# Q7.16     From a uniform disk of radius $R$ , a circular hole of radius $R/2$  is cut out. The centre               of the hole is at $R/2$ from the centre of the original disc. Locate the centre of gravity              of the resulting flat body.

Let the mass per unit area of the disc be $\sigma$.

So total mass is   $=\ \Pi r^2\sigma\ =\ m$

Mass of the smaller disc is given by :

$=\ \Pi \left ( \frac{r}{2} \right )^2\sigma\ =\ \frac{m}{4}$

Now since the disc is removed, we can assume that part to have negative mass with respect to the initial condition.

So, the centre of mass of the disc is given by the formula :

$x\ =\ \frac{m_1r_1\ +\ m_2r_2}{m_1\ +\ m_2}$

or                                                                    $=\ \frac{m\times 0\ -\ \frac{m}{4}\times \frac{r}{2}}{m\ -\ \frac{m}{4}}$

$=\ \frac{-r}{6}$

Hence the centre of mass is shifted   $\frac{r}{6}$   leftward from point O.

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