TEST BANK FOR Dynamics with Dynamics of Rigid Bodies By S L Loney (Solution Manual)

If r be the perpendicular distance from any given line of any
element m of the mass of a body, then the quantity åmr2 is called
the moment of inertia of the body about the given line.
In other words, the moment of inertia is thus obtained; take each
element of the body, multiply it by the square of its perpendicular
distance from the given line; and add together all the quantities thus
obtained.
If this sum be equal to Mk2, where M is the total mass of the body,
then k is called the Radius of Gyration about the given line. It has
sometimes been called the Swing-Radius.
If three mutually perpendicular axes Ox;Oy;Oz be taken, and if the
coordinates of any element m of the system referred to these axes be
x;y and z, then the quantities åmyz;åmzx; and åmxy are called the
products of inertia with respect to the axes y and z; z and x; and x
and y respectively.
Since the distance of the element from the axis of x is
p
y2+z2;
the moment of inertia about the axis of x
=åm(y2+z2):
1
2 Chapter 11: Moments and Products of Inertia: Principal Axes
145. Simple cases of Moments of Inertia.
I. Thin uniform rod of mass M and length 2a: Let AB be the rod,
and PQ any element of it such that AP = x and PQ =d x: The mass
of PQ is
d x
2a
:M:
Hence the moment of inertia about an axis through A perpendicular
to the rod
=åd x
2a
:M:x2 =
M
2a
Z 2a
0
x2dx =
M
2a
:
1
3
[2a]3 = M:
Ma2
3
:
Similarly, if O be the centre of the rod, OP = y and PQ = d y; the
moment of inertia of the rod about an axis through O perpendicular
to the rod
=åd y
2a
:M:y2 =
M
2a
Z +a
¡a
y2dy =
M
2a
:
1
3
[y3]+a
¡a = M:
a2
3
:
II. Rectangular lamina. Let ABCD be the lamina, such that AB = 2a
and AD=2b; whose centre is O: By drawing a large number of lines
parallel to AD we obtain a large number of strips, each of which is
ultimately a straight line.
The moment of inertia of each of these strips about an axis through
O parallel to AB is (by I) equal to its mass multiplied by
b2
3
: Hence
the sum of the moments of the strips, i.e. the moment of inertia of
the rectangle, about the same line is M
b2
3
:
So its moment of inertia, about an axis through O parallel to the
side 2b; is M
a2
3
:
If x and y be the coordinates of any point P of the lamina referred
to axes through O parallel to AB and AD respectively, these results
give
LONEY’S DYNAMICS OF RIGID BODIES WITH SOLUTION MANUAL (Kindle edition) 3
åmy2 = moment of inertia about Ox = M
b2
3
and åmx2 = M
a2
3
:
The moment of inertia of the lamina about an axis through O perpendicular
to the lamina
=åm:OP2 =åm(x2+y2) = M
a2+b2
3
:
III. Rectangular Parallelepiped. Let the lengths of its sides be
2a;2b; and 2c: Consider an axis through the centre parallel to the
side 2a; and conceive the solid as made up of a very large number
of thin parallel rectangular slices all perpendicular to this axis;
each of these slices has sides 2b and 2c and hence its moment of
inertia about the axis is its mass multiplied by
b2+c2
3
: Hence the
moment of inertia of the whole body is the whole mass multiplied
by
b2+c2
3
; i:e:; M
b2+c2
3
:
IV. Circumference of a circle. Let Ox be any axis through the centre
O, P any point of the circumference such that xOP = q ;PQ an
element adq ; then the moment of inertia about Ox
=å
·
adq
2pa
¸
a2 sin2q =
Ma2
2p
Z 2p
0
sin2q dq
=4£
Ma2
2p
Z 2=p
0
sin2q dq =
2Ma2
p :
1
2
:
p
2
= M
a2
2
:
V. Circular disc of radius a. The area contained between concentric
circles of radii r and r +d r is 2prd r and its mass is thus
2prdr
pa2 :M; its moment of inertia about a diameter by the previous
article =
2rdr
a2 M:
r2
2
:
4 Chapter 11: Moments and Products of Inertia: Principal Axes
Hence the required moment of inertia
=
M
a2
Z a
0
r3dr =
M
a2 :
a4
4
= M:
a2
4
:
So for the moment about a perpendicular diameter.
The moment of inertia about an axis through the centre perpendicular
to the disc = (as in II) the sum of these = M:
a2
2
:
Elliptic disc of axes 2a;2b: Taking slices made by lines parallel to
the axis of y, the moment of inertia about the axis of x clearly
=2
Z 0
p=2
·
2bsinf d(acosf )
pab
M
¸
:
b2 sin2f
3
=
4
3
Mb2
p
Z p=2
0
sin4f df
=M:
b2
4
:
So the moment of inertia about the axis of y = M:
a2
4
:
VI. Hollow sphere. Let it be formed by the revolution of the circle
of IV about the diameter. Then the moment of inertia about the
diameter
=å
·
adq :2pasinq
4pa2 M
¸
a2 sin2q
=
Ma2
2
Z p
0
sin3q :dq
=2:
Ma2
2
:
2
3:1
=
2Ma2
3
:
VII. Solid sphere. The volume of the thin shell included between
spheres of radii r and r+d r is 4pr2d r; and hence its mass is
LONEY’S DYNAMICS