Topic 6 – Angular Momentum
Chapter 10
I.
Introduction.
a. As with conservation of energy and conservation of linear
momentum, conservation of angular momentum is one of the
basic principles of physics.
b. Experimental evidence shows that angular momentum is
never created nor destroyed.
c. We extend our study of rotational motion to situations in
which the direction of the axis of rotation may change.
Angular velocity, angular acceleration, and torque are
presented in Chapter 9. Here we begin by introducing the
vector nature of these quantities and of angular momentum,
which is the rotational analog of linear momentum. We then
show that the net torque acting on a system equals the time
rate of change of its angular momentum. Angular momentum
is conserved in systems that have zero net external torque.
Like conservation of linear momentum, conservation of
angular momentum is a fundamental law of nature, relating
to even atoms, molecules, subatomic particles, and photons.
II.
The Vector Nature of Rotation.
a. In Chapter 9, we indicated the direction of rotation about an
axis by assigning plus and minus signs to indicate the
direction of the angular velocity, just as in Chapter 2 we
used plus and minus signs to indicate the direction of the
velocity in one-dimensional motion.
b. However, plus and minus signs are not adequate to specify
the direction of the angular velocity if the direction of the
rotation axis is not fixed in space.
c. This inadequacy is overcome by treating the angular velocity
as a vector quantity directed along the rotation axis.
d. If the rotation is directed as shown,
is directed as shown; if the
rotation direction is reversed, so is
the direction of .
e. The convention relating the
direction of with the direction of
rotation is specified by a convention
called the right-hand rule.
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Topic 6 – Angular Momentum
Chapter 10
f. You can obtain the direction of by curling the fingers of
your right hand in the direction of rotation ( Figure 10-2);
your thumb then points along the rotation axis in the
direction of .
g. In Chapter 9, we indicated the direction of torque about an
axis by assigning plus and minus signs to indicate the
direction of the torque. In this chapter, we define the torque
about a point as a vector quantity, and, as with , the
direction of is specified by a right hand rule.
i. Figure 10-3 shows a force
acting on a particle at some
position relative to the origin
.
ii. The torque about exerted by
this force is defined as a vector
that is perpendicular to both
and
and has magnitude
where
,
is the angle between the directions of
and
.
iii. If and are both perpendicular to the z axis, as in
Figure 10-3, the torque vector is parallel to the z
axis.
iv. If is applied to the rim of a
disk of radius r as shown in
Figure 10-4, the torque
vector has the magnitude
and is along the axis of
rotation in the direction shown.
h. The Vector Product.
i. Review.
1. The Cross Product Method.
2. The cross product method produces a vector
when two vectors are multiplied together.
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Topic 6 – Angular Momentum
Chapter 10
3. The magnitude is the area of a parallelogram
formed by two vectors.
4. The resultant direction is defined as
perpendicular to the other two and is determined
by the right-hand rule when two vectors are
placed tail to tail.
a. Align your right hand along the first vector
so that the palm of your hand is at the tail
of the vector and your fingertips are
pointing toward the head.
b. Then, curl your fingers via the small angle
toward the second vector.
c. The direction your thumb points in making
the angle from the first to the second
vector is the direction of the resultant.
5. Two vectors that are parallel or antiparallel will
have a cross product of 0.
ii. Torque is expressed mathematically as the vector
product of
and
.
iii. Because of the used to indicate this type of
multiplication, the vector product is also called the
cross product.
iv. The vector product of two vectors
and is defined to be a vector
whose magnitude
equals the area of the parallelogram
formed by
v. The vector
and .
is perpendicular
to both and in the direction
of the thumb of your right
hand if you curl your fingers
from the direction of
the direction of .
toward
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Topic 6 – Angular Momentum
Chapter 10
vi. If
is the angle between
and
,
*
and
vector that is perpendicular to both
direction of
, the vector product of
is a unit
and in the
and is:
vii. Note that the order in which two vectors are multiplied
in a vector product makes a difference.
i. Finding the Vector Product of Two Vectors.
1. PICTURE
a. At times it is easier to find a vector
product of two vectors by using the
equation
.
b. At other times it is easier to find the vector
product using the Cartesian components of
the two vectors.
2. SOLVE
a. The vector product obeys a distributive
law under addition:
b. If
and are functions of some variable
such as t, the derivative of
follows
the usual product rule for derivatives:
c. The unit vectors , , and , which are
mutually perpendicular, have vector
products given by
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Topic 6 – Angular Momentum
Chapter 10
i. Reversing the order of multiplication
gives
and
.
ii. A tool for remembering this is shown
in Figure 10 - 8.
iii. Furthermore,
d. CHECK.
i. Make sure that your vector products
make sense.
ii. For example, the vector product of
two vectors is a vector and is
perpendicular to each of the two
vectors.
iii. In addition, check your work to
make certain you did not
inadvertently reverse the order of
the two vectors being multiplied, and
thus create a sign error.
e. TAKING IT FURTHER.
i. Any coordinate system for which the
equations above hold is called a
right-handed coordinate system.
ii. Only right-handed coordinate
systems are used in this book.
j. Example 10 – 1: Vector and Dot Products - If
and
= 12 find
.
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Topic 6 – Angular Momentum
Chapter 10
III.
Torque and Angular Momentum.
a. The linear momentum of the particle is
momentum
of the particle relative to the origin
defined to be the vector product of
b. If
. The angular
and
is
:
and
are both perpendicular to the z axis, as in Figure 10-10,
is parallel to the z axis and is given by
. Like torque, angular momentum is
defined relative to a point in space; in this case the angular
momentum is defined about the origin.
c. Figure 10-11 shows a particle of mass m attached to a
circular disk of negligible mass moving in a circle in the xy
plane with its center at the origin. The disk is spinning about
the z axis with angular speed ω The speed of the particle
and its angular speed are related by
.
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Topic 6 – Angular Momentum
Chapter 10
d. The angular momentum of the particle relative to the center
of the disk is
e. Note:
In this example the angular momentum vector is in the same
direction as the angular velocity vector.
f. The angular momentum of this particle about a general point
on the z axis is not parallel to the angular velocity vector.
i. Figure 10-12 shows the angular momentum vector
for the same particle attached to the same disk, but
with
computed about a point on the z axis that is not
at the center of the circle. In this case, the angular
momentum is not parallel to the angular velocity vector
, which is parallel to the z axis.
ii. In Figure 10-13, we attach a second particle of equal
mass to the spinning disk at a point diametrically
opposite this first particle.
1. The angular-momentum vectors
shown relative to the same point
and
.
are
2. The total angular momentum
of
the two-particle system is again parallel to the
angular velocity vector .
3. In this case, the axis of rotation, the z axis,
passes through the center of mass of the twoparticle system, and the mass distribution is
symmetric about this axis.
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Topic 6 – Angular Momentum
Chapter 10
4. Such an axis is called a symmetry axis.
5. For any system of particles that rotates about a
symmetry axis, the total angular momentum
(which is the sum of the angular momenta of the
individual particles) is parallel to the angular
velocity and is given by
g. Example 10 – 2: Angular Momentum About the Origin. Find
the angular momentum about the origin for the following
situations. (a) A car of mass 1200 kg moves in a circle of
radius 20 m with a speed of 15 m/s. The circle is in the xy
plane, centered at the origin. When viewed from a point on
the positive z axis, the car moves counterclockwise. (b) The
same car moves in the xy plane with velocity
along the line y = y0 = 20 m parallel to the
x axis. (c) A uniform disk in the xy plane of
radius 20 m and mass 1200 kg rotates at
0.75 rad/s about its axis, which is also the z
axis. When viewed from a point on the
positive z axis, the disk rotates
counterclockwise. Model the car as a point
particle and the disk as a uniform disk.
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Topic 6 – Angular Momentum
Chapter 10
h. There are several additional results concerning torque and
angular momentum for a system of particles. The first of
these is
i. In Equation 10-10, the net external torque about the point
is the vector sum of the external torques about that point
acting on the system. Integrating both sides of this equation
with respect to time gives
j. Equation 10-11 is the rotational analog of
k. It is often useful to split the total angular momentum of a
system about an arbitrary point O into orbital angular
momentum and spin angular momentum.
l. Earth has spin angular momentum due to its spinning motion
about its rotational axis, and it has orbital angular
momentum about the center of the Sun due to its orbital
motion around the Sun (Figure 10-17).
i. The total angular momentum of Earth relative to the
center of the Sun is the vector sum of the spin and
orbital angular momenta.
is the angular
momentum of a system about its center of mass, and
is the angular momentum that a point particle of
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Topic 6 – Angular Momentum
Chapter 10
mass M located at the center of mass and moving at
the velocity of the center of mass would have. That is,
m. In
Chapter 9, torques are computed about axes instead of
about points.
i. The relation between the torque about an axis and the
torque about a point is straightforward.
ii. If point O is the origin and if force exerts torque
about O, then (the z component of ) is the torque
of about the z axis.
iii. Do not confuse torque about a point with torque about
an axis. The torque of a force about the z axis is the z
component of the torque of the force about any point
on the z axis.
iv. Taking components of vector products requires some
care. If
then
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Topic 6 – Angular Momentum
Chapter 10
v.
(see Figure 10-18) are the vector
components of , , and . The vector component in a
given direction is the scalar component in that direction
times the unit vector in that direction. For example,
Here,
is the vector component of in the
positive radial direction (directly away from the z axis),
and
is the component of
perpendicular to the z
axis, and thus parallel to the xy plane
.
n. The relation between angular momentum about an axis and
angular momentum about a point is also straightforward.
i. If the angular momentum of a point particle about the
origin is
then the angular momentum of the
particle about the z axis is
where
is the component of the linear momentum
perpendicular to the z axis
ii. Taking the z vector components of both sides of
Equation 10-10 gives
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Topic 6 – Angular Momentum
Chapter 10
For a symmetric rigid object rotating about the z axis,
where
is the moment of inertia about the z
axis.
iii. For a system of particles, the total angular momentum
about the z axis equals the sum of the angular
momenta about the z axis. In addition, the total torque
about the z axis is the sum of the external torques
about the z axis acting on the system.
iv. Example 10 – 3: The Atwood‘s Machine Revisited - An
Atwood‘s machine has two blocks with masses m1 and
m2 (m1 > m2) connected by a string of negligible mass
that passes over a pulley with frictionless bearings. The
pulley is a uniform disk of mass M and radius R. The
string does not slip on the pulley. Apply Equation 1016 to the system consisting of the two blocks, the
string, and the pulley, to find the angular acceleration
of the pulley and the linear acceleration of the blocks.
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Topic 6 – Angular Momentum
Chapter 10
IV.
Conservation of Angular Momentum.
a. When the net external torque acting on a system about some
point remains zero, we have
b. Equation 10-20 is a statement of the law of conservation
of angular momentum.
c. This law is the rotational analog of the law of conservation of
linear momentum.
i. If a system is isolated from its surroundings, so that
there are no external forces or torques acting on it,
three quantities are conserved: energy, linear
momentum, and angular momentum.
ii. The law of conservation of angular momentum is a
fundamental law of nature.
d. Although conservation of angular momentum is a law,
independent of Newton‘s laws of motion, the fact that the
internal torques of a system cancel is suggested by Newton‘s
third law.
e. Consider the two particles shown in Figure 10-26.
i. Let
and
be the force exerted by particle 1 on particle 2
be the force exerted by particle 2 on particle 1.
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Topic 6 – Angular Momentum
Chapter 10
ii. By Newton‘s third law,
. The sum of the
torques exerted by these forces about the origin O is
iii. The vector
particles.
iv. If
is along the line joining the two
acts parallel to the line joining m1 and m2,
and
are either parallel or antiparallel and
v. If this is true for all the internal forces, the internal
torques cancel in pairs.
f. Example 10 – 4: A Rotating
Disk - Disk 1 is rotating freely
and has an angular velocity
about a vertical axis that
coincides with its symmetry
axis, as shown in Figure 1027. Its moment of inertia about
this axis is I1. It drops onto disk
2, of moment of inertia I2, that
is initially at rest. Disk 2 is
centered on the same axis as disk 1 and is free to rotate
about that axis. Because of kinetic friction, the two disks
eventually attain a common angular velocity . Find .
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Topic 6 – Angular Momentum
Chapter 10
g. In the collision of the two disks in Example 10-4,
mechanical energy is dissipated.
i. We can see this by writing the energy in terms of the
angular momentum.
ii. An object rotating with an angular velocity ω has
kinetic energy.
iii. This result is analogous with that for linear motion
Equation 8-25.
1. The
initial
kinetic
energy in Example 10-4 is
2. The final kinetic energy is
3. Because
the ratio of the final to the initial
kinetic energy is
which is less than one.
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Topic 6 – Angular Momentum
Chapter 10
4. This interaction of the disks is analogous to a
one-dimensional perfectly inelastic collision of
two objects.
iv. Example 10 – 5: Mud in Your Eye - You and three of
your friends have been bullied for many years by Gene,
who has avoided taking physics classes. So you and
your three friends who are
taking advanced-placement
physics decide to teach him a
lesson using conservation of
angular momentum. Here is
your plan. A nearby park has a
small merry-go-round
(Figure 10-28) with a 3.0-m2
diameter turntable that has a 130-kg · m moment of
inertia. You initially get all five of you to stand on the
merry-go-round next to the rim while the merry-goround is rotating at a modest 20 rev/min. When the
signal is given, you and your friends will quickly walk to
the center of the merry-go-round, leaving Gene near
the rim. The merry-go-round will speed up, throwing
Gene off and into the mud. (You plan to do this after a
heavy rain.) Gene is very quick and very strong, so
throwing him off will require that the centripetal
acceleration of the rim be at least 4.0 gs. Will this plan
work? (Assume that each person has a mass of 60 kg.)
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Topic 6 – Angular Momentum
Chapter 10
1. The moment of inertia of the students–merry-goround system decreases as the friends walk
toward the center.
2. Thus, the system‘s moment of inertia decreases
while its angular momentum remains constant.
3. As a result, we can see from Equation 10-21
that the kinetic energy of the students–merrygo-round system increases.
4. The energy for this kinetic energy increase
comes from the internal energy of the friends.
Walking radially inward, like walking up a steep
incline, requires the expenditure of internal
energy.
v. Example 10 – 6: Another Ride on the Merry-Go-Round
- A 25-kg child in a playground runs
with an initial speed of 2.5 m/s
along a path tangent to the rim of a
merry-go-round, whose radius is
2.0 m. The merry-go-round, which
is initially at rest, has a moment of
2
inertia of 500 kg · m The child then
jumps on (Figure 10-29). Find the
final angular velocity of the child
and the merry-go-round together.
vi. PRACTICE PROBLEM 10-1 Calculate the initial and
final kinetic energies of the child–merry-go-round
system.
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Topic 6 – Angular Momentum
Chapter 10
vii. Example 10 – 7: Spinning the Wheel
Conceptual - You are sitting on a stool
on a frictionless turntable holding a
bicycle wheel ( Figure 10-30).
Initially, neither the wheel nor the
turntable is spinning. Following
instructions from your teacher, you
hold the spin axis of the wheel vertical
with one hand, and with your other
hand you set the wheel spinning
counterclockwise (as viewed from
above). Surprise! When you start the
wheel spinning one way, the
turntable, the stool, you, and the axis of the wheel
start rotating in the opposite direction. After a few
seconds, you use your free hand to brake the spinning
motion of the wheel. You are surprised again when
you, the stool, and the wheel axis cease rotating as the
wheel ceases spinning. Explain.
viii. Example 10 – 8: Pulling Through a Hole - A particle of
mass m moves with speed
in a circle of radius on a
frictionless tabletop. The particle is attached to a string
that passes through a hole in the table, as shown in
Figure 10-31. The string is slowly pulled downward
until the particle is a distance from the hole, after
which the particle moves in a circle of radius . (a)
Find the final velocity in terms of ,
and . (b) Find
the tension when the particle is moving in a circle of
radius r in terms of m, r, and the angular momentum
. (c) Calculate the work done on the particle by the
tension force by integrating
answer in terms of r and
. Express your
.
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Topic 6 – Angular Momentum
Chapter 10
ix. In Figure 10-33 a puck on a
frictionless plane is given an initial
speed
The puck is attached to a
string that wraps around a vertical
post.
1. This situation looks similar to
Example 10-8, but it is not
the same.
2. There is no agent that does work on the puck,
nor is there any mechanism for energy
dissipation.
3. Thus, mechanical energy must be conserved.
4. Because
, where L is the magnitude
of the angular momentum about the axis of the
post, is constant and I decreases as decreases,
L must also decrease.
a. Note that the tension force does not act
toward the axis of the post.
b. The tension force on the puck produces a
torque vector about the axis of the post
in the downward direction, which reduces
the puck‘s angular-momentum vector
about the axis, which is in the upward
direction.
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Topic 6 – Angular Momentum
Chapter 10
x. Example 10 – 9: The Ballistic Pendulum Revisited - A
thin rod of mass M and length d hangs
vertically from a pivot attached to one
end. A piece of clay of mass m and
moving horizontally at speed hits the
rod a distance x from the pivot and
sticks to it (Figure 10-34). Find the
ratio of the clay–rod system‘s kinetic
energy just after the collision to its
kinetic energy just before the collision.
xi. Example 10 – 10: Tipping the Wheel Conceptual - A
student sitting on a stool that rests on a
turntable with frictionless bearings
(Figure 10-35a) is holding a rapidly
spinning bicycle wheel. The rotation axis
of the wheel is initially horizontal, and
the magnitude of the spin-angularmomentum vector of the spinning wheel
is
. What will happen if the student
suddenly tips the axle of the wheel
(Figure 10-35b) so that after the
rotation the spin axis of the wheel is
vertical and the wheel is spinning
counterclockwise (when viewed from
above)?
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Topic 6 – Angular Momentum
Chapter 10
V.
Proofs (If time).
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Topic 6 – Angular Momentum
Chapter 10
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