Applying law of conservation of energy to rotating rod 
Can anyone solve this question using law of conservation of energy??
actually I tried to solve this by taking a part of rod dx  and then integrating it to the whole of the rod to find potential energy. But I am not getting the correct answer . can anyone correct me?]

 A: 
Can anyone solve this question using law of conservation of energy?

Of course. Conservation of energy requires that the sum of potential and kinetic energy remain constant. The potential energy is the height of the center of mass relative some arbitrary point. Choosing the rod being horizontal as that arbitrary point, the potential energy as a function of $\theta$ is $P = \frac 1 2 mgl \cos \theta$. The kinetic energy is purely rotational from the perspective of a frame with origin at the pivot point: $T = \frac12 \mathrm{I}\, {\dot\theta}^2$ where $\mathrm I = \frac13 ml^2$ is the moment of inertia of a slender rod about one end. Given that the rod is not moving initially when $\theta = 0$, conservation of energy dictates that
$$
\frac 1 2 mgl \cos \theta + \frac12 \frac13 ml^2\,{\dot\theta}^2 = \frac12 mgl
$$
Solving for $\dot\theta$ yields
$${\dot\theta}^2 = 3\frac g l (1-\cos\theta)$$
Differentiating with respect to time yields
$$2{\dot\theta}{\ddot\theta} = 3 \frac g l \sin\theta \dot\theta$$
or
$$\ddot \theta = \frac 3 2 \frac g l \sin\theta$$
A: You need not use conservation of energy to solve this problem. You can do this by merely writing the expression for torque on the rod.

Note that the force acts on the centre of mass, which is why I have calculated the perpendicular distance to be lsin∅/2.
