Question

In the figure a uniform disc of radius $R$, from which a hole of radius $R / 2$ has been cut out from left of the centre and is placed on right of the centre of disc. Find the centre of mass of the resulting disc.

• A. $\frac{ R }{3}, 0$
• B. $\frac{R}{2}, 0$
• C. $\frac{ R }{4}, 0$
• D. $R , 0$

Answer 1

Answer: (C)

$x_{c m}=\frac{-m_{1} \times\left(-\frac{R}{2}\right)+m_{2} \times 0+m_{3} \frac{R}{2}}{-m_{1}+m_{2}+m_{3}}$
$x_{c m}=\frac{\sigma \frac{\pi R^{2}}{4} \times \frac{R}{2}+\sigma \frac{\pi R^{2}}{4} \times \frac{R}{2}}{-\sigma \frac{\pi R^{2}}{4}+\sigma \pi R^{2}+\sigma \frac{\pi R^{2}}{4}}$
$=\frac{R}{8}+\frac{R}{8}=\left(\frac{R}{4}, 0\right)$