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Byju's Answer

Standard XII

Physics

Race of Rollers

One end of a ...

Question

One end of a thin uniform rod of length $L$ and mass $M_{1}$ is rivetted to the centre of a uniform circular disc of radius $r$ and mass $M_{2}$ 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)

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

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

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

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

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Solution

The correct option is A $\frac{M_{1} L}{2 (M_{1} + M_{2})}$
$\begin{matrix} r_{c o m} = \frac{M_{1} (\frac{L}{2}) + M_{2} (0)}{M_{1} + M_{2}} = \frac{M_{1} L}{2 (M_{1} + M_{2})} \end{matrix}$
$∴$ Option $A$ is correct.

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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)

The correct option is A M1L2(M1+M2)rcom=M1(L2)+M2(0)M1+M2=M1L2(M1+M2)∴ Option A is correct.

One end of a thin uniform rod of length $L$ and mass $M_{1}$ is rivetted to the centre of a uniform circular disc of radius $r$ and mass $M_{2}$ 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)

The correct option is A $\frac{M_{1} L}{2 (M_{1} + M_{2})}$
$\begin{matrix} r_{c o m} = \frac{M_{1} (\frac{L}{2}) + M_{2} (0)}{M_{1} + M_{2}} = \frac{M_{1} L}{2 (M_{1} + M_{2})} \end{matrix}$
$∴$ Option $A$ is correct.