One end of a thin uniform rod of length L and mass M₁ is rivetted to the centre of a uniform circular disc of radius 'r' and mass M₂ so that both are coplanar. The centre of mass of the combination from the centre of the disc is (Assume that the point of attachment is at the origin)

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$\frac{L (M_{1} + M_{2})}{2 M_{1}}$

$\frac{{LM}_{1}}{2 (M_{1} + M_{2})}$

$\frac{2 (M_{1} + M_{2})}{{LM}_{1}}$

$\frac{2 {LM}_{1}}{(M_{1} + M_{2})}$

answer is D.

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Detailed Solution

The rod's mass can be thought of as being concentrated at its centre if the origin is in the centre of the disc and the x-axis runs along the length of the rod.

From the question,

$R_{cm} = \frac{M_{1} \times \frac{L}{2} + M_{2} \times 0}{M_{1} + M_{2}}$

$R_{cm} = \frac{{LM}_{1}}{2 (M_{1} + M_{2})}$

Hence the correct answer is $\frac{{LM}_{1}}{2 (M_{1} + M_{2})} .$

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