# Moving center of mass, torque and axis of rotation

1. Jul 26, 2015

### Electric to be

Hello. Suppose I have a coordinate system with the origin at the center of mass at rest of some rigid object and I wish to calculate the total change in angular momentum of the object with respect to it's center of mass.

If I apply a torque to the object, that also happens to be an unbalanced force, the object will move in the direction of the force and as a result the center of mass will no longer be on the origin.

So now the torques will be calculated with respect the original origin and not the c.o.m. . I can't see a way to reconcile this because following the center of mass for my reference system would result in a non inertial frame.

Is the only way to do this calculation is to use this frame and then switch to the center of mass frame after the calculation is over to determine it's final angular momentum?

2. Jul 26, 2015

### Staff: Mentor

The easiest way to do that is to use coordinates in which the center of mass is at rest.

Only if you use it to describe the object's motion while the force is being applied. If you use the inertial frame in which the center of mass is at rest after the force is finished, you don't have that problem.

3. Jul 26, 2015

### Electric to be

So would a solution be calculating the torque and change in angular mom. about some point and then changing to the center of mass frame afterwards?

4. Jul 27, 2015

### Staff: Mentor

You could do that, yes.

5. Jul 27, 2015

### Electric to be

Alright thanks. My last question is how would being in a non inertial frame of reference affect say, the calculation of change in angular momentum about the center of mass? How would it be different than pretending that the center of mass is simply fixed in space?

6. Jul 27, 2015

### Staff: Mentor

You would have to include a "fictitious force" in the non-inertial frame that would oppose the action of the applied force, in order to explain why the center of mass of the object remains at rest.

I'm not sure what you mean by "pretending that the center of mass is simply fixed in space". You can't just "pretend" that; you have to calculate its consequences, which is what adopting the above non-inertial frame does.

Last edited: Jul 27, 2015
7. Jul 27, 2015

### Electric to be

Does this force have to be applied at a location that would apply an opposite torque, or would it just need to cancel out the force causing translation? Essentially keeping the 'rotational' part of the reference frame inertial but accounting for the acceleration of the center of mass? Sorry if this doesn't make sense.

8. Jul 27, 2015

### Staff: Mentor

It's present everywhere in the non-inertial frame; it's not "applied" at any particular point. Think of the Newtonian "gravitational force" in a frame that's at rest with respect to the Earth (and small enough that tidal effects are negligible); it's present everywhere, and with the same magnitude and direction (since a rock released anywhere in the frame will accelerate downwards). The key point is that the system's center of mass is at rest in the frame.

Yes. See above.

9. Jul 27, 2015

### Electric to be

Thanks again. So if I model this as a fictitious force going through the object's center of mass (like a gravitational field), would the overall calculation differ at all from one in which there is a fixed axis in the object's center of mass? In both cases the net force is zero and there is a torque acting at the same distance.

10. Jul 27, 2015

### Staff: Mentor

I'm not sure I understand. There has to be some axis for the torque to be about, and that axis will pass through the center of mass, regardless of which frame you use.

11. Jul 27, 2015

### Electric to be

Yes, I understand. But it seems that applying fictitious forces to the non inertial frame and doing calculations in that frame for change in angular momentum would have the same results as 'pretending' that the axis is fixed if a torque is applied on the same location for the same period of time.

12. Jul 27, 2015

### Electric to be

Essentially what I mean is adding the fictitious force in this case seems almost unnecessary since it doesn't factor into the torque/impulse calculation anyways ?

13. Jul 27, 2015

### Staff: Mentor

Doesn't it? Remember that the fictitious force is present everywhere in the non-inertial frame, not just at the center of mass. In particular, it's present at the point where the torque is applied.

14. Jul 27, 2015

### Electric to be

I'm having trouble understanding. So the fictitious force of -ma is applied to every single particle? Wouldn't that just create a net force in the center of mass (and resulting torque of 0), like your gravitational field analogy?

15. Jul 27, 2015

### Staff: Mentor

Yes, just like in a gravitational field.

Yes, that's exactly what it has to do in order to keep the center of mass at rest.

What you may not have realized is that the magnitude of this "fictitious force" will not be the same as the magnitude of the total force applied to a point off center on the object. In other words, the "fictitious force" does not cancel the entire applied force; it only cancels enough of it to keep the center of mass of the object at rest. There will still be "extra" applied force left over to create a torque about the center of mass.

In other words, if we look in the original frame (the one in which the object as a whole starts out at rest), the applied force is not entirely a torque; it's part torque and part linear force. If it were all torque, the center of mass would not move; the object would just rotate about it.

16. Jul 27, 2015

### Electric to be

If I'm not mistaken, don't all forces act like they move through the center of mass in terms of translational motion (ignoring rotation)?

So I understand that there still would be a torque about the center of mass, but this would be because one of the forces (the real one not going through the center of mass) points in a direction off center and has a lever arm and not because 'some' of the applied force goes into rotation and some into translation, correct?

So in this case, the fictitious force would have to be the same magnitude, and opposite direction. The net torque would just come from the fact that the fictitious force acts at the center of mass (like gravity) and the real force has a lever arm.

Then as a result it would appear that using this new 'inertial frame' would have the exact same calculation for change in angular momentum as if the axis was fixed at the center of mass and the same force was acting with the same lever arm, so basically the need for these fictious forces is removed. However if it was about some other axis not through the center of mass, then fictitious forces would become part of the calculation (since then they would have a lever arm) . At least this is what I have gathered.

17. Jul 27, 2015

### Staff: Mentor

If you're viewing the object as a single object, yes, you can treat translational forces as if they were exerted directly at the center of mass. But that doesn't mean the are actually, physically applied there; it just means you can treat them that way at that level of analysis.

If you're viewing the object as a collection of interacting particles, then the "fictitious force" in the non-inertial frame is applied to each particle equally; actual applied forces are applied at, well, the particle to which they are applied. But then you have to model the interaction forces between the particles as well in order to see how the object as a whole moves. That doesn't seem like the model you have in mind.

Yes.

No. Some of the applied force has to go into translation; otherwise the center of mass of the object would not move. So the full applied force is split, for purposes of analysis, into two pieces, a translational force and a torque. The former moves the center of mass; the latter induces a rotation about the center of mass.

Same magnitude as the translational force; not same magnitude as the total force (including torque). If it were the same magnitude as the total force, the center of mass would not be at rest in the non-inertial frame.

If this were the case, the center of mass would move (see above).

No. See above.

I don't understand what you mean by this. You don't get to pick the axis about which a torque is applied: the direction of the force plus the vector from the center of mass to the point of application together define a plane, and the axis must be perpendicular to that plane and passing through the center of mass.

It looks to me like using a non-inertial frame is confusing you; in that case you probably shouldn't do that. The analysis can be done in either the inertial frame in which the object is originally at rest, or the inertial frame in which the object's center of mass is at rest after the force is applied. The first of those two is probably the easiest to grasp intuitively: you just apply the off-center force and see what happens to the object.

18. Jul 27, 2015

### Electric to be

By total force I mean the actual physical quantity of force that is equal to ma, not the sum of ''translational forces'' and torque. But yeah we both mean the same thing (I guess we just use different terminology).

I still don't understand why. In both cases the net force acting on the object is zero, as both the center of mass of the fixed axis frame and the center of mass frame with fictitious forces doesn't move. In addition, the only torque acting on both is the same since the only forces acting on the fictitious force frame are the actual applied force which creates a torque and the fictitious force which does not. In the fixed frame essentially the same forces act which is the applied force which exerts a torque, and the forces at the center of mass at the fixed axis which exert a centripetal force that creates no torque.

Also, why can't I choose an axis? Can't I calculate angular momentum about the origin of any coordinate system, not necessarily going through the center of mass. ( though that is very convenient)

Last edited: Jul 27, 2015
19. Jul 27, 2015

### Staff: Mentor

I don't understand what you mean by "the fixed axis frame", but if it's any frame other than the non-inertial center of mass frame with fictitious forces, then the center of mass must move in that frame. The center of mass only remains at rest in the non-inertial center of mass frame; that's how that frame is defined. Obviously, since the center of mass undergoes non-inertial motion when the force is applied, the center of mass cannot remain at rest in any inertial frame.

I didn't say you couldn't choose an axis about which to calculate angular momentum. I said you couldn't choose the axis about which the torque is applied. You can calculate angular momentum about any axis you like, but if that axis doesn't match the axis about which the torque is applied (which, as I said, is determined by the physics of the problem), then the angular momentum you calculate won't answer the question you appear to want to answer.

20. Jul 27, 2015

### Electric to be

Ok, but why does the torque have to be calculated about the center of mass? Can't I calculate torque about any point?

edit: I understand. You can have the axis go through any point but i should make sure to have the axis of the angular momentum and torque go along the same point.

Last edited: Jul 27, 2015