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?)

enter image description here


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!

  • $\begingroup$ You cannot expect the energy to be conserved in a system with time-dependent forces (as the friction in this case). $\endgroup$ – yuggib Nov 3 '14 at 21:16
  • $\begingroup$ I do not say it is conserved, but it is changing in the wrong direction. There is an increase in kinetic energy (instead of a decrease) that could be extracted back as work to return to the initial condition, and also heat generation from which a small proportion can also be converted into work. $\endgroup$ – Wolphram jonny Nov 3 '14 at 21:20
  • $\begingroup$ In addition to my previous comment, I do not understand the system very well. The "friction" you utilize is strange...kinetic friction opposes to the motion (and causes dissipation), but in your case FR1 does not seem to oppose the motion. If it is an external force applied, then as I said you cannot expect energy conservation because it is time-dependent, and energy increase is possible. $\endgroup$ – yuggib Nov 3 '14 at 21:38
  • $\begingroup$ to the kind gentleperson who voted to close, this is not a personal theory, it is about finding what mistake I am doing in a standard mechanical reasoning. I do not believe that perpetual mobile devices could exist. $\endgroup$ – Wolphram jonny Nov 3 '14 at 22:58
  • $\begingroup$ @yuggib 1) but kinetic friction is opposing the motion! think it this way: imagine an arrow pointing down when the disk is at t2 is printed on the disk. Without friction, the arrow will still be pointing down after a small clockwise displacement. So the rails sees the disk moving counterclockwise, and that is why the friction forces point that way. $\endgroup$ – Wolphram jonny Nov 3 '14 at 23:04

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.

  • $\begingroup$ I see. So if the forces were equal there would be a torque that will also slow down the center of mass, plus in this case angular momentum would not even be conserved. Correct? $\endgroup$ – Wolphram jonny Nov 4 '14 at 1:02
  • $\begingroup$ Well - angular momentum would be transferred to "the world". Because it is always conserved - you just may have to look at the bigger picture and the external forces. $\endgroup$ – Floris Nov 4 '14 at 1:18
  • $\begingroup$ Just one more question. If the frictions were applied in a direction parallel to the arm (by using some adequately shaped pliers attached to the arm instead of the annular rail) then there would be no net force on the disk that would cause the center of mass to decelerate (or I am still wrong?). What would be the explanation in this case? $\endgroup$ – Wolphram jonny Nov 4 '14 at 2:10
  • $\begingroup$ You would have to do work with the pliers to move the disk... $\endgroup$ – Floris Nov 4 '14 at 2:22
  • 1
    $\begingroup$ I know, I just completely forgot about it! my mind is deteriorating faster than I thought. I would have been able to spot this in a split second when I was in my 20's. See you! $\endgroup$ – Wolphram jonny Nov 5 '14 at 17:16

1) When there is friction at the disk there is a force on the arm that is transmitted to the central axle that torques the central axle to rotate around the disk in the opposite sense (and if floating in space would keep the total of angular momentum constant). If the central axis is fixed at this time, say to the ground, then there is a force from the earth that is changing the energy and the total angular momentum.

2) After the disk is phase locked to the axis, the system is rotating as a solid figure and there is no rubbing and no dissipation at the disk.

  • $\begingroup$ 1) Is this still true if both friction forces differ in magnitude so the total torque is zero? (I mentioned that at point 2b) ? 2) I agree with that is why i mentioned that friction was switched off (either explicitly or because of the internal dynamics. I chose to turn them off explicitly to be on the safest side) $\endgroup$ – Wolphram jonny Nov 3 '14 at 21:11
  • $\begingroup$ The friction at the disk torques the rail and acts to reduce $\omega_1$ and compensates for increase of $\omega_2$, thus keeping total angular momentum constant. The friction at the axis acts in the same direction (reduces $\omega_1$) so the total torque there cannot be zero. Floating in space there is no friction at the axis but locked down on the Earth, the system looses angular momentum due to friction at the axis and gives it up to the Earth. Energy is lost from friction to heat. To do this properly for floating in space one needs to consider $\omega_3$, angular velocity of the c.o.m. $\endgroup$ – eshaya Nov 4 '14 at 18:58

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