then it will speed up (in a precise way whichwe will soon determine). There is nothing mysterious going on here. If you pushon the record, it goes faster. L points in the same direction as before, but it nowsimply has a larger magnitude. In fact, in this type of problem, you can completelyforget that L is a vector; you can just deal with its magnitude L, and everythingwill be fine. This first case will be the subject of the present chapter.

In contrast, the second case, where L changes direction, can get rather confusing.This will be the subject of the following chapter, where we will talk about gyroscopes,tops, and other such spinning objects that have a tendency to make one’s head spin

VII-1

VII-2 CHAPTER 7. ANGULAR MOMENTUM, PART I (CONSTANT L)

also. In this case, the entire point is that L is actually a vector. And unlike the firstcase, you really have to visualize things in three dimensions to see what’s going on.1

The angular momentum of a point mass is given by the simple expression in eq.(7.1). But in order to deal with setups in the real world, which invariably consistof many particles, we must learn how to calculate the angular momentum of anextended object. This is the task of the Section 7.1. (But we’ll only deal withmotion in the x-y plane in this chapter. Chapter 8 deals with general 3-D motion.)

7.1 Pancake object in x-y plane

Consider a flat, rigid body undergoing arbitrary motion in the x-y plane (seeFig. 7.1). What is the angular momentum of this body, relative to the origin

y

x

ω

V

CM

Figure 7.1

of the coordinate system?2

If we imagine the body to consist of particles of mass mi, then the angularmomentum of the entire body is the sum of the angular momenta of each mi (whichare Li = ri × pi). So the total angular momentum is

L =∑

i

ri × pi. (7.2)

(For a continuous distribution of mass, we’d have an integral instead of a sum.)L depends on the locations and momenta of the masses. The momenta in turndepend on how fast the body is translating and spinning. Our goal here is to findthis dependence of L on the distribution and motion of its constituent masses. Thisresult will involve the geometry of the body in a specific way, as we will show.

In this chapter, we will deal only with pancake-like objects which move in thex-y plane (or simple extensions of these). We will find L relative to the origin, andwe will also derive an expression for the kinetic energy.

Note that since both r and p for our pancake-like objects always lie in the x-yplane, the vector L = r×p always points in the z direction. This fact is what makesthese pancake cases easy to deal with; L changes only because its length changes,not its direction. So when we eventually get to the τ = dL/dt equation, it will takeon a simple form.

Let’s first look at a special case, and then we will look at general motion in thex-y plane.

7.1.1 Rotation about the z-axis

The pancake in Fig. 7.2 rotates with angular speed ω around the z-axis, in the

y

x

ω

Figure 7.2

counterclockwise direction (as viewed from above). Consider a little piece of the1The difference between these two cases is essentially the same as the difference between the

two basic F = dp/dt cases. The vector p can change simply because its magnitude changes, inwhich case we have F = ma. Or, p can change because its direction changes, in which case we endup with the centripetal acceleration statement, F = mv2/r. (Or, there could be a combination ofthese effects). The former case seems a bit more intuitive than the latter.

2Remember, L is defined relative to a chosen origin (since it has the vector r in it), so it makesno sense to ask what L is, without specifying what origin you’ve chosen.

7.1. PANCAKE OBJECT IN X-Y PLANE VII-3

body, with mass dm and position (x, y). Let r =√

x2 + y2. This little piece travelsin a circle around the origin. Its speed3 is v = ωr. Therefore, the angular momentumof this piece (relative to the origin) is equal to L = r × p = r(v dm)z = dmr2ωz.The z direction arises from the cross product of the (orthogonal) vectors r and p.The angular momentum of the entire body is therefore

L =∫

r2ωz dm

=∫

(x2 + y2)ωz dm, (7.3)

where the integration runs over the area of the body. (If the density of the objectis constant, as is usually the case, then we have dm = ρ dx dy.) If we define themoment of inertia around the z-axis to be

Iz ≡∫

r2 dm =∫

(x2 + y2) dm, (7.4)

then the z-component of L isLz = Izω, (7.5)

and Lx and Ly are both equal to zero. In the case where the rigid body is made upof a collection of point masses, mi, in the x-y plane, the moment of inertia in eq.(7.4) simply takes the discretized form,

Iz ≡∑

i

mir2i . (7.6)

Given any rigid body in the x-y plane, we can calculate Iz. And given ω, wecan then multiply it by Iz to find Lz. In Section 7.2.1, we will get some practicecalculating many moments of inertia.

What is the kinetic energy of our object? We need to add up the energies ofall the little pieces. A little piece has energy dmv2/2 = dm(rω)2/2. So the totalkinetic energy is

T =∫ 1

2r2ω2 dm. (7.7)

With our definition of Iz in eq. (7.4), we have

T =12Izω

2. (7.8)

This is easy to remember, because it looks a lot like the kinetic energy of a pointmass, (1/2)mv2.

7.1.2 General motion in x-y plane

How do we deal with general motion in the x-y plane? For the motion in Fig. 7.3,

y

x

ω

V

CM

Figure 7.3

3The velocity is actually given by v = ~ω× r, which reduces to v = ωr in our case. The vector ~ωis the angular velocity vector, which is defined to point along the axis of rotation, with magnitude ω(so ~ω = ωz here). There is no great need to use the vector ~ω in the constant L case in this chapter,so we won’t. But don’t worry, in the next chapter you’ll get all the practice with ~ω that you couldpossibly hope for.


the various pieces of mass are not traveling in circles about the origin, so we cannotwrite v = ωr, as we did above.

It turns out to be highly advantageous to write the angular momentum, L, andthe kinetic energy, T , in terms of the center-of-mass (CM) coordinates and thecoordinates relative to the CM. The expressions for L and T take on very nice formswhen written this way, as we now show.

Let the coordinates of the CM be R = (X, Y ), and let the coordinates relativeto the CM be r′ = (x′, y′). Then r = R + r′ (see Fig. 7.4). Let the velocity of the

y

x

CM

r

r'

R

Figure 7.4

CM be V, and let the velocity relative to the CM be v′. Then v = V + v′. Let thebody rotate with angular speed ω′ around the CM (around an instantaneous axisparallel to the z-axis, so that the pancake remains in the x-y plane at all times).4

Then v′ = ω′r′.Let’s look at L first. The angular momentum is

L =∫

r× v dm

=∫

(R + r′)× (V + v′) dm

= MR×V +∫

r′ × v′ dm (cross terms vanish; see below)

= MR×V +(∫

r′2ω′dm

)z

≡ MR×V +(ICMz ω′

)z. (7.9)

where M is the mass of the pancake. In going from the second to the third lineabove, the cross terms,

∫r′×V dm and

∫R×v′ dm, vanish because of the definition

of the CM (which says that∫

r′ dm = 0, and hence∫

v′ dm = d(∫

r′ dm)/dt = 0).The quantity ICM

z is the moment of inertia around an axis through the CM, parallelto the z-axis.

Eq. (7.9) is a very nice result, and it’s important enough to be called a theorem.In words, it says:

Theorem 7.1 The angular momentum (relative to the origin) of a body can befound by treating the body as a point mass located at the CM and finding the angularmomentum of this point mass (relative to the origin), and by then adding on theangular momentum of the body, relative to the CM. 5

Note that if we have the special case where the CM travels around the originin a circle, with angular speed Ω (so that V = ΩR), then eq. (7.9) becomes L =(MR2Ω + ICM

z ω′)z.

4What we mean here is the following. Consider a coordinate system whose origin is the CM andwhose axes are parallel to the fixed x- and y-axes. Then the pancake rotates with angular speed ω′

in this reference frame.5This theorem only works if we use the CM as the location of the imagined point mass. True,

in the above analysis we could have chosen a point P other than the CM, and then written thingsin terms of the coordinates of P and the coordinates relative to P (which could also be describedby a rotation). But then the cross terms in eq. (7.9) wouldn’t vanish, and we’d end up with anunenlightening mess.

7.1. PANCAKE OBJECT IN X-Y PLANE VII-5

Now let’s look at T . The kinetic energy is

T =∫ 1

2v2 dm

=∫ 1

2|V + v′|2 dm

=12MV 2 +

∫ 12v′2 dm (cross term vanishes; see below)

=12MV 2 +

∫ 12r′2ω′2dm

≡ 12MV 2 +

12ICMz ω′2. (7.10)

In going from the second to third line above, the cross term∫

V ·v′ dm vanishes bydefinition of the CM. Again, eq. (7.10) is a very nice result. In words, it says:

Theorem 7.2 The kinetic energy of a body can be found by treating the body as apoint mass located at the CM, and by then adding on the kinetic energy of the bodydue to motion relative to the CM.

7.1.3 The parallel-axis theorem

Consider the special case where the CM rotates around the origin at the same rateas the body rotates around the CM. (This may be achieved, for example, by gluing astick across the pancake and pivoting one end of the stick at the origin; see Fig. 7.5.)

y

x

ω

CMglue

stick

R

Figure 7.5

This means that we have the nice situation where all points in the pancake travelin circles around the origin. Let their angular speed be ω.

In this situation, the speed of the CM is ωR, so eq. (7.9) says that the angularmomentum around the origin is

Lz = (MR2 + ICMz )ω. (7.11)

In other words, the moment of inertia around the origin is

Iz = MR2 + ICMz . (7.12)

This is the parallel-axis theorem. It says that once you’ve calculated the moment ofinertia of an object relative to the CM (namely ICM

z ), then if you want to calculateIz around an arbitrary point in the plane of the pancake, you simply have to addon MR2, where R is the distance from the point to the CM, and M is the mass ofthe pancake.

The parallel-axis theorem is simply a special case of the more general result, eq.(7.9), so it is valid only with the CM, and not with any other point.

Likewise, in this situation, eq. (7.10) gives

T =12(MR2 + ICM

z )ω2. (7.13)


Example (A stick): Let’s verify the parallel-axis theorem for a stick of mass mand length `, in the case where we want to compare the moment of inertia about anend with the moment of inertia about the CM. (Both of the relevant axes will beperpendicular to the stick, and parallel to each other, of course.)For convenience, let ρ = m/` be the density. The moment of inertial about an end is

Iend =∫ `

0

x2 dm =∫ `

0

x2ρ dx =13ρ`3 =

13(ρ`)`2 =

13m`2. (7.14)

The moment of inertial about the CM is

ICM =∫ `/2

−`/2

x2 dm =∫ `/2

−`/2

x2ρ dx =112

ρ`3 =112

m`2. (7.15)

This is consistent with the parallel axis theorem, eq. (7.12), because

Iend = m

(`

2

)2

+ ICM. (7.16)

Remember that this only works with the CM. If you instead want to compare Iend

with the I around a point, say, `/6 from that end, then you cannot say they differby m(`/6)2. But you can compare each of them to ICM and say that they differ by(`/2)2 − (`/3)2 = 5`2/36.

7.1.4 The perpendicular-axis theorem

This theorem is valid only for pancake objects. Consider a pancake object in thex-y plane (see Fig. 7.6). Then the perpendicular-axis theorem says that

y

x

pancake

Figure 7.6

Iz = Ix + Iy, (7.17)

where Ix and Iy are defined analogously to eq. (7.4). (That is, to find Ix, youimagine spinning the object around the x-axis at angular speed ω, and then defineIx ≡ Lx/ω. Similarly for Iy.) In other words,

Ix ≡∫

(y2 + z2) dm, Iy ≡∫

(z2 + x2) dm, Iz ≡∫

(x2 + y2) dm. (7.18)

To prove the theorem, we simply use the fact that z = 0 for our pancake object.Hence Iz = Ix + Iy.

In the limited number of cases where this theorem is applicable, it may save yousome trouble. A few examples are given in the next section.

7.2 Calculating moments of inertia

7.2.1 Lots of examples

Let’s now compute the moments of inertia for a few objects, around specified axes.We will use ρ to denote mass density (per unit length, area, or volume, as appro-priate). We will assume that this density is uniform throughout the object. For

7.2. CALCULATING MOMENTS OF INERTIA VII-7

the more complicated of the objects below, it is generally a good idea to slice theobject up into pieces for which I is already known. The problem then reduces tointegrating over these known I’s. There is usually more than one way to do thisslicing. For example, a sphere may be looked at as a series of concentric shells ora collection of disks stacked on top of each other. In the examples below, you maywant to play around with slicings other than the ones given.

Consider at least a few of these examples to be problems and try to work themout for yourself.

1. A ring of mass M and radius R (axis through center, perpendicular to plane; Fig. 7.7):

R

R

Figure 7.7I =∫

r2 dm =∫ 2π

0

R2ρR dθ = (2πRρ)R2 = MR2 , (7.19)

as it should be (since every bit of the mass is a distance R from the axis).

2. A ring of mass M and radius R (axis through center, in plane; Fig. 7.7):

The distance from the axis is (the absolute value of) R sin θ. Therefore,

I =∫

r2 dm =∫ 2π

0

(R sin θ)2ρR dθ =12(2πRρ)R2 = 1

2MR2 , (7.20)

where we have used sin2 θ = (1−cos 2θ)/2. You can also do this via the perpendicularaxis theorem. In the notation of section 7.1.4, we have Ix = Iy, by symmetry. Hence,Iz = 2Ix. Using Iz = MR2 from Example 1 gives the proper result.

3. A disk of mass M and radius R (axis through center, perpendicular to plane; Fig. 7.8):

R

R

Figure 7.8I =∫

r2 dm =∫ 2π

0

∫ R

0

r2ρr dr dθ = (R4/4)2ρπ =12(ρπR2)R2 = 1

2MR2 . (7.21)

You can save one (trivial) integration step by considering the disk to be made upof many concentric rings, and invoke Example 1. The mass of each ring is ρ2πr dr.Integrating over the rings gives I =

∫ R

0(ρ2πr dr)r2 = πR4ρ/2 = MR2/2, as before.

Slicing the disk up is fairly inconsequential in this example, but it will save a gooddeal of effort in others.

4. A disk of mass M and radius R (axis through center, in plane; Fig. 7.8):

Slice the disk up into rings, and use Example 2.

I =∫ R

0

(1/2)(ρ2πr dr)r2 = (R2/4)ρπ =14(ρπR2)R2 = 1

4MR2 . (7.22)

Or, just use Example 3 and the perpendicular axis theorem.

5. A spherical shell of mass M and radius R (any axis through center; Fig. 7.9):

R

R

Figure 7.9

Choose the z-axis. We may slice the sphere into a large number of horizontal ring-likestrips. In spherical coordinates, the radii of the rings are given by (the absolute value


of) R sin θ, where θ is the angle down from the north pole. The area of a strip is then2π(R sin θ)Rdθ. Using

∫sin3 θ =

∫sin θ(1− cos2 θ) = − cos θ + cos3 θ/3, we have

I =∫

(x2 + y2) dm =∫ π

0

(R sin θ)2ρ2π(R sin θ)Rdθ = 2πρR4

∫ π

0

sin3 θ

= 2πρR4(4/3) =23(4πR2ρ)R2 = 2

3MR2 . (7.23)

6. A sphere of mass M and radius R (any axis through center; Fig. 7.9):A sphere is made up of concentric spherical shells. The volume of a shell is 4πr2dr.Using Example 5, we have

I =∫ R

0

(2/3)(4πρr2dr)r2 = (R5/5)(8πρ/3) =25(4/3πR3ρ)R2 = 2

5MR2 . (7.24)

7. A thin uniform rod of mass M and length L (axis through center, perpendicular torod; Fig. 7.10):

L

L

Figure 7.10I =

∫x2dm =

∫ L/2

−L/2

x2ρ dx =112

(ρL)L2 = 112ML2 . (7.25)

8. A thin uniform rod of mass M and length L (axis through end, perpendicular to rod;Fig. 7.10):

I =∫

x2dm =∫ L

0

x2ρ dx =13(ρL)L2 = 1

3ML2 . (7.26)

9. An infinitesimally thin triangle of mass M and length L (axis through tip, perpen-dicular to plane; Fig. 7.11):

L

L

L

2β

Figure 7.11

Let the base have length a (we will assume a is infinitesimally small). Then a slice ata distance x from the tip has length a(x/L). If the slice has thickness dx, then it isessentially a point mass of mass dm = ρax dx/L. So

I =∫

x2dm =∫ L

0

x2ρax/Ldx =12(ρaL/2)L2 = 1

2ML2 , (7.27)

because aL/2 is the area of the triangle. This of course has the same form as the diskin Example 3, because a disk is made up of many of these triangles.

10. An isosceles triangle of mass M , vertex angle 2β, and common-side length L (axisthrough tip, perpendicular to plane; Fig. 7.11):Let h be the altitude of the triangle (so h = L cos β). Slice the triangle into thin stripsparallel to the base. Let x be the distance from the vertex to a thin strip. Then thelength of a strip is ` = 2x tan β, and its mass is dm = ρ(2x tanβ dx). Using Example7 above, along with the parallel axis theorem, we have

I =∫ h

0

dm

(`2

12+ x2

)=

∫ h

0

(ρ2x tanβ dx)(

(2x tan β)2

12+ x2

)

=∫ h

0

2ρ tan β

(1 +

tan2 β

3

)x3 dx = 2ρ tanβ

(1 +

tan2 β

3

)h4

4. (7.28)

But the area of the whole triangle is h2 tanβ, so we have I = (Mh2/2)(1+ tan2 β/3).In terms of L, this is

I = (ML2/2)(cos2 β + sin2 β/3) = 12ML2

(1− 2

3 sin2 β)

. (7.29)

7.2. CALCULATING MOMENTS OF INERTIA VII-9

11. A regular N -gon of mass M and “radius” R (axis through center, perpendicular toplane; Fig. 7.12):

N= 6

a

b

R

Figure 7.12

The N -gon is made up of N isosceles triangles, so we can use Example 10, withβ = π/N . The masses of the triangles simply add, so if M is the mass of the wholeN -gon, we have

I = 12MR2(1− 2

3 sin2 πN ) . (7.30)

Let’s list the values of I for a few N . We’ll use the shorthand notation (N, I/MR2).Eq. 7.30 gives (3, 1

4 ), (4, 13 ), (6, 5

12 ), (∞, 12 ). These values of I form a nice arithmetic

progression.

12. A rectangle of mass M and sides of length a and b (axis through center, perpendicularto plane; Fig. 7.12):Let the z-axis be perpendicular to the plane. We know that Ix = Mb2/12 andIy = Ma2/12, so the perpendicular axis theorem tells us that

Iz = Ix + Iy = 112M(a2 + b2) . (7.31)

7.2.2 A neat trick

For some objects with certain symmetries, it is possible to calculate I without doingany integrals. All that is needed is a scaling argument and the parallel-axis theorem.We will illustrate this technique by finding I for a stick (Example 7, above). Otherapplications can be found in Problems 4 and 5.

In the present example, the basic trick is to compare I for a stick of length Lwith I for a stick of length 2L. A simple scaling argument shows the latter is eighttimes the former. This is true because the integral

∫x2 dm =

∫x2ρ dx has three

powers of x in it. So a change of variables, y = 2x, brings in a factor of 23 = 8.In other words, if we imagine expanding the smaller stick to create the larger one,then a corresponding piece will now be twice as far from the axis, and also twice asmassive.

The technique is most easily illustrated with pictures. If we denote a moment ofinertia of an object by a picture of the object (with a dot signifying the axis), thenwe have:

L

L LL

M

=

=

=

8

2

2

2( )+ __

The first line comes from the scaling argument, the second is obvious (moments ofinertia simply add; the left-hand side is two copies of the right-hand side, attachedat the pivot), and the third comes from the parallel-axis theorem. Equating theright-hand sides of the first two gives


= 4

Plugging this expression for into the third equation gives the desiredresult,

LM= 2__112

Note that sooner or later you must use real live numbers (which enter herethrough the parallel axis theorem). Using only scaling arguments isn’t enough,because they only provide linear equations homogeneous in the I’s, and thereforegive no way to pick up the proper dimensions.

Once you’ve mastered this trick and applied it to the fractal objects in Problem5, you can impress your friends by saying that you can “use scaling arguments,along with the parallel-axis theorem, to calculate moments of inertia of objects withfractal dimension.” (And you never know when that might come in handy.)

7.3 Torque

We will now show that (under certain conditions, stated below) the rate of changeof angular momentum is equal to a certain quantity, τ , which we call the torque.That is, τ = dL/dt. This is the rotational analog of our old friend F = dp/dtinvolving linear momentum. The basic idea here is straightforward, but there aretwo subtle issues. One deals with internal forces within a collection of particles.The other deals with origins (the points relative to which the angular momentumis calculated) that are not fixed. To keep things straight, we’ll prove the generaltheorem by dealing with three increasingly complicated situations.

Our derivation of τ = dL/dt here holds for completely general motion; we cantake the result and use it in the following chapter, too. If you wish, you can constructa more specific proof of τ = dL/dt for the special case of a pancake object in thex-y plane. But since the general proof is no more difficult, we’ll present it here inthis chapter and get it over with.

7.3.1 Point mass, fixed origin

Consider a point mass at position r relative to a fixed origin (see Fig. 7.13). The

y

x

r

Figure 7.13

time derivative of the angular momentum, L = r× p, is

dLdt

=d

dt(r× p)

=drdt× p + r× dp

dt= v × (mv) + r× F

= 0 + r× F, (7.32)

7.3. TORQUE VII-11

where F is the force acting on the particle. (This is the same proof as in Theorem6.1, except that here we are considering an arbitrary force instead of a central one.)Therefore, if we define the torque on the particle as

τ ≡ r× F, (7.33)

then we have

τ =dLdt

. (7.34)

7.3.2 Extended mass, fixed origin

In an extended object, there are internal forces acting on the various pieces of theobject, in addition to whatever external forces exist. For example, the external forceon a given atom in a body might come from gravity, while the internal forces comefrom the adjacent atoms. How do we deal with these different types of forces?

In what follows, we will deal only with internal forces that are central forces,that is, where the force between two objects is directed along the line joining them.This is a valid assumption for the pushing and pulling forces between atoms in asolid. (It isn’t valid, for example, when dealing with magnetic forces. But we won’tbe interested in such things here.) We will invoke Newton’s third law, which saysthat the force that particle 1 applies to particle 2 is equal and opposite to the forcethat particle 2 applies to particle 1.

For concreteness, let us assume that we have a collection of N discrete particleslabeled by the index i (see Fig. 7.14). (In the continuous case, we’d need to replace

y

x

r1

r4

r2

r3

4N=

Figure 7.14

the following sums with integrals.) Then the total angular momentum of the systemis

L =N∑

i=1

ri × pi. (7.35)

The force acting on each particle is Fexti + Fint

i = dpi/dt. Therefore,

dLdt

=d

dt

∑

i

ri × pi

=∑

i

dri

dt× pi +

∑

i

ri × dpi

dt

=∑

i

vi × (mvi) +∑

i

ri × (Fexti + Fint

i )

= 0 +∑

i

ri × Fexti =

∑

i

τ exti . (7.36)

The last line follows because vi×vi = 0, and also∑

i ri×Finti = 0, as you can show

in Problem 15. (This is fairly obvious. It basically says that a rigid object with noexternal forces won’t spontaneously start rotating.) Note that the right-hand sideinvolves the total torque acting on the body, which may come from forces acting atmany different points.


7.3.3 Extended mass, non-fixed origin

Let the position of the origin be r0 (see Fig. 7.15). Let the positions of the particles

y

x

r1

r2

r0

r1 r0

r2 r0

-

-

Figure 7.15

be ri (r0, ri, and all other vectors below are measured with respect to a given fixedcoordinate system). Then the total angular momentum of the system, relative tothe (possibly moving) origin r0, is

L =∑

i

(ri − r0)×mi(ri − r0). (7.37)

Therefore,

dLdt

=d

dt

(∑

i

(ri − r0)×mi(ri − r0)

)

=∑

i

(ri − r0)×mi(ri − r0) +∑

i

(ri − r0)×mi(ri − r0)

= 0 +∑

i

(ri − r0)× (Fexti + Fint

i −mir0), (7.38)

because miri is the net force (namely Fexti + Fint

i ) acting on the ith particle. But aquick corollary of Problem 15 is that the term involving Fint

i vanishes (show this).And since

∑miri = MR (where M =

∑mi is the total mass, and R is the position

of the center-of-mass), we have

dLdt

=∑

i

(ri − r0)× Fexti −M(R− r0)× r0. (7.39)

The first term is the external torque, relative to the origin r0. The second term issomething we wish would go away. And indeed, it usually does. It vanishes if anyof the following three conditions is satisfied.

1. R = r0. That is, the origin is the CM.

2. r0 = 0. That is, the origin is not accelerating.

3. (R− r0) is parallel to r0. This condition is rarely invoked.

If any of these conditions is satisfied, then we are free to write

dLdt

=∑

i

(ri − r0)× Fexti ≡

∑

i

τ exti . (7.40)

That is, we can equate the total torque with the rate of change of the total angularmomentum. An immediate corollary of this result is:

Corollary 7.3 If the total torque on a system is zero, then its angular momentumis conserved. In particular, the angular momentum of an isolated system (one thatis subject to no external forces) is conserved.

7.4. ANGULAR IMPULSE VII-13

In the present chapter, we are dealing only with cases where L is constant.Therefore, dL/dt = d(LL)/dt = (dL/dt)L. But L = Iω, so dL/dt = Iω ≡ Iα.Taking the magnitude of each side of eq. (7.40) therefore gives

τ = Iα, (7.41)

if L is constant.Invariably, we will calculate angular momentum and torque around either a fixed

point or the CM. These are “safe” origins, in the sense that eq. (7.40) holds. Aslong as you vow to always use one of these safe origins, you can simply apply eq.(7.40) and basically ignore its derivation.

Remark: There is one common situation where the third condition above is applicable.Consider a wheel rolling on the ground. Mark a point on the rim. At the instant this pointin in contact with the ground, it is a valid choice for the origin. This is true because (R−r0)points vertically. And r0 also points vertically. (A point on a rolling wheel traces out acycloid. Right before the point hits the ground, it is moving straight downward; right afterit hits the ground, it is moving straight upward.) But never mind, it’s still a good idea topick your origin to be the CM or a fixed point, even if the third condition holds. ♣

For conditions that number but three,We say, “Torque is dL by dt.”But though they’re all true,I’ll stick to just two;It’s CM’s and fixed points for me.

7.4 Angular impulse

In Section 4.5.1, we defined the impulse, I, to be the time integral of the forceapplied to an object (which is the net change in linear momentum). That is,

I ≡∫ t2

t1F(t) dt = ∆p. (7.42)

We now define the angular impulse, Iθ, to be the time integral of the torqueapplied to an object (which is the net change in angular momentum). That is,

Iθ ≡∫ t2

t1τ (t) dt = ∆L. (7.43)

These are just definitions, devoid of any content. The place where the physicscomes in is the following. Consider a situation where F(t) is always applied at thesame position relative to the origin around which τ (t) is calculated. Let this positionbe R. Then we have τ (t) = R×F(t). Plugging this into eq. (7.43), and taking theconstant R outside the integral, gives Iθ = R× I. That is,

∆L = R× (∆p) (for F(t) applied at one position). (7.44)


This is a very useful result. It deals with the net changes in L and p, and notwith their changes at any particular instant. Hence, even if the magnitude of F ischanging in some arbitrary manner as time goes by, and we have no idea what ∆pand ∆L are, eq. (7.44) is still true. And note that Eq. (7.44) holds for generalmotion, so we can apply it in the next chapter, too.

In many cases, you don’t have to worry about the cross product in eq. (7.44),because the lever arm, R, is perpendicular to the change in momentum, ∆p. Also,in many cases the object starts at rest, so you don’t have to bother with the ∆’s. Thefollowing example is a classic application of this type of angular impulse situation.

Example (Striking a stick): A stick, initially at rest, is struck with a hammer.The blow is made is perpendicular to the stick, at one end. Let the stick have mass mand length `. Let the blow occur quickly, so that the stick doesn’t move much whilethe hammer is in contact. If the CM of the stick ends up moving at speed v, whatare the speeds of the ends, right after the blow?

Solution: We have no idea exactly what F (t) looks like, or for how long it isapplied, but we do know from eq. (7.44) that ∆L = (`/2)∆p, where L is calculatedrelative to the CM (so that the lever arm is `/2). Therefore, (m`2/12)ω = (`/2)mv.Hence, the final v and ω are related by ω = 6v/`.The speeds of the ends are obtained by adding (or subtracting) the rotational motionto the CM’s translational motion. The rotational speeds of the ends are ±ω(`/2) =±(6v/`)(`/2) = 3v. Therefore, the end that was hit moves with speed v + 3v = 4v,and the other end moves with speed v − 3v = −2v (that is, backwards).

What L was, he just couldn’t tell.And p? He was clueless as well.But despite his distress,He wrote down the right guessFor their quotient: the lever-arm’s `.

Impulse is also useful for “collisions” that occur over extended times (see, forexample, Problem 17).

7.5. EXERCISES VII-15

7.5 Exercises

Section 7.3: Torque

1. Maximum frequency *A pendulum is made of a uniform stick of length `. A pivot is place somewherealong the stick, which is allowed to swing in a vertical plane. Where shouldthe pivot be placed on the stick so that the frequency of (small) oscillations ismaximum?

Section 7.4: Impulse

2. Not hitting the pole *A (possibly non-uniform) stick of mass m and length ` lies on frictionless ice.Its midpoint (which is also its CM) touches a thin pole sticking out of the ice.One end of the stick is struck with a quick blow perpendicular to the stick, sothat the CM moves away from the pole. What is the minimum value of thestick’s moment of inertia which allows the stick to not hit the pole?

3. Up, down, and twisting **A uniform stick is held horizontally and then released. At the same instant,one end is struck with a quick upwards blow. If the stick ends up horizontalwhen it returns to its original height, what are the possible values for themaximum height to which the stick’s center rises?

4. Repetitive bouncing *Using the result of Problem 18, what must the relation be between vx and ω,so that a superball will continually bounce back and forth between the sametwo points of contact on the ground?

5. Bouncing under a table **You throw a superball so that it bounces off the floor, then off the undersideof a table, then off the floor again. What must the initial relation between vx

and Rω be, so that the ball returns to your hand?6 (Use the result of Problem18, and modifications thereof.)

6. Bouncing between walls **A stick of length ` slides on frictionless ice. It bounces between two walls, adistance L apart, in such a way that only one end touches the walls, and thestick hits the walls at an angle θ each time. What should θ be, in terms of Land `? What does the situation look like in the limit L ¿ `?

What should θ be, in terms of L and `, if the stick makes an additional n fullrevolutions between the walls? Is there a minimum value of L/` for this to bepossible?

6You are strongly encouraged to bounce a ball in such a manner and have it magically comeback to your hand. It turns out that the required value of ω is rather small, so a natural throwwith ω ≈ 0 will essentially get the job done.


7.6 Problems

Section 7.1: Pancake object in x-y plane

1. Leaning rectangle ***A rectangle of height 2a and width 2b rests on top of a fixed cylinder of radiusR (see Fig. 7.16). The moment of inertia of the rectangle around its center is

b ba2

R

Figure 7.16

I. The rectangle is given an infinitesimal kick, and then ‘rolls’ on the cylinderwithout slipping. Find the equation of motion for the tilting angle of therectangle. Under what conditions will it fall off the cylinder, and under whatconditions will oscillate back and forth? Find the frequency of these smalloscillations.

2. Leaving the sphere **A small ball with moment of inertia ηmr2 rests on top of a sphere. There isfriction between the ball and sphere. The ball is given an infinitesimal kickand rolls downward without slipping. At what point does it lose contact withthe sphere? (Assume that r is much less than the radius of the sphere.) Howdoes your answer change if the size of the ball is comparable to, or larger than,the size of the sphere? (Assume that the sphere is fixed.)

You may want to solve Problem 4.3 first, if you haven’t already done so.

3. Sliding ladder ***A ladder of length ` and uniform mass density per unit length leans againsta frictionless wall. The ground is also frictionless. The ladder is initially heldmotionless, with its bottom end an infinitesimal distance from the wall. Theladder is then released, whereupon the bottom end slides away from the wall,and the top end slides down the wall (see Fig. 7.17).

Figure 7.17

A long time after the ladder is released, what is the horizontal component ofthe velocity of its center of mass?

Section 7.2: Calculating moments of inertia

4. Slick calculations of I **In the spirit of section 7.2.2, find the moments of inertia of the following objects(see Fig. 7.18).

l

l

Figure 7.18

(a) A uniform square of mass m and side ` (axis through center, perpendic-ular to plane).

(b) A uniform equilateral triangle of mass m and side ` (axis through center,perpendicular to plane).

5. Slick calculations of I for fractal objects ***In the spirit of section 7.2.2, find the moments of inertia of the following fractalobjects. (Be careful how the mass scales.)

7.6. PROBLEMS VII-17

(a) Take a stick of length `, and remove the middle third. Then remove themiddle third from each of the remaining two pieces. Then remove themiddle third from each of the remaining four pieces, and so on, forever.Let the final object have mass m (axis through center, perpendicular tostick; see Fig. 7.19).7

l

Figure 7.19

(b) Take a square of side `, and remove the ‘middle’ square (1/9 of thearea). Then remove the ‘middle’ square from each of the remaining eightsquares, and so on, forever. Let the final object have mass m (axisthrough center, perpendicular to plane; see Fig. 7.20).

l

Figure 7.20

(c) Take an equilateral triangle of side `, and remove the ‘middle’ triangle(1/4 of the area). Then remove the ‘middle’ triangle from each of theremaining three triangles, and so on, forever. Let the final object havemass m (axis through center, perpendicular to plane; Fig. 7.21).

l

Figure 7.21

6. Minimum I

A moldable blob of matter of mass M is to be situated between the planesz = 0 and z = 1 (see Fig. 7.22). The goal is to have the moment of inertia

1=z

2=z

Figure 7.22

around the z-axis be as small as possible. What shape should the blob take?

Section 7.3: Torque

7. Removing a support

(a) A rod of length ` and mass m rests on supports at its ends. The rightsupport is quickly removed (see Fig. 7.23). What is the force on the left

l

d dr r

Figure 7.23

support immediately thereafter?

(b) A rod of length 2r and moment of inertia ηmr2 (where η is a numericalconstant) rests on top of two supports, each of which is a distance d awayfrom the center. The right support is quickly removed (see Fig. 7.23).What is the force on the left support immediately thereafter?

7This object is the Cantor set, for those who like such things. It has no length, so the densityof the remaining mass is infinite. If you suddenly develop an aversion to point masses with infinitedensity, simply imagine the above iteration being carried out only, say, a million times.


8. Oscillating ball *A small ball (with uniform density) of radius r rolls without slipping near thebottom of a fixed cylinder of radius R (see Fig. 7.24). What is the frequency

R

Figure 7.24 of small oscillations about the bottom? (Assume r ¿ R.)

9. Ball hitting stick *A ball of mass M hits a stick with moment of inertia I = ηm`2. The ball isinitially traveling with velocity V0, perpendicular to the stick. The ball strikesthe stick at a distance d from its center (see Fig. 7.25). The collision is elastic.

V0d

l

m

M

I= mlη 2

Figure 7.25

Find the resulting translational and rotational speeds of the stick, and alsothe resulting speed of the ball.

10. A ball and stick theorem *A ball of mass M hits a stick with moment of inertia I. The ball is initiallytraveling with velocity V0, perpendicular to the stick. The ball strikes the stickat a distance d from its center (see Fig. 7.25). The collision is elastic.

Prove that the relative speed of the ball and the point of contact on thestick is the same before and immediately after the collision. (This theorem isanalogous to the ‘relative speed’ theorem of two balls.)

11. A triangle of circles ***Three circular objects with moments of inertia I = ηMR2 are situated in atriangle as in Fig. 7.26. Find the initial downward acceleration of the topRR

R

Figure 7.26

circle, if:

(a) There is friction between the bottom two circles and the ground (so theyroll without slipping), but there is no friction between any of the circles.

(b) There is no friction between the bottom two circles and the ground, butthere is friction between the circles.

Which case has a larger acceleration?

12. Lots of sticks ***This problem deals with rigid ‘stick-like’ objects of length 2r, masses Mi, andmoments of inertia ηMir

2. The center-of-mass of each stick is located at thecenter of the stick. (All the sticks have the same r and η. Only the massesdiffer.) Assume M1 À M2 À M3 À · · ·.The sticks are placed on a horizontal frictionless surface, as shown in Fig. 7.27.

....

M1

M2

M3

M4

M5

2r

Figure 7.27

The ends overlap a negligible distance, and the ends are a negligible distanceapart.

The first (heaviest) stick is given an instantaneous blow (as shown) whichcauses it to translate and rotate. (The blow comes from the side of stick #1on which stick #2 lies; the right side, as shown in the figure.) The first stick

7.6. PROBLEMS VII-19

will strike the second stick, which will then strike the third stick, and so on.Assume all collisions among the sticks are elastic.

Depending on the size of η, the speed of the nth stick will either (1) approachzero, (2) approach infinity, or (3) be independent of n, as n →∞.

What is the special value of η corresponding to the third of these three sce-narios? Give an example of a stick having this value of η.

(You may work in the approximation where M1 is infinitely heavier than M2,which is infinitely heavier than M3, etc.)

13. Falling stick *A massless stick of length b has one end attached to a pivot and the other endglued perpendicularly to the middle of a stick of mass m and length `.

(a) If the two sticks are held in a horizontal plane (see Fig. 7.28) and then

b

b

l

l

g

g

Figure 7.28

released, what is the initial acceleration of the CM?

(b) If the two sticks are held in a vertical plane (see Fig. 7.28) and thenreleased, what is the initial acceleration of the CM?

14. Falling Chimney ****A chimney initially stands upright. It is given a tiny kick, so that it topplesover. At what point along its length is it most likely to break?

In doing this problem, work with the following two-dimensional simplifiedmodel of a chimney. Assume the chimney consists of boards stacked on top ofeach other; and each board is attached to the two adjacent ones with stringsat each end (see Fig. 7.29). Assume that the boards are slightly thicker at

Figure 7.29

their ends, so that they only touch each other at their endpoints. The goal isto find where the string has the maximum tension.

(In doing this problem, you may work in the approximation where the widthof the chimney is very small compared to its height.)

15. Zero torque from internal forces **Given a collection of particles with positions ri, let the force on the ith particle,due to all the others, be Fint

i . Assuming that the force between any twoparticles is a central force, use Newton’s third law to show

∑i ri × Fint

i = 0.

16. Lengthening the string **A mass hangs from a string and swings around in a circle, as shown in Fig. 7.30.

h

r

l

θ

Figure 7.30

The length of the string is very slowly increased (or decreased). Let θ, `, r,and h be defined as in the figure.

(a) Assuming θ is very small, how does r depend on `?

(b) Assuming θ is very close to π/2, how does h depend on `?


Section 7.4: Impulse

17. Sliding to rolling **A ball initially slides, without rotating, on a horizontal surface with friction(see Fig. 7.31). The initial speed of the ball is V0, and the moment of inertia

RV0

Figure 7.31about its center is I = ηmR2.

(a) Without knowing anything about how the friction force depends on po-sition, find the speed of the ball when it begins to roll without slipping.Also, find the kinetic energy lost while sliding.

(b) Now consider the special case where the coefficient of sliding friction is µ,independent of position. At what time, and at what distance, does theball begin to roll without slipping?Verify that the work done by friction equals the loss in energy calculatedin part (a) (be careful on this).

18. The superball **A ball with radius R is thrown in the plane of the paper (the x-y plane), whilealso spinning around the axis perpendicular to the page. The ball bounces offthe floor. Assuming that the collision is elastic, show that the v′x and ω′ afterthe bounce are related to the vx and ω before the bounce by

(v′x

Rω′

)=

(3/7 4/710/7 −3/7

) (vx

Rω

), (7.45)

where our convention is that positive vx is to the right, and positive ω isclockwise.8

19. Many bounces *Using the result of Problem 18, describe what happens over the course of manysuperball bounces.

20. Rolling over a bump **A ball with radius R (and uniform density) rolls without slipping on theground. It encounters a step of height h and rolls up over it. Assume thatthe ball sticks to the corner of the step briefly (until the center of the ballis directly above the corner). And assume that the ball does not slip withrespect to the corner.

Show that the minimum initial speed, V0, required for the ball to climb overthe step, is given by

V0 ≥ R√

14gh/57R/5− h

. (7.46)

8Assume that there is no distortion in the ball during the bounce, which means that the forcesin the x- and y-directions are independent, which then means that the kinetic energies associatedwith the x- and y-motions are separately conserved.

7.7. SOLUTIONS VII-21

7.7 Solutions

1. Leaning rectangleWhen the rectangle has rotated through an angle θ, the position of its CM is (relativeto the center of the cylinder)

(x, y) = R(sin θ, cos θ) + Rθ(− cos θ, sin θ) + a(sin θ, cos θ), (7.47)

where we have added up the distances along the three shaded triangles in Fig. 7.32(note

b

a

a

2

2

R

Rθ

θ

Figure 7.32

that the contact point has moved a distance Rθ along the rectangle).We’ll use the Lagrangian method to find the equation of motion and the frequency ofsmall oscillations. Using eq. (7.47), the square of the speed of the CM is

v2 = x2 + y2 = (a2 + R2θ2)θ2. (7.48)

(There’s an easy way to see this clean result. The CM instantaneously rotates aroundthe contact point with angular speed θ, and from Fig. 7.32 the distance to the contactpoint is

√a2 + R2θ2.)

The Lagrangian is

L = T − V =12M(a2 + R2θ2)θ2 +

12Iθ2 −Mg

((R + a) cos θ + Rθ sin θ

). (7.49)

The equation of motion is

(Ma2 + MR2θ2 + I)θ + MR2θθ2 = Mga sin θ −MgRθ cos θ. (7.50)

Consider small oscillations. Using the small-angle approximations, and keeping termsonly to first order in θ, we obtain

(Ma2 + I)θ + Mg(R− a)θ = 0. (7.51)

Therefore, oscillatory motion occurs for a < R (note that this is independent of b).The frequency of small oscillations is

ω =

√Mg(R− a)Ma2 + I

. (7.52)

Some special cases: If I = 0 (i.e., all the mass is located at the CM), we have ω =√g(R− a)/a2. If the rectangle is a uniform horizontal stick, so that a ¿ R, a ¿ b, and

I ≈ Mb2/3, we have ω ≈√

3gR/b2. If the rectangle is a vertical stick (satisfying a < R),

so that b ¿ a and I ≈ Ma2/3, we have ω ≈√

3g(R− a)/4a2. If in addition a ¿ R, then

ω ≈√

3gR/4a2.

Remarks:

(a) Without doing much work, there are two ways that we can determine the conditionunder which there is oscillatory motion. The first is to look at the height of theCM. Using small-angle approximations in eq. (7.47), the height of the CM is y ≈(R + a) + (R − a)θ2/2. Therefore, if a < R, the potential increases with θ, so therectangle wants to decrease its θ and fall back down to the middle. If a > R, thepotential decreases with θ, so the rectangle wants to keep increasing its θ, and thusfalls off the cylinder.

The second way is to look at the horizontal positions of the CM and the contact point.Small-angle approximations in eq. (7.47) show that the former equals aθ and the latterequals Rθ. Therefore, if a < R then the CM is to the left of the contact point, so thetorque from gravity makes θ decrease, and the motion is stable. If a > R then thetorque from gravity makes θ increase, and the motion is unstable.


(b) The small-angle equation of motion, eq. (7.51), can also be derived using τ = dL/dt,using the instantaneous contact point, P , on the rectangle as the origin around whichwe calculate τ and L. (From section (7.3.3), we know that it is legal to use this pointwhen θ = 0.)

However, point P cannot be used as the origin to use τ = dL/dt to calculate the exactequation of motion, eq. (7.50), because for θ 6= 0 the third condition in section (7.3.3)does not hold.

It is possible to use the CM as the origin for τ = dL/dt, but the calculation is rathermessy. ♣

2. Leaving the sphere

In this setup, the ball still leaves the sphere when the normal force becomes zero; soeq. (4.91) is still applicable, from the solution to Problem 4.3. The only change comesin the calculation of v. The ball has rotational energy, so conservation of energy givesmgR(1− cos θ) = mv2/2 + Iω2/2 = mv2/2 + ηmr2ω2/2. Using rω = v, we have

v =

√2gR(1− cos θ)

1 + η. (7.53)

Plugging this into eq. (4.91), we see that the ball leaves the sphere when

cos θ =2

3 + η. (7.54)

For, η = 0, this is 2/3, of course. For a uniform ball with η = 2/5, we have cos θ =10/17, so θ ≈ 54. For, η →∞, we have cos θ → 0, so θ ≈ 90 (v will be very small,because most of the energy will take the form of rotational energy.)If the size of the ball is comparable to, or bigger than, the size of the sphere, we haveto take into account the fact that the CM of the ball does not move along a circle ofradius R. It moves along a circle of radius R + r. So eq. (4.91) becomes

mv2

R + r= mg cos θ. (7.55)

Also, the conservation-of-energy equation takes the form mg(R + r)(1 − cos θ) =mv2/2+ηmr2ω2/2. But rω still equals v (prove this). So we have the same equationsas above, except that R is replaced everywhere by R + r. But R didn’t appear in theoriginal answer, so the answer is unchanged.

Remark: Note that the method of the second solution to Problem 4.3 will not work in

this problem, because there is a force available to make vx decrease, namely the friction

force. And indeed, vx does decrease before the rolling ball leaves the sphere. (The v in this

problem is simply 1/√

1 + η times the v in Problem 4.3, so the maximum vx is still achieved

at cos θ = 2/3, and the angle in eq. (7.54) is larger than this.) ♣3. Sliding ladder

The key to this problem is the fact that the ladder will lose contact with the wallbefore it hits the ground. The first thing we must do is calculate exactly where thisloss of contact occurs.Let r = `/2, for convenience. It is easy to see that while the ladder is in contact withthe wall, the CM of the ladder will move in a circle of radius r. (The median to thehypotenuse of a right triangle has half the length of the hypotenuse.) Let θ be the


angle between the wall and the radius from the corner to the CM of the ladder; seeFig. 7.33. (This is also the angle between the ladder and the wall.)

r

rr

θ

θ

Figure 7.33

We will solve the problem by assuming that the CM always moves in a circle, andthen determining the point at which the horizontal CM speed starts to decrease (i.e.,the point at which the normal force from the wall becomes negative, which it of coursecan’t do).

By conservation of energy, the kinetic energy of the ladder is equal to the loss inpotential energy, which is mgr(1 − cos θ), where θ is defined above. This kineticenergy may be broken up into the CM translational energy plus the rotation energy.The CM translational energy is simply mr2θ2/2 (since the CM travels in a circle).The rotational energy is Iθ2/2. (The same θ applies here as in the CM translationalmotion, because θ is the angle between the ladder and the vertical.) Letting I ≡ηmr2, to be general (η = 1/3 for our ladder), we have, by conservation of energy,(1 + η)mr2θ2/2 = mgr(1− cos θ). Therefore, the speed of the CM, v = rθ, is

v =√

2gr

1 + η

√(1− cos θ) . (7.56)

The horizontal speed is therefore

vx =√

2gr

1 + η

√(1− cos θ) cos θ. (7.57)

Taking the derivative of√

(1− cos θ) cos θ, we see that the speed is maximum atcos θ = 2/3. (This is independent of η.)

Therefore the ladder loses contact with the wall when

cos θ = 2/3. (7.58)

Using this value of θ in eq. (7.57) gives a horizontal speed of (letting η = 1/3)

vx =√

2gr

3≡√

g`

3. (7.59)

This is the horizontal speed just after the ladder loses contact with the wall, and thusis the horizontal speed from then on, because the floor exerts no horizontal force.

You are encouraged to compare various aspects of this problem with those in Problem2.

4. Slick calculations of I

(a) We claim that the I for a square of side 2` is 16 times the I for a square of side` (where the axes pass through any two corresponding points). The factor of 16comes in part from the fact that dm goes like the area, which is proportional tolength squared. So the corresponding dm’s are increased by a factor of 4. Thereare therefore four powers of 2 in the integral

∫r2 dm =

∫r2 dx dy.

With pictures, we have:


lm

=

=

16

2

2

2

( )+ ___

ll

= 4

The first line comes from the scaling argument, the second is obvious (momentsof inertia add), and the third comes from the parallel axis theorem. Equatingthe right-hand sides of the first two, and then using the third to eliminate

gives

ll

m=2_1

6

This agrees with the result of example 12 in section 7.2.1, with a = b = `.

(b) This is again a two-dimensional object, so the I for a triangle of side 2` is16 times the I for a triangle of side ` (where the axes pass through any twocorresponding points).Again, with pictures, we have:

lm

=

=

16

2

3

2

(

(

)

)

+

+

___

l l

= 3

The first line comes from the scaling argument, the second is obvious, and thethird comes from the parallel axis theorem. Equating the right-hand sides of

the first two, and then using the third to eliminate gives

l

l

m=2__1

12

This agrees with the result of example 11 in section 7.2.1, with N = 3 (becausethe ‘radius’, R, used in that example equals `/

√3).

5. Slick calculations of I for fractal objects

(a) The scaling argument here is a little trickier than that in section 7.2.2. Ourobject is self-similar to an object 3 times as big, so let’s increase the length by


a factor of 3 and see what happens to I. In the integral∫

x2dm, the x’s pickup a factor of 3, so this gives a factor of 9. But what happens to the dm? Well,tripling the size of our object increases its mass by a factor of 2 (since the newobject is simply made up of two of the smaller ones, plus some empty space inthe middle), so the dm picks up a factor of 2. Thus the I for an object of length3` is 18 times the I for an object of length ` (where the axes pass through anytwo corresponding points).

With pictures, we have (the following symbols denote our fractal object):

lm

=

=

18

2

3

( )

+

l l

l= 2

2/

The first line comes from the scaling argument, the second is obvious (momentsof inertia add), and the third comes from the parallel axis theorem. Equating theright-hand sides of the first two, and then using the third to eliminategives

ll m 2_18=

This is larger than the I for a uniform stick (m`2/12), because the mass isgenerally further away from the center.

Remark: When we increase the length of our object by a factor of 3 here, the factor

of 2 in the dm is between the factor of 1 relevant to a zero-dimensional object, and

the factor of 3 relevant to a one-dimensional object. So in some sense our object has a

dimension between 0 and 1. It is reasonable to define the dimension, d, of an object as

the number for which rd is the increase in ‘volume’ when the dimensions are increased

by a factor of r. In our example, we have 3d = 2, so d = log3 2 ≈ 0.63. ♣

(b) Again, the mass scales in a strange way. Let’s increase the dimensions of ourobject by a factor of 3 and see what happens to I. In the integral

∫x2dm, the

x’s pick up a factor of 3, so this gives a factor of 9. But what happens to thedm? Tripling the size of our object increases its mass by a factor of 8 (since thenew object is made up of eight of the smaller ones, plus an empty square in themiddle), so the dm picks up a factor of 8. Thus the I for an object of side 3`is 72 times the I for an object of side ` (where the axes pass through any twocorresponding points).

Again, with pictures, we have (the following symbols denote our fractal object):


l

l

m

m

=

=

=

72

2

2

3

(

(

)

)+

+

+

l l

= 4 ( )4

2

The first line comes from the scaling argument, the second is obvious, and thethird and fourth come from the parallel axis theorem. Equating the right-handsides of the first two, and then using the third and fourth to eliminate

and gives

ll

m=2__3

16

This is larger than the I for the uniform square in problem 4, because the massis generally further away from the center.

Note: Increasing the size of our object by a factor of 3 increases the ‘volume’ by a

factor of 8. So the dimension is given by 3d = 8; hence d = log3 8 ≈ 1.89.

(c) Again, the mass scales in a strange way. Let’s increase the dimensions of ourobject by a factor of 2 and see what happens to I. In the integral

∫x2dm, the

x’s pick up a factor of 2, so this gives a factor of 4. But what happens to thedm? Doubling the size of our object increases its mass by a factor of 3 (sincethe new object is simply made up of three of the smaller ones, plus an emptytriangle in the middle), so the dm picks up a factor of 3. Thus the I for anobject of side 2` is 12 times the I for an object of side ` (where the axes passthrough any two corresponding points).Again, with pictures, we have (the following symbols denote our fractal object):

lm

=

=

12

2

3

2

(

(

)

)

+ ___

l l

= 3

The first line comes from the scaling argument, the second is obvious, and thethird comes from the parallel axis theorem. Equating the right-hand sides of

the first two, and then using the third to eliminate gives

l

l

m=2_1

9


This is larger than the I for the uniform triangle in problem 4, because the massis generally further away from the center.

Note: Increasing the size of our object by a factor of 2 increases the ‘volume’ by a

factor of 3. So the dimension is given by 2d = 3; hence d = log2 3 ≈ 1.58.

6. Minimum I

The shape should be a cylinder with the z-axis as its symmetry axis. This is fairlyobvious, and a quick proof (by contradiction) is the following.Assume the optimal blob is not a cylinder, and consider the surface of the blob. Ifthe blob is not a cylinder, then there exist two points on the surface, P1 and P2, thatare located at different distances, r1 and r2, from the z-axis. Assume r1 < r2 (seeFig. 7.34). Then moving a small piece of the blob from P2 to P1 will decrease the

z

z

x

= 1

rP1

r2

1

P2

Figure 7.34

moment of inertia,∫

r2ρ dV . Hence, the proposed blob was not the one with smallestI.In order to avoid this contradiction, we must have all points on the surface be equidis-tant from the z-axis. The only blob with this property is a cylinder.

7. Removing a support

(a) First Solution: Let the desired force on the left support be F . Let theacceleration of the CM of the stick be a. Then (looking at torques around theCM, to obtain the second equation; see Fig. 7.35),

l

mgF

Figure 7.35mg − F = ma,

F`

2=

m`2

12α,

a =`

2α. (7.60)

Solving for F gives F = mg/4. (So the CM accelerates at 3g/4, and the rightend accelerates at 3g/2.)Second Solution: Looking at torques around the CM, we have

F`

2=

m`2

12α. (7.61)

Looking at torques around the fixed end, we have

mg`

2=

m`2

3α. (7.62)

These two equations give F = mg/4.

(b) First Solution: As in the first solution above, we have (see Fig. 7.36)

d r

mgF

Figure 7.36mg − F = ma,

Fd = (ηmr2)α,

a = dα. (7.63)

Solving for F gives F = mg(1 + d2/ηr2)−1. For d = r and η = 1/3, we get theanswer in part (a).Second Solution: As in the second solution above, looking at torquesaround the CM, we have

Fd = (ηmr2)α. (7.64)


Looking at torques around the fixed pivot, we have

mgd = (ηmr2 + md2)α. (7.65)

These two equations give F = mg(1 + d2/ηr2)−1.

Some limits: If d = r, then: in the limit η = 0, F = 0; if η = 1, F = mg/2; and in thelimit η = ∞, F = mg; these all make sense. In the limit d = 0, F = mg. And in thelimit d = ∞, F = 0. (More precisely, we should be writing d ¿ √

η r or d À √η r.)

8. Oscillating ballLet the angle from the bottom of the cylinder be θ (see Fig. 7.37). Let Ff be the

R

θ

Ff

Figure 7.37

friction force. Then F = ma gives

Ff −mg sin θ = ma. (7.66)

Looking at torque and angular momentum around the CM, we have

−rFf =25mr2α. (7.67)

Using rα = a, this equation gives Ff = −2ma/5. Plugging this into eq. (7.66), andusing sin θ ≈ θ, yields mgθ + 7ma/5 = 0. Under the assumption r ¿ R, we havea ≈ Rθ, so we finally have

θ +(

5g

7R

)θ = 0. (7.68)

This is the equation for simple harmonic motion with frequency

ω =

√5g

7R. (7.69)

This answer is slightly smaller than the√

g/R answer if the ball were sliding. (Therolling ball effectively has a larger inertial mass, but the same gravitational mass.)This problem can also be done using the contact point as the origin around which τand L are calculated.

Remark: If we get rid if the r ¿ R assumption, we leave it to you to show that rα = a

still holds, but a = Rθ changes to a = (R−r)θ. Therefore, the exact result for the frequency

is ω =√

5g/7(R− r). This goes to infinity as r gets close to R. ♣9. Ball hitting stick

Let V be the speed of the ball after the collision. Let v be the speed of the CMof the stick after the collision. Let ω be the angular speed of the stick after thecollision. Conservation of momentum, angular momentum (around the initial centerof the stick), and energy give (see Fig. 7.38)

V0d

l

m

M

I= mlη 2

Figure 7.38

MV0 = MV + mv,

MV0d = MV d + ηm`2ω,

MV 20 = MV 2 + mv2 + ηm`2ω2. (7.70)

We must solve these three equations for V , v, and ω. The first two equations quicklygive vd = η`2ω. Solving for V in the first equation and plugging it into the third, andthen eliminating ω through vd = η`2ω gives

v = V02

1 + mM + d2

η`2

, and thus v = V0

2 dη`2

1 + mM + d2

η`2

. (7.71)


Knowing v, the first equation above gives V as

V = V0

1− mM + d2

η`2

1 + mM + d2

η`2

. (7.72)

Another solution is of course V = V0, v = 0, and ω = 0. Nowhere in eqs. (7.70) doesit say that the ball actually hits the stick.

The reader is encouraged to check various limits of these answers.

10. A ball and stick theorem

Let V be the speed of the ball after the collision. Let v be the speed of the CMof the stick after the collision. Let ω be the angular speed of the stick after thecollision. Conservation of momentum, angular momentum (around the initial centerof the stick), and energy give (see Fig. 7.38)

MV0 = MV + mv,

MV0d = MV d + Iω,

MV 20 = MV 2 + mv2 + Iω2. (7.73)

The speed of the contact point on the stick right after the collision is v + ωd. So thedesired relative speed is (v + ωd) − V . We can solve the three above equations forV , v, and ω and obtain our answer (i.e., use the results of problem 9), but there’s aslightly more appealing method.

The first two equations quickly give mvd = Iω. The last equation may be written inthe form (using Iω2 = (Iω)ω = (mvd)ω)

M(V0 − V )(V0 + V ) = mv(v + ωd). (7.74)

Dividing this by the first equation, written in the from M(V0 − V ) = mv, givesV0 + V = v + ωd, or

V0 = (v + ωd)− V, (7.75)

as was to be shown.

11. A triangle of circles

(a) Let the normal force between the circles be N . Let the friction force from theground be Ff (see Fig. 7.39). If we consider torques around the centers of the

FfFf

N

N N

N

Figure 7.39

bottom balls, then the only force we have to worry about is Ff (since N , gravity,and the normal force from the ground point through the centers).Let ax be the initial horizontal acceleration of the right bottom circle (so α =ax/R is its angular acceleration). Let ay be the initial vertical acceleration ofthe top circle (downward taken to be positive). Then

N cos 60 − Ff = Max,

Mg − 2N sin 60 = May,

FfR = (ηMR2)(ax/R). (7.76)

We have four unknowns, N , Ff , ax, and ay. So we need one more equation.Fortunately, ax and ay are related. The ‘surface’ of contact between the topand bottom circles lies at an angle of 30 with the horizontal. Therefore, if a


bottom circle moves a distance d to the side, then the top circle moves a distanced tan 30 downward. So

ax =√

3ay. (7.77)

We now have four equations and four unknowns. Solving for ay, by your methodof choice, gives

ay =g

7 + 6η. (7.78)

(b) Let the normal force between the circles be N . Let the friction between thecircles be Ff (see Fig. 7.40). If we consider torques around the centers of the

Ff FfN

N N

N

Figure 7.40

bottom balls, then the only force we have to worry about is Ff .Let ax be the initial horizontal acceleration of the right bottom circle. Letay be the initial vertical acceleration of the top circle (downward taken to bepositive). From the same reasoning as in part (a), we have ax =

√3ay. Let α

be the angular acceleration of the right bottom circle (counterclockwise takento be positive). Note that α is not equal to ax/R, because the bottom circlesslip. The four equations analogous to eqs. (7.76) and (7.77) are

N cos 60 − Ff sin 60 = Max,

Mg − 2N sin 60 − 2Ff cos 60 = May,

FfR = (ηMR2)α,

ax =√

3ay. (7.79)

We have five unknowns, N , Ff , ax, ay, and α. So we need one more equation.The tricky part is relating α to ax. To do this, it is easiest to ignore the ymotion of the top circle and imagine the bottom right circle to be rotating upand around the top circle, which is held fixed. If the bottom circle moves aninfinitesimal distance d to the right, then its center moves a distance d/ cos 30

up and to the right. So the angle through which the bottom circle rotates isd/(R cos 30). Bringing back in the vertical motion of the balls does not changethis result. Therefore,

α =2√3

ax

R. (7.80)

We now have five equations and five unknowns. Solving for ay, by your methodof choice, gives

ay =g

7 + 8η. (7.81)

Remark: If η 6= 0, this result is smaller than that in part (a). This is not all that

intuitive. The basic reason is that the bottom circles in part (b) have to rotate a bit

faster, so they take up more energy. Also, one can show that the N ’s are equal in (a)

and (b), which is likewise not obvious. Since there’s an extra force (from the friction)

holding the top ball up in part (b), the acceleration is smaller. ♣12. Lots of sticks

Consider the collision between two sticks. Let the speed of the end of the heavy onebe V . Since this stick is essentially infinitely heavy, we may consider it to be aninfinitely heavy ball, moving at speed V . (The rotational degree of freedom of theheavy stick is irrelevant, as far as the light stick is concerned.)We will solve this problem by first finding the speed of the contact point on the lightstick, and then finding the speed of the other end of the light stick.


• Speed of contact point:We can invoke the result of problem 10 to say that the point of contact onthe light stick picks up a speed of 2V . But let’s prove this from scratch herein a different way: In the same spirit as the (easier) problem of the collisionbetween two balls of greatly disparate masses, we will work things out in therest frame of the infinitely heavy ball right before the collision. The situationthen reduces to a stick of mass m, length 2r, moment of inertia ηmr2, and speedV , approaching a fixed wall (see Fig. 7.41). To find the behavior of the stick

m

V 2r

Figure 7.41

after the collision, we will use (1) conservation of energy, and (2) conservationof angular momentum around the contact point.Let u be the speed of the center of mass of the stick after the collision. Let ωbe its angular velocity after the collision. Since the wall is infinitely heavy, itwill acquire zero kinetic energy. So conservation of E gives

12mV 2 =

12mu2 +

12(ηmr2)ω2. (7.82)

The initial angular momentum around the contact point is L = mrV , so con-servation of L gives (breaking L after the collision up into the L of the CM plusthe L relative to the CM)

mrV = mru + (ηmr2)ω. (7.83)

Solving eqs. (7.82) and (7.83) for u and rω in terms of V gives

u = V1− η

1 + η, and rw = V

21 + η

. (7.84)

(The other solution, u = V and rω = 0 represents the case where the stickmisses the wall.) The relative speed of the wall (i.e., the ball) and the point ofcontact on the light stick is

rw − u = V, (7.85)

as was to be shown.Going back to the lab frame (i.e., adding V onto this speed) shows that thepoint of contact on the light stick moves at speed 2V .

• Speed of other end:Consider a stick struck at an end, with impulse I. The speed of the CM isthen vCM = I/m. The angular impulse is Ir, so Ir = ηmr2ω, and hencerω = I/mη = vCM/η.The speed of the struck end is vstr = rω + vCM. The speed of the other end(taking positive to be in the reverse direction) is voth = rω − vCM. The ratio ofthese is

voth

vstr=

vCM/η − vCM

vCM/η + vCM

=1− η

1 + η. (7.86)

In the problem at hand, we have vstr = 2V . Therefore,

voth = V2(1− η)1 + η

. (7.87)

The same analysis works in all the other collisions. Therefore, the bottom ends of thesticks move with speeds that form a geometric progression with ratio 2(1−η)/(1+η).If this ratio is less than 1 (i.e., η > 1/3), then the speeds go to zero, as n →∞. If it


is greater than 1 (i.e., η < 1/3), then the speeds go to infinity, as n →∞. If it equals1 (i.e., η = 1/3), then the speeds are independent of n, as n →∞. Therefore,

η =13

(7.88)

is the desired answer. A uniform stick has η = 1/3 (usually written in the formI = m`2/12, where ` = 2r).

13. Falling stick

(a) It is easiest to calculate τ and L relative to the pivot point. The torque is dueto gravity, which effectively acts on the CM. It has magnitude mgb.The moment of inertia of the stick around a horizontal axis through the pivot(and perpendicular to the massless stick) is simply mb2. So when the stick startsto fall, τ = dL/dt gives mgb = (mb2)α. Therefore, the initial acceleration of theCM, bα, is

bα = g, (7.89)

independent of ` and b.This makes sense. This stick initially falls straight down, and the pivot providesno force because it doesn’t know right away that the stick is moving.

(b) The only change from part (a) is the moment of inertia of the stick arounda horizontal axis through the pivot (and perpendicular to the massless stick).From the parallel axis theorem, this moment is mb2 + m`2/12. So when thestick starts to fall, τ = dL/dt gives mgb = (mb2 + m`2/12)α. Therefore, theinitial acceleration of the CM, bα, is

bα =g

1 + `2

12b2

. (7.90)

As ` → 0, this goes to g, as it should. As ` → ∞, it goes to 0, as it should (a tinymovement in the CM corresponds to a very large movement in the points far out alongthe stick).

14. Falling Chimney

Let the height of the chimney be `. Let the width be 2r. The moment of inertia aroundthe pivot point is m`2/3 (if we ignore the width). Let the angle with the vertical beθ. Then the torque (around the pivot point) due to gravity is τ = mg(`/2) sin θ. Soτ = dL/dt gives mg(`/2) sin θ = (1/3)m`2θ, or

θ =3g sin θ

2`. (7.91)

Consider the chimney to consist of a chimney of height a, with another one of height`−a placed on top of it. We will find the tensions in the strings connecting these two‘sub-chimneys’; then we will maximize one of the tensions as a function of a.The forces on the top piece are gravity and the forces on each end of the bottomboard. Let us break these latter forces up into transverse and longitudinal forcesalong the chimney. Let T1 and T2 be the two longitudinal components, and let Fbe the sum of the transverse components. These are shown in Fig. 7.42. We have

T2

T1

rr

F

Figure 7.42

picked the positive directions for T1 and T2 such that positive T1 corresponds to anormal force, and positive T2 corresponds to a tension in the string. This is the case


we will be concerned with. (If, for example, T1 happens to be negative, then it simplycorresponds to a tension instead of a normal force.) It turns out that if r ¿ `, thenT2 À F (as we will see below), so the tension in the right string is essentially equalto T2. We will therefore be concerned with maximizing T2.In writing down the force and torque equations for the top piece, we have threeequations (Fx = max, Fy = may, and τ = dL/dt around its center-of-mass), andthree unknowns (F , T1, and T2). Using the fact that the top piece has length (`− a),its CM travels in a circle of radius (` + a)/2, and its mass is m(` − a)/`, our threeequations are, respectively,

(T1 − T2) sin θ + F cos θ =m(`− a)

`

(` + a

2θ cos θ

),

(T1 − T2) cos θ − F sin θ − mg(`− a)`

= −m(`− a)`

(` + a

2θ sin θ

),

(T1 + T2)r − F`− a

2=

m(`− a)`

((`− a)2

12θ

). (7.92)

We can solve for F by multiplying the first equation by cos θ, the second by sin θ, andsubtracting. Using (7.91) to eliminate θ gives

F =mg sin θ

4(−1 + 4f − 3f2), (7.93)

where f ≡ a/` is the fraction of the way along the chimney.We may now solve for T2. Multiplying the second of eqs. (7.92) by r and subtractingfrom the third gives (to leading order in the large number `/r)9

T2 ≈ F`− a

4r+

m(`− a)`

(`− a)2

24rθ. (7.94)

Using eqs. (7.91) and (7.93), this may be written as

T2 ≈ mg` sin θ

8rf(1− f)2. (7.95)

As stated above, this is much greater than F (since `/r À 1), so the tension in theright string is essentially equal to T2. Taking the derivative with respect to f , we seethat T2 is maximum at

f ≡ a

`=

13. (7.96)

So our chimney is most likely to break at a point one-third of the way up (assumingthat the width is much less than the height).

15. Zero torque from internal forcesLet Fint

ij be the force that the ith particle feels due to the jth particle (see Fig. 7.43.

r

r

r r

F

F

i

i

j

j

ij

ji

-

Figure 7.43

ThenFint

i =∑

j

Fintij , (7.97)

and Newton’s third law says that

Fintij = −Fint

ji . (7.98)

9This result is simply the third equation with T1 set equal to T2. Basically, T1 and T2 are bothvery large and are essentially equal; the difference between them is of order 1.


Therefore,τ int ≡

∑

i

ri × Finti =

∑

i

∑

j

ri × Fintij . (7.99)

But if we change the indices (which were labeled arbitrarily), we have

τ int =∑

j

∑

i

rj × Fintji = −

∑

j

∑

i

rj × Fintij . (7.100)

Adding the two previous equations gives

2τ int =∑

i

∑

j

(ri − rj)× Fintij . (7.101)

But our central-force assumption says that Fintij is parallel to (ri − rj). Therefore,

each cross-product in the sum is zero.

16. Lengthening the pendulum

Consider the angular momentum, L, around the support point, P . The forces on themass are the tension in the string and gravity. The former provides no torque aroundP , and the latter provides no torque in the z-direction. Therefore, Lz is constant.Let ω` be the frequency of the circular motion, when the string has length `. Then

mr2ω` = Lz (7.102)

is constant.The frequency ω` is obtained by using F = ma for the circular motion. The tensionin the string is mg/ cos θ, so the horizontal radial force is mg tan θ. Therefore,

mg tan θ = mrω2` = m(` sin θ)ω2

` =⇒ ω` =√

g

` cos θ. (7.103)

plugging this into eq. (7.102) gives

mr2

√g

` cos θ= mr2

√g

h= Lz. (7.104)

(a) For θ ≈ 0, we have h ≈ `, so eq. (7.104) gives r2/√

` ≈ C. Therefore,

r ∝ `1/4. (7.105)

So r grows very slowly with `.

(b) For θ ≈ π/2, we have r ≈ `, so eq. (7.104) gives `2/√

h ≈ C. Therefore,

h ∝ `4. (7.106)

So h grows very quickly with `.

17. Sliding to rolling

(a) Let the ball travel to the right. Define all linear quantities to be positive to theright, and all angular quantities to be positive clockwise, as shown in Fig. 7.44.

RV

ω

Figure 7.44(Then, for example, the friction force Ff is negative.) The friction force slows


down the translational motion and speeds up the rotational motion, accordingto (looking at torque around the CM)

Ff = ma,

−FfR = Iα. (7.107)

Eliminating Ff , and using I = ηmR2, gives a = −ηRα. Integrating this overtime, up to the time when the ball stops slipping, gives

∆V = −ηR∆ω. (7.108)

(This is the same statement as the impulse equation, eq. (7.44).) Using ∆V =Vf − V0, and ∆ω = ωf − ω0 = ωf , and also ωf = Vf/R (the non-slippingcondition), we find

Vf =V0

1 + η, (7.109)

independent of how Ff depends on position. (For that matter, Ff could evendepend on time or speed. The relation a = −ηRα would still be true at alltimes, and hence also eq. (7.108).)

Remark: You could also calculate τ and L relative to the instantaneous point ofcontact on the ground (which is a fixed point). There is zero torque relative to thispoint. The motion around this point is not a simple rotation, so we have to add the Lof the CM plus the L relative to the CM. τ = dL/dt gives 0 = (d/dt)(mRv + ηmR2ω).Hence, a = −ηRα.

Note that it is not valid to calculate τ and L relative to the instantaneous point of

contact on the ball. The ball is slowing down, so there is a horizontal component to

the acceleration, and hence the third condition in section 7.3.3 does not hold. ♣

The loss in kinetic energy is given by (using eq. (7.109), and also the relationωf = Vf/R)

∆KE =12mV 2

0 −(

12mV 2

f +12Iω2

f

)

=12mV 2

0

(1− 1

(1 + η)2− η

(1 + η)2

)

=12mV 2

0

(η

1 + η

). (7.110)

For η → 0, no energy is lost, which makes sense. And for η →∞, all the energy is lost,

which also makes sense (this case is essentially like a sliding block which can’t rotate).

(b) Let’s find t. The friction force is Ff = −µmg. So F = ma gives −µg = a(so a is constant). Therefore, ∆V = at = −µgt. But eq. (7.109) says that∆V ≡ Vf − V0 = −V0η/(1 + η). So we find

t =η

µ(1 + η)V0

g. (7.111)

For η → 0, we have t → 0, which makes sense. And for η → ∞, we havet → V0/(µg) which is exactly the time a sliding block would take to stop.Now let’s find d. We have d = V0t + (1/2)at2. Using a = −µg, and plugging int from above gives

d =V 2

0

g

η(2 + η)2µ(1 + η)2

. (7.112)


The two extreme cases for η check here.To calculate the work done by friction, one might be tempted to take the productFfd. But the result doesn’t look much like the loss in kinetic energy calculatedin eq. (7.110). What’s wrong with this? The error is that the friction force doesnot act over a distance d. To find the distance over which Ff acts, we must findhow far the surface of the ball moves relative to the ground.The relative speed of the point of contact and the ground is Vrel = V (t)−Rω(t) =(V0 + at)−Rαt. Using a = −ηRα and a = −µg, this becomes

Vrel = V0 − 1 + η

ηµgt. (7.113)

Integrating this from t = 0 to the t given in eq. (7.111) yields

drel =∫

Vreldt =V 2

0 η

2µg(1 + η). (7.114)

The work done by friction is Ffdrel = µmgdrel, which does indeed give the ∆KEin eq. (7.110).

18. The superball

The y-motion of the ball is irrelevant in this problem, because the y-velocity simplyreverses direction, and the vertical impulse from the floor provides no torque aroundthe CM of the ball.With the positive directions for x and ω as stated in the problem, eq. (7.44) may beused to show that the horizontal impulse from the floor changes vx and ω accordingto

I(ω′ − ω) = −Rm(v′x − vx). (7.115)

The conservation-of-energy statement is

12mv′x

2 +12Iω′2 =

12mv2

x +12Iω2. (7.116)

Given vx and ω, eqs. (7.115) and (7.116) are two equations in the two unknowns v′xand ω′. They can be solved in a messy way using the quadratic formula, but it ismuch easier to use the standard trick of rewriting eq. (7.116) as

I(ω′2 − ω2) = −m(v′x2 − v2

x), (7.117)

and then dividing this by eq. (7.115) to obtain

R(ω′ + ω) = (v′x + vx). (7.118)

Eqs. (7.118) and (7.115) are now two linear equations in the two unknowns v′x and ω′.Using I = (2/5)mR2 for a solid sphere, you can easily solve the equations to obtainthe desired result, eq. (7.45).

Remark: The other solution to eqs. (7.115) and (7.116) is of course v′x = vx and ω′ = ω.This corresponds to the ball bouncing off a frictionless floor (or even just passing throughthe floor). Eq. (7.115) is true for any ball, but the conservation-of-energy statement in eq.(7.116) is only true for two special cases. One is the case of a frictionless floor, where thereis “maximal” slipping at the point of contact. The other is the case of zero slipping, whichis the case with the superball. If there is any intermediate amount of slipping, then energy


is not conserved, because the friction force does work and generates heat. (Work is forcetimes distance, and in the first special case, the force is zero; while in the second specialcase, the distance is zero). Therefore, in addition to being made of a very bouncy material,a superball must also have a surface that won’t slip while in contact with the floor.

Note that eq. (7.118) may easily be used to show that the relative velocity of the point of

contact and the ground exactly reverses direction during the bounce. ♣19. Many bounces

Eq. (7.45) gives the result after one bounce, so the result after two bounces is(

v′′xRω′′

)=

(3/7 4/710/7 −3/7

)(v′x

Rω′

)

=(

3/7 4/710/7 −3/7

)2 (vx

Rω

)

=(

1 00 1

)(vx

Rω

)

=(

vx

Rω

). (7.119)

The square of the matrix turns out to be the identity. Therefore, after two bounces,both vx and ω return to their original values. The ball then repeats the motion ofthe previous two bounces (and so on, after every two bounces). The only differencebetween successive pairs of bounces is that the ball may shift horizontally. You arestrongly encouraged to experimentally verify this strange periodic behavior.

20. Rolling over a bumpWe will use the fact that the angular momentum of the ball with respect to the cornerof the point (call it point P ) is unchanged by the collision. This is true because anyforces exerted at point P provide zero torque around P .10 This fact will allow us tofind the energy of the ball right after the collision, which we will then require to begreater than Mgh.Breaking L into the contribution relative to the CM, plus the contribution from theball treated like a point mass located at its CM (eq. (7.9)), gives an initial angularmomentum equal to L = (2/5)MR2ω0 + MV0(R − h), where ω0 is the initial rollingangular speed. But the non-slipping condition requires that V0 = Rω0. Hence, L maybe written as

L =25MRV0 + MV0(R− h) = MV0

(7R

5− h

). (7.120)

Let ω′ be the angular speed of the ball around point P immediately after the collision.The parallel-axis theorem says that the ball’s moment of inertia around P is equalto (2/5)MR2 + MR2 = (7/5)MR2. Conservation of L (around point P ) during thecollision then gives

MV0

(7R

5− h

)=

75MR2ω′, (7.121)

which gives ω′. The energy of the ball right after the collision is therefore

E =12

(75MR2

)ω′2 =

12

(75MR2

)(MV0(7R/5− h)

(7/5)MR2

)2

=MV 2

0 (7R/5− h)2

(14/5)R2.

(7.122)10The torque from gravity will be relevant once the ball rises up off the ground. But during the

(instantaneous) collision, L will not change.


The ball will climb up over the step if E ≥ Mgh, which gives

V0 ≥ R√

14gh/57R/5− h

. (7.123)

Remarks: It is indeed possible for the ball to rise up over the step, even if h > R (as longas the ball sticks to the corner, without slipping). But note that V0 →∞ as h → 7R/5. Forh ≥ 7R/5, it is impossible for the ball to make it up over the step. (The ball will actuallyget pushed down into the ground, instead of rising up, if h > 7R/5.)

For an object with a general moment of inertia I = ηMR2 (so η = 2/5 in our problem), youcan easily show that the minimum initial speed is

V0 ≥R

√2(1 + η)gh

(1 + η)R− h. (7.124)

This decreases as η increases. It is smallest when the “ball” is a wheel with all the mass on

its rim (so that η = 1), in which case it is possible for the wheel to climb over the step even

if h approaches 2R. ♣

Ch7 angular momentum

Technology

Transcript of Ch7 angular momentum