## The eccentric wheel.
I found this picture years ago on a website that is now dead. I removed distractions from it, reducing it to its simplest form.
Comments: To satisfy the conditions of equilibrium, the system must not only have a net force of zero and a net torque of zero acting on it, but also be properly held or constrained on its supports. The more stable the supports the more stable is the equilibrium, that is the equilibrium is achived more easily. If the system has continual torque imbalance, why can't it turn continually?
## Solution.The rotatable system does not have torque imbalance. Several observations all lead to that conclusion.Of course we see that as the wheel is turned, every roller gains the same potential energy on one side as it loses on the other during a full rotation. I said "is turned" not "turns", for it won't turn by itself. Case closed. We also see that rotation of the wheel by 1/8 turn gives the very same configuration, so it won't make that move by itself. Again, case closed.
But the most basic reason this won't work has to do with a fundamental principle of physics—the definition of work. Work is defined to be W = Fs(cosθ) where θ is the angle between and F. When two objects are in contact they exert forces on each other that are perpendicular to the tangent of the surfaces at the point of contact. Therefore the motion of one body sliding on another is always in the direction of this tangent. The force at the contact point is perpendicular to the displacement and zero work is done by this force. No work is done on or by bodies sliding along a frictionless unmoving surface. sThe wall is a rigid, unmoving frictionless surface. It does no work on the rotatable system. This is true for any shape of wall.
## Looking deeper.The system has positions of equilibrium. If the picture shows one of them, there are seven more since rotation by 1/8 revolution looks the same. In the position shown the center of mass of the wheel is on a level line through the axle, and is to the left of the axle.
Now imagine turning the wheel 1/16 revolution, so the rollers are between the positions shown. Where is the new center of mass? Take pairs of rollers on the same vertical. Their center of mass is at the same level as the axle. Add all pairs, and the center of mass of the system is again on the same level as the axle. Is it possible this cockeyed device has the same center of mass in
Torques present an interesting slant on this, and are the final nail in its coffin. While the net torque on the If the curved wall weren't there, the springs and rollers would be extended to their limit and the system would be balanced in all positions. The net torque on the system would be zero. This is situation (A). Now put the curved wall in place and you have situation (B) shown in the diagram. What has changed? We have added forces of the ramp acting on the rollers. But these forces all have zero torque for they all pass through a common point. So the net torque on the wheel is still zero. Also, they are paired with equal and oppositely directed forces due to the springs. These are all internal forces in the system, and by Newton's third law such internal forces always sum to zero. We use here an important fact about torques. We can choose any point as a center of torques, and if the net torque is zero about that point it will also be zero for any other point we choose. In this case we make our task easy by choosing the center of the wall's circular arc as the center of torques. Any force whose line of action passes through the center of torques has zero torque because its lever arm is zero.
This process for reaching this result was subtle, and perhaps devious. Here's an elaboration.What about the springs? They are irrelevant to our conclusion, for each spring compresses, then decompresses by the same amount when the wheel is rotated a full circle. Net work done: zero, assuming the springs are perfectly elastic, obeying Hooke's law. If the springs are not perfectly elastic, there will be dissipative heating of them, losing energy and slowing the wheel. Also, the springs exert forces all passing through a common point, the axle, so they all exert zero torque on the system. The springs' forces are internal to the system and in a force analysis they add nothing to the net force on the system. This analysis is overkill, but instructive. There's a whole class of unworkable perpetual motion proposals that use internal or external fixed constraints to move masses on a wheel as it is rotated. Similar analysis applies to them all. Case finally closed for good.
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