From a uniform circular disc of radius 2 ,cm | vedclass

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(D) Let $R = 2 \,cm$ be the radius of the original disc and $r = 1 \,cm$ be the radius of the removed portion.The area of the original disc is $A_1 =

Vedclass · Accepted Answer

$120$

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(D) Let $R = 2 \,cm$ be the radius of the original disc and $r = 1 \,cm$ be the radius of the removed portion.The area of the original disc is $A_1 = \pi R^2 = 4\pi \,cm^2$ and the area of the removed portion is $A_2 = \pi r^2 = \pi \,cm^2$.Let $\sigma$ be the surface mass density. The mass of the original disc is $M = \sigma A_1 = 4\sigma\pi$ and the mass of the removed portion is $m = \sigma A_2 = \sigma\pi$.The remaining mass is $M' = M - m = 3\sigma\pi = \frac{3}{4}M$.To maximize the shift in the centre of mass, the removed portion must be tangent to the original disc at the edge. The distance of the centre of the removed portion from $O$ is $d = R - r = 2 - 1 = 1 \,cm$.The shift in the centre of mass $x$ is given by $x = \frac{m d}{M'} = \frac{(\sigma\pi)(1)}{3\sigma\pi} = \frac{1}{3} \,cm$.When the disc is rotated by an angle $	heta$, the new centre of mass moves along a circular arc of radius $x = \frac{1}{3} \,cm$. The displacement $PQ$ between the initial and final positions of the centre of mass is given as $\frac{1}{\sqrt{3}} \,cm$.In the isosceles triangle formed by the centre $O$ and the two positions of the centre of mass, the angle at $O$ is $	heta$. Using the chord length formula $PQ = 2x \sin(\frac{	heta}{2})$:$\frac{1}{\sqrt{3}} = 2(\frac{1}{3}) \sin(\frac{	heta}{2})$$\sin(\frac{	heta}{2}) = \frac{3}{2\sqrt{3}} = \frac{\sqrt{3}}{2}$$\frac{	heta}{2} = 60^{\circ} \Rightarrow 	heta = 120^{\circ}$.

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