Question

A cavity of radius b is made in a disc of mass M, radius R, as shown in Fig. Find the new COM :

• A. $\frac{-b^2}{R+b}$
• B. $\frac{-b^2}{R-b}$
• C. $\frac{-R}{2R+b}$
• D. $\frac{-R}{3R+b}$

Answer 1

The correct option is A $\frac{-b^2}{R+b}$

Let the mass per unit area of ring be $\gamma$

First we need to calculate the mass removed:

$m^{\prime }=\gamma \left(AreaRemoved\right)=\gamma \left(\pi b^2\right)=\frac{M}{\pi R^2}\left(\pi b^2\right)=M\frac{b^2}{R^2}$

COM of the part cut is at $(R-b)$ along X-axis.

New Center of mass is given by:

$x^{\prime }=\frac{M\left(0\right)-m^{\prime }(R-b)}{M-m^{\prime }}=\frac{-M\frac{b^2}{R^2}(R-b)}{M-M\frac{b^2}{R^2}}=\frac{-b^2(R-b)}{R^2-b^2}=\frac{-b^2}{R+b}$