A rod of length $l$ is standing vertically on a frictionless surface.

It is disturbed slightly from this position. Let $\omega$ and $\alpha$

be the angular speed and angular acceleration of the rod when the rod

turns through an angle $\theta$ with the vertical, then find the value

of the acceleration of the centre of mass of rod.

Attempt:

Using conservation of energy,

$\dfrac 12 mv^2 + \dfrac 1 2 I_{cm}\omega^2 = \dfrac{mlg}{2}(1-\cos \theta)$

$I_{cm}= \dfrac {ml^2}{12}$

Substituting this and then differentiating wrt time,

we get:

$2a v = g\omega\sin \theta - \dfrac{2}{3}l \omega \alpha$ \where $a=$ acceleration.

Now, somehow I have to eliminate $v$. I am unable to do that. How do I proceed?

The answer given is:

$a= \dfrac l 2 \sin\theta \alpha + \dfrac l 2 \omega^2 \cos \theta $