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total moment of inertia about any axis will be
I=I+ MRậu. (19.7)
This theorem ¡s called the parailel-azis theorem, and may be easily proved.
The moment of inertia about any axis is the mass times the sum of the z¿'s and
the z;'s, each squared: 7 = Y)(z‡ + ÿ)m¿. We shall concentrate on the #'s, but
of course the 's work the same way. Now z is the distance of a particular point
mass from the origin, but let us consider how it would look iŸ we measured z
trom the CM, instead of z from the origin. To get ready for this analysis, we
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HP 1 + XeM:
Then we just square this to fnd
3? =0 +2XecM#; + XêM:
So, when this is multiplied by rm¿ and summed over all ;, what happens? Taking
the constants outside the summation sign, we get
Ty — » ma? + 2X*eM » m4 + XếM » Tạ.
The third sum is easy; it is just Äf Xổ. In the second sum there are two pieces,
one of them is Ð }m;z;, which is the tobal mass times the #-coordinate of the
center oŸ mass. But this contributes nothing, because # is rmeasured from the
center of mass, and in these axes the average position of all the particles, weighted
by the masses, is zero. The first sum, of course, is the #ø part of ?„. Thhus we
arrive at Bd. (19.7), jusE as we guessed.
Let us check (19.7) for one example. Let us just see whether it works for the
rod. For an axis through one end, the moment of inertia should be Ä# 2/3, for we
calculated that. The center of mass of a rod, of course, is in the center of the rod,
at a distance L/2. Therefore we should find that MfL”/3 = ML2/12+ M(L/2).
Since one-quarter plus one-twelfth is one-third, we have made no fundamental
©TTOT.
Incidentally, we did not really need to use an integral to ñnd the moment of
inertia (19.5). If we simply assume that it is A2? times +y, an unknown coefficient,
and then use the argument about the two halves to get z+ for (19.6), then from
our argument about transferring the axes we could prove that + = 1 + + SO +
must be 1/3. There is always another way to do itl
In applying the parallel-axis theorem, it is oŸ course Important to remember
that the axis for Ï¿ musứ be parallel to the axis about which the moment of inertia
is wanbed.
One further property of the moment of inertia is worth mentioning because it
1s often helpful in ñnding the moment of inertia of certain kinds of objects. This
property is that if one has a pÏane figure and a set of coordinate axes with origin
in the plane and z-axis perpendicular to the plane, then the moment of inertia of
this fñgure about the z-axis is equal to the sum of the moments of inertia about
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the zø- and -axes. This is easily proved by noting that
1„ = m(wŸ + z2) = À ) mu
(since z¿ =0). Similarly,
lạ = À m,(x; +z7)= » m7,
1= m(x2 + 2) = À mịư? + ` mịy;
= l¿ + ly.
As an example, the moment ofinertia ofa uniform rectangular plate of mass Ä,
width +, and length b, about an axis perpendicular to the plate and through
1ts center is simply
I= M(u + L”)/12,
because its moment of inertia about an axis in its plane and parallel to its length
is Mu2/12, i.e., just as for a rod of length +, and the moment of inertia about
the other axis in its plane is MƒL”/12, just as for a rod of length L.
To summarize, the moment of inertia of an object about a given axis, which
we shall call the z-axis, has the following properties:
(1) The moment of inertia is
1= 3 mi(eỆ + uệ) = [GẺ +) dm,
(2) T the object is made of a number of parts, each of whose moment of inertia
1s known, the total moment of inertia is the sum of the moments of inertia
of the pieces.
(3) The moment of inertia about any given axis is equal to the moment of
inertia about a parallel axis through the ƠM plus the total mass times the
square of the distance from the axis to the ƠM.
(4) LÝ the object is a plane fñgure, the moment of inertia about an axis perpen-
dicular to the plane is equal to the sum of the moments of inertia about
any two mutually perpendicular axes lying in the plane and intersecting at
the perpendicular axis.
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The moments of inertia of a number of elementary shapes having uniform
mass densities are given in Table 19-1, and the moments of inertia of some other
objects, which may be deduced from 'Table 19-1, using the above properties, are
given in Table 19-2.
Table 19-1
Thin rod, length U | .L rod at center ML?/12
Thin concentric
circular ring, radii | .L ring at center | Mr +r3)/2
r1 and 7a
Sphere, radius r through center 2Mr2 /5
Table 19-2
Rect. sheet, sides ø, b | || b at center Ma?/12
Rect. sheet, sides a, b | _L sheet at M(a2 + b2)/12
center
Thin annular ring, any diameter Mf(r?+r3)/4
radii r1, ra
Rect. parallelepiped, || c, through M(a2 + b2)/12
sides a, Ù, € center
lRt. circ. cyL, radius || L, through Mr?/2
r, length L center
lRt. circ. cyL, radius .L L, through | A(r?/4+ L2/12)
r, length L center
19-4 Rotational kinetic energy
Now let us go on to discuss dynamics further. In the analogy between linear
motion and angular motion that we discussed in Chapter 18, we used the work
theorem, but we did not talk about kinetic energy. What is the kinetic energy of
a rigid body, rotating about a certain axis with an angular velocity œ¿? We can
immediately guess the correct answer by using our analogies. The moment of
inertia corresponds to the mass, angular velocity corresponds to velocity, and so
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the kinetic energy ought to be j1 œ2, and indeed it is, as will now be demonstrated.