Question

The figure shows a disc of radius $3R$ from which a circular hole of radius $R$ is cut as shown in the figure. The distance of the centre of mass of the remaining object from the point $O$ is

• A. $R/4$
• B. $R/5$
• C. $R/3$
• D. $R/6$

Answer 1

Answer: (A)

The centre of mass of the remaining object is
$x_{cm}=\frac{m_1{x}_1-m_2{x}_2}{m_1-m_2}=\frac{\sigma A_1{x}_1-\sigma A_2{x}_2}{\sigma A_1-\sigma A_2}$
where $m_1$ and $A_1$ are the mass and area of the total disc and $m_2$ and $A_2$ are the mass and area of the cut-out portion. $\sigma$ is the mass per unit length of the disc.
$A_1=\pi (3R^2$ , $A_2=\pi R^2$
$x_1=0$ , $x_2=2R$
$\Rightarrow x_{cm}=-R/4$