Question

A flat thin circular disc has a radius $4$ cm and circular hole of radius $\frac{1}{2}$ cm is made in it with its centre at a distance of $1$ cm from the centre of the disc. The mass of the disc is $10$ kg. If the moment of inertia of the system about an axis passing through the centre of the hole is $n\times 10$${}^{-3}$ kg-m${}^2$, find n.

• A. 9
• B. 4
• C. 16
• D. 81

Answer 1

The correct option is A 9

The moment of inertia of the whole disc about the axis of rotation through O'
$I_1=\frac{1}{2}M{a}^2+M{c}^2$
The moment of inertia of the disc removed portion is
$I_2=\frac{1}{2}m{b}^2=\frac{1}{2}M\frac{b^2}{a^2}{b}^2$
$(\because m=\frac{M}{\pi a^{2}p}.\pi b^{2}p)=\frac{1}{2}\frac{M{b}^4}{a^2}$
Moment of inertia of the system
$1=I_1-1_2=M[\frac{a^2}{2}+c^2-\frac{b^4}{2a^e}]$
$=10[\frac{16}{2}\times {10}^{-4}+1\times 104-\frac{105}{2\times 16\times 16}\times {10}^4]$
= $10(8\times 10$${}^{-4}$ + $10$${}^{-4}$ $-0.002\times 10$${}^{-4}$)
= $9\times 10$${}^{-3}$ kg-m${}^2$ $\Rightarrow$ $n=9$