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. $\frac{\text{R(} 2 - \pi \left.\right)}{2 \left(3 \pi + 2\right)}$
• B. $\frac{\text{R(} 1 - \pi \left.\right)}{2 \left(2 \pi + 1\right)}$
• 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 $=R cos 4 5^{^\circ } = \frac{\text{R}}{\sqrt{2}}$
Area of square $= \frac{\text{R}^{2}}{2}$
$\text{X}\left(\text{COM}\right)=\frac{\left(\pi \times \left(\text{R}\right)^{2} \times \sigma \times 0 + \pi \times \frac{\left(\text{R}\right)^{2}}{4} \left(- \sigma \right) \times \frac{\text{R}}{2} + \frac{\left(\text{R}\right)^{2}}{2} \times \sigma \times \frac{\text{R}}{2}\right)}{\left(\pi \times \left(\text{R}\right)^{2} \times \sigma + \pi \times \frac{\left(\text{R}\right)^{2}}{4} \left(- \sigma \right) + \frac{\left(\text{R}\right)^{2}}{2} \times \sigma \right)}$
$=\frac{\text{R(} 2 - \pi \left.\right)}{2 \left(\right. 3 \pi + 2 \left.\right)}$
$\therefore $ The centre of mass of the system is at a distance of $\frac{\text{R(} 2 - \pi \left.\right)}{2 \left(\right. 3 \pi + 2 \left.\right)}$ from the centre $O$ towards the plate as shown in the figure