A particle of mass $m$ is attached to an end of a uniform rod of mass $M = 2 m$ and length $l$ which is suspended through its mid-point by an inextensible string as shown. Initially the rod is in horizontal position and at rest. The system is released from this position. Just after the release.

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The angular acceleration of the system is $\frac{6 g}{5 l}$

The angular acceleration of the system is $\frac{2 g}{5 l}$

The tension in the string is $\frac{12}{5} m g$

The tension in the string is $\frac{2}{5} m g$

answer is A, C.

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Detailed Solution

The FBD of the rod and the ball are shown. Applying $τ = I α$ about the C.M. of the rod,

we have, $N (\frac{l}{2}) = (\frac{M l^{2}}{12}) α$
Writing Newton’s $I I^{n d}$ law in the vertical direction on the CM of the rod we have
$T - N - 2 m g = 0$ and writing Newton’s II law in the vertical direction on the ball we have,
$m g - N = m (\frac{l}{2}) α$