Question

There is a thin uniform disc of radius $R$ and mass per unit area $\sigma$ , in which a hole of radius $R/2$ has been cut out as shown in the figure. Inside the hole, a square plate of same mass per unit area $\sigma$ is inserted so that its corners touch the periphery of the hole. The distance of the centre of mass of the system from the origin is

• A. $\frac{\text{R(}2-\pi )}{2\left(3\pi +2\right)}$
• B. $\frac{\text{R(}1-\pi )}{2\left(2\pi +1\right)}$
• C. $\frac{2\text{R}\pi }{2\left(3\pi +2\right)}$
• D. $\frac{3\text{R}\pi }{2\left(2\pi +1\right)}$

Answer 1

Answer: (A)

Side of square $=Rcos45^{{}^{\circ }}=\frac{\text{R}}{\sqrt{2}}$
Area of square $=\frac{{\text{R}}^2}{2}$
$\text{X}\left(\text{COM}\right)=\frac{\left(\pi \times {\left(\text{R}\right)}^2\times \sigma \times 0+\pi \times \frac{{\left(\text{R}\right)}^2}{4}\left(-\sigma \right)\times \frac{\text{R}}{2}+\frac{{\left(\text{R}\right)}^2}{2}\times \sigma \times \frac{\text{R}}{2}\right)}{\left(\pi \times {\left(\text{R}\right)}^2\times \sigma +\pi \times \frac{{\left(\text{R}\right)}^2}{4}\left(-\sigma \right)+\frac{{\left(\text{R}\right)}^2}{2}\times \sigma \right)}$
$=\frac{\text{R(}2-\pi )}{2(3\pi +2)}$
$\therefore$ The centre of mass of the system is at a distance of $\frac{\text{R(}2-\pi )}{2(3\pi +2)}$ from the centre $O$ towards the plate as shown in the figure