# 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 $$3g/2$$?

• The line of action of the tension force is along the rod. There must be a force producing the centripetal acceleration of the particles which make up the rod. – Farcher May 13 '19 at 7:28
• Even when the angular velocity is zero? (I only wanted to know the vertical component of the acceleration though). – Random May 13 '19 at 7:29

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.