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When there are 2 forces normal to each end of a uniform pole, if they had the same force, the rotation would be cancelled out and all force would be acting in one direction. But if one force was smaller, some force would act to rotate the pole. This is easy to calculate with moments and torque, but how would you calculate the force pushing it in a line? I'm rather sure it won't be the whole force acting in that direction, as that would suggest a force on one end with no force on the other would not spin around the CoM but that the CoM would spin around a point in space.

Another way to put it would be: can the force that is "used" in torque be directly used to calculate linear force (this would make it not spin around the CoM)?

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  • $\begingroup$ Found this similar question posted just now link $\endgroup$
    – Charlie Parker
    Commented Jun 7 at 12:33
  • $\begingroup$ It seems to me that the "accurate and reliable" way to do this is to calculate the net force (as a vector) to find the acceleration of the CM, and calculate the net torque to find the angular acceleration about the CM. This business about "rotational force" is what's tripping you up; torque is the important concept. $\endgroup$
    – Michael Seifert
    Commented Jun 7 at 14:05
  • $\begingroup$ The thing I'm thinking a about is that can the force 'used' in torque be 'used' in linear movement or is it a fraction if the total force $\endgroup$
    – Charlie Parker
    Commented Jun 7 at 15:49
  • $\begingroup$ The force which determines the linear acceleration of the CM is the total force, no matter which direction it points. You take each individual force, add them as vectors, and divide by the mass. For example, in your "all rotational" diagram, the force indicated at the bottom of the object would cause an acceleration up and to the left in the absence of gravity. $\endgroup$
    – Michael Seifert
    Commented Jun 7 at 16:44
  • $\begingroup$ @MichaelSeifert I understand how it works with only one force and when it is in line with the CoM but it doesnt work with one force aplied at the end of a pole because what youhave said suggests it would not rotate around the CoM (as said in my updated question to make it more general) $\endgroup$
    – Charlie Parker
    Commented Jun 7 at 16:58

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I'm rather sure it won't be the whole force acting in that direction, as that would suggest a force on one end with no force on the other would not spin around the CoM but that the CoM would spin around a point in space.

A force on one end would simultaneously produce a torque (causing the object to rotate), and cause the center of mass to accelerate. The entire net force contributes to net acceleration. It does not matter if the force is in line with the center, the net acceleration of the object is identical.

can the force that is "used" in torque be directly used to calculate linear force

Yes.

(this would make it not spin around the CoM)?

Why do you think that would be so? Strike one end of a ruler rapidly. It will both accelerate away from you and it will rotate around its center of mass.

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  • $\begingroup$ Interesting, the bit about the ruler made me understand it as I was imagining the force was like velocity in that it will accelerate in rotational speed only. Just to be sure, say the ruler had a constant acceleration and was in space, would it still move? This would mean the CoM would be moving until the moment the acceleration stops, in which it would then continue to spin but around the CoM with its gained velocity? $\endgroup$
    – Charlie Parker
    Commented Jun 8 at 9:31
  • $\begingroup$ I'm not sure I understand the comment. The CoM stays moving even after the acceleration. The velocity is constant when there are no forces. The CoM would be accelerating until the forces/(acceleration) stops. Yes it would continue to move and spin after that. $\endgroup$
    – BowlOfRed
    Commented Jun 8 at 17:27
  • $\begingroup$ sorry, was just a little confused but it makes a lot of sense now, thanks $\endgroup$
    – Charlie Parker
    Commented Jun 9 at 18:39

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