Question

A circular disc of radius R has a uniform thickness. A circular hole of diameter equal to the radius of the disc has been cut out as shown in figure. Find the centre of mass of the remaining disc.

• A. $\frac{R}{6}$ units towards left
• B. $\frac{R}{6}$ units towards Right
• C. $\frac{R}{3}$ units towards left
• D. $\frac{R}{3}$ units towards right

Answer 1

The correct option is A

$\frac{R}{6}$ units towards left

If $\sigma$ be the mass/unit area of the disc, Mass of the whole disc, $m_1=\pi R^2\sigma$
Radius of the hole = $\frac{R}{2}$
Mass of the hole, $m_2$ = $\pi (\frac{R}{2}{)}^{2}P$
Distance of centre of mass of hole from O, i.e., $x_2$ = $\frac{R}{2}$
Since the hole has been taken out, its mass is taken as negative.
If O is the origin, $x_{cm}$ = $\frac{m_1{x}_1+m_2{x}_2}{m_1+m_2}$ = $\frac{\left(\pi R^2\sigma \right)0-\left(\pi \right(\frac{R}{2}{)}^2\sigma \left)\right(\frac{R}{2})}{\pi R^2\sigma -\pi (\frac{R}{2}{)}^2}\sigma$ = $\frac{-\pi R^3\frac{\sigma }{8}}{\left(\frac{3}{4}\right)\pi R^2\sigma }$ = $\frac{-R}{6}$
The negative sign indicates that the CM is to be the left of O.