I am trying to analyse the following situation using classical mechanical concepts. Consider a a straight rod $AB$ of mass $M$ and length $L$ placed on a frictionless horizontal surface. A force $F$ acts at the end $A$ perpendicular to the rod. The direction of $F$ is fixed. I am trying to find out the initial acceleration of end $B$ just after force $F$ is applied. But I cannot figure out which point on the rod should be taken as the instantaneous centre. If I consider the mid point of $AB$ (call it $O$) as centre. Then, torque about $O$,

$$T=\frac{FL}{2}=\frac{ML^2a}{12}$$ where $a$ is the angular acceleration of rod. Hence the linear acceleration of end B would be $\frac{3F}{M}$

But I can also assume end $B$ to be the centre. Then it's linear acceleration would be zero. Can someone please help me with this situation?

  • $\begingroup$ An unaided rotation will alway happen around the center of mass, so yes the midpoint of the rod. Can you show how you found the linear acceleration of end B to be $3F/M$? $\endgroup$ – Steeven Sep 20 '15 at 14:09
  • $\begingroup$ From the torque equation about O we get angular acceleration a=6F/ML. Now linear acceleration of B=(L/2)*a=3F/M . But this is incorrect. $\endgroup$ – Irrational 3.14 Sep 21 '15 at 17:20
  • $\begingroup$ See calculating the center of percussion. $\endgroup$ – Mike Dunlavey Sep 21 '15 at 18:00

Irrational 3.14

It seems to me that you are using the equation torque = moment of inertia * angular acceleration. This particular equation ONLY holds under three conditions:

  1. The rotational center is fixed.
  2. The rotational center is the center of mass.
  3. The rotational center is moving at a constant velocity.

In your scenario, the center of the rod satisfies condition 2, and thus is a "safe“ origin. The point B, however, is accelerating and thus does not satisfy any of the three conditions. That is why point B is not a valid choice.

I refer you to chapter 8 of David Morin's Introduction to Classical Mechanics, for a complete derivation of the three conditions.

  • $\begingroup$ Thank you for clearing the doubt regarding origin. But even after considering O as a valid origin, I am getting an incorrect value for the linear acceleration of B (The correct value is 2F/M). Why is it so? $\endgroup$ – Irrational 3.14 Sep 21 '15 at 17:12
  • $\begingroup$ Ok so the 3F/M part is just the linear acceleration of point B due to the torque that you applied. There is also the force contribution right? Since F=Ma, the force contributes an acceleration of a= F/M forward. While the torque contributes 3F/M backward. So the net acceleration is 2F/M backward. $\endgroup$ – Zhengyan Shi Sep 21 '15 at 17:20

The linear acceleration of $B$ due to the rotational movement: $$ a_r = {L \over 2} {FL/2 \over ML^2/12} = {3F \over M} $$ The linear acceleration of $B$ due to the linear acceleration of the center of mass of the rod (which is anti-parallel to the rotational acceleration): $$ a_t = {F \over M} $$ Since these two accelerations are on opposite directions to each other, the total linear acceleration of point $B$ is their difference, $2F/M$.


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