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} $
Show Answer
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} $
BETA
