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Suppose we have a rigid uniform bar AB of length L slipping from its vertical position on a frictionless floor. At a certain instant of time the angle made by the bar with the vertical is $\theta$. Now, looking from ground frame is it logically correct to say that the instantaneous torque about point B is $mg\frac{L}{2}\sin(\theta)$?

However, some people told me that to find toque about $B$ we need to consider torque due to the pseudo force acting at the COM of the rod as well. But, I believe that torque due to pseudo force should not be included when just asked for toque about B. We should take only toque due to the real force $mg$ as the acceleration terms are anyway taken care while writing the equations of rotational motion. Any suggestions? This is perhaps just a terminology/semantics problem?

  • $\begingroup$ Not sure what you mean by a pseudo force? There are two forces acting on the rod: the force of gravity acting downwards at the center of mass and the normal force acting at B. Torque is $\vec{r}\times\vec{F}$ for each of those forces, with $\vec{r}$ measured from the point at which you want to know the torque. $\endgroup$ – WAH May 27 '17 at 16:17
  • $\begingroup$ The rod (and the point B) is accelerating in horizontal direction as well. If you consider point B as the frame of reference then a pseudo force acts at the COM. @WAH $\endgroup$ – user139621 May 27 '17 at 16:19
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    $\begingroup$ While the end of the rod in contact with the ground is moving horizontally, I wouldn't say that the rod (as represented by its COM) is accelerating in the horizontal direction: there are no linear forces acting in the horizontal direction. Again, I don't know what you mean by a pseudo force. Rather, there is the real force of gravity acting at the center of mass. Or by pseudo force do you mean that the fact that all the little bits of gravity acting all along the rod is equivalent to $mg$ acting at the COM? In any case, yes, the torque on the rod about B is given by $mgL\sin(\theta)/2$. $\endgroup$ – WAH May 28 '17 at 1:32

Your advisers are correct. Whenever you are using a non-inertial frame of reference you need to apply pseudo-forces to compensate.

You are using an accelerating frame of reference attached to B. In the inertial ground frame of reference the COM does not move horizontally, because there are no horizontal forces; end B accelerates to the right. In the B-frame end B is at rest; in addition to the gravity force $mg$ acting downwards there is also a pseudo-force $ma$ acting on the COM to the left, where $a=\frac12 L\ddot \theta$. This pseudo-force also exerts a torque about B.

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  • $\begingroup$ The fact is no frame of reference was mentioned explicitly in the question. It just asked for torque about B. If I am using the ground frame the only torque should be $mgL/2\sin(\theta)$ as WAH pointed out in the comments. Here are the two schools of thought: 1) i.stack.imgur.com/McQEA.png (considers pseudo torque and non-inertial frame) and 2) i.stack.imgur.com/3AaDm.jpg(considers inertial frame and only real torque). Now, I guess it is just a semantics problem as to which interpretation is correct. $\endgroup$ – user139621 May 28 '17 at 12:38
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    $\begingroup$ Your question asks for the instantaneous torque about point B, which is the end of the bar. The linked questions ask for the instantaneous torque about the point in contact with the floor which is again B because the point of contact is moving. The question does not ask for torque about a fixed point on the floor. Hence Q8 is incorrect and Q13 is correct. I agree that the wording of the question could have been clearer. $\endgroup$ – sammy gerbil Jan 26 at 8:39

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