Rotational Motion Question 78

Question: A rod PQ of mass M and length L is hinged at end P. The rod is kepts horizontal by a massless string tied to point Q . When string is cut, the initial angular acceleration of the rod is

Options:

A) $ \frac{3g}{2L} $

B) $ \frac{g}{L} $

C) $ \frac{2g}{L} $

D) $ \frac{2g}{3L} $

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Answer:

Correct Answer: A

Solution:

  • Torque on the rod = Moment of weight of the rod about P

    $ \tau =mg\frac{L}{2} $ ….(i)

    $ \because $ Moment of inertia of rod about

    $ P=\frac{ML^{2}}{3} $ ….(ii)

    As $ \tau =la $

    From Eqs.(i) and (ii), we get

    $ Mg\frac{L}{2}=\frac{ML^{2}}{3}\alpha $

    $ \therefore $ $ \alpha =\frac{3g}{2L} $



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