Finding relation between angular acceleration and liner acceleration directed in one direction 
In the above given configuration what could be the relation between the angular acceleration of the rod $\alpha$ and the acceleration $a$ of the $2m$ block. 
My attempt : What I thought is that the string connected to the rod will move with the same acceleration as that of the block (using constraint relation or simple intuition). The only thing I can think of after that was to take the component of acceleration perpendicular to the rod. As we have in usual cases $\alpha L=a$ so we can have here $\alpha L=acos37^{\circ}$ but that is not getting me to the answer provided. Where am I wrong. Do the component of acceleration along the rod has any role here ?
 A: You are almost correct.  The linear acceleration of the end of the rod is $\alpha L$ perpendicular to the rod.  The component of this acceleration in the direction of the string is $\alpha L\sin A$ where $A$ is the angle which the string makes with the rod (not the angle which the rod makes with the horizontal).  So at the instant shown in the diagram the component along the string is $\alpha L\sin(90-37)^\circ=\alpha L\cos 37^\circ$, which is also the acceleration of the $2m$ block.  
There is no component of acceleration along the rod, because the length of the rod is not changing.  There is no need to consider forces - it only confuses the situation.
A: The rod will only rotate upwards due to the force that is normal to it's length, acting from the tether on the rod, 
$$F_{t,\perp} = F_t\cos\theta$$ 
Here $F_t = (2m)g$ is the force from the tether, equal to the weight of the block, and $\theta$ is the angle. The rod also feels a downward force, gravity, that has a normal component 
$$F_{G,\perp} = mg\cos\theta$$
The net force is then 
$$ F_\perp = F_{t,\perp} - F_{G,\perp} =2mg\cos\theta - mg\cos\theta = mg\cos\theta$$
We then have the link $a_\perp = F_\perp/m =  L\alpha$, where $a_\perp$ is the acceleration perpendicular to the length of the rod, $L$ is the length and $\alpha$ is the angular acceleration. We then have
$$\alpha = \frac{a_\perp}{L} = \frac{g}{L}\cos\theta$$
In the mean time, the acceleration of the tether (and then also the block) is the vertical component of $a_\perp$: $a_y = a_\perp\cos\theta$, giving 
$$a_{block} = g cos^2\theta$$ 
Which in turn gives the final relation:
$$ \alpha = \frac{a_{block}}{L\cos\theta}$$
