Question

A uniform thin stick of length ( and mass m is held horizontally
with its end B hinged at a point B on the edge of a table. Point
A is suddenly released. The acceleration of the centre of mass
of the stick at the time of release, is :- [AIPMT (Mains) 2007]​

Answer 1

Answer:

The weight of the stick works through its midpoint,

That is: $\sf{\frac{l}{2}}$

Thus,

$\boxed{\sf{Torque\:on\:the\:hinge = {\frac{mgl}{2}}}}$

But:

Torque = Moment of inertia × Angular acceleration moment of inertia of a rod free to rotate at one axis

$\implies$ $\boxed{\sf{\frac{1}{3}ml^{2}}}$

That is:

$\implies$ $\boxed{\sf{\frac{mgl}{2}=\frac{1}{3}ml^{2}\alpha}}$

Now:

$\implies$ $\sf{\alpha{=\frac{3g}{2l}}}$

$\implies$ $\sf{a=\alpha l}$

$\implies$ $\sf{a=\frac{3g}{4}}$

Therefore:

Correct option: (a) $\sf{\frac{3}{4}g}$

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Answered by: Niki Swar, Goa❤️