Question

A circular hole of radius $1 \, cm$ is cut off from a disc of radius $6 \, cm$ . The centre of the hole is $3 \, cm$ from the centre of the disc. Then the distance of the centre of mass of the remaining disc from the centre of the disc is

• A. $\frac{3}{35}cm$
• B. $\frac{1}{35}cm$
• C. $\frac{3}{10}cm$
• D. none of these

Answer 1

Answer: (A)

For the calculation of the position of centre of mass, cut off mass is taken as negative. The mass of disc is

$\text{m}_{1} = \pi \text{r}_{1}^{2} \sigma $ $ = \pi \text{6}^{2} \sigma = \text{36} \pi \sigma $
where, $\sigma $ = surface mass density. The mass of cut out portion is
$ \left(\text{m}\right)_{\text{2}} = \pi \left(\text{1}\right)^{2} \sigma = \pi \sigma $
$ \text{x}_{\text{cm}} = \frac{\text{m}_{1} \text{x}_{\text{1}} - \text{m}_{2} \text{x}_{2}}{m_{1} - m_{2}}$
Taking origin at the centre of disc,
$x_{1}=0,x_{2}=3cm$
$\text{x}_{\text{cm}} = \frac{\text{36} \pi \sigma \times 0 - \pi \sigma \times 3}{3 6 \pi \sigma - \pi \sigma }$
$=\frac{- 3 \pi \sigma }{3 5 \pi \sigma }=-\frac{3}{3 5}\text{cm}$