Chapter 9 Angular Momentum

42 Pages·2004·0.7 MB·English

by  Bruce Sherwood

Copyright 2003 John Wiley & Sons. Adopters of Matter & Interactions by Ruth Chabay and Bruce Sherwood
may provide this revised version of Chapter 9 to their students.
Chapter 9
Angular Momentum
9.1 Rotational kinetic energy and moment of inertia 296
9.2 Angular momentum 297
9.2.1 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 301
9.4.1 An ice skater’s spin 301
9.4.2 The dumbbell demonstration 301
9.4.3 A high dive 302
9.5 Translational angular momentum 303
9.5.1 Visualizing the magnitude of translational angular momentum 303
9.5.2 Another view of rotational angular momentum 304
9.5.3 Another example of conservation of angular momentum 304
9.5.4 An alternative way to calculate translational angular momentum306
9.6 Direction of angular momentum 306
9.6.1 The vector cross product 308
9.7 Angular momentum of multiparticle systems 310
9.7.1 Rotational angular momentum as a vector 311
9.7.2 More exercises on angular momentum 312
9.8 The angular momentum principle 313
9.8.1 Multiparticle systems 314
9.8.2 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 318
9.11.1 The Bohr model of the hydrogen atom 318
9.11.2 Particle spin 322
9.11.3 Consequences of angular momentum quantization 322
9.12 Calculating torque 323
9.12.1 Example: Falling object 324
9.12.2 More exercises on the angular momentum principle 325
9.13 Applications to systems with nonzero torques 325
9.13.1 A meter stick on the ice 325
9.13.2 A puck with string wound around it 327
9.13.3 Static equilibrium 327
9.14 *Gyroscopes 328
9.14.1 *Magnetic Resonance Imaging (MRI) 330
9.14.2 *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
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 “angular
momentum,” the rotational analogue of the ordinary “linear” momentum
with which we are already familiar. Angular momentum is conserved, just as
energy and ordinary momentum are. In the atomic world angular momen-
tum is quantized, which has far-reaching consequences for the structure of
atoms 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 “angular
speed” in radians per second but with different linear speeds in meters per
second, depending on their distances from the axis (Figure 9.1). This is the
m
4
situation for a rigid object.
v4 = ωr4 v3 = ωr3 As we saw in Chapter 2, angular speed, normally denoted by ω (lowercase
r
4 r3 m3 ω = 2--T--π-- Gmraekeeks oomnee gcao)m, ipsl eat me teuarsnu roef o3f6 h0o dwe gfarsete sso (m2eπt hraindgia niss )ro itna tain tgim. Ief aTn, wobe jseacyt
m that its angular speed is ω = 2π⁄T radians per second. The time T is called
1 r1 the period. We measure angular speed ω in the “natural” units of radians
r
2 per second rather than degrees per second, because the fundamental geo-
v1 = ωr1 v2 = ωr2 metrical relationship between angle and arc length (arc length = rθ) is valid
m2 only if the angle θ is measured in radians. The speed of an object at a per-
Figure 9.1 A case of rigid rotation about an pendicular distance r⊥ from the center is the circumference divided by the
axis with angular speed ω. Note the differ- period:
ent speeds, with vi = ωri. ⎛ ⎞
v = -2---π----r--⊥-- = ⎝2----π--⎠r⊥ = ωr⊥
T T
The kinetic energy of a rotating rigid object like the one shown in Figure 9.1
is 12---m1v12+12---m2v22+12---m3v32 + .... Since the speed of each mass is v = ωr⊥, we
can write the rotational kinetic energy like this:
Krot = 12---[m1(r⊥1ω)2+m2(r⊥2ω)2+m3(r⊥3ω)2 + ...]
= 12---[m1r⊥21 +m2r⊥22 +m3r⊥23 + ...]ω2
The quantity in brackets is called the “moment of inertia” and is usually de-
noted by the letter I:
MOMENT OF INERTIA
I = m1r⊥21 +m2r⊥22 +m3r⊥23 +m4r⊥24 + ...
With this definition of moment of inertia we have the following convenient
way to calculate rotational kinetic energy:
ROTATIONAL KINETIC ENERGY
K = 1---Iω2
rot 2
9.2: Angular momentum 297
Review the derivation of this formula and notice that this is not some new
kind of energy. It is ordinary kinetic energy of the atoms that are moving
with respect to the axle. The new formula K = 1---Iω2 is simply a compact
rot 2
and convenient way of expressing the total kinetic energy of all the atoms
in 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:
K = K +K , where K = 1---M v 2
trans rel trans 2 tot cm
For the system shown in Figure 9.1, the center of mass isn’t moving, so
K = 0, and K = K since there is no vibrational kinetic energy. In
trans rel rot
Chapter 7 we often calculated rotational kinetic energy by using the energy
principle. Now we have an additional way of evaluating rotational kinetic
energy, in terms of how fast the object is spinning.
Example: Find the moment of inertia of a “linear” triatomic
molecule such as CO (three atoms in a row) consisting of a central
2
atom of mass M flanked by two atoms each of mass m (Figure 9.2).
The distance between the centers of adjacent atoms is d. d d
Remember that almost all of the mass in an atom is concentrated
in the tiny nucleus, whose diameter is much smaller than d.
Solution: Using the definition of moment of inertia, we have
I = M(02)+m(d2)+m(d2) = 2md2 m M m
Even if the nucleus of the central atom has a large mass, it
Figure 9.2 A “linear” triatomic molecule,
doesn’t contribute to the moment of inertia. This reflects the
such as CO , with the three atoms in a row.
2
fact that if the molecule rotates, nucleons in the tiny central nu-
The masses are concentrated in the tiny
cleus have almost no rotational kinetic energy.
nuclei (drawn much larger than the nuclei
Ex. 9.1 What is the moment of inertia of a diatomic molecule such really are!).
as N around its center of mass if the mass of each atom is M and
2
the distance between nuclei is d? It helps to draw a diagram.
Remember that the electrons hardly contribute to the moment of
inertia because their mass is so small.
Ex. 9.2 A bicycle wheel has almost all its mass M located in the
outer rim at radius R. If it rotates on a stationary axle with angular
speed 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 made
of.
Ex. 9.3 The moment of inertia of a uniform-density disk rotating
about an axle through its center can be shown to be 1---MR2. This
2
result is obtained by using integral calculus to add up the
contributions of all the atoms in the disk (see Problem 9.3). The
factor of 1/2 reflects the fact that some of the atoms are near the
center and some are far from the center; the factor of 1/2 is an
average of the square distances. A uniform-density disk whose mass
is 10 kg and radius is 0.4 m makes one complete rotation every 0.2
s. What is its kinetic energy?
9.2 Angular momentum
A single particle moving at a slow speed (v << c) has translational kinetic en-
ergy 1---m v 2 (a scalar) and momentum mv (a vector). We have seen in the
2
298 Chapter 9: Angular Momentum
previous chapters that both energy and momentum are useful concepts,
and the energy principle and the momentum principle tell how the energy
and 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 M v , and the center
tot cm
of mass of the device is located at the axle, which is stationary. But does the
device 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 an
object that is rotating but not translating (stationary center of mass), is “ro-
tational angular momentum,” with the symbol L :
rot
ROTATIONAL ANGULAR MOMENTUM
L = Iω
rot
The relationship between rotational kinetic energy 1---Iω2 and rotational an-
2
gular momentum Iω is very similar to the relationship between translation-
al kinetic energy 1---mv 2 and ordinary momentum mv. In fact, L = Iω is
2 rot
really the z component (into or out of the page) of an angular momentum
vector, which makes the analogy to ordinary momentum even stronger. This
lets us account for the two different possible directions of rotation of the de-
vice:
DIRECTION OF ROTATIONAL ANGULAR MOMENTUM
Counterclockwise: ω positive, angular momentum vector in +z direction
Clockwise: ω negative, angular momentum vector in –z direction
It may seem odd to say that angular momentum is a vector pointing in the z
direction when the rotation is in the x-y plane. It’s simply a convenient way
to 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 describe
ω
whether the rotation is clockwise or counterclockwise (viewed from some-
where along the +z axis, looking back toward the x-y plane) by saying wheth-
Rotational angular momentum er your descriptive vector points in the +z or –z direction.
counterclockwise, out of page (+z) Because it is difficult to sketch a situation in three dimensions on paper,
whenever possible we will work with two-dimensional projections onto the
x-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 as
Rotational angular momentum
showing the tip of an arrow (a vector) pointing out at you, while the symbol
clockwise, into page (–z)
represents the feathers of an arrow (a vector) pointing away from you.
ω
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.
Figure 9.3 2-D projections: symbols for
rotational angular momentum out of page
and into page.
Example: A bicycle wheel of mass M and radius R has almost all its
mass concentrated in the rim. It spins on its axis, making one
revolution in T seconds. What is the magnitude of its rotational
angular momentum?
Solution: L = Iω = (MR2)(2π⁄T)
rot
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 L ?
rot
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 spins
with angular speed 300 radians/s. What is the magnitude of its
rotational angular momentum?
Ex. 9.8 The moment of inertia of a uniform-density sphere can be
shown to be 2---MR2. Approximating the Earth as a sphere of mass
6×1024 kg an5d radius 6.4×106 m, what is the magnitude of the
Earth’s rotational angular momentum? Remember that the Earth
takes 24 hours to rotate once.
Ex. 9.9 The Earth rotates from west to east. That is, if you look
down on the Earth from above the North Pole, the Earth rotates
counterclockwise. What is the direction of the rotational angular
momentum 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 different
ways:
K = 1---mv 2 or --p-----2-, since p = mv
trans 2 2m
The second formula is particularly useful in situations where we happen to
know 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
L2
K = 1---Iω2 or -----r-o---t, since L = Iω
rot 2 2I rot
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 5 kg⋅m2 and the
magnitude of its rotational angular momentum is 20 kg⋅m2/s,
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 why
it makes sense to define something called “the rotational angular momen-
tum” of a rotating object as Iω, 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 a
m
distance r from the axis, and turn the object through a small angle ∆θ. We 4
v = ωr
show the force split into two components F|| and F⊥, parallel and perpen- 4 4 v3 = ωr3
dicular to the arm.
m
3
? Which of these force components does no work on the system, and F⊥
m
why? 1 r
F
F will do no work, because it is at right angles to the movement of the point
of|| application of the force; there is no displacement in the direction of this v1 = ωr1 F|| v2 = ωr2
m
force component and therefore it does no work. 2
We will calculate the amount of work in terms of the angle through which ∆θ
we apply our force. We can express the distance ∆l through which the force Figure 9.4 Apply a force to change the
F⊥ acts as r∆θ, where ∆θ is the angle through which the system rotates. It rotational kinetic energy.
300 Chapter 9: Angular Momentum
is a fact of geometry that the length of an arc of a circle is r∆θ when the an-
gle is measured in radians. For example, when the angle is 2π radians, the
length of the arc is 2πr, the circumference of the circle.
Given the geometrical fact that we apply a force through a distance r∆θ,
here is the energy principle applied to the current situation:
∆Krot = F⊥(r∆θ) = rF⊥∆θ
The quantity rF⊥ is called the “torque,” which is Latin for “twist”. The units
of torque are N⋅m, which technically is the same as joules, but since torque
is physically not the same thing as energy, we say that a torque has units of
N⋅m rather than joules.
We’ll see why rotational angular momentum is defined to be Iω 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:
∆K ∆θ
-----∆----tr-o---t = rF⊥-∆----t--
In the limit of infinitesimal time interval we have derivatives:
dK dθ
--------r--o--t = rF⊥------
dt dt
-d----(1---Iω2) = rF⊥d----θ--
dt 2 dt
? See if you can do these derivatives with respect to time yourself.
Note in particular that dθ⁄dt is simply the angular speed ω, and that
ω 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:
Iωd----ω--- = rF⊥ω
dt
Cancel the common factor ω, and since the moment of inertia I of our ro-
tating object isn’t changing, write the result like this:
d
-----(Iω) = rF⊥
dt
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 similar
to 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 all
the torques due to all the forces:
THE ANGULAR MOMENTUM PRINCIPLE
FOR A ROTATING OBJECT
d
-----(Iω) = net torque
dt
To keep the derivation simple, we assumed that the moment of inertia
wasn’t changing. However, a more complete treatment shows that the angu-
lar momentum principle as stated here does correctly describe the motion
of a rotating object even if the moment of inertia changes. Our result is
more general than our simplified derivation.
In reality the angular momentum principle is a vector principle, and what
we have dealt with in this particular case is the z component of the angular
momentum (and the z component of the torque, which is also a vector).
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 on
the tip of one skate, with her arms and one leg
outstretched, then pull her leg in and bring her
arms to a vertical position above her head. She
then spins much faster.
What’s going on? There is some frictional
force of the ice on the tip of the skate, but this
force is applied so close to the axis of rotation
that the torque is small. That is, rF⊥ is very small
because r is small and the frictional force is small.
? During the short time when the skater
quickly changes her configuration, what can
you say about the rotational angular
momentum? Why?
Small torque implies small change in angular
momentum per unit time, so in a short time the
angular momentum of the skater will hardly
change.
? But if the rotational angular momentum hardly changes, how can
the 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 the
rotational angular momentum not to change, the angular velocity must in-
crease to compensate for the decreased moment of inertia: I ω = I ω ,
1 1 2 2
and therefore ω = I ω ⁄I .
2 1 1 2
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 maybe
hold your nose) and say, “Well, we did all that general analysis of the effect
of torques on the angular momentum of a rotating object, so if that’s what
we get when we apply these general principles, I guess that’s that.” But at
another 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 skating
maneuver. The angular momentum of the skater doesn’t change, so as her
moment of inertia decreases, the kinetic energy of the skater L2 ⁄(2I) ac-
rot
tually increases.
? Where does this increased energy come from?
Evidently the skater has to expend chemical energy in order to increase her
kinetic energy. In fact, at high spin rates it takes a noticeable effort to pull
her arms and leg toward the spin axis. This effect is even more dramatic
when 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 a
dumbbell in each hand. Start spinning slowly with the arms held out, then
pull your hands in toward your chest. You find yourself spinning much fast-
302 Chapter 9: Angular Momentum
er, because your smaller moment of inertia I is compensated by a larger an-
gular speed ω in order that the product Iω stay constant in the absence of
significant torques. This works only if the frictional torques exerted on the
stool 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 the
dumbbells 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 radial
component 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 a
tight 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 the
surface (Figure 9.5).
? What can you say about the diver’s rotational angular momentum
while he is in the air? Why?
We can probably neglect air resistance, because the diver is very far from
reaching terminal speed. Then the only significant force acting is the
Earth’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 angular
momentum is constant during the dive, where did that rapid spin
come from?
When the diver jumps off the board, he must thrust with his feet in such a
way that the force of the board on his feet has a sizable lever arm about his
center of mass, thereby exerting a torque around the center of mass. By the
time his feet lose contact with the board, the diver already has acquired a
sizable rotational angular momentum due to the torque.
? When the diver pulls out of the tucked position, why does he stop
spinning rapidly?
Figure 9.5 A diver spins, then straightens There is a large increase in the moment of inertia, because many atoms are
out. now much farther from the center of mass than they were when the diver
was 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 rotational
angular momentum. But his angular speed may be so small that you hardly
notice it, especially in comparison with the very rapid spin in the preceding
tucked position. Moreover, his body could approximately follow the curving
path of his center of mass, with the body rotating just enough to stay tangent
to 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.
9.5: Translational angular momentum 303
9.5 Translational angular momentum
r
h ⊥
In Figure 9.6 a piece of clay starting from rest falls from rest a height h and
sticks 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 the
A
same momentum of the clay just before it hits and sticks. (a)
? 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 r⊥, where r⊥ is the perpendicular distance from the
axle to the line of motion of the falling clay as shown In Figure 9.6. The fast- h
r = 0
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? A
(b)
As you probably expect, if the impact parameter is zero (the line of motion
of the clay passes through the axle) the device won’t rotate; case (b).
? What is the difference in the rotation in case (a) compared with
r
case (c)? ⊥ h
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-
A
(c)
tate clockwise (rotational angular momentum into the page, negative z
component).
? If we increase the mass of the clay, how will that affect the rotation
of the device? If the clay is going faster just before it sticks to the ball, r
⊥
how will that affect the rotation of the device?
As you probably expect, the device will rotate faster if the clay has more mass h
or hits with a higher speed. In other words, the magnitude p of the mo- A
(d)
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 both
the impact parameter r⊥ and also to the magnitude of the momentum p .
A particle moving in a straight line has angular momentum because it can Figure 9.6 A falling piece of clay could hit
make something else rotate. Both straight-line motion and rotation involve and stick to the wheel in different loca-
angular momentum. tions, with different impact parameters
We define the magnitude LA of the angular momentum of the clay rel- (r⊥).
ative to location A (the location of the axle) to be the product of the impact
parameter and the magnitude of the momentum, LA = r⊥ p , and we call
this “translational” angular momentum. Although translational angular
momentum is a vector, at the moment we’re just interested in its magni-
tude.
TRANSLATIONAL ANGULAR MOMENTUM
RELATIVE TO LOCATION A r⊥
LA = r⊥ p
p LA = r⊥p
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:
(cid:127) Draw a vector representing the momentum p of the particle.
(cid:127) Draw a line through location A, parallel to p,
(cid:127) Draw the impact parameter r⊥ from this line to the particle. A
(cid:127) Draw a rectangle bounded by r⊥ and p.
(cid:127) The area of the rectangle is the magnitude of the translational angular Figure 9.7 Visualizing the magnitude of
momentum, LA = r⊥ p . angular momentum for the falling clay.
304 Chapter 9: Angular Momentum
Example: A dog whose mass is 10 kg runs at 20 m/s along the left
side of a highway that is 5 meters wide. What is the magnitude of its
translational angular momentum relative to location A on the right
side of the highway (Figure 9.8)? Relative to location B on the right
side of the highway? Relative to location C on the left side of the
p = (10 kg)(20 m/s) LA = r⊥p highway? Include appropriate units.
dog r⊥ = 5 m Solution: In Figure 9.8 we show the relevant rectangle for
location A. The magnitude of the translational angular
B
momentum relative to location A is
C L = (10 kg)(20 m/s)(5 m) = 1000 kg⋅m2/s
A
A
The same rectangle is valid for location B, so L is also equal to
B
1000 kg⋅m2/s. Relative to location C, the impact parameter is
Figure 9.8 The translational angular
zero, and the rectangle has zero area: L = 0.
C
momentum of a dog running along the
left side of a highway. Ex. 9.11 In Figure 9.9 are seven particles each with the same
magnitude of momentum p = 10 kg⋅m/s but with different
positions relative to location A, and different directions of ordinary
5 m 5 m
linear momentum. Calculate the magnitude LA = r⊥ p in each
(1) (2) (3) case. In each case, draw the appropriate rectangle, showing r⊥ and
p that bound the rectangle.
8 m
p = 10 kg⋅m/s
9.5.2 Another view of rotational angular momentum
(5) (6)
Our new formula for calculating translational angular momentum can be
A
(4) used to rederive the formula for rotational angular momentum, thereby il-
luminating the relationship between these two kinds of angular momen-
7 m tum. At a particular instant the translational angular momenta of one of the
atoms in a rotating object is this:
(7) r⊥ p = r⊥(mv) = r⊥(mωr⊥) = (mr⊥2)ω
Figure 9.9 Draw the translational angular
Add up the contributions of all of the atoms:
momentum rectangles and calculate the
mmaogmneitnutduem oinf etahceh ctarasen.slational angular Lrot = (mr⊥2)1ω+(mr⊥2)2ω+(mr⊥2)3ω+ .....
Lrot = [(mr⊥2)1+(mr⊥2)2+(mr⊥2)3+ .....]ω
L = Iω
rot
From this you can see that rotational angular momentum is just the sum of
the translational angular momenta of all the atoms in the rotating object.
The formula for the translational angular momentum of a single particle,
LA = r⊥ p , is really the fundamental definition of angular momentum.
Nevertheless it is often very useful to express the angular momentum of an
object that rotates about a fixed axis as rotational angular momentum, eval-
uated with the convenient formula L = Iω.
rot
9.5.3 Another example of conservation of angular momentum
Angular momentum, whether rotational or translational or a combination
of both kinds of angular momentum, is a conserved quantity, like linear mo-
mentum and energy. That is, the change in angular momentum of a system
plus 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 the
combined system (device plus clay) will stay constant from just before the