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we move it in a circle or in some other curve, then it generates a rather peculiar
volume. For a curved path, the outside goes around farther, and the inside goes
around less, and these efects balance out. So If we want to locate the center
of mass of a plane sheet oŸ uniform density, we can remember that the volune
generat©ed by spinning ¡i% about an axis is the distance that the center of mass
goes around, times the area oŸ the sheet.
For example, If we wish to fnd the center of mass of a right triangle of
base D and height (Eig. 19-2), we might solve the problem in the following
way. Imagine an axis along H, and rotate the triangle about that axis through
a full 360 degrees. This generates a cone. The distance that the #-coordinate
of the center of mass has moved is 2rz. The area which is beïing moved is the
area of the triangle, sH D. So the z-distance of the center of mass tỉimes the
area, of the triangle is the volume swept out, which is of course x!22//3. Thus
--- Trang 352 ---
2S XI» --`
“ mà `
f ` D \
N —_— - ` r
Fig. 19-2. A right triangle and a right circular cone generated by
rotating the triangle.
(2xz)(3HD) = xD?H/3, or z = D/3. In a similar manner, by rotating about
the other axis, or by symmetry, we lnd = H/3. In fact, the center oŸ mass
of any uniform triangular area is where the three medians, the lines from the
vertices through the centers of the opposite sides, all meet. That point is 1/3
of the way along each median. (C?ue: Slice the triangle up into a lot of little
pleces, each parallel to a base. Note that the median line bisects every piece, and
therefore the center of mass must lie on this line.
Now let us try a more complicated figure. Suppose that ït is desired to fñnd
the position of the center of mass of a uniform semicircular disc—a disc sliced in
half. Where is the center of mass? Eor a full disc, i is at the center, of course,
but a half-dise is more difficult. Let r be the radius and z be the distance of the
center of mass from the straight edge of the disc. Spin it around this edge as axis
to generate a sphere. Khen the center of mass has gone around 27rz, the area
is Zr2/2 (because it is only half a circle). The volume generated is, of course,
4r3/3, from which we fnd that
(2xz)(šmr?) = 4mr3/3,
+ = Ar/3n.
'There is another theorem of Pappus which is a special case of the above one,
and therefore equally true. Suppose that, instead of the solid semicircular disc,
we have a semicircular piece of wire with uniform mass density along the wire,
and we want to find its center of mass. In this case there is no mass in the
Interior, only on the wire. 'Then it turns out that the area which is swept by
a plane curved line, when it moves as before, is the distance that the center of
mass moves times the iength of the line. (The line can be thought of as a very
narrow area, and the previous theorem can be applied to it.)
--- Trang 353 ---
19-3 Finding the moment of inertia
Now let us discuss the problem of finding the zmomen‡s oƒ inertia of various
objects. The formula for the moment of inertia about the z-axis of an object 1s
T=) mị(zỶ + tý)
I= Jú2+2) đm = J2+20sat (19.4)
That is, we must sum the masses, each one multiplied by the square of its
distance (z‡ + ÿ) from the axis. Note that it is not the three-dimensional
distance, only the two-dimensional distance squared, even for a three-dimensional
object. For the most part, we shall restrict ourselves to two-dimensional objeects,
but the formula for rotation about the z-axis is just the same in three dimensions.
L——————
x ——| |-gx
Fig. 19-3. A straight rod of length L rotating about an axis through
one end.
As a simple example, consider a rod rotating about a perpendicular axis
through one end (Eig. 19-3). Now we must sum all the masses times the z-
distances squared (the s being all zero in this case). What we mean by “the
sum,” of course, is the integral of z2 times the little elements of mass. lÝ we
divide the rod into small elements of length dz, the corresponding elements of
mass are proportional to đz, and if dz were the length of the whole rod the mass
would be Mĩ. Therefore
đưmm = M dr/L
and so
¬- '“—=. q95
= øˆ—>———=— “dư = ———. :
0 TL Lo b)
The dimensions of moment of inertia are always mass times length squared, so
all we really had to work out was the factor 1/3.
--- Trang 354 ---
Now what is ƒ If the rotation axis is at the center of the rod? We could just
do the integral over again, letting # range from —šL to +§1. But let us notice
a few things about the moment of inertia. We can imagine the rod as two rods,
cach of mass ă/2 and length 7/2; the moments of inertia of the two small rods
are equal, and are both given by the formula (19.5). Thherefore the moment of
Inertia 1s 2 2
]= 2(M/2)L/2)ˆ = HE (19.6)
'Thus it is much easier to turn a rod about its center, than to swing it around an
Of course, we could go on to compute the moments oŸ inertia of various other
bodies of interest. However, while such computations provide a certain amount of
important exercise in the calculus, they are not basically of interest to us as such.
'There is, however, an interesting theorem which is very useful. Suppose we have
an object, and we want to fnd its moment of inertia around some axis. hat
means we want the inertia needed to carry it by rotation about that axis. Now if
we support the object on pivots at the center of mass, so that the obJect does not
turn as iE rotates about the axis (because there is no torque on it from inertial
effects, and therefore it will not turn when we start moving it), then the forces
needed to swing it around are the same as though all the mass were concentrated
at the center of mass, and the moment of inertia would be simply 71 = MHệu,,
where em 1s the distance from the axis to the center of mass. But of course
that is not the right formula for the moment of inertia of an object which is really
beïng rotated as it revolves, because not only is the center of it moving ïn a circle,
which would contribute an amount 1¡ to the moment of inertia, but also we must
turn it about its center of mass. So it is not unreasonable that we must add to 1
the moment oŸ inertia ?„ about the center of mass. So it is a good guess that the