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ABC is a part...

Question

$A B C$ is a part of ring having radius $R_{2}$ and $A D C$ is a part of disc having inner radius $R_{1}$ and outer $R_{2}$ . Part $A B C$ and $A D C$ have same mass. Then center of mass will be located, from the centre $O$ .

\frac{(R_{2} - R_{1}) (2 R_{1} + R_{2})}{3 π (R_{1} + R_{2})}

(above)

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\frac{(R_{2} - R_{1}) (2 R_{1} + R_{2})}{3 π (R_{1} + R_{2})}

(below)

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\frac{2 R_{1} + R_{2}}{3 π}

(above)

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\frac{2 R_{1} + R_{2}}{3 π}

(below)

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Solution

The correct option is B $\frac{(R_{2} - R_{1}) (2 R_{1} + R_{2})}{3 π (R_{1} + R_{2})}$ (below)
Center of mass of lower given disc is

$y_{c o m .} = \frac{\int x d m}{\int d m} = \frac{\int x \frac{(2 π x) d x}{π (R_{2}^{2} - R_{1}^{2})}}{m} = \frac{2 \int x^{2} d x}{(R_{2}^{2} - R_{1}^{2}) m} = \frac{{[2 \frac{x^{3}}{3}]}_{R_{1}}^{R_{2}}}{(R_{2}^{2} - R_{1}^{2}) m} = \frac{2}{3} \frac{(R_{2}^{2} - R_{1}^{2})}{(R_{2}^{2} - R_{1}^{2}) m} = \frac{2}{3} \frac{(R_{2} - R_{1}) (R_{1}^{2} + R_{2}^{2} + R_{1} R_{2})}{(R_{2} - R_{1}) (R_{2} + R_{1})} = \frac{2 (R_{1}^{2} + R_{2}^{2} + R_{1} R_{2})}{3 (R_{2} + R_{1})} b e l o w$

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