Object moving radially outward on Rotating Flywheel

1. Dec 25, 2011

julienl07

Hello. This has been bothering me.

A point mass is on a rotating flywheel that has a constant initial angular velocity, ω0. The object (point mass), initially at some distance r0 from the axis of rotation, now moves out to a further distance rf, and then stops. Say the wheel has a moment of inertia Iw, and the mass is m.

If the system includes the wheel and object, its total moment of inertia is

I = Iw+mr2

By conservation of angular momentum

ωf0$\frac{I_{0}}{I_{f}}$

If the initial rotational kinetic energy of the system was $\frac{1}{2}I_{0}ω_{0}^{2}$, the final is $\frac{1}{2}I_{f}ω_{f}^{2}$ = $\frac{1}{2}I_{0}ω_{0}^{2}\frac{I_0}{I_f}$ = $E_{rk_0}\frac{I_0}{I_f}$. Ignoring the gory details, the rotational kinetic energy of the system has decreased, but no external work has been done.

In my ideal situation, I imagine that the object is attached to the axis of rotation by a cord, and that there are 'guide walls' forming an 'aisle' that goes radially outward from the axis. The tension in the cord is momentarily lowered, and the object moves radially outward through this 'aisle', before the cord is held still. The only interaction that I can think of (other than from the cord) is a normal force between the guide wall and the object. This force both increases the speed of the object, and slows the angular speed of the wheel by a torque, so that the angular velocity of the object and wheel remain equal as it moves outward.

So, sure internal work has been done from the torque, but no external work has been done.
How is it that the total rotational/kinetic energy of the system has changed, but no external work has been done? No heat was generated or anything.

Is there something that I have not seen? Can anyone explain this?

Thanks.

Last edited: Dec 25, 2011
2. Dec 26, 2011

AlephZero

You forgot the work done by the cord when the particle moves radially. That's where the change of KE comes from.

3. Dec 26, 2011

rcgldr

Imagine that a coiled like spring (like a clock spring) is being wound up as the cord lenthens. The decrease in kinetic energy of the system would equal the increase in potential energy of the spring. If there was no friction or any losses, such a system could oscillate between a minimum and maximum radius for the object, exchanging kinetic for potential energy and vice versa.

Last edited: Dec 26, 2011
4. Dec 26, 2011

julienl07

Thanks for the replies.

Although I didn't mention it, I know that work is done by the cord. The cord pulls towards the axis while the object moves away from the axis, thus the work is negative.

But again, this is internal work. By definition, if no external work has been done, then the total energy of the system (kinetic, potential, heat) should remain constant, right?

@rcgldr:

Are you suggesting that the cord is elastic? (You say that it would oscillate between maximum and minimum radius.) I meant that the cord does not wind around the axis, and that the object moves "infinitely slowly" outward before stopping (and remaining) at some new, greater radius (theoretical ideal system). Unless I misunderstood you, Is there some sort of potential gain by being at a greater radius?

Thanks.

5. Dec 26, 2011

AlephZero

If you define "the system" as including whatever changes the tension in the string, then yes this is internal work.

But conservation of total energy is not the same as conservation of kinetic energy. Some other form of energy is being converted into KE when the string changes length. For example if the length of the string was being controlled by an electric motor/generator. you would need to supply electrical energy to the motor to wind the string in and reduce the radius of the mass (and increase the KE of the rotating system), or the generator would supply electrical energy as the string moved out and the KE of the system decreased.

6. Dec 26, 2011

rcgldr

Not the cord, but a wind up coiled spring attached to post that the cord is wrapped around on, so that the cord can move inwards and unwind the spring, or outwards and wind the spring. Since there's no dampening (friction) the system would tend to oscillate. If there was some type of latch mechanism to release and then stop the spring without any losses, then that would allow the radius to be changed without oscillation.

The potential gain only exists because the spring winds up as the radius increases, which increases the potential energy in the spring. The point of using the spring was to create an example where an internal potential energy increases as the radius increases, so that total energy of the system would remain constant.

If you allowed external work to be done, then the cord could pass through a hole at the center of the plane, with the object sliding around the hole at various distances (radius) on a frictionless surface. Then external work could be done by pulling the cord inwards through the hole, or negative work done by allowing the tension to pull the cord outwards. The work done would be equal to the integral of tension as a function of distance over the distance the cord moved. Tension will equal some constant / r3 (if the radius is doubled, the tension is 1/8th of what is was before).