# Is the Change in Rotational Kinetic Energy Frame Invariant?

1. Oct 18, 2009

### e2m2a

Is the Change in Rotational Kinetic Energy Frame Invariant?

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I know the translational kinetic energy of an object is frame dependent. That is, in the center of mass frame of the object, the kinetic energy is zero. In any other frame it is not. In fact, work itself is frame dependent. A force applied through a distance will give a change in kinetic energy of the object with respect to one frame, but in another frame the same force could go through a different displacement, hence, with respect to the other frame, a different work or change in kinetic energy will be observed. However, impulse seems to be frame invariant (if we are dealing with non-relativistive speeds). That is, a force applied to an object for a given time increment will cause a change in momentum of the object in one frame and will give that SAME change in momentum with respect to any other frame. (Remember, I am assuming sub-relativistic speeds.)
However, I don't see how this could apply to angular motion. If an observer in one frame applies a torque through an angular displacement on an object that is pivoted at one end, not through its center of mass, there will be an increase in the rotational kinetic energy equal to:

∆KE = 1/2 I ω²

Will the same change in rotational kinetic energy be observed in any other frame? That is, is angular work frame invariant at sub-relativistic speeds?

2. Oct 18, 2009

### Staff: Mentor

That is an interesting question. My gut feeling is that it is not, but I don't really know. If I were you and were interested in this question I would imagine two massive particles attached to each other with a massless rigid rod and analyze that. I would have them start on the x axis and rotate in the xy plane and then boost them in z and x and calculate the angular momentum.

3. Oct 19, 2009

### e2m2a

To be honest I don't have the capability to do a Lorentz transformation for the hypothetical problem you suggested. Also, I am dealing with non-relativistic speeds, so Galilean transformation and Newtonian mechanics is adequate and simplies the problem. Furthermore, I think the question can be simplified as follows. First of all I think anyone interested in this question would agree that for a symmetric body rotating around an axis through the center of mass would be frame invariant for both rotational kinetic energy and any change in rotational kinetic energy. I think it is intuitive and easy to prove that a disk rotating around an axis through its center of mass will have the same rotational kinetic energy with respect to any reference frame. The real question deals with bodies that rotate around an axis which is not through the center of mass of the body. I think the results would be the same for this reason. The angular velocity vector is a free vector. A body rotating around an axis not through its center of mass would still have an angular velocity vector determined by the right-hand rule. This vector is not affected by translation of the axis. How could it be?
You can't sum an angular velocity vector and a linear translation vector. Dimensional analysis would show they do not even have the same units. This alone forbids addition. The only thing that would impact the angular velocity is a torque. If an observer initially at rest observes the rotating object, he or she will see a rotational kinetic energy with respect to his frame. If the frame of the observer then accelerates, and then afterwards moves at a constant velocity, the magnitude of the angular velocity vector will not have changed due to the acceleration of this frame because no torque was manifested on the object during the acceleration of the frame. This is because relative to the observer on the accelerating frame, the rotating object would be "falling" in a gravitational field. This WOULD NOT impact the angular velocity vector. One way to prove this would be to do the following. Rotate an object around an axis connected to a second object. Insure the axis of rotation is not through the center of mass of the rotating object, and choose the ratio of the masses of the rotating object and the other object such that the centrifugal reactive force acting on the axis will be inconsequential. Drop the rotating object-second object combination and measure the angular velocity of the rotator as the combination falls to the earth. Neglecting friction, there will be no change in the angular velocity of the rotating object. If I am wrong about my conclusions, I welcome any input and discussion.

4. Oct 19, 2009

### Staff: Mentor

Yes, I agree.
I am not sure about this at all. That is what I would investigate with the "2 mass object" that I described. I think that this assumption is begging the question.
I would most strongly suggest that you not do this analysis in an accelerating reference frame. That will simply add completely unnecessary complications and make it difficult to convince anyone that your analysis is correct. Use two different inertial reference frames, for your purpose Galilean is fine.

5. Oct 19, 2009

### pervect

Staff Emeritus
As long as your rotating system is isolated, there isn't any ambiguity about the total energy it contains. And the rotational energy does indeed contribute to the energy of the system.

If your system is interacting with the rest of the world, there are some issues with the relativity of simultaneity that come into play. However, you can _always_ say that the density of energy and momentum transforms as a rank 2 tensor, the stress-energy tensor. I'm not sure how helpful this will be to you, but it's the easiest way I know of explaining the general rules.

You can also say that the energy and momentum of a system will transform as a four-vector, IF the system is isolated. However, if the system is NOT isolated, you might be in for an unpleasant surprise. Again, I'm not sure how helpful this will be to you, but if you don't know about four-vectors, it would be very useful for you to find out more :-).

So, rotating or not, as long as your system is isolated, it will have a definite energy and a definite momentum (assuming SR, i.e. a flat space-time) that transforms in a well known and standard manner.

6. Oct 19, 2009

### e2m2a

The assumption I made about a symmetric body rotating around an axis through its center of mass is actually taught as a standard problem in first-year physics courses. For example, suppose we have a long, thin rod of mass m and length L that rotates around an axis through its center of mass with angular velocity ω. Furthermore, suppose the center of mass of the rod moves with a translational velocity Vcom. Then the total energy of the rod with respect to a laboratory frame is:

Etotal = 1/2 m Vcom² + 1/2 Icom ω² (1)

Where,
Icom = the moment of inertia of the rod with respect to an
axis through its center of mass, which is 1/12 m L²

Again, this is a standard equation, easily derived in an introductory physics course. Now, examing equation 1, no where in the expression do we see anything explicit or implied that the rotational energy expressed in the second term on the right of the equation is a function of Vcom. Hence, the second term must be invariant with respect to any other inertial frame, unless what is taught in introductory physics courses about this is wrong. However, it WOULD be a function of Vcom if we approached relativistic velocities, but I am only concerned with the energy dynamics within the Newtonian domain.
The problem for me is knowing what the total energy is for the case where the rod rotates around an axis which is at one end of the rod, everything else being equal. I proposed the following expression as the correct one in a previous thread and was told it was correct. I want to make sure it is correct. Here is the expression.

Etotal = 1/2 m Vcom² + 1/2 Iend ω² (2)

Where,
Iend = the moment of inertia of the rod with respect to an
axis through one end of the rod, which is 1/3 m L²

Equation 2 is derived from the simple fact, as far as I understand, that any planar motion of an object can be expressed as the sum of a translational motion and a rotational motion around a fixed point. And if the above is the correct expression, for non-relativistic velocities, then the second term on the right for the rotational kinetic energy of the rod should be frame invariant. Is equation 2 the correct expression?

7. Oct 19, 2009

### Staff: Mentor

Oh, that is a much easier approach than mine. I like it.