2
$\begingroup$

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 $3g/2$?

$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – Farcher May 13 '19 at 7:28
  • $\begingroup$ Even when the angular velocity is zero? (I only wanted to know the vertical component of the acceleration though). $\endgroup$ – Random May 13 '19 at 7:29
0
$\begingroup$

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.

$\endgroup$
  • 1
    $\begingroup$ The question doesn't mention a string, just a hinged rod. $\endgroup$ – PM 2Ring May 13 '19 at 9:04
  • $\begingroup$ I have edited it @PM 2Ring $\endgroup$ – sawan kumawat May 13 '19 at 11:45
  • $\begingroup$ But what is wrong with my reasoning? I analysed only the end point not the entire rod. $\endgroup$ – Random May 13 '19 at 11:53
  • $\begingroup$ Please do note that, Tension always has a pulling effect on body, so it will act along the axis of rod, not in horizontal direction $\endgroup$ – sawan kumawat May 13 '19 at 11:55
  • $\begingroup$ I assumed the rod to be in the horizontal direction and the axis I was talking about was the axis about which the rod is rotating. $\endgroup$ – Random May 13 '19 at 12:20

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.