First slide

Location of centre on mass for multiple point masses

Question

From a circular disc of radius R, a square is cut out with a radius as its diagonal. The centre of mass of remaining portion is at a distance (from the centre)

Moderate

A

$\frac{R}{(4 π - 2)}$
B

$\frac{R}{(2 π)}$
C

$\frac{R}{(π - 2)}$
D

$\frac{R}{(2 π - 2)}$

Solution

Here, $A_{1} ({CC}_{1}) = A_{2} ({CC}_{2})$
where, A = area of square and A₂ = area of remaining portion.
Side of square will be $\frac{R}{\sqrt{2}}$
$∴ C C_{2} = \frac{A_{1}}{A_{2}} (C C_{1}) = \frac{(R / \sqrt{2})^{2}}{π R^{2} - (R / \sqrt{2})^{2}} (\frac{R}{2}) = (\frac{R}{4 π - 2})$

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