Question

A disc (of radius $r cm$ ) of uniform thickness and uniform density $\sigma$ has a square hole with sides of length $l=\frac{r}{\sqrt{2}} cm .$ One corner of the hole is located at the centre of the disc and centre of the hole lies on $y$ -axis as shown. Then the $y$ -coordinate of position of centre of mass of disc with hole (in $cm$ ) is

• A. $-\frac{r}{2(\pi-1 / 4)}$
• B. $-\frac{r}{4(\pi-1 / 4)}$
• C. $-\frac{r}{4(\pi-1 / 2)}$
• D. $-\frac{3 r}{4(\pi-1 / 4)}$

Answer 1

Answer: (C)

This disc can be assumed to be made of a complete uniform disc and a square plate with same negative mass density
$ Y_{c m} =\frac{m_{1} y_{1}+m_{2} y_{2}}{m_{1}+m_{2}}=\frac{\left(\pi r^{2}\right) \sigma(0)+\ell^{2}(-\sigma)(r / 2)}{\pi r^{2} \sigma+\ell^{2}(-\sigma)} $
$=\frac{-\ell^{2} r}{2\left(\pi r^{2}-\ell^{2}\right)}=\frac{-\frac{r^{3}}{2}}{2\left(\pi r^{2}-\frac{r^{2}}{2}\right)}=\frac{-r}{4\left(\pi-\frac{1}{2}\right)} $