Question

From a uniform circular disc of mass $M$ and radius $R$ a small circular disc of radius $R/2$ is removed in such a way that both have a common tangent. Find the distance of centre of mass of remaining part from the centre of original disc.

Answer 1

Let $\sigma$ is mass per unit area of the circular disc.

Given $\sigma \times \pi R^2=M$ $\Rightarrow \sigma =\frac{M}{\pi R^2}$,

Now mass of removed disc$=\sigma \times \frac{\pi R^2}{4}=\frac{M}{4}$

Now calculate the centre of mass from the Point $A$ as shown in figure.

$X_{cm}=\frac{M\times R-\frac{M}{4}\times \frac{R}{2}}{M-\frac{M}{4}}=\frac{7}{6}R$

Hence, Centre of mass from the centre of disc$=\frac{7}{6}R-R=\frac{1}{6}R$ right to the centre.