Question

There is a uniform bar of mass m and length / hinged at centre so that it can turn in a vertical plane about
a horizontal axis. Two point like masses 'm' and '2m' are rigidly attached to the ends and the arrangement
is released from rest.



Initial angular acceleration of the bar is
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Answer 1

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Answer 2

Answer:

Initial angular acceleration of the bar is α = $\frac{3g}{5l}$

Explanation:

The net torque will be due to the extra mass m,

τ = (2m - m)g$\frac{l}{2}$

τ = mgl/2

now moment of inertia of a solid rod about its horizontal axis $I_{rod}$ = $\frac{ml^{2} }{12}$

now moment of inertia of point masses m and 2m at a distance $\frac{l}{2}$

$I_{m} =\frac{ml^{2} }{4}$

$I_{2m}=2m\frac{l^{2} }{4}=\frac{ml^{2} }{2}$

as masses are lying on the axis

$I=I_{rod} +I_{m} +I_{2m}$

$I=\frac{ml^{2} }{12}+\frac{ml^{2} }{4}+\frac{ml^{2} }{2}$

$I=\frac{5ml^{2} }{6}$

Again,   τ = Iα

α =  τ/I

α =$\frac{\frac{mgl}{2} }{\frac{5ml^{2} }{ 6} }$

α = $\frac{3g}{5l}$