question_answer

A rod of length \[\ell \] is pivoted about a horizontal, frictionless pin through one end. The rod is released from rest in a vertical position shown in Figure. Find velocity of the C.M. of the rod, when rod is inclined at an angle \[\theta \] from the vertical.

A) \[\sqrt{\frac{gl(1-\cos \theta )}{4}}\]        

B) \[\sqrt{\frac{gl(1-\sin \theta )}{4}}\]

C) \[\sqrt{\frac{3gl(1-\sin \theta )}{4}}\]

D) \[\sqrt{\frac{3gl(1-\cos \theta )}{4}}\]

Correct Answer: D

Solution :

The fall in position of C.M of the rod, \[h=\frac{\ell }{2}(1-\cos \theta )\]

In the process, decrease in P.E. is equal to the increase in rotational K.E. of the rod, so \[mgh=\frac{1}{2}I{{\omega }^{2}}\]

Or \[mg\frac{\ell }{2}(1-\cos \theta )=\frac{1}{2}\left( \frac{m{{\ell }^{2}}}{3} \right) &  &  &  &  &  &  & _{{}}^{{{\omega }^{2}}}\]

\[\therefore \omega =\sqrt{\frac{3g}{\ell }(1-\cos \theta )}\]

The velocity of C.M. of the rod \[V{{ & }_{cm}}=\omega r\]

\[=\sqrt{\frac{3g}{\ell }(1-\cos \theta )}\times \frac{\ell }{2}=\sqrt{\frac{3g\ell (1-cos\theta )}{4}}\].