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Physics For Scientists And Engineers 6E - part 88

S E C T I O N 1 1 . 4 • Conservation of Angular Momentum

349

We use the fact that radians are dimensionless to ensure
consistent units for each term.

Finally, the elastic nature of the collision tells us that ki-

netic energy is conserved; in this case, the kinetic energy
consists of translational and rotational forms:

In solving Equations (1), (2), and (3) simultaneously,

we find that v

and ) #

To finalize the problem, note that these values seem rea-

sonable. The disk is moving more slowly after the collision
than it was before the collision, and the stick has a small
translational speed. Table 11.1 summarizes the initial and fi-
nal values of variables for the disk and the stick, and verifies
the conservation of linear momentum, angular momentum,
and kinetic energy.

What If?

What if the collision between the disk and the stick

is perfectly inelastic? How does this change the analysis?

Answer In this case, the disk adheres to the end of the
stick upon collision. The conservation of linear momentum
principle leading to Equation (1) would be altered to

2.0 rad/s.

1.3 m/s,

2.3 m/s,

(3)

18 m

# 2.0 v

(1.33 m

))

(1.33 kg&m

))

(2.0 kg)(3.0 m/s)

(2.0 kg)v

(1.0 kg)v

# K

(2)

9.0 rad/s % (3.0 rad/m)v

)

(1.33 kg&m

))

12 kg&m

/s # $(4.0 kg&m)v

(1.33 kg&m

))

(2.0 m)(2.0 kg)(3.0 m/s) # $(2.0 m)(2.0 kg)v

For the rotational part of this question, we need to find the
center of mass of the system of the disk and the stick. Choos-
ing the center of the stick as the origin, the y position of the
center of mass along the vertical stick is

Thus, the center of mass of the system is 2.0 m $ 1.33 m
#

0.67 m from the upper end of the stick.

The conservation of angular momentum principle leading

to Equation (2) would be altered to the following, evaluating
angular momenta around the center of mass of the system:

The moment of inertia of the stick around the center of
mass of the system is found from the parallel-axis theorem:

Thus, Equation (4) becomes

Evaluating the total kinetic energy of the system after the colli-
sion shows that it is less than that before the collision because
kinetic energy is not conserved in an inelastic collision.

) #

4.0 kg&m

# $

1.0 rad/s

4.0 kg&m

/s # (4.0 kg&m

))

(3.1 kg&m

))

(0.67 m)(2.0 kg)(3.0 m/s) # [(2.0 kg)(0.67 m)

])

1.33 kg&m

(1.0 kg)(1.33 m)

3.1 kg&m

(4)

(0.67 m)m

(0.67 m)

]) % I

)

) %

)

(2.0 kg)(2.0 m) % (1.0 kg)(0)

(2.0 kg % 1.0 kg)

1.33 m

2.0 m/s

(2.0 kg)(3.0 m/s) # (2.0 kg % 1.0 kg)v

v (m/s)

(rad/s)

p (kg · m/s)

L (kg · m

/s)

trans

( J)

rot

( J)

Before
Disk

3.0

—

6.0

9.0

—

Stick

Total for

—

6.0

9.0

System

After
Disk

2.3

—

4.7

9.3

5.4

—

Stick

1.3

2.0

1.3

2.7

0.9

2.7

Total for

—

6.0

6.3

2.7

System

Comparison of Values in Example 11.10 Before and After the Collision

Table 11.1

Notice that linear momentum, angular momentum, and total kinetic energy are conserved.

At the Interactive Worked Example link at http://www.pse6.com, you can adjust the speed and position of the disk and
observe the collision.

11.5 The Motion of Gyroscopes and Tops

A very unusual and fascinating type of motion you probably have observed is that of a
top spinning about its axis of symmetry, as shown in Figure 11.14a. If the top spins very
rapidly, the symmetry axis rotates about the z axis, sweeping out a cone (see Fig.
11.14b). The motion of the symmetry axis about the vertical—known as

precessional

motion—is usually slow relative to the spinning motion of the top.

It is quite natural to wonder why the top does not fall over. Because the center of

mass is not directly above the pivot point O, a net torque is clearly acting on the top
about O—a torque resulting from the gravitational force M

g. The top would certainly

fall over if it were not spinning. Because it is spinning, however, it has an angular mo-
mentum

L directed along its symmetry axis. We shall show that this symmetry axis

moves about the z axis (precessional motion occurs) because the torque produces a
change in the direction of the symmetry axis. This is an excellent example of the impor-
tance of the directional nature of angular momentum.

The essential features of precessional motion can be illustrated by considering the

simple gyroscope shown in Figure 11.15a. The two forces acting on the top are the
downward gravitational force M

g and the normal force n acting upward at the pivot

point O. The normal force produces no torque about the pivot because its moment
arm through that point is zero. However, the gravitational force produces a torque
! #

r " Mg about O, where the direction of ! is perpendicular to the plane formed

r and Mg. By necessity, the vector ! lies in a horizontal xy plane perpendicular to

the angular momentum vector. The net torque and angular momentum of the gyro-
scope are related through Equation 11.13:

From this expression, we see that the nonzero torque produces a change in angular
momentum d

L—a change that is in the same direction as !. Therefore, like the torque

vector, d

L must also be perpendicular to L. Figure 11.15b illustrates the resulting pre-

cessional motion of the symmetry axis of the gyroscope. In a time interval dt, the
change in angular momentum is d

L # L

dt. Because d

L is perpendicular to

L, the magnitude of L does not change ("L

" # "L

"). Rather, what is changing is the

direction of

L. Because the change in angular momentum d L is in the direction of !,

which lies in the xy plane, the gyroscope undergoes precessional motion.

! #

350

C H A P T E R 1 1 • Angular Momentum

∆L

(a)

(b)

Figure 11.15 (a) The motion of a simple gyroscope pivoted a distance h from its center

of mass. The gravitational force Mg produces a torque about the pivot, and this torque

is perpendicular to the axle. (b) Overhead view of the initial and final angular momen-

tum vectors. The torque results in a change in angular momentum d L in a direction

perpendicular to the axle. The axle sweeps out an angle d

in a time interval dt.

Figure 11.14 Precessional motion

of a top spinning about its symme-

try axis. (a) The only external

forces acting on the top are the

normal force n and the gravita-

tional force Mg. The direction of

the angular momentum L is along

the axis of symmetry. The right-

hand rule indicates that

r " F # r " Mg is in the xy

plane. (b). The direction of .L is

parallel to that of

in part (a). The

fact that L

# .

L % L

indicates

that the top precesses about the z

axis.

(a)

(b)

d L

Precessional motion

To simplify the description of the system, we must make an assumption: The total

angular momentum of the precessing wheel is the sum of the angular momentum I$
due to the spinning and the angular momentum due to the motion of the center of
mass about the pivot. In our treatment, we shall neglect the contribution from the
center-of-mass motion and take the total angular momentum to be just I$. In practice,
this is a good approximation if $ is made very large.

The vector diagram in Figure 11.15b shows that in the time interval dt, the angular

momentum vector rotates through an angle d!, which is also the angle through which
the axle rotates. From the vector triangle formed by the vectors

, and d

L, we see

that

where we have used the fact that, for small values of any angle ", sin "

& ". Dividing

through by dt and using the relationship L # I), we find that the rate at which the axle
rotates about the vertical axis is

(11.20)

The angular speed )

is called the

precessional frequency. This result is valid only

when )

)

. Otherwise, a much more complicated motion is involved. As you can

see from Equation 11.20, the condition )

)

is met when ) is large, that is, the

wheel spins rapidly. Furthermore, note that the precessional frequency decreases as )
increases—that is, as the wheel spins faster about its axis of symmetry.

As an example of the usefulness of gyroscopes, suppose you are in a spacecraft in

deep  space  and  you  need  to  alter  your  trajectory.  You  need  to  turn  the  spacecraft
around  in  order  to  fire  the  engines  in  the  correct  direction.  But  how  do  you  turn  a
spacecraft  around  in  empty  space?  One  way  is  to  have  small  rocket  engines  that  fire
perpendicularly out the side of the spacecraft, providing a torque around its center of
mass. This is desirable, and many spacecraft have such rockets.

Let us consider another method, however, that is related to angular momentum

and does not require the consumption of rocket fuel. Suppose that the spacecraft car-
ries a gyroscope that is not rotating, as in Figure 11.16a. In this case, the angular mo-
mentum of the spacecraft about its center of mass is zero. Suppose the gyroscope is set
into rotation, giving the gyroscope a nonzero angular momentum. There is no exter-
nal torque on the isolated system (spacecraft % gyroscope), so the angular momentum
of this system must remain zero according to the principle of conservation of angular
momentum. This principle can be satisfied if the spacecraft rotates in the direction op-
posite to that of the gyroscope, so that the angular momentum vectors of the gyro-
scope and the spacecraft cancel, resulting in no angular momentum of the system. The
result of rotating the gyroscope, as in Figure 11.16b, is that the spacecraft turns
around! By including three gyroscopes with mutually perpendicular axles, any desired
rotation in space can be achieved.

This effect created an undesirable situation with the Voyager 2 spacecraft during its

flight. The spacecraft carried a tape recorder whose reels rotated at high speeds. Each
time the tape recorder was turned on, the reels acted as gyroscopes, and the spacecraft
started an undesirable rotation in the opposite direction. This had to be counteracted
by Mission Control by using the sideward-firing jets to stop the rotation!

11.6 Angular Momentum as a

Fundamental Quantity

We have seen that the concept of angular momentum is very useful for describing
the motion of macroscopic systems. However, the concept also is valid on a submi-
croscopic scale and has been used extensively in the development of modern

)

Mgh

sin(d!)

& d! #

dt
L

(Mgh)dt

S E C T I O N 1 1 . 6 • Angular Momentum as a Fundamental Quantity

351

Figure 11.16 (a) A spacecraft car-

ries a gyroscope that is not spin-

ning. (b) When the gyroscope is set

into rotation, the spacecraft turns

the other way so that the angular

momentum of the system is

conserved.

(a)

(b)

Gyroscope rotates

counterclockwise

Spacecraft

rotates

clockwise

theories  of  atomic,  molecular,  and  nuclear  physics.  In  these  developments,  it  has
been found that the angular momentum of a system is a fundamental quantity. The
word  fundamental in  this  context  implies  that  angular  momentum  is  an  intrinsic
property  of  atoms,  molecules,  and  their  constituents,  a  property  that  is  a  part  of
their very nature.

To explain the results of a variety of experiments on atomic and molecular systems,

we rely on the fact that the angular momentum has discrete values. These discrete val-
ues are multiples of the fundamental unit of angular momentum # h/2,, where h is
called Planck’s constant:

Fundamental unit of angular momentum # # 1.054 ' 10

kg · m

Let us accept this postulate without proof for the time being and show how it can

be used to estimate the angular speed of a diatomic molecule. Consider the O

mole-

cule as a rigid rotor, that is, two atoms separated by a fixed distance d and rotating
about the center of mass (Fig. 11.17). Equating the angular momentum to the funda-
mental unit , we can find the order of magnitude of the lowest angular speed:

In Example 10.3, we found that the moment of inertia of the O

molecule about this

axis of rotation is 1.95 ' 10

kg · m

. Therefore,

Actual angular speeds are found to be multiples of a number with this order of
magnitude.

This simple example shows that certain classical concepts and models, when prop-

erly modified, are useful in describing some features of atomic and molecular systems.
A wide variety of phenomena on the submicroscopic scale can be explained only if we
assume discrete values of the angular momentum associated with a particular type of
motion.

The Danish physicist Niels Bohr (1885–1962) accepted and adopted this radical

idea of discrete angular momentum values in developing his theory of the hydrogen
atom. Strictly classical models were unsuccessful in describing many of the hydrogen
atom’s properties. Bohr postulated that the electron could occupy only those circular
orbits about the proton for which the orbital angular momentum was equal to n ,
where n is an integer. That is, he made the bold claim that orbital angular momentum
is quantized. One can use this simple model to estimate the rotational frequencies of
the electron in the various orbits (see Problem 42).

)

1.054 ' 10

kg&m

1.95 ' 10

kg&m

( 10

rad/s

)

& /

)

352

C H A P T E R 1 1 • Angular Momentum

⊕

Figure 11.17 The rigid-rotor

model of a diatomic molecule. The

rotation occurs about the center of

mass in the plane of the page.

The

torque ! due to a force F about an origin in an inertial frame is defined to be

! r " F

(11.1)

Given two vectors

A and B, the cross product A " B is a vector C having a magnitude

! AB sin "

(11.3)

where " is the angle between A and B. The direction of the vector

C # A " B is per-

pendicular to the plane formed by

A and B, and this direction is determined by the

right-hand rule.

The

angular momentum L of a particle having linear momentum p # mv is

L ! r " p

(11.10)

where

r is the vector position of the particle relative to an origin in an inertial frame.

S U M M A R Y

Take a practice test for

this chapter by clicking on
the Practice Test link at
http://www.pse6.com.

Content .. 86 87 88 89 ..