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(B) Let the mass of the semi-circular ring $ADC$ be $m$. Then the mass of the semi-circular ring $ABC$ is $3m$.The center of mass of a semi-circular r

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$\frac{1}{\pi}$

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(B) Let the mass of the semi-circular ring $ADC$ be $m$. Then the mass of the semi-circular ring $ABC$ is $3m$.The center of mass of a semi-circular ring of radius $R$ is at a distance $\frac{2R}{\pi}$ from its diameter.Let $A$ be the origin $(0,0)$. The center of mass of $ADC$ (mass $m$) is at $x_1 = \frac{2R}{\pi}$ and the center of mass of $ABC$ (mass $3m$) is at $x_2 = -\frac{2R}{\pi}$.The $x$-coordinate of the center of mass of the system is $X_{com} = \frac{m_1 x_1 + m_2 x_2}{m_1 + m_2} = \frac{m(\frac{2R}{\pi}) + 3m(-\frac{2R}{\pi})}{m + 3m} = \frac{-\frac{4mR}{\pi}}{4m} = -\frac{R}{\pi}$.In equilibrium,the center of mass lies directly below the hinge $A$. If $	heta$ is the angle the line $AC$ makes with the vertical,then $	an 	heta = \frac{|X_{com}|}{Y_{com}}$. Since the vertical distance $Y_{com}$ of the center of mass from $A$ is $R$,we have $	an 	heta = \frac{R/\pi}{R} = \frac{1}{\pi}$.

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