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)

$x_{cm}=\frac{-m_1\times \left(-\frac{R}{2}\right)+m_2\times 0+m_3\frac{R}{2}}{-m_1+m_2+m_3}$
$x_{cm}=\frac{\sigma \frac{\pi R^2}{4}\times \frac{R}{2}+\sigma \frac{\pi R^2}{4}\times \frac{R}{2}}{-\sigma \frac{\pi R^2}{4}+\sigma \pi R^2+\sigma \frac{\pi R^2}{4}}$
$=\frac{R}{8}+\frac{R}{8}=\left(\frac{R}{4},0\right)$