Kinetic energy of a rotating and translating body?

[Chris L]

Homework Statement

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.

Homework 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

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

Which gives a time-invariant answer

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!

 

[Orodruin]

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 ...

 

[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?

 

[Orodruin]

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?

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.

 

[HalfLostAndSearching]

I would approach this problem by considering the conservation of energy. In this scenario, the rock is experiencing both rotational and translational motion, so we need to consider the total kinetic energy of the rock, which is the sum of its rotational and translational kinetic energies.

The rotational kinetic energy of the rock is given by (1/2)Iω², where I is the moment of inertia of the rock and ω is its angular velocity. The translational kinetic energy is given by (1/2)mv², where m is the mass of the rock and v is its linear velocity.

In this scenario, the rock is experiencing a constant angular velocity and a constant linear velocity, so its kinetic energy will also be constant. As the rock rotates and translates, the energy is being constantly transferred between the rotational and translational components, but the total energy remains the same.

So, to determine the total kinetic energy of the rock in the observer's frame, we can simply sum the rotational and translational kinetic energies:

E_k = (1/2)Iω² + (1/2)mv²

This approach accounts for the fact that the two motions can cancel each other out and result in a constant total kinetic energy. It also takes into consideration the conservation of energy, as the energy is being constantly transferred between the two components, but the total energy remains the same.

In regards to the second solution, which considers the motion parametrically, it is important to note that the equations for the linear and angular velocities are not independent of each other. The linear velocity (u) is dependent on the angular velocity (ω) and the radius of rotation (R), as seen in the equations x'(t) and y'(t). So, while the kinetic energy may appear to change with time in this approach, it is actually a result of the changing angular velocity, which is dependent on the linear velocity.

Overall, the first approach is a more comprehensive and accurate way to determine the total kinetic energy of the rotating and translating body. It considers the conservation of energy and takes into account the interdependence of the linear and angular velocities.