Q:

I need help figuring out what is wrong in this aspiring perpetuum mobile

Credits: My question is motivated from a question from another user (https://physics.stackexchange.com/q/143377/), I just reformulated what I think he tried to ask into, what seem to me, simpler terms.
The system starts with a disk, attached to an arm, whose center revolves around another axis, as shown in the figure. The kinetic energy of the system can be expressed as $$E_k=\frac{1}{4}mr^2\omega_2^2+\frac{1}{2}md^2\omega_1^2$$ Where r is the radius of the disk, $d$ is the length of the arm,  $\omega_1$ is the angular speed of revolution of the arm and $\omega_2$ is the angular speed of rotation of the disk in the lab's reference frame. 
In addition, there is no gravity here.
Note: This expression for the kinetic energy seems to be correct (see question Is this expression for the kinetic energy of a spinning disk revolving about a second axis correct?) 

Dynamics: 
1) At $t=t_1$
a) The disk does not rotate around its center of mass (as it revolves around the center), a a vertical arrow drawn on the disk will remain vertical as the disk revolves.
b) There is no friction between rail and disk
c) The Kinetic energy of the system will be $$E_k= \frac{1}{2}md^2\omega_1^2$$
2) At $t=t_2$
a) The disk does not rotate around its center of mass (as it revolves around the center)--
a vertical arrow drawn on the disk will remain vertical as the disk revolves.
b) Friction between rail and disk is switched on (the arrows in the figure show the directions of the force felt by the disk). Friction at the top and bottom are made slightly different so the total torque relative to the center is zero and the revolving speed stays constant at $\omega_1$
c) The Kinetic energy of the disk system (not including the annular rail) will start to increase to $$E_k=\frac{1}{4}mr^2\omega_2^2+\frac{1}{2}md^2\omega_1^2$$
3) At $t=t_3$
a) The disk reached a rotating speed $\omega_2=\omega_1$, friction stops and the disk rotation and revolution are locked:  the same point on the disk will keep facing the center as it revolves around it)--an arrow drawn on the disk will rotate and remain pointing parallel to the arm as the disk revolves.
b) Friction is switched off.
c) The Kinetic energy of the disk system becomes $$E_k=\frac{1}{4}mr^2\omega_1^2+\frac{1}{2}md^2\omega_1^2$$
My guess is that the circular ring stays at rest because the torque remains zero, so the work from the friction forces it feels from the disk will be dissipated as heat. 
Conclusion: 
The kinetic energy of the system seems to increase from $$E_k= \frac{1}{2}md^2\omega_1^2$$ to $$E_k=\frac{1}{4}mr^2\omega_1^2+\frac{1}{2}md^2\omega_1^2$$ In addition, heat is generated and dissipated, both without any apparent source. But this is not possible (assume that this system is floating isolated in space), so, what is going on here? Thanks!

A:

so the total torque relative to the center is zero and the revolving speed stays constant at $\omega_1$

This is not possible. When the two forces are unequal, there is a net force on the disk which means its center of mass will decelerate. 
The net angular momentum of the system (disk plus arm) remains constant - because there is no net torque on the system according to the constraint you gave. But you transfer angular momentum from the center of mass to the rotation of the disk - $\omega_1$ will decrease. And since the energy stored can be written as $\frac12 I \omega^2$, when angular momentum is transferred from the enter of mass and shared between two components, each with smaller $L$, then less energy is stored in the motion.

