Question 298: Centre of Mass and Rotational Motion

Question

A square shaped hole of side l = ${a \over 2}$ is carved
out at a distance d = ${a \over 2}$ from the centre ‘O’ of
a uniform circular disk of radius a. If the
distance of the centre of mass of the
remaining portion form O is
$ - {a \over X}$ , value of X (to
the nearest integer) is :
![](https://imagex.cdn.examgoal.net/jgy2xukexykx2z/badf58e4-51c5-4ddd-9ddb-fd03c8944954/f52ea4b0-f405-11ea-8293-3b6cc591631c/file-jgy2xukexykx30-352w.png)

Accepted Answer

X
com
= ${{\sigma \pi {a^2} 	imes 0 - \sigma .{{{a^2}} \over 4}.d} \over {\sigma \pi {a^2} - \sigma .{{{a^2}} \over 4}}}$
= ${{ - {{{a^2}} \over 4}.d} \over {\pi {a^2} - {{{a^2}} \over 4}}}$
= ${{ - d} \over {4\pi  - 1}}$ = ${{ - a} \over {2\left( {4\pi  - 1} ight)}}$
$ 	herefore $ $ - {a \over X}$ = ${{ - a} \over {2\left( {4\pi  - 1} ight)}}$
$ \Rightarrow $ X = ${2\left( {4\pi  - 1} ight)}$ = ${\left( {8\pi  - 2} ight)}$ = 23.12
So, nearest integer value of X = 23

Centre of Mass and Rotational Motion

A square shaped hole of side l = {a \over 2} is carved out at a distance d = {a \over 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 form O is - {a \over X} , value of X (to the nearest integer) is :

Integer / Numeric Answer Type