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How can we calculate torque in a non-inertial frame? Take for instance a bar in free fall with two masses, one on either end, $M_1$ and $M_2$. Taking the point of rotation to not be the center of mass, i.e. $M_1\neq M_2$ and take the point of rotation to be the center, what is the proper way of analyzing the situation to come to the conclusion that there is no rotation?

  • If you want to force two masses at a constant distance to rotate around anything but the center of mass, then you have to exert a force on them (you can already do it with one mass on a string). Mechanical engineers know and dread these forces from imbalanced rotating machine parts very well... they like to destroy bearings in no time. If there is no such force, on the other hand, any point that you pick on the line connecting the two masses that is not the center of mass will exhibit a circular motion (in 2d and a screw-like motion in 3d). Is that your question? – CuriousOne May 27 '15 at 17:59
  • @CuriousOne No, I forgot to say a bar in free fall. You actually commented on the last question I asked- looking at the definition of torque, $\tau = dL/dt$, this is only true for a fixed reference point, right? So if I have a bar in free-fall, a non-inertial reference frame, how do I come to conclusion that there is no rotation? –  May 27 '15 at 18:04
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    In a non-inertial system I would expect to see a pseudo-torque similar to pseudo-forces for the case of linear motion, but both are non-physical. Having said that, I have never done the non-inertial calculations, so I can't assure you that that's exactly how it works out. – CuriousOne May 27 '15 at 18:16
  • possible duplicate of What is the proof that a force applied on a rigid body will cause it to rotate around its center of mass? – John Alexiou May 27 '15 at 18:26

1 Answers1

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Follow the rules of motion:

  1. Sum of forces equals mass times acceleration of the center of mass: $$ \sum_i \vec{F}_i = m \vec{a}_{cm} $$
  2. Sum of torques about the center of mass equals change in angular momentum: $$ \sum_i (M_i + \vec{r}_i \times \vec{F}_i) = I_{cm} \dot{\vec{\omega}} + \vec{\omega} \times I_{cm} \vec{\omega}$$ where $\vec{r}_i$ is the relative location of force $\vec{F}_i$ to the center of mass.

So for an accelerating rigid body that is not rotating $\dot{\vec{\omega}} = \vec{\omega} = 0$ the right hand side of the last equation must be zero.

See https://physics.stackexchange.com/a/80449/392 for a complete treatment of how you go from linear/angular momentum to the equations of motion.

Also see https://physics.stackexchange.com/a/82494/392 for a similar situation where a force is applied away from the center of mass.

The rule that comes out of the above equations of motion are:

  1. If the net torque about the center of mass is zero then the body will purely translate
  2. If the sum of the forces on a body are zero (but not the net torque) then the body will purely rotate about its center of mass.
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John Alexiou
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  • Did you conclude there is no rotation, or assume there is no rotation? –  May 27 '15 at 18:08
  • I cannot conclude anything unless I know the details. – John Alexiou May 27 '15 at 18:21
  • The problem I was trying to convey is having a bar, with two unequal masses, in free fall in a a vacuum, experiencing gravity. We do not expect the bar to rotate, but if the two masses are uneven, then they will product an uneven torque about the center of the bar (not the COM). Why doesn't this uneven torque generate rotation? –  May 27 '15 at 18:23
  • Because it is the net torque about the center of mass that counts, not the geometric center. – John Alexiou May 27 '15 at 18:25
  • See the 2nd link in my edited post. – John Alexiou May 27 '15 at 18:26
  • (In response to your first comment) I know that is true, but we should be able to discuss the torque about the geometric center, and still come to the conclusion that there is no rotation.

    I'll read the link.

     –  May 27 '15 at 18:26
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    @Anthony, in the accelerating reference frame, you get a fictitious force opposite the direction of acceleration applied through the COM. If you calculate torque through some other axis, you must apply this fictitious force as well. – BowlOfRed May 27 '15 at 18:28
  • Then you need to add the torque due to the inertial loads of the body. A $d m g$ term where $d$ is the center of mass distance to the center of the bar. – John Alexiou May 27 '15 at 18:28
  • Can you elaborate on that? I assumed there should be something like that, but I feel like I've already taken into account the gravity by saying that there are two gravitational forces on either end of the bar. –  May 27 '15 at 18:29
  • In the 1st link you see the equation $\sum \vec{M}_A = I_C \vec{\alpha} + m \vec{c} \times \vec{a}_A + \ldots$? The $m \vec{c} \times \vec{a}_A$ term is what I am talking about. – John Alexiou May 27 '15 at 18:31
  • @BowlOfRed If I include that fictitious force, do I still get to include the two torques from gravity on the two masses? I feel like that's double counting the forces. –  May 27 '15 at 18:55
  • Yes. The force comes from the acceleration of the frame, not from "gravity" itself. – BowlOfRed May 27 '15 at 18:56
  • I see. Do you derive that through coordinate transforms? –  May 27 '15 at 18:57
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    @Anthony, yes. The details of the cross terms (torque due to linear acceleration, and force due to angular acceleration) are again in link 1. – John Alexiou May 27 '15 at 19:09