(^626) „„„„„ A Textbook of Engineering Mechanics

3
0
2
r
mxdxπ∫

44
0
2
42
r
xmr
m
⎡⎤ π
π=⎢⎥
⎣⎦

2
0.5^2
2
Mr
= Mr ...(Q M = πr^2 m)
Now as per theorem of perpendicular axis, the mass moment of inertia about X-X or Y-Y axis
IXX=
2
0.5 0.25 2
22
ZZ
YY
I Mr
IMr== =
Note: The above formula holds good for the mass, moment of inertia of a solid cylinder
also. In this case, M is taken as the mass of the solid cylinder.
31.11.MASS MOMENT OF INERTIA OF A SOLID SPHERE
Consider a solid sphere of radius r with O as centre.
Let m= Mass per unit volume of the sphere
∴ Total mass of the sphere
M=
4 3
3
mrπ
Now consider an elementary plate PQ of thickness dx and at a distance x from O as shown
in Fig. 31.6.
We know that the radius of this plate
y = rx^22 –
∴ Mass of this plate = mπy^2 dx = mπ (r^2 – x^2 ) dx
and moment of inertia of this plate about X-X axis

(Radius)^2
Mass ×
2

22
(–)^22 (–)
2
rx
mr xdxπ×
= (–)^222
2
m
rxdx
π
= (–2)44 22
2
m
rx rxdx
π

• The mass moment of inertia of the whole sphere may now be found out by integrating the
  above equation from – r to + r. Therefore
  I =
  44 22

(–2)

2

r

r

m

rx rxdx

+

π

∫ +

=

44 22

(–2)

2

r

r

m

rx rxdx

+

π

∫ +

Fig. 31.6. Sphere.