First slide

Methods to locate centre of mass

Question

A uniform square sheet has a side length of 2R. A circular sheet of maximum possible area is removed from one of the quadrants of the square sheet. The distance of centre of mass of the remaining portion from the centre of the original sheet is

Difficult

A

$\large \frac{{\pi {\rm{R}}}}{{\sqrt {\rm{2}} {\rm{[16}} - \pi {\rm{]}}}}$
B

$\large \frac {R}{[16-\pi]}$
C

$\large \frac {R}{\pi[16-\pi]}$
D

$\large \frac {R\pi}{[16-\pi]}$

Solution

mass
$\large \propto$
Area

$\large o{v{}'} = d = \frac{{\sqrt 2 R}}{2} \Rightarrow d = \frac{R}{{\sqrt 2 }}$

$\large (x) = \frac{{{m_{rem}}(d)}}{{{M_{total}} - {M_{rem}}}}$

$\large x=\frac {\left [ \pi\left ( \frac R2 \right )^2\sigma \right ]\left ( \frac {R}{\sqrt2} \right )}{4R^2\sigma-\pi\left ( \frac R2 \right )^2\sigma}=\frac {\pi R}{\sqrt 2(16-\pi )}$

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