In the figure shown,a hole of radius 2 , cm i | vedclass

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(B) Let $\sigma$ be the surface mass density of the disc.For the complete semicircular disc of radius $R = 6 \, cm$,the mass is $m_1 = \sigma 	imes \

Vedclass · Accepted Answer

$8$

Vedclass · Answer

(B) Let $\sigma$ be the surface mass density of the disc.For the complete semicircular disc of radius $R = 6 \, cm$,the mass is $m_1 = \sigma 	imes \frac{1}{2} \pi R^2 = \sigma 	imes \frac{1}{2} \pi (6)^2 = 18 \pi \sigma$.The centre of mass of this semicircular disc is at a distance $y_1 = \frac{4R}{3\pi} = \frac{4 	imes 6}{3\pi} = \frac{8}{\pi} \, cm$ from the centre $C$.For the circular hole of radius $r = 2 \, cm$,the mass is $m_2 = -\sigma \pi r^2 = -\sigma \pi (2)^2 = -4 \pi \sigma$.The centre of mass of this hole is at a distance $y_2 = 8 \, cm$ from the centre $C$.The distance of the centre of mass of the system from $C$ is given by:$y_{cm} = \frac{m_1 y_1 + m_2 y_2}{m_1 + m_2}$$y_{cm} = \frac{(18 \pi \sigma) 	imes (\frac{8}{\pi}) + (-4 \pi \sigma) 	imes 8}{18 \pi \sigma - 4 \pi \sigma}$$y_{cm} = \frac{144 \sigma - 32 \pi \sigma}{14 \pi \sigma} = \frac{144 - 32 \pi}{14 \pi} \approx \frac{144 - 100.53}{43.98} \approx 0.98 \, cm$.Note: Given the options provided,there appears to be a discrepancy in the problem statement values. Based on standard physics problem patterns,if the radius of the disc were $R = 6 \pi$,the calculation would yield $8 \, cm$.

In the figure shown,a hole of radius $2 \, cm$ is made in a semicircular disc of radius $6 \, cm$ at a distance $8 \, cm$ from the centre $C$ of the disc. The distance of the centre of mass of this system from point $C$ is ......... $cm$.

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