For instance, consider a weight on one end of the ring. Assume that the ring has negligible mass compared to the weight. When the weight splits into two, moves around the ring and recombines at the opposite end of the ring, is it true that the center of mass moves without the ring moving in opposite direction?
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$\begingroup$ How does a ring have and end to attached a weight to? Maybe a diagram is in order here. $\endgroup$– John AlexiouNov 3, 2016 at 19:04
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$\begingroup$ Is there anything in contact with the ring other than the weights? $\endgroup$– garypApr 26, 2018 at 20:31
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$\begingroup$ If the weight splits, it is no longer a rigid body. $\endgroup$– John AlexiouApr 27, 2018 at 3:15
3 Answers
No. The center of mass will not move, the ring will.
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$\begingroup$ Yes, there is force on it due to the sliding bodies. $\endgroup$ Jan 24, 2015 at 5:35
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$\begingroup$ But aren't the forces acting on the ring balanced each other? When the split objects moving around the ring, isn't the centrifugal force on the ring during the first half of the ring equal to that on the ring for the second half of the ring? $\endgroup$– tzw101Jan 24, 2015 at 5:53
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$\begingroup$ Forces could cancel their effect on the ring if they acted at the same time. In this case, sum of forces by the two bodies on the ring is oriented oppositely to the velocity of the bodies' center of mass during motion in the first quarter of the circle and is oriented along that velocity in the second quarter of the circle. During both periods the net force gives the ring momentum of same magnitude and opposite direction. But, because the second period happens after the first, the action of forces do not cancel at any time and the ring moves. $\endgroup$ Jan 24, 2015 at 15:59
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$\begingroup$ When the bodies meet again on the other side of the ring, the bodies and the ring cease to move but both the ring and the bodies have moved in the opposite directions. So the two periods cancel in the sense that ring stops when the bodies meet again, but not in the sense that ring does not move at all. $\endgroup$ Jan 24, 2015 at 16:00
In general ,you can say that the center of mass of a system tends to remain in its state of uniform velocity or of rest unless the whole system is being acted upon by an external force. In this case the center of mass of the whole ring and the weight system was initially at rest . So when the weight splits there has been no external force on the whole system . As such the two fragments of weight and the ring will always move in such a way that the velocity of the center of mass remains zero and at any instant of time the coordinates of the two fragments and the center of the ring will remain in such a manner that the center of mass stays intact at its initial coordinate . Although I must say that the neglecting of the mass of the ring is a bit fishy cause that is dictating you to conclude that the center of mass was initially at the initial position the weight and hence after the collision again it gives you the impression of shifting to the new position of the pieces of the weight.
By Newton's first law, the center of mass of any body set can moves at any uniform velocity without any force internal or external.