Question 297: 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 :

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