Question

A square shaped hole of side $l=\frac{a}{2}$is carved out at a distance $d=\frac{a}{2}$from the centre ‘$O$’ of a uniform circular disk of radius $a$. If the distance of the centre of mass of the remaining portion from $O$ is– $\frac{a}{X}$, value of $X$ (to the nearest integer) is _________.

Answer 1

Step 1. Position of center of mass when some mass is removed is

$\overrightarrow{\mathbf{r}}\mathit{c}\mathit{m}\mathbf{=}\frac{{\mathbf{m}}_{\mathbf{1}}\overrightarrow{{\mathbf{r}}_{\mathbf{1}}}\mathbf{-}{\mathbf{m}}_{\mathbf{2}}\overrightarrow{{\mathbf{r}}_{\mathbf{2}}}}{{\mathbf{m}}_{\mathbf{1}}\mathbf{-}{\mathbf{m}}_{\mathbf{2}}}$

where, $m_1=$whole mass

$m_2=$removed mass

$\overrightarrow{r_1}=$position vector of center of mass for whole mass

$\overrightarrow{r_2}=$position vector of center of mass for removed mass

Step 2. Let the density be $\rho$

So, $m_1={\mathrm{\pi a}}^2\mathrm{\sigma }$

And $m_2=\frac{a^2}{4}\sigma$

Also, $r_1=0,r_2=\frac{a}{2}$

$\therefore$$X_{cm}=\frac{\left(\sigma \pi a^2\times 0\right)-\left(\sigma {\displaystyle \frac{a^2}{4}}\times {\displaystyle \frac{a}{2}}\right)}{\sigma \pi a^2-{\displaystyle \frac{\sigma a^2}{4}}}$

$\Rightarrow$$X_{cm}=\frac{{\displaystyle \frac{-a^3}{8}}}{\left(\mathrm{\pi }-{\displaystyle \frac{1}{4}}\right)a^2}$

$\Rightarrow$$X_{cm}=\frac{-a}{2\left(4\mathrm{\pi }-1\right)}$

$\Rightarrow$$X_{cm}=\frac{-a}{8\mathrm{\pi }-2}$

$\Rightarrow$$X_{cm}=-\frac{a}{23}$

Thus, the value of $\mathit{X}\mathbf{=}\mathbf{23}$