I know people will say it violates many laws of motion and conservation but could anyone explain why it is so? It's NOT a question about free energy.

Imagine a motor in space. It has an arm (rod) attached to it. At the end of the arm there's a ball (weight) attached to it. The motor produces a circular motion. So the motor is the center and the rod is the radius. To the ball there is a rope attached which connects to the rod aswell, but the rod keeps the ball in place. The rope is longer than the radius. The rod has a realease mechanism to let the ball go. When the ball is released, it disconnects from the rod and flies away in a tangent. Then the ball flies for some distance until it is stopped by the rope.

Does the sudden stopping of the ball by the rope make the motor move off its place and head in a single direction?

Is there any opposite reaction force to the direction of the moving ball on the moment that it is released?

Is there an opposite reaction force somewhere that makes this impossible and cancels out any potential movement causing forces all together?

Is it possible that this can make the system as a whole (motor, ball, rope and rod) move in a single direction without reaction mass being expelled and without external forces acting upon it? And if so, will this system stop moving when it catches up with its ball?

I appreciate help a lot.

  • 1
    $\begingroup$ "The motor produces a circular motion. So the motor is the center and the rod is the radius." Well, it depends on how heavy the motor is compared to the ball at the end of the rod. The system will (I think?) rotate about its center of mass, which in general will not be located at the motor, but will be somewhere between the motor and the ball. So the motor will already be moving around as is. To simplify matters, you can consider the system to be a dumbbell (motor and ball) and see how the math works out. $\endgroup$ – DumpsterDoofus Apr 15 '14 at 19:21
  • $\begingroup$ Ok I think the motor should have the largest mass, thus making the center of mass as close to the motor as possible. That way, I can call the motor the center turning point where the dumbell rotates around. $\endgroup$ – Arundel Apr 15 '14 at 19:34

As soon as the ball is released from the rod, it starts going in a tangential direction. Since there are no external forces on the system, the linear momentum of the system (ball + motor) must be conserved and the motor starts moving in the opposite direction of the ball with some velocity (depending upon the masses of ball and motor), so that the linear momentum of the system remains zero.

When the rope becomes fully stretched, the tension force in the rope acts to change the momenta of each individual subsystem (ball and motor) back to zero.

During all this, the center of mass of the system remains fixed.

  • $\begingroup$ When the ball is released from the rod, the motor starts moving in opposite direction of the ball, you say. But doesn't that opposite force just push the rod so the rod starts turning in the opposite direction (so that it's a torque?) and leave the motor at fixed position? $\endgroup$ – Arundel Apr 16 '14 at 19:07
  • $\begingroup$ It comes back to the fact that the motor and ball will be rotating around a common center of mass. Even if the motor is much larger than the ball so that the center of mass is close to the motor, it won't be centered exactly on the center of mass and so the motor will always have some motion in the opposite direction as the motion of the ball. $\endgroup$ – NeutronStar May 22 '14 at 3:47

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