# Kinetic energy of a rotating and translating body?

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1. Oct 28, 2014

### Chris L

1. The problem statement, all variables and given/known data
Not a homework or coursework question, but given the simplicity of the problem I feel that this is an appropriate subforum.

Consider a person spinning a rock on a string above their head at a constant angular velocity, walking away from the observer at a constant linear velocity (ignore gravity for simplicity - consider it a 2D problem as viewed from above). The goal is to determine the total kinetic energy of the rock in the observer's frame.

2. Relevant equations

Typing this on my phone so apologies for the lack of LaTeX

I = mR² (ignoring the string)
Angular kinetic energy = (1/2)Iω²
Linear kinetic energy = (1/2)mv²

3. The attempt at a solution

In my mind there are two ways to solve this, but they don't agree with each other:

The first way is to simply consider the problem as a superposition of the rotational and linear components, and sum their energies:

E_k = (1/2)mv² + (1/2)mR²ω²

The second way is to consider the motion parametrically. The motion could look something along the lines of (letting the walker's velocity wrt the observer be u and the velocity of the rock in the observer's frame be v):

x(t) = Rcos(ωt) + ut
y(t) = Rsin(ωt)

Taking time derivatives of each coordinate yields

x'(t) = -Rωsin(ωt) + u
y'(t) = Rωcos(ωt)

The kinetic energy would then be

E_k = (1/2)m(x'(t)² + y'(t)²)

Which for u ≠ 0 is not time invariant. After considering how linear and angular velocity components can "cancel" each other if the linear velocity is not normal to the plane of rotation, a different answer is reached.

I can't convince myself which should be correct (if either). The problem I have with the first solution is that it doesn't consider how the two motions can cancel, and the problem I have with the second one is that the energy shouldn't change with time. For the energy to change, work would have to be done (and the energy would have to go somewhere) but the only force acting on the rock is the tension of the string, which doesn't move relative to the rock and shouldn't be doing any work.

I think that the second solution is not considering some sort of virtual force which would explain how energy is periodically removed and then restored, but I'm not sure why a virtual force would appear in what I think is an intertial frame (but it's possible that my understanding of inertial frames is wrong as well). The rock is of course accelerating with respect to the observer, but if you choose u = 0 then the two solutions yield the same answer and no virtual force is required to describe this scenario (despite the rock still accelerating in the observer's frame).

I'm in 3rd year engineering (and ashamed that I can't answer this) so I have a reasonably thorough math background - if an adequate answer requires some university-level math then don't hold back. Thanks in advance for any responses!

2. Oct 28, 2014

### Orodruin

Staff Emeritus
You are not allowed to split the kinetic energy into translational and rotational motion like that. You can split it into translational energy of the center of mass and due to motion relative to the center of mass (which will be rotation only for a rigid body), but the center of mass in your case is where the stone is. It is a useful and instructive exercise to prove this.

As the person is walking, the force in the observer system is not orthogonal to the velocity of the stone. Therefore, work is performed and the energy of the stone must change.

Edit: Translational and rotational, not rotational and rotational as I wrote first .......

3. Oct 28, 2014

### Chris L

Ah, that makes sense. I was considering the work done from the frame of the walking person (which would be zero), not the observer. The net force on the rock still points along the string but F•d isn't zero because of the non-circular path the rock takes in the observer's frame.

Followup question: where does the discrepency in the energy go? Is it supplied /absorbed by the person holding the rock?

4. Oct 28, 2014

### Orodruin

Staff Emeritus
Yes. If the stone does positive work on the rock, the rock does negative work on the person. If the person was passing by in outer space they would both circle around their common center of mass (much like the Earth and the Moon). If the person is walking on an (infinitely massive) object like the Earth, the work lost to the stone can be supplied by the force between the feet and the ground so the person could keep a constant velocity.

Anyway, I strongly recomment the exercise of showing that the kinetic energy can be split into translational energy of the center of mass plus motion relative to the center of mass. It is very similar to proving the parallel axis theorem for moment of inertia.