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Imagine a Stanford torus rotating with 1 rpm so that centripetal/reactive centrifugal acceleration provides about 1.0g of artificial gravitational acceleration inside the ring. The picture below shows a Stanford torus in a cutaway view.

Stanford torus

The rotating torus has a specific rotational energy (let's say 100 J). A mass of 1 kg is located 1 m above the ground inside the torus.

Step 1:

The mass is dropped. The torus' moment of inertia increases, its angular velocity decreases. The rotational energy of the torus does not change.

Rotational energy of the torus after step 1: 100 J.

Step 2:

¾ of the torus' rotational energy is converted into electric energy. We assume no heat losses in the process. The electric energy is temporarily stored.

Rotational energy of the torus after step 2: 25 J.
Stored electric energy: 75 J.

Step 3:

The mass is lifted again. The torus' moment of inertia decreases, its angular velocity increases. The rotational energy of the torus does not change.

Rotational energy of the torus after step 3: 25 J.
Stored electric energy: 75 J.

Step 4:

The stored electric energy is converted into rotational energy of the torus. Again, we assume no heat losses in the process.

Rotational energy of the torus after step 4: 100 J.
Stored electric energy: 0 J.

The sum of rotational energy and electric energy is equal in all steps. The critical actions are accelerating/decelerating the torus. When rotational energy is invested in or taken out of the torus, not only does the rotational energy of the torus change but also the potential energies of the masses inside the torus. In step 2 potential energy "disappears" and in step 4 potential energy is "created". Energy is gained in this example since the mass is lifted under less acceleration and is dropped when acceleration is greater. But the process can be done the other way around.

This would indicate a contradiction of classical mechanics and energy conservation. Do you see any mistakes or something that I missed?

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    $\begingroup$ don't see in your energy budget the energy required to 'lift' the 1kg mass. Did I miss it? $\endgroup$
    – tom
    Dec 12, 2014 at 18:09

5 Answers 5

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In step 1 you lower the mass and this generates energy. Let's say you store this energy is a spring, and for the sake of argument let's say the energy stored is 1J. The energy has to come from somewhere, and of course it comes from the rotational energy of the torus so the rotational energy of the torus is now 99J.

In step 2 you slow the torus, perhaps by spinning a counterwheel to conserver angular momentum, so if you store ¾ of the energy the stored energy is 74¼J and the torus energy is 24¾J.

Now you lift the mass back up using some of the energy you stored when you lowered it. This energy goes into rotational energy of the disk. Rather than do the sum properly let me guestimate that it takes a quarter of the energy you got from dropping the mass to lift it again. The mass is back where it was, you have ¾J left in your spring and the torus energy is now 24¾ + ¼ = 25J.

Finally you release the 74¼J you have in your spinning counterweight to spin up the torus again. The torus now has 25 + 74¼ = 99¼J, and you have ¾J left in your spring. The total energy is 99¼ + ¾ = 100J.

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  • $\begingroup$ @ John Rennie. You did a circular argument in my view. You start by assuming that energy is conserved ("The energy has to come from somewhere") to prove that energy is conserved. I think the disagreement comes down to this: "When a mass falls inside of a Stanford torus (and does work while doing so), does angular momentum of the torus change?" My answer is No, your answer is Yes. $\endgroup$
    – Chris
    Dec 13, 2014 at 11:11
  • $\begingroup$ My answer is No - Your answer disagrees with experiment $\endgroup$
    – Christoph
    Dec 13, 2014 at 12:08
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    $\begingroup$ @Chris: you're quite correct that I assume energy is conserved, but then if it wasn't all of modern physics would come tumbling down around our ears. Energy can't not be conserved without violating some pretty fundamental symmetries of the universe. The point of my answer is to show how the energy is moving around your system and explain why the initial impression that energy isn't conserved is wrong. If you want to start from the position that energy isn't conserved then, well, you're on your own because for a start we couldn't use Newtonian mechanics to analyse the system. $\endgroup$
    – John Rennie
    Dec 13, 2014 at 14:04
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Actually, if the mass comes to rest relative to the torus after landing, the energy of the system goes down. Let $I$ be the moment of inertia of the torus, $r$ be the radius of the mass from the axis of rotation before it is dropped, $R$ be its radius after it lands, $\omega_1$ be the angular velocity before the mass is dropped, and $\omega_2$ be the angular velocity after. Then $I_1 = I + mr^2$ and $I_2 = I + mR^2$.

Conservation of angular momentum (no external torque applied to the system) says: $L_1 = I_1 \omega_1 = L_2 = I_2 \omega_2$.

Therefore, $\omega_2 = \dfrac{I_1}{I_2}\omega_1$. Meanwhile, $E_1 = \dfrac{1}{2}I_1 \omega_1^2$, and

$E_2 = \dfrac{1}{2}I_2 \omega_2^2 = \dfrac{1}{2}I_2\dfrac{I_1^2}{I_2^2}\omega_1^2 = \dfrac{1}{2}I_1 \omega_1^2 \dfrac{I_1}{I_2} = \left(\dfrac{I_1}{I_2} \right)E_1$.

Since $I_1 < I_2$, some energy is lost. Where does it go? Into heat; the collision of the mass with the outer radius of the torus is inelastic because the mass sticks to the outer radius. (This is the old sand-on-the-conveyor-belt paradox in disguise.)

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  • $\begingroup$ @ pwf. I agree, in reality heat losses would slow down the system. But the focus here is the relationship of rotational kinetic energy and potential energies. I.e. when the torus is accelerated from standstill two energy forms develop: (a) rotational kinetic energy of the torus, (b) potential energies of the masses inside the torus. But you only have to "invest" the rotational energy, the potential energies of the masses are created out of nothing (or from a source we don't know). $\endgroup$
    – Chris
    Dec 13, 2014 at 11:28
  • $\begingroup$ @Chris, There isn't really any potential energy in this situation, as seen from an inertial frame of reference. In the rotating frame there is a sort of fictitious potential energy (associated with the fictitious centrifugal force), but analyzing the problem in those terms masks where the energy is actually coming from and going to. (For example, if an electric motor lifts the weight back up in Step 3, the energy of the torus-and-mass system is restored, but it comes from whatever is fueling the electric motor; mechanical energy is not conserved.) $\endgroup$
    – pwf
    Dec 15, 2014 at 17:28
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Step 1: the rotational energy $E=1/2 \omega L$ does increase because $\omega$ increases (bacuse I decreases) and L is a constant ( $L=\omega I$). This is wrong again in step 3:

The rotational energy of the torus does not change

So the potential energies of the lifted objects is not the same in both situations, because they will also convert rotational into potential energy. The more rotational energy the more the potential energy acquired by the lifted objects when they are raised.

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  • $\begingroup$ @ Wolphram jonny. You did the formula wrong. Rotational kinetic energy = 0.5 * moment of inertia * (angular velocity)². When the mass falls, moment of inertia increases and angular velocity decreases, keeping angular momentum constant. When angular momentum does not change, rotational kinetic energy does not change. $\endgroup$
    – Chris
    Dec 13, 2014 at 11:17
  • $\begingroup$ @Chris but if you replace in your equation $L=\omega I$ you get mine. My equation is not wrong is just another way to write it in terms of L intesad of I. $\endgroup$
    – user65081
    Dec 13, 2014 at 16:37
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Moving the mass inside the rotating object takes energy - when it is "lowered" against the artificial gravity it does work against the force resisting this motion so the final kinetic energy of the system with "lowered" mass is not the same as before. The equation:

$$E=\frac12I\omega^2=\frac{L^2}{2I}$$

is the clearest way to see why this is so - when angular momentum is conserved but moment of inertia goes up, stored energy goes down.

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Step 1 The mass is dropped ... The rotational energy of the torus does not change.

Going to stop you right there. Although total energy is indeed conserved, rotational energy does not need to be conserved. Remember that the potential energy of dropped objects is generally converted to heat when they go thud on the ground. Angular momentum is however conserved in this step.

In fact, it is simply impossible for rotational energy ($L^2/2I$) to stay the same when a rotating object adjusts its mass distribution without interacting with other objects (changing moment of inertia $I$ while keeping constant angular momentum $L$). When the mass drops, energy is released and if not captured that energy goes to heat.

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  • $\begingroup$ @ Nanite. "In fact, it is simply impossible for rotational energy (L2/2I) to stay the same when a rotating object adjusts its mass distribution without interacting with other objects". This is not correct. The adjustment comes from changing angular velocity. A good real world example is a figure skater doing a pirouette. Remember that angular momentum is the product of moment of inertia and angular velocity. $\endgroup$
    – Chris
    Dec 13, 2014 at 12:10
  • $\begingroup$ Indeed the classic skater example illustrates what is going on. Suppose the skater starts a spin with moment of inertia I = 2 kg m^2 and angular momentum L = 10 kg m^2/s. He then pulls in his arms, decreasing to I = 1 kg m^2 and maintaining constant L. His rotational speed doubles. In this case the rotational energy L2/2I increases from 25 J to 50 J. The energy is provided by the arms pulling inwards, i.e., chemical energy has been converted to rotational energy. $\endgroup$
    – Nanite
    Dec 13, 2014 at 13:39

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