5.3.1. Uniform slender rod; axis perpendicular to length rod
Rod has
mass
and length
.
We wish to compute its moment of inertia about an axis through O, at
an arbitrary distance from one end. Using Eq. (70), we choose
as an element of mass a short section having length
at
a distance
from
point
O.
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The ratio
of the mass
of
this element to the total mass
is equal to the ratio of its length
to
the total length
.Thus
Where
Using (70) we obtain:
.
From
this general expression we can find the moment
of inertia about an axis through any point of the
rod. For example, if the axis is at the left end,
and
If the axis
is the right end
and
As would be expected, if the axis passes though the center,
5.3.2. Hollow or solid cylinder
Fig.
38 shows a hollow cylinder of length
and
inner and outer radii
and
.
We
choose as the most convenient volume element a thin cylindrical sheet
of radius
,
thickness
and length
.
The volume of this shell is very nearly equal to that
of flat sheet of thickness
,
length
,
and width
.
Then
.
The moment of inertia is given by
Volume
Hence
and the moment of inertia is
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If
the cylinder is solid
letting outer radius
be
we
find that the moment of inertia of a solid
cylinder of radius
is
If
the cylinder is very thin,
and
,
are
nearly
equal, if
represents
this common radius
(76)
Note, that the moment of inertia of cylinder does not depend on the length . It depends only on the radial distribution of mass, not on distribution along the axis.
5.3.3. Uniform sphere of radius , axis through center
Divide the sphere into thin disks. The radius of the disk shown in Fig. 7 is
Its
volume is
and
its mass is
.
Hence
from Eq. ()
Integrating this expression from 0 to gives the moment of inertia of the right hemisphere. From symmetry, the total for the entire sphere is just twice this:
Carrying out the integration, we obtain
The
mass
of
the sphere is
.
Hence
.
11-8 Torque
A doorknob is located as far as possible from the door's hinge line for a good reason. If you want to open a heavy door, you must certainly apply a force; that alone, however, is not enough. Where you apply that force and in what direction you push are also important. If you apply your force nearer to the hinge line than the knob, or at any angle other than 90° to the plane of the door, you must use a greater force to move the door than if you apply the force at the knob and perpendicular to the door's plane.
Figure 11-15a shows a cross section of a body that is free to rotate about an axis passing through О and perpendicular to the cross section. A force is applied at point P, whose position relative to О is defined by a position vector . The directions of vectors and make an angle with each other. (For simplicity, we consider only forces that have no component parallel to the rotation axis; thus, is in the plane of the page.)
To
determine how
results
in a rotation of the body around the rotation axis, we resolve
into
two components (Fig. 11-15b).
One
component, called the radial
component
,
points
along
.
This
component does not cause rotation, because it acts along a line that
extends through O.
(If
you pull on a door parallel to the plane of the door, you do not
rotate the door.) The other component of
tangential component
,
is
perpendicular to
and
has magnitude
.
This
component does
cause
rotation. (If you pull on a door perpendicular to its plane, you can
rotate the door.)
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The ability
of
to
rotate the body depends not only on the magnitude of its tangential
component
,
but
also on just how far from О
the
force is applied. To include both these factors, we define a quantity
called torque
as the product of the two factors and write it as
equivalent ways of computing the torque are
And
,
where
is
the perpendicular distance between the rotation axis at О
and
an extended line running through the vector
(Fig.
11-15c). This extended line is
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called the line of action of , and is called the moment arm of . Figure 11-15b shows that we can describe , the magnitude of , as being the moment arm of the force component .
Torque, which comes from the Latin word meaning "to twist," may be loosely identified as the turning or twisting action of the force . When you apply a force to an object - such as a screwdriver or torque wrench - with the purpose of turning that object, you are applying a torque. The SI unit of torque is the newton-meter (N • m). Caution: The newton-meter is also the unit of work.
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Torque and work, however, are quite different quantities and must not be confused. Work is often expressed in joules (1 J = IN- m), but torque never is.
In the next chapter we shall discuss torque in a general way as being a vector quantity. Here, however, because we consider only rotation around a single axis, we do not need vector notation. Instead, a torque has either a positive or negative value depending on the direction of rotation it would give a body initially at rest: If the body would rotate counterclockwise, the torque is positive. If the object would rotate clockwise, the torque is negative. (The phrase "clocks are negative" from Section 11-2 still works.)
Torques
obey the superposition principle that we discussed in Chapter 5 for
forces: When several torques act on a body, the net
torque
(or
resultant
torque)
is
the sum of the individual torques. The symbol for net torque is
.
11-9 Newton's Second Law for Rotation
A torque
can cause rotation of a rigid body, as when you use a torque to
rotate a door. Here we want to relate the net torque
on a rigid body to the angular acceleration
it
causes about a rotation axis. We do so by analogy with Newton's
second law (
)
for
the acceleration a of a body of mass
due
to a net force
along a coordinate axis. We replace
with
,
with
,
and
with
,
writing
(Newton's
second law for rotation), 11-34
where must be in radian measure.
Proof of Equation 11-34
We prove Eq. 11-34 by first considering the simple situation shown in Fig. 11-16.
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The rigid body there consists of a particle of mass on one end of a massless rod of length . The rod can move only by rotating about its other end, around a rotation axis (an axle) that is perpendicular to the plane of the page. Thus, the particle can move only in a circular path that has the rotation axis at its center.
A force
acts
on the particle. However, because the particle can move only along
the circular path, only the tangential component
of
the force (the component that is tangent to the circular path) can
accelerate the particle along the path. We can relate
to
the particle's tangential acceleration
along
the path with Newton's second law, writing
.
The torque acting on the particle is, from Eq. 11-32,
.
From Eq.
11-22 (
)
we
can write this as
11-35
The quantity in parentheses on the right side of Eq. 11-35 is the rotational inertia of the particle about the rotation axis (see Eq. 11-26). Thus, Eq. 11-35 reduces to
For the situation in which more than one force is applied to the particle, we can generalize Eq. 11-26 as
which we set out to prove. We can extend this equation to any rigid body rotating about a fixed axis, because any such body can always be analyzed as an assembly of single particles.
11-10 Work and Rotational Kinetic Energy
As we discussed in Chapter 7, when a force causes a rigid body of mass to accelerate along a coordinate axis, it does work on the body. Thus, the body's kinetic energy ( ) can change. Suppose it is the only energy of the body that changes. Then we relate the change in kinetic energy to the work with the work-kinetic energy theorem (Eq. 7-10), writing
(work-kinetic
energy theorem). (11-41)
For motion confined to an axis, we can calculate the work with Eq. 7-32,
This
reduces to
when
is
constant and the body's displacement is d.
The rate at which the work is done is the power, which we can find with Eqs. 7-43
(power,
one-dimensional motion). (11-43)
Now let us
consider a rotational situation that is similar. When a torque
accelerates a rigid body in rotation about a fixed axis, it does work
on
the body. Therefore, the body's rotational kinetic energy (
)
can
change. Suppose that it is the only energy of the body that changes.
Then we can still relate the change
in
kinetic energy to the work
with
the work-kinetic energy theorem, except now the kinetic energy is a
rotational kinetic energy:
(work-kinetic
energy theorem). (11-44)
Here,
is the rotational inertia of the body about the fixed axis and
and
are the angular speeds of the body before and after the work is done,
respectively. Also, we can calculate the work with a rotational
equivalent of Eq. 11-42,
(work,
rotation about fixed axis),(11-45)
where
is the torque doing the work
,
and
and
are
the body's angular positions before and after the work is done,
respectively. When
is constant, Eq. 11-45 reduces to
(work,
constant torque). (11-46)
The rate at which the work is done is the power, which we can find with the rotational equivalent of Eq. 11-43,
power,
rotation about fixed axis).
(11-47)
Proof of Eqs. 11-44 through 11-47
Let us again consider the situation of Fig. 11-16, in which force rotates a rigid body consisting of a single particle of mass fastened to the end of a massless rod. During the rotation, force does work on the body. Let us assume that the only energy of the body that is changed by is the kinetic energy. Then we can apply the work-kinetic energy theorem of Eq. 11-41
.
11-48
Using and Eq 11-18 ( ) we can rewrite Eq. 11-48 as
(11-
From Eq.
11-26, the rotational inertia for this one-particle body is
.
Substituting
this into Eq. 11-49 yields
,
which is Eq. 11-44. We derived it for a rigid body with one particle, but it holds for any rigid body rotated about a fixed axis.
We
next relate the work
done
on the body in Fig. 11-16 to the torque
оп
the
body due to force
.
When
the particle moves a distance
along
its circular path,
only the tangential component Ft
of
the force accelerates the particle along the path.
Therefore, only
does
work on the particle. We write that work
.
However,
we can replace
with
,
where
is
the angle through which the particle
moves. Thus we have
. (11-50)
From Eq.
11-32, we see that the product
is
equal to the torque
,
so we can rewrite
Eq. 11-50 as
(11-51)
The work done during a finite angular displacement from to
which is Eq. 11-45. It holds for any rigid body rotating about a fixed axis. Equation 11-46 comes directly from Eq. 11-45.
We can find the power P for rotational motion from Eq. 11-51:
which is Eq. 11-47
Sample Problem 11-10
A rigid sculpture consists of a thin hoop (of mass and radius = 0.15 m) and a thin radial rod (of mass and length = 2.0/ ), arranged as shown in Fig. 11-19. The sculpture can pivot around a horizontal axis in the plane of the hoop, passing through its center.
(a) In terms of and , what is the sculpture's rotational inertia about the rotation axis?
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A Key Idea here is that we can separately find the rotational inertias of the hoop and the rod and then add the results to get the sculpture's total rotational inertia . The hoop has rotational inertia hoop = mR2/2 about its diameter. The rod has rotational inertia com = mL2/12 about an axis through its center of mass and parallel to the sculpture's rotation axis. To find its rotational inertia rod about that rotation axis, we use Eq. 11-29, the parallel-axis theorem:
where we
have used the fact that
and where the perpendicular distance between the rod's center of mass
and the rotation axis is
.
Thus, the rotational inertia /
of
the sculpture about the rotation axis is
(b) Starting from rest, the sculpture rotates around the rotation axis from the initial upright orientation of Fig. 11-19. What is its angular speed about the axis when it is inverted?
SOLUTION: Three Key Ideas are required here:
We can relate the sculpture's speed to its rotational kinetic energy with Eq. 11-27 (
).We can relate to the gravitational potential energy of the sculpture via the conservation of the sculpture's mechanical energy during the rotation. Thus, during the rotation, does not change (
)
as energy is transferred from
to
.For the gravitational potential energy we can treat the rigid sculpture as a particle located at the center of mass, with the total mass
concentrated there.
We can write the conservation of mechanical energy ( ) as
As the sculpture rotates from its initial position at rest to its inverted position, when the angular speed is , the change in its kinetic energy is
From Eq.
8-7
,
the
corresponding change
in the gravitational
potential energy is
where
is the sculpture's total mass, and
is the vertical displacement of its center of mass during the
rotation.
To find
,
we
first find the initial location
of the center of mass in Fig. 11-19. The hoop (with mass
)
is
centered at
.
The rod (with mass
)
is centered at
.
Thus,
from Eq. 9-5, the sculpture's center of mass is at
When the
sculpture is inverted, the center of mass is this same distance
from the rotation axis but below
it.
Therefore, the vertical displacement of the center of mass from the
initial position to the inverted position is
Now let's pull these results together. Substituting Eqs. 11-55 and 11-56 into 11-54 gives us
Substituting
from (a) and
from above and solving for со,
we
find
(Answer)
Q. 2.01. Do the internal forces affect the motion of a system under the effect of some external force ?
Ans. No. Torques acting on the system due to internal forces cancel out.
Q. 2.02. What do you mean by a rigid body?
Ans. A body, whose constituent particles remain at their respective positions, when a body is in translational or rotational motion is called a rigid body.
Q. 2.01. Torque and work are both defined as force times distance. Explain, how do they differ.
Ans. (i) Whereas work is a scalar quantity, torque is a vector quantity.
(ii) Work done is measured as the product of the applied force and the distance, which the body covers along the direction of the force. On the other hand, torque is measured as the product of the force and its perpendicular distance from the axis of rotation.
Q.2.02. Why is a ladder more apt to slip, when you are high up on it than when you just begin to climb?
Ans. When a person is high up on a ladder, then torque produced due to his weight about the point of contact between the ladder and the floor becomes quite large. On the other hand, when he starts climbing up, the torque is small. Due to this reason, the ladder is more apt to slip, when one is high up on it.
Q. 2.03. A planet moves around the sun under the effect of gravitational force exerted by the sun. Why is the torque on the planet due to the gravitational force zero?
Ans. The torque on the planet due to the sun,
,
where
is
the position vector of the planet w.r.t. the sun and
is the gravitational force on the planet. Since the gravitational
force on the planet acts along the line joining the planet to the
sun, the vectors
and
are always parallel and hence
.
Ans. An object will not acquire angular momentum, if no external torque acts on it. During its flight, a torque acts on the projectile due to gravity and hence it acquires angular momentum.
Q. 2.06. A planet revolves around a massive star in a highly elliptical orbit. Is its angular momentum constant
1.
Starting from Newton's second law of motion, derive the equation
of motion of a particle (capable of rotation about an
axis) on which a torque
is
acting. Assume that the motion
of the particle is in a plane.
2 Derive expression for torque on a system in cartesian coordinates.
3, Derive expression for the torque acting on a system of -particles.
Derive expression for the angular momentum of a system in cartesian co-ordinates.
Prove that the time rate of change of the angular momentum of a particle is equal to the torque acting on it.
Derive a relation between angular momentum and torque.
Derive the relation between the torque and the angular momentum. Hence obtain and state the law of conservation of angular momentum.
State and prove the principle of conservation of angular momentum.
B. On Torque
9. A disc of radius 0.5 m is rotating about an axis passing through its centre and perpendicular to its plane.A tangential force of 2,000 N is applied to bring the disk to rest in 2 s. Calculate its angular momentum.
[Ans. 2,000 kg m2/s
10. A torque of 20 N m is applied on a wheel initially at rest. Calculate the angular momentum of the wheel after 3 s.
[Ans. 60 kg m2/s
Type C. On Angular momentum
11. In hydrogen atom, electron revolves in a circular orbit of radius 0.53 A with a velocity of 2.2 x 106 m/s. If the mass of the electron is 9.0 x 10-31 kg, find its angular momentum.
[Ans. 1.05 x 10-34 kg m2/s
12. Find the angular momentum of Neptune about the sun. Given distance of Neptune from the sun is 5 x 1012 m. period of revolution about the sun = 5 x 109s and mass of Neptune = 1027 kg. [Ans. 3.14 x 1043 kg m2I