Classical Mechanics contradicts Conservation of energy? 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.

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?
 A: 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.)
A: 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.
A: 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.
A: 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.
A: 
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.
