A meterstick of length $L$ is set on the very edge of a table. Almost all of the meter sticks length is off the table. The meterstick is then released from rest.

I may be wrong in my thinking but here is how it goes:

Originally I chose the very edge of the meterstick (the one on the table) as my pivot point. If you choose the center of mass as the pivot the normal force will exert less torque because the meterstick falls. My only explanation would be that the rotational inertia about the center of mass would be less than that about the edge. Would these two effects cancel out?

  • $\begingroup$ It'd be nice if you could make this question more 'to the point' $\endgroup$ – Buraian Apr 18 '20 at 19:52
  • $\begingroup$ @DDD4C4U I tried to shorten it. I didn't know how to add a picture so I wanted to make sure people understood my question. $\endgroup$ – Jordan E Apr 18 '20 at 22:40
  • $\begingroup$ Click edit then then on the row where there is BOLD and ITALICS , the second part of that row has the add image button $\endgroup$ – Buraian Apr 19 '20 at 8:36

You can choose any pivot point to write down the equations of motion, only some choices are easier to analyze. In any case, the motion of the rod does not depend on that choice.

If there is a point of the body that remains at rest during the process, then this point may be useful point of reference for writing down torques and equation they obey.

But if we are not sure that the point of contact is at rest at all times (in your case, the rod will eventually fall down and lose contact) then it may be useful to write down torques around the center of mass and an equation these torques obey.

  • $\begingroup$ Would the net torque/angular acceleration still be the same if we chose the pivot point to be the center of mass the instant after release? $\endgroup$ – Jordan E Apr 18 '20 at 22:54
  • $\begingroup$ Torque depends on the reference point, so no for torque. Angular acceleration is absolute (when referred to the inertial frame), so this will be the same. $\endgroup$ – Ján Lalinský Apr 18 '20 at 23:04
  • $\begingroup$ Okay, thank you. So that means that though the torques will be different, the moment of inertia will also change, which causes it to have no effect on the angular acceleration? $\endgroup$ – Jordan E Apr 18 '20 at 23:28
  • $\begingroup$ Indeed. Angular velocity and angular acceleration are objective quantities (if observed from inertial frame), they do not refer to any particular point of reference. $\endgroup$ – Ján Lalinský Apr 19 '20 at 0:38

Case1 choosing edge as point Torque due to mg is given by T=mgl/2
Which is equated with Ialpha Mgl/2 = ml^2/3 alpha 1 Case 2 choosing centre of mass as reference Nl/2=ml^2/12 (4N)l/2= ml^2/3alpha 2 Equation 1 and 2 as alpha has to be same 4N=mg N=mg/4 But note this will be very initial condition can after that com will have an acceleration Mg-N=ma So not normal reaction has to be less


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.