Calculating acceleration of end point of rod Suppose I have a rod hinged at its end, of length $\ell$ free to rotate about an horizontal axis; initially in the horizontal position and I have to calculate the acceleration of its other end point. I know the derivation using torque but what if I consider the free body diagram of the end point. Two forces, gravity and tension act on it but since tension can only act in the horizontal direction only gravitational force acts in the vertical direction and hence its acceleration is $g$(in the downward direction), which is incorrect. What is wrong in this? Which force is responsible for increasing its acceleration to $3g/2$?
 A: The other parts of the rod exerts a tangential force on the end which contributes to its downward acceleration.  If you get a chance to watch a video of a tall brick chimney being demolished, notice that it breaks before hitting the ground.  It is not designed to withstand the lateral forces required to match the angular acceleration of the upper end with that of the lower sections.
A: The component of $mg$ force, ($mgcos\theta$) will be equal to tension in the rod ( if string is inextensible, and $\theta$ is the angle of string with vertical) and therefore only $mgsin\theta$ will be left, which is responsible for restoring force in the string. And by dividing restoring force by  its mass, you can calculate its acceleration.
EDIT
In case of hinged rod, the same direction of tension and $mg$ force will be there, but they will act at Centre of mass of Rod, so again the formula for calculating acceleration will be same.
A: 
Two forces, gravity and tension act on it but since tension can only act in the horizontal direction only gravitational force acts in the vertical direction and hence its acceleration is g

You are missing a major force called the hinge reaction . I've drawn it

*

*you can yourself check that your fbd must be wrong because in your case there is only one horizontal force on rod which is tension which makes it unstable (I.e there must be an accelerationin x direction). So there has to be a horizontal force balancing it and this force is provided by hinge


*similarly there is a normal hinge reaction in y direction too since you know acceleration is 3g/2 the hinge reaction must be in negative y direction . Here I've taken it in +y but on balancing forces you'll get it negative

