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Chapter 7
Rotational Motion
GOALS
When you have mastered the contents of this chapter, you will be able to achieve the
following goals:
Definitions
Define each of the following terms, and use it in an operational definition:
angular displacement
center of mass
angular velocity
moment of inertia
angular acceleration
rotational kinetic energy
uniformly accelerated angular
angular momentum
motion
torque
Equilibrium
State the conditions for static equilibrium.
Rotational Motion
Write the equations for rotational motion with constant angular acceleration.
Rotational Kinematics
Solve problems for systems with a fixed axis of rotation using the principles of
rotational kinematics.
Rotational Dynamics
Solve problems using the principles of rotational dynamics, for systems with fixed
axes of rotation, including conservation of energy and conservation of angular
momentum.
Equilibrium Problems
Solve problems involving conditions of static equilibrium.
PREREQUISITES
Before beginning this chapter you should be familiar with Chapter 4,
Forces and Newton's Laws, and Chapter 5, Energy. The quantitative
aspects of rotational motion are very similar to those of kinematics
(Chapter 3).
Chapter 7
Rotational Motion
OVERVIEW
This chapter may seem long because it contains both Rotational Kinematics and
Rotational Dynamics. The essential features of each of these concepts are
summarized and compared to the linear case in the table on Rotational
Kinematics and table 7.1 on Rotational Dynamics. The chapter will be easier to
handle if you keep these subdivisions in mind.
SUGGESTED STUDY PROCEDURE
When you begin to study this chapter, read the following Chapter Goals:
Definitions, Equilibrium, Rotational Dynamics, and Equilibrium. An expanded
discussion of each of the terms listed as Definitions can be found in the next
section of this Study Guide chapter. Next, read Chapter Sections 7.1-7.9. As you
read, be sure to note the parallels between rotational motion and linear motion.
The table on rotational kinematics summarizes the expressions for rotational
kinematics and table 7.1 summarizes rotation dynamics. As you read, remember
that the answers to all questions posed in the text sections are answered in the
second section of this Study Guide chapter.
At the end of the chapter, read the Chapter Summary and complete Summary
Exercises 3, 9, 10, 11, 12, and 15. Next, do Algorithmic Problems 1, 3, 4, 5, and 6
and complete Problems and Exercises 5, 6, 7, 8, 9, 15, 16, 17, and 23. For
additional practice on rotational motion problems, see section three of this Study
Guide for more Examples. Finally, attempt the Practice Test. If you have
difficulties with any part of the test, seek extra work from the appropriate text
section. This study procedure is outlined below.
-------------------------------------------------------------------------------------------------------------------Chapter Goals
Suggested
Summary
Algorithmic
Exercises
Text Readings
Exercises
Problems
& Problems
--------------------------------------------------------------------------------------------------------------------Definitions
7.1, 7.2, 7.3
3,10,11,12
1,3,4
Equilibrium
7.4
9
6,7,8
Rotational
7.5, 7.6, 7.7,
5,6
9,23
Dynamics
7.8, 7.9
Equilibrium
,
Problems
7.4
15
5,6,7,8,
15,16,17
Chapter 7
Rotational Motion
DEFINITIONS
ANGULAR DISPLACEMENT
Angle between two positions of a rigid body rotating about a fixed axis. Usually
given in radians.
You are probably most likely to give angular displacements in degrees, such as
"Make a 45ø turn," rather than "Turn p/4
radians." So you need to practice the conversion from radians to degrees and vice
versa so that both ways of measuring
angles are familiar to you.
ANGULAR VELOCITY
Time rate of change of angular displacement. Usually given in radians per second.
The most common way of talking about angular velocity in daily life is probably in
turns of revolutions per second or per
minute. Do you know what the 33 1/3 for a phonograph record means? It stands for
33 1/3 r.p.m. (revolutions per minute). Do
you know how to convert that to an angular speed measured in radians per second?
ANGULAR ACCELERATION
Time rate of change of angular velocity. Usually given in radians per second per
second.
When the rate of rotation of an object is changing then it has an angular acceleration.
If the rate of rotation is increasing, then
the angular acceleration is positive.
UNIFORMLY ACCELERATED ANGULAR MOTION
Rotational motion with a constant angular acceleration.
Common examples of this kind of motion seem scarce except for systems that roll
under the action of gravity, such as a
yo-yo rolling down its string.
TORQUE (moment of force)
A vector product of a force and its perpendicular distance to the point of rotation. A
right-hand rule can be used to determine
its direction.
You can note that the direction of the torque vector is exactly the direction
wherenothing is happening since the acting force
and the distance to the rotation point form a plane perpendicular to the torque.
CENTER OF MASS
The balance point of an object; the force of gravity (weight) produces zero torque
about the center of mass.
You have noticed since your early childhood that even large objects can be
supported at one point. There are several kinds of
children's toys that make use of an unusual shape to stably balance on a point. What
can you deduce about the location of
the center of mass of such toys?
MOMENT OF INERTIA
Measure of the ability of a body to resist a change in rotation. It depends upon the
distribution of its mass relative to an axis
of rotation.
If most of the mass of an object is far from the axis of rotation, then the moment of
inertia is large. Hence, a large baseball
bat is easier to swing if you choke up on the handle. You are moving more of the
mass of the bat closer to the center of
rotation, your hands, so that you reduce its moment of inertia for the swing.
ROTATIONAL KINETIC ENERGY
Energy of rotation equal to one-half the product of moment of inertia and angular
velocity squared.
Objects that rotate about an axis fixed in space have this kind of kinetic energy.
ANGULAR MOMENTUM
The product of moment of inertia and angular velocity gives the magnitude of the
angular momentum. A right-hand rule can be
used to determine its direction.
It is a straight forward matter to show that a system may have a constant linear
momentum while its angular momentum
changes, but the reverse case is not possible. Consider the case of motion of a system
of two equal forces acting in
opposite directions but separated by some distance. What properties of such a
system are constant? What properties
change?
ANSWERS TO QUESTIONS FOUND IN THE TEXT
SECTION 7.1 Introduction
The most common way of talking about rotation is in terms of the number of
complete rotations or revolutions an object or system makes. This is often done in
terms of revolutions per minutes or rotations per second.
You can start an object rotating by applying some forces to the object that are
some distance away from the point of rotation, or axis, and do not point exactly
toward the axis.
We can specify the angular motion of a system in terms of its angular
displacement, angular velocity, angular acceleration, moment of inertia,
rotational kinetic energy, and angular momentum.
Photographs, page 154
Notice the line of action of the applied torque with respect to the point of
rotation; i.e., the point of contact between the spool and the table.
EXAMPLES
ROTATIONAL KINEMATICS
1. A physics student is earning some extra money for his college expenses by serving
as a teacher's aide at a nursery school. One day on the playground he notices a child
(24 kg) playing on the merry-go-round which has an off-center set of hand and foot
bars so that it can be started from rest by a person sitting on the merry-go-round. He
makes the following observations: the child starts the merry-go-round from rest and
pumps it up to a constant rotation rate in 12 seconds, after which time the child
passes the observer each 1.4 seconds. The diameter of the merry-go-round is 5
meters. Quantitatively describe the child's rotation.
What Data Are Given?
The mass of the child is 24 kg. The period of angular acceleration is 12 sec and
the time required for one complete rotation of the child is 1.4 sec. If the child is
assumed to be sitting on the rim of the merry-go-round, then the child is 2.5
meters from the axis of rotation.
What Data Are Implied?
In order to calculate the child's angular acceleration, let us assume that the
motion is uniformly accelerated angular motion (u.a.a.m.) until the merry- goround reaches a constant speed of rotation. It then is a special case of u.a.a.m.;
i.e., the angular acceleration is zero.
What Physics Principles Are Involved?
You can begin the problem by using the concepts of u.a.a.m. as given on page
150, or summarized in the right-hand column of the table on page 153. You can
also calculate the work done by the child and the kinetic energy of the child
using concepts from Section 7.7.
What Equations Are to be Used?
Angular velocity; ω = 2π/T where T is the period of rotation.
For u.a.a.m.; α = ω/t where t is the time of acceleration
(7.3)
the angle turned during acceleration = (½) α t2
(7.8)
2
2 2
Kinetic Energy = (Iω )/2 = (mr ω )/2
for a point mass m located at a distance r from the axis of rotation
(7.26)
Work = change in kinetic energy
(7.25)
Algebraic Solutions
The equations above are all written with terms we may wish to calculate on the
left-hand side of each equation.
Numerical Solutions
Final angular velocity = (2π rad.)/(1.4 sec.) = 45 rad/sec
Angular acceleration = (4.5 rad/sec.)/(12 sec.) = 0.37 rad/s2
Angle of rotation during acceleration = (½) (0.37)(12)2
θ = 27 radians
Angular displacement of the child at any time at 12 seconds where t is the time
from rest,
θ = 27 rad. + (4.5 rad/s)(t - 12) where t > 12 s.
Distance travelled in meters = rθ = (2.5 m) θ
Kinetic energy of the child = (½) (24 kg)(2.5 m)2(4.5 rad/s)2 = 1.5 x 103 J
Work done to rotate the child = 1.5 x 103 J, this neglects the rotational kinetic
energy of the merry-go-round.
ROTATIONAL DYNAMICS
2. Consider a playground merry-go-round at rest and a child (24 kg) runs at a speed
of 6.2 m/s tangential to the rim of the merry-go-round on 3.0 meter radius and
jumps on it. If the moment of inertia of the merry-go-round is 1.44 x 102 kgm2, what
is the final angular speed of the system? What is the final kinetic energy of the
system? Is any energy lost? If so, what happens to it?
What Data Are Given?
The mass, speed, and distance of the child from the axis of rotation are given.
The moment of inertia of the merry-go-round is given.
What Data Are Implied?
The visual arrangement of the problem is implied by the setting of this problem.
Can you picture it in your mind?
See Fig. 7.1
This problem also implies that friction be neglected.
What Physics Principles Are Involved?
The concept of the conservation of angular momentum can be used to analyze
the motion of the combined system of child and merry-go-round. From the
instant the child jumps from the ground to land on the merry-go-round, the total
angular momentum is constant since there are then no external torques acting on
the combined child - merry-go-round system if the frictional forces on the system
are neglected. It seems as if conservation of energy can also be assumed from the
instant the child jumps up to the merry-go-round. There are no external forces
acting on the system.
What Equations Are to be Used?
The angular momentum is a constant
Iiωi = Ifωf
(7.30)
Total energy is a constant, perhaps
(½) mvi2 = (½) Ifωf2
To use these equations you need to know that the magnitude linear velocity v
can be related to the magnitude angular velocity ω about a fixed axis is the
perpendicular distance from the velocity to the axis is v by the following
equation.
v = rω
(7.12)
Algebraic Solutions
The initial angular momentum of the child - merry-go-round system is equal to
the angular momentum of the child just as she jumps on the merry-go-round
because the merry-go-round is initially at rest and so has an angular momentum
of zero.
Initial angular momentum = Li = mcvcrc
(1)
where the subscript c stands for the child of mass, speed, and distance from axis
of rotation of m, v, and r respectively.
Final angular momentum = Lf = ImWm + IcWc
(2)
where the subscript m applies to the merry-go-round. Since the child and the
merry-go-round are rotating together in the final state the angular speed of the
child and the merry-go-round are the same.
Wm = WC
(3)
The child can be treated as a rotating point mass, so
Ic = mcrc2
Then, from conservation of angular momentum,
Lf = Li
Wm(Im + mcrc2) = mcvcrc
(4)
so Wm = (mcvcrc) / (Im + mcrc2)
Is energy conserved?
The initial energy = KEi = (½) mcvc2
(5)
The final energy = KEf = (½) ImWm2 + (½) IcWc2 = (½) Wm2 (Im + Ic)
(6)
We can substitute the above expression for Wm, then
KEf = (½) [(mcvcrc) / (Im + mcrc2)]2 = [Im + mcrc2] = (½) [(mc2vc2rc2) / (Im + mcrc2)]
Let us set the initial and final kinetic energies equal
(½) mcvc2 = (½) [(mc2vc2rc2) / (Im + mcrc2)]
(5)
Divide by (½) mcvc2
1 = (mcrc2)/(Im + mcrc2) which will be true if Im = 0.
The energy will be conserved only if the moment of inertia of the merry- goround is neglected. What happens to the energy when Im is not treated as zero?
What explanation can you give of the failure of the law of conservation of energy
to be true for this system?
Numerical Solutions
Initial angular momentum = (24 kg)(6.2 m/s)(3.0 m) = 4.5 x 102 kgm2/s
Final angular speed = Wm = (4.5 x 102 kgm2/s) / (1.44 x 102 kgm2 + 24 kg (3.0 m)2)
= (4.5 x 102)/(3.6 x 102) rad/sec = (5.0 rad/s)/4.0 = 1.3 rad/s.
Initial kinetic energy = (1/2)(24 kg)(6.2 m/s)2 = 4.6 x 102 J
Final kinetic energy = (½) (1.44 x 102)(5/4)2 + (½) (24)(3.0)2(5/4)2 = 2.8 x 102 J
Fraction of kinetic energy lost = (4.6 x 102 J - 2.8 x 102J)/(4.6 x 102 J) = 0.39
Thinking About the Answer
Notice how just the simple statement that the child jumps on the merry-go-round,
with its implication of a completely inelastic collision means that 39% of the initial
kinetic energy is lost. What happens to it? You are invited to postulate a different
final situation. Assume the child makes a completely elastic collision with the merrygo-round. What then is the final state of the child and of the merry-go-round? This
means that the condition that the final angular velocity of the child and the merrygo-round must no longer be equal, so Equation (3) is no longer true. Can you solve
the completely elastic collision case? Hint: Equations (1), (2), (5), and (6) are still
valid, but they must be solved simultaneously for Wm and Wc.
[Answers: rotation speed of the merry-go-round = 2.5 rad/s. speed of the child = 1.1 m/s
still along the tangent line] If you wish you can extend the elastic collision
calculations and compute the impulsive torque on the merry-go-round and the
impulse on the child. Would he be hurt by an elastic collision with the merry-goround?
EQUILIBRIUM PROBLEMS
3. What is the force applied by the quadriceps tendon to the lower portion of the
human leg to hold it up at a 45ø angle from horizontal? Assume the values of the
variables as shown in the following mechanical analog to the human leg.
What Data Are Given?
See Figure 7.2
What Data Are Implied?
Use the strength of the earth's gravitational field to be g = 9.8 m/s2. Then a
generalized force diagram can be drawn as follows
See Figure 7.3
What Physics Principles Are Involved?
The system must satisfy the conditions for equilibrium, p. 157.
Sum of forces = zero
Sum of torques = zero
What Equations Are to be Used?
There is an unknown force L exerted by the upper leg bone of the hinge at the
knee. If we sum the torques about point A, then we can eliminate the force L and
solve for F.
Sum of torques = 0 = -Fd sinα + W2 (1/2)cosθ + WF (l cosθ + F/2)
(6)
where a negative sign indicates a counter-clockwise rotation about the point A.
Algebraic Solution
Solve for F in Equation 6
F = [WL (½) cos θ + WF (l cos θ + f/2)]/(dsinα)
(7)
Numerical Solution
F = [((4 kg)(9.8 m/s2)(0.22 m) cos 45ø + (2 kg)(9.8 m/s2) (0.44 cos 45ø + 0.09))]/
((0.10 m) sin (arctan 0.5))
F = (6.1 + 7.9) / (4.5 x 10-2) = 31 N
Thinking About the Answer
You can see from Equation (7) that the smaller you make the angle q; i.e., the
more nearly you hold your lower leg out horizontally, the greater is the force
applied by the quadriceps tendon. Try lifting your leg and feeling of this tendon
as you do. What can you infer from what you feel? Does our mechanical analog
seem to be a good approximation for the way your leg really works?
PRACTICE TEST
1. The six meter uniform bar has a weight of 20 Newtons. It is placed so that one end
rests on the side of a table, and a 15 Newton weight is placed 1 meter from its center.
a. What force if applied to point D will keep the bar in equilibrium?
b. With the bar in equilibrium, what force acts on the bar at point A?
2. The bicycle wheel shown below has a mass of 5 kilograms and a moment of
inertia of .5 kg m2 about its center. The wheel is supported above the ground so that
it can rotate freely and be driven by the chain C. The driving sprocket has a radius of
10 cm.
a. If the chain is producing an angular acceleration a = 5 rad/sec2, what is the
tension in the driving chain?
b. If the wheel starts at rest, what will be its angular velocity after 3 sec?
ANSWERS:
1. 20 N, 15 N
2. 2.5 N, 15 rad/sec