# rotating mass

by imanator
Tags: action/reaction, force and torque, mass, momentum and energy, rotating
 P: 8 Hi there, I'm new to the forum, but hopefully this question has a simply answer. my question is, suppose you have a mass (a) on the end of a string or rod attached to a vertical support which is grounded. The mass is rotating about that support. Now assume you have another mass (b) traveling towards the rotating mass along a path, so that at one point, it is tangential to the radius of mass (a)'s rotation. when mass (b) reaches the rotating mass (a) and collides with it there will be an action reaction. my Question is will the reaction felt by mass (a) be through the centre support and in the same direction as mass (b) was going? or will it just induce a torque at the centre point, in other words if there was a force sensor at the centre point that was directed back (in the direction that mass (b) was coming from, would it register the same reaction force that object (b) felt when it hit object (a)? and what about the momentum of the system? I hope this question is not too difficult to follow. I Attached a sketch for guidance. I have done high-school physics and university physics, so I should be able to follow most replies. Thanks for your help. Regards. Attached Thumbnails
 PF Gold P: 5,707 attachment is unreadable even at max magnification
 P: 8 Thanks for that, I uploaded a better one.
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## rotating mass

A force sensor at the center won't feel the collision itself as there are no radial momentum changes. Afterwards, the changed rotation speed leads to a different radial force.
 P: 8 thanks mfb, not sure what you mean by afterwards the changed speed leads to different radial force though. are you referring to the fact the fact that the rotating mass has slowed down will result in a lower v^2/r value?
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P: 10,846
 Quote by imanator thanks mfb, not sure what you mean by afterwards the changed speed leads to different radial force though. are you referring to the fact the fact that the rotating mass has slowed down will result in a lower v^2/r value?
Right.
 P: 8 does everyone else agree with this?
 Sci Advisor PF Gold P: 11,383 More or less, I think. The pivot, holding the string will be experiencing a radial force (m vsquared/ radius), which constantly changes direction as the mass orbits it. I assume that the pivot is on a massive enough base to prevent any significant motion (wobble). If the new mass hits the orbiting mass tangentially and the two values of momentum add to zero then the two masses will end up at rest and there will be no tension in the string. If the resulting collision changes the tangential velocity to any other value, the tension in the string will be m newvsquared / radius. This tension will, initially, be at right angles to the path of the arriving mass. The force on the pivot can only ever be radial, along the string.

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