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Q.

A disc of radius $R$ is cut off from a uniform thin sheet of metal. A circular hole of radius $\frac{R}{2}$ is now cut out from the disc, with the hole being tangent to the rim of the disc. Find the distance of the centre of mass from the centre of the original disc.

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a

$\frac{R}{6}$

b

$\frac{R}{4}$

c

$\frac{R}{2}$

d

$\frac{R}{3}$

answer is D.

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Detailed Solution

Mass of original uncut circle $m_{1} = m$ & Location of $C M = (0, 0)$

Mass of hole of negative mass : $m_{2} = \frac{m}{4}$ ; Location of $C M = (0, \frac{R}{2})$

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Thus, $y_{C M} = \frac{m_{1} y_{1} + m_{2} y_{2}}{m_{1} + m_{2}} = \frac{m (0) + (- \frac{m}{4}) \frac{R}{2}}{m + (- \frac{m}{4})} = - \frac{R}{6}$

So the centre of mass is at the point $(0, - \frac{R}{6})$ .

Thus, the required distance is $\frac{R}{6}$ .

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A disc of radius R is cut off from a uniform thin sheet of metal. A circular hole of radius R2 is now cut out from the disc, with the hole being tangent to the rim of the disc. Find the distance of the centre of mass from the centre of the original disc.