Two semicircular rings of linear mass densiti | vedclass

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(B) Let the mass of the semicircular ring with density $\lambda$ be $M_1 = \lambda (\pi R) = M$. Then,the mass of the semicircular ring with density $

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$\frac{R}{\pi}$

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(B) Let the mass of the semicircular ring with density $\lambda$ be $M_1 = \lambda (\pi R) = M$. Then,the mass of the semicircular ring with density $3\lambda$ is $M_2 = 3\lambda (\pi R) = 3M$. The center of mass of a semicircular ring of radius $R$ is at a distance $\frac{2R}{\pi}$ from its center along the axis of symmetry. Let the geometrical center be at the origin $(0,0)$. Place the first ring (mass $M$) in the left half,so its center of mass is at $x_1 = -\frac{2R}{\pi}$. Place the second ring (mass $3M$) in the right half,so its center of mass is at $x_2 = \frac{2R}{\pi}$. The center of mass of the combined system is $X_{cm} = \frac{M_1 x_1 + M_2 x_2}{M_1 + M_2}$. $X_{cm} = \frac{M(-\frac{2R}{\pi}) + 3M(\frac{2R}{\pi})}{M + 3M} = \frac{\frac{4MR}{\pi}}{4M} = \frac{R}{\pi}$. Thus,the distance from the geometrical center is $\frac{R}{\pi}$.

Two semicircular rings of linear mass densities $\lambda$ and $3\lambda$ and of radius $R$ each are joined to form a complete ring. The distance of the center of mass of the complete ring from its geometrical center is

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