Chapter 31 : Kinetics of Motion of Rotation „„„„„ 625

Fig. 31.5. Circular lamina.

The mass moment of inertia of the whole rod may be found out by integrating the above
equation for the whole length of the rod i.e. from 0 to 2l. Therefore

I=

2 332

2

(^00)
(2 )
–0
33
l l
xl
mx dx m m
⎡⎤ ⎡ ⎤
==⎢⎥ ⎢ ⎥
⎣⎦ ⎣ ⎦
∫

8432
33
ml Ml
= ...(QM = 2ml)
31.9.MASS MOMENT OF INERTIA OF A THIN CIRCULAR RING
Consider a thin circular ring of radius r and O as centre as shown in Fig. 31.4.
Let m= Mass per unit length of the ring.
∴ Total mass of the ring
M=2 πrm
Now consider a small element of length dx as shown in
Fig. 31.4. We know that mass moment of inertia of the strip about the
central axis
=(m dx) r^2
The mass moment of inertia of the whole ring about central
axis (i.e. at right angles to the plane)
IZZ= (2 πrm) r^2 = Mr^2
...(Q M = 2πrm)
Now as per theorem of perpendicular axis, the mass moment
of inertia about X-X or Y-Y axis
IXX=
2
0.5^2
22
ZZ
YY
I Mr
IMr== =
31.10.MASS MOMENT OF INERTIA OF A CIRCULAR LAMINA
Consider a circular lamina of radius r with O as centre.
Let m = Mass per unit area of the lamina.
∴ Total mass of the lamina M = πr^2 m
Now consider an elementary ring of thickness dx at a radius x as shown in Fig. 31.5. We know
that mass of this thin ring
=m 2 πx dx
We know that the mass moment of inertia of a thin ring
(of mass m.2πx.dx) about central axis
=(m 2 πx dx) x^2
=m 2 πx^3 dx
The mass moment of inertia of the whole section about the
central axis, may be found out by integrating the above equation
for the whole radius of the circle, i.e. from 0 to r. Therefore
IZZ=
3
0
2
r
∫mxdxπ
Fig. 31.4. Circular ring.