Question

A circular hole of radius $\frac{R}{2}$ is cut from a circular disc of radius $R$. The disc lies in the XY-plane and its centre coincides with the origin. If the remaining mass of the disc is $M$ then,

(1)determine the initial mass of the disc.

(2)determine its moment of inertia about the z-axis.

Answer 1

(A) Let, surface density of the disc be=$\sigma$

It can be written as, $\sigma =\frac{M}{\pi R^2-\pi {\left(\frac{R}{2}\right)}^2}=\frac{4}{3}\frac{M}{\pi R^2}$

So, initial mass of the disc is $m_1=\sigma \left(\pi R^2\right)=\frac{4}{3}M$

Mass of the disc, that removed, $m_2=\sigma \left(\pi \frac{R^2}{4}\right)=\frac{M}{3}$

(B) Moment of inertia of the disc, $I=\frac{m1R^2}{2}$

Moment of inertia of the smaller disc of radius $\frac{R}{2}$ about $O^{/}$ and parallel to z-axis is,

$I_2=\frac{m_2{\left(\frac{R}{2}\right)}^2}{2}=\frac{m_2{R}^2}{8}$

(Using parallel theorem), the moment of inertia of smaller disc about O,

$I_{O_2}=\frac{m2R^2}{8}+\frac{m2R^2}{4}=\frac{3m2R^2}{8}$

As the part of the smaller disc of mass $m_2$ is removed from the disc $m_1$.

Hence the net moment of inertia is $I_o=I-I_{O_2}$

$I=\frac{m1R^2}{2}-\frac{3m2R^2}{8}$

Substituting the value of $m_1$ and $m_2$ from (A), in the above equation,

$I=\frac{13}{24}M{R}^2$