A. Center of the Mass
B. Torque and Equilibrium of a Rigid Body
C. The Second Condition of Equilibrium
D. Rigid Bodies in Equilibrium
E. Levers
F. Stability of Equilibrium
G. Moment of Inertia: The Rotational Analog to Mass
H. Angular momentum
I. Rotational Kinetic Energy
Problems
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In Chapter 6 we saw that
the rotational equations of motion were exactly the same as the corresponding
equations for linear motion. In this
chapter we will continue to see that linear quantities and laws, such as Newton’s
second law, kinetic energy, and momentum, have rotational counterparts.
The transitional and
rotational motion of rigid bodies can be analyzed most easily with the help of
a concept called the center of mass.
The center of mass is just the average position of the mass distribution
of an object.
We make use if the
center of mass because it has the following important property:
All
the mass of a rigid body may be assumed to be at the center of mass of the
object when considering the transitional behavior of that object under the
action of external forces.
This means that the
rigid body is equivalent to an equally massive point mass placed at the mass
center insofar as its transitional motion under a given external force is
concerned. For example, consider the
projectile motion of a bowling pin or lopsided softball. Their center of mass follows a smooth curve,
the same curve a point object or spherically symmetric object would follow if
subjected to the same external forces.
As we shall see, the importance of center of mass is that it allows us
to separate the transitional and rotational motion of an object. The object will be translated as though all
its mass were concentrated at its mass center.
B. Torque
and Equilibrium of a Rigid Body
When an object is in
equilibrium its state of motion is not changing. An obvious condition for equilibrium is that the net force acting
must be zero. Then from Newton’s second
law, Fnet = ma, we have that the acceleration is zero and the
velocity is constant. Although this is
the only requirement necessary for a point mass to be in equilibrium it is not
sufficient to guarantee the equilibrium of a rigid body.
Consider three cases below where a stick is subjected to two forces with the same magnitude.
In all three cases the
net force on the stick is zero. Yet we
know that in the last case the stick does not remain at rest. It rotates clockwise. We are therefore forced to conclude that
while a necessary condition for equilibrium is that the net force is zero, this
is not a sufficient condition for a rigid body. A second condition is required to insure rotational
equilibrium. There must be no net turning
effect of forces rotating an object about a pivot point if the rigid object is
to be in equilibrium.
Experience shows that
the action causing a rotation depends not only on the magnitude and direction
of the applied force but also on the point of application. When you open a door, for example, you push
with your hand near the doorknob. You
know that a much larger force is required to turn the door if you push near its
hinges at the axis of rotation. Two
factors determine the rotational impetus:
the perpendicular component of the force and the distance from the point
of application to the axis of rotation.
The turning effect is given the name torque. The definition to torque is
Torque = (perpendicular force)(distance)
Or in symbols
(7.1)
Where the symbol t (Greek “tau”) has been used for
torque. Note that torque has the units
of force times length (e.g., N • m, lb • ft).
Examples:
1.
Find the torque exerted by the wrench on the bolt in
the diagram below.
Using Eq. (7.1) we have
2.
A student with an arm of length 2 ft holds a weight of
20 lb in her hand. Calculate the torque
exerted on her shoulder by the weight if she holds her arm a) outstretched,
b) at her side, and c) at an angle of 60° below horizontal.
a) First
draw a schematic diagram and then determine the torque using Eq. (7.1).
b) We
see that when the weight is held at her side there is no component of the force
perpendicular to d. This force gives no
turning effect and thus the torque is zero.
c) In this case the perpendicular component of
the force is (20 lb) (cos 60°) which is equal to 10 lb. Thus by Eq. (7.1) we have
C. The
Second Condition for Equilibrium
We have now acquired enough
insight so that we can now make a precise statement of the second condition for
equilibrium.
For
an object to be in equilibrium there must be no change in the rotational
motion. Thus the net torque must be
zero. This is equivalent to the requirement
that the sum of all clockwise (CW) torques must equal the sum of all
counterclockwise (CCW) torques.
D. Rigid
Bodies in Equilibrium
Now that we know the
second condition for equilibrium, we are able to treat all ordinary situations
in which a rigid body is at equilibrium.
To analyze such situations we follow the simple procedure outlined
below:
1.
Isolate the body for discussion.
2.
Draw all the forces acting on the body.
3.
Choose a pivot point about which you will determine
torques.
4.
Determine the torques.
5.
Solve the equations Fnet = 0 , and tCW = tCCW
Before we look at some
examples consider the choice of the axis of rotation for solving the torque
equation. When a rigid body is in
equilibrium it is not rotating about any axis. As a result, the axis for solving the torque equation may be
taken anywhere and is completely arbitrary.
We are therefore justified in taking the axis anywhere we think is
convenient.
Examples:
1.
A 16 ft seesaw is pivoted in the center. At what distance from the center would a 200
lb person sit to balance a 150 lb person on the opposite end?
A
convenient point to calculate the torques would be at the fulcrum. With this choice the upward force exerted by
the fulcrum does not contribute to the torques.
2.
A 10 ft, 25 lb seesaw is balanced by a little girl (50
lb) and her father (200 lb) at opposite ends as shown below. How far from the seesaw’s center of mass
must the fulcrum be placed?
Calculate
the torques about the fulcrum.
3.
A 6 ft tall 180 lb student is instructed to lie on a
board as shown below. The board is 8 ft
long, weighs 10 lbs and is supported on the left end by a scale that reads 80
lbs. Use this information to determine
the distance from his feet to his center of mass.
Calculate
the torques about the right pivot.
Note: In the last example we found the 6 ft man to
have a center of mass 3.33 ft from his feet.
The ratio of these two values (3.33 ft/6 ft = 0.56) tells us the position
of the center of mass of the student relative to his height. For an average man the center of mass is
approximately 56% of his height. This
value drops to approximately 54% for a woman.
A lever is a rigid bar
free to rotate about a fixed point called the fulcrum. The position of the fulcrum is fixed so that
it is not free to move with respect to the bar. Levers are used to lift the loads in an advantageous way and to
transfer movement from one point to another.
There are three basic
parts to any lever: the fulcrum, the
applied force, and the load. Levers are
classified by which one of these parts is in the middle, as shown below.
In a class 1 lever the
fulcrum is placed between the applied force and the load. A crowbar, an automobile jack, and a pair of
scissors are examples of class 1 levers.
In a class 2 lever the load is in the middle. A wheelbarrow is an example of a class 2 lever. A class 3 lever has the applied force in the
middle. As we shall see, some human
limb movements are performed by class 3 levers.
Examples:
1.
Determine the force that must be applied to end of the
class 1 lever shown below to lift the 500 lb load.
In
this example we will neglect the weight of the lever. To solve this calculate the torques about the fulcrum.
2.
The bones and muscles in the arm can be examined with a
very simple model of a class 3 lever.
Consider what happens when you hold a weight outstretched in your hand
with your upper arm against your body. The
elbow acts like a fulcrum with the lower arm as a lever. The load is the weight of your lower arm and
the weight being held in your hand. The
biceps apply an upward force to keep the system in equilibrium. Calculate the tension in the biceps if you hold
a 15 lb weight in your hand as shown below.
The dimensions given are typical.
We assume the lower arm weighs 4 lb.
Before
starting this problem you should first convince yourself that this is indeed an
example of a third class lever. To find
the applied force calculate the torques about the elbow.
Note: We recognize first and second class levers
as being very advantageous in lifting weights.
However, as we see in the last example, in a class 3 lever the applied
force is much greater than the weight of the load. Can you think of any particular advantage that a class 3 lever
gives?
The principles we have
discussed in previous sections allow us to examine what happens when a body is
tipped or knocked over. Consider the
motion of the glass of orange juice shown below. Close observation shows us that the glass is rotating about point
A. The rotation is a result of an
unbalanced torque on the glass.
Example: Suppose you are given the job of moving a
100 lb refrigerator. Being basically
lazy you decide to slide the refrigerator instead of lifting it up and carrying
it. Suppose further that this task occurs
in typical student ghetto housing where the linoleum is usually warped. If the edge of the refrigerator catches a
piece of linoleum as it is pushed as shown below, determine the force F that
will tip it over.
This type of problem
is easily solved if we recognize three important features. First we see that the refrigerator will
rotate about point A if it tips over.
Then we realize that it will only tip over if the torque about point A
due to the external force exceeds the torque due to the weight of the
refrigerator. Finally remember that
only the perpendicular component of the force gives a torque. If we extend the lines of the applied force
and the weight of the refrigerator, then torques are easily calculated as shown
below.
If the clockwise
torque is to balance the counterclockwise torque we have
Thus
if the applied force exceeds 50 lb the refrigerator will be tipped over.
Problem: If the refrigerator is 6 ft tall and you
push on the top of the refrigerator, what applied force will tip it over?
Question: In view of the previous discussion, if you
were given the task of moving your own refrigerator where would you push it?
G. Moment
of Inertia: The Rotational Analog to
Mass
The most important
equation for linear motion was given to us by Newton in his second law, F =
ma. We now seek a rotational analog to
the second law. It is easy to see that
for rotational motion angular acceleration must replace linear
acceleration. Furthermore, since an
unbalanced torque is required to change rotational motion, just as an
unbalanced force is required to change linear motion, torque will replace
force. However the rotational analog to
mass is not as obvious.
To find the rotational
analog to mass first consider how mass is viewed in the second law. Mass is a measure of the inertia of an
object. It tells us how hard it is to
set an object into motion or to stop it if it is already moving. With this concept of linear inertia in mind
we define the moment of inertia of an object. The moment of inertia tells us how hard it is to set an object
into rotational motion, or how difficult it is to stop a rotating object. Just as inertia depends on the mass of an
object, the moment of inertia depends on the mass of an object. In addition, how this mass is distributed
will also affect the moment of inertia.
We may now state the
rotational analog too Newton’s second law as
t = Ia (7.2)
Where I is the moment
of inertia of the body, and α is the angular acceleration which must be in
units of rad/sec2. In this
equation I is viewed as resisting angular acceleration by an external
torque. Moments of inertia for various
objects are given below. It is
important to note that objects with large moments of inertia have more mass
concentrated away from the axis of rotation. This makes them more difficult to set into rotational
motion. As a result, a hollow cylinder
will have a larger I than an equally massive solid cylinder. And I is four times greater for a rod
rotating about its end than about its middle.
Examples:
1.
Find the torque necessary to give a bike wheel of
radius 0.4 m and mass 2 kg an angular acceleration of 10 rad/sec2.
We
need to solve the Eq. 7.2. First use
the fact that a bike wheel is a hollow cylinder to find its moment of inertia.
Now
solve Eq. 7.2
t = I a = (0.32 kg m2)(10 rad /sec2) =3.2 N m
2.
A typical bowling ball has a mass of 5 kg and a radius
of 10 cm. What torque is necessary to increase
the angular velocity of a bowling ball from rest to 40 rad/sec in a time of 2
seconds?
Once
again we must solve Eq. 7.2, but in this case both I and a
must be determined from the information given.
First determine I. Neglecting
the holes in a bowling ball we use the moment of inertia of a solid sphere.
Now
use angular kinematics to find a.
Noticing
that the angular acceleration is in the appropriate units, we are now ready to
determine the torque.
In dealing with
transitional motion we found it useful to introduce the quantity linear
momentum (p = mv). This allowed us to
state Newton’s law F = ma in a convenient way.
Namely, that force is the rate at which linear momentum changes. We saw that an external force was necessary
to change momentum. Furthermore, when
no external force was present we found that linear momentum was conserved. This gave us a powerful tool in treating
many physical problems such as collisions.
The rotational analog
to linear momentum is called angular momentum. As one might expect from the fact that linear momentum is given
by mv, the defining equation for angular momentum is
Angular momentum = Iw (7.3)
The angular momentum
of an object or a system of objects obeys a conservations law much like the one
obeyed by linear momentum. It may be
stated as follows:
If no
external torques act on a system, its angular momentum will remain constant.
Example: An ice skater with arms outstretched is
spinning with an angular velocity if 0.5 rev/sec. In this position her moment of inertia is 12 kg • m2. Suppose she brings her arms close to her
body and reduces her moment of inertia to 3 kg • m2. What is her new angular velocity?
To solve this we
simply use the law of conservation of angular momentum.
The center of mass of
a rotating wheel mounted on a stationary axle has zero velocity, and thus the
wheel has no kinetic energy of translation.
Yet we know that a rotating flywheel can do work because of its
rotation. For example, it could wind up
a rope with a weight hanging from it.
Therefore, it does possess kinetic energy, that is energy of motion.
Once again we can use
the definition of translational kinetic energy given by ˝ mv2 to
find an expression for rotational kinetic energy. We replace mass by the moment of inertia and linear velocity by
angular velocity and obtain:
(7.4)
Remembering out
definition of kinetic energy, this equation tells us what amount of work must
be done on a wheel to set it in motion, or what amount of work it can do by
virtue of its rotation.
In many cases an object
will be rotating and translating. In
such instances, the total kinetic energy is the sum of translational
and rotational kinetic energies, ˝ mv2 + ˝ Iw2. If the object rolls without slipping we have
a further restriction that v = rw, where v is the
velocity if the center of mass. Examine
the object below which rolls without slipping and summarize these equations.
For all objects that
rotate, we see that a rotational kinetic energy exists due to the rotational
motion of the object. This rotational
kinetic energy is interconvertible with work, potential energy, and translational
energy. So we can again use the law of
conservation of energy.
Example: A flywheel of radius 0.2 meters and mass 40
kg is rotating at 50 rad/sec. Find the
moment of inertia of the wheel and its rotational kinetic energy.
We assume the flywheel
is a solid cylinder so we first calculate the moment of inertia.
Now use Eq 7.4 to find
the rotational kinetic energy
Question: In the previous example how much work is
required to set the flywheel in motion?
Summary
Center of Mass Moment
of Inertia
Equilibrium Angular
Momentum
Torque Conservation
of Angular Momentum
Levers Rotational
Kinetic Energy
Stability of
Equilibrium
1.
Calculate the torque exerted by the applied force in
the situations described below:
a.
A 200 lb force acting perpendicularly at a distance of
3 ft.
b. A 70
kg person 2 m from a seesaw fulcrum.
c.
A uniform 12 ft ladder of 50 lb suspended at one end on
a horizontal position.
d. A uniform
12 ft ladder of 50 lb suspended at one end on a vertical position.
2.
A 40 lb little boy would like to ride on the seesaw but
there is no one else to play with. The
seesaw is 12 ft long and weighs 20 lb.
Where must the fulcrum be placed to balance the boy alone at one end?
3.
A 20 ft long, 200 lb plant rests on two points as shown
below. The plank is not fastened at
either point. What is the furthest
distance x that the 100 lb girl can walk before the plank tips over?
4.
The position of the foot when standing with the tip of
the feet on the edge of a step is shown below.
The floor pushes upward at the toes with a force equal to the weight of
the person. Calculate the tension on
the Achilles tendon and the compressional force on the tibia when a 200 lb
person stands in this position.
5.
A dolly is very useful instrument to have when you need
to move massive objects. It is another
example of a class 2 lever. Consider
the task of moving a 4 ft, 300 lb sleeper sofa with a 6 foot dolly as
diagrammed below. What force must you
apply to lift the sofa?
6.
What weights must be placed on a 4 foot hurdle if it is
to be just knocked over with a horizontal force of 7 lb? Assume the bottoms of the hurdle are each 2
ft long and weigh 2 lb each.
7.
In the two situations shown below identical objects are
shown in different positions. In which
position are the objects more stable against a toppling force?
8.
A 200 lb student stands 6 ft tall. What horizontal force is required to knock
him over in the following circumstances?
a.
The force is applied 3 ft from the ground and his feet
are spread 1 ft apart.
b. The
force is applied 3 ft from the ground and his feet are spread 2 ft apart.
c.
The force is applied 5 ft from the ground and his feet
are spread 1 ft apart.
9.
A beer keg has a radius of 20 cm (.2 m) and a mass of
25 kg. What is its moment of
inertia? What torque is necessary to
give it an angular acceleration of 8 rad/sec2?
10. A
high diver is rotating in the tuck position with an angular velocity of 2
rev/sec. His mass is 80 kg.
a.
Assume that in the tuck position he has a moment of
inertia similar to a solid cylinder of radius 0.5 meters. Find his moment of inertia.
b. Before
entering the water a diver stretches out of the tuck position and assumes the
shape of a long rigid rod. Assuming he
stretches out to a length of 2 meters calculate his moment of inertia in this
position.
c.
What is his angular velocity in this new position?
11. Why
is it necessary to have a second propeller of a helicopter?
12. A
merry-go-round of radius 2 m and moment of inertia 400 kg•m2 is
rotating with an angular velocity of 0.30 rev/sec. If a 50 kg student steps onto the edge of the merry-go-round
find:
a.
The final moment of inertia of the system.
b. The
final angular velocity if the system.
13. A
bowling ball has a mass of 5 kg and a radius of 0.15 m. As it rolls down the bowling alley without
slipping its center of mass travels with a velocity of 4 m/sec.
a.
What is the moment of inertia of the bowling ball?
b. What
is the angular velocity of the bowling ball?
c.
What is the rotational kinetic energy of the bowling
ball?
d. What
is the translational kinetic energy of the bowling ball?
e.
What is the total kinetic energy of the bowling ball?
14. A
hollow cylinder is rolling with a speed of 4 m/sec on a horizontal surface as
it approaches an incline. What height
above the horizontal will it rise?
