Question

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

• A. $R/20$
• B. $R/16$
• C. $R/6$
• D. $\frac{3}{4}R$

Answer 1

The correct option is C $R/6$

Let the mass of the disc $+M$

Therefore, the mass of the removed part of disc, $m=\left(\frac{M}{R^2}\right)\times {\left(\frac{R}{2}\right)}^2=\frac{M}{4}$

Now, the center of gravity of the resulting flat body.

$\begin{array}{c}R=\left[\frac{M\times 0-\left(\frac{M}{4}\right)\times \left(\frac{R}{2}\right)}{\left(M-\frac{M}{4}\right)}\right]\\ =-\frac{\left(\frac{MR}{8}\right)}{\left(\frac{3M}{4}\right)}\\ =\frac{-R}{6}\end{array}$

Negative sin shows that the center of garavity lies at opposite direction of original $COM$ at a distance $R/6$.