Question

A circular hole of radius $R/2$ 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
(a) determine the initial mass of the disc, and
(b) determine its moment of inertia about the z-axis.

Answer 1

Let $\sigma$ be the surface mass density of the plate given by $\sigma =\frac{M}{\pi R^2-\pi {\left(\frac{R}{2}\right)}^2}=\frac{4}{3}\frac{M}{\pi R^2}$

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

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

The given arrangement may be considered as a combination of a circular plate of mass $\left(M/3\right)$

The moment of inertia of the big disc of radius $R$ and mass $m_1$ about $O$ and parallel to the z-axis is

$I_{o_1}=\frac{m_1{R}^2}{2}$

The moment of inertia of the small disc of mass $m_2$ and radius $R/2$ about $O^{\prime }$ and parallel to z-axis

$I_{o_1}=\frac{m_2{\left(R/2\right)}^2}{2}=\frac{m_2{R}^2}{8}$

The moment of inertia of the small disc about $O$ (using parallel axis theorem)

$I_{o_1}=\frac{m_2{R}^2}{8}+\frac{m_2{R}^2}{4}=\frac{{3m}_2{R}^2}{8}$

As the part of the disc of mass $m_2$ is removed from disc $m_1$, hence the net moment of inertia of structure is

$I_o=I_{o_1}-I_{o_2}=\frac{m_1{R}^2}{2}-\frac{{3m}_2{R}^2}{8}$

Substituting the values of $m_1$ and $m_2$, we get

$I_0=\frac{13}{24}M{R}^2$