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Minimum velocity for particle to lose contact

  1. May 30, 2015 #1
    1. The problem statement, all variables and given/known data


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    3. The attempt at a solution

    Honestly I have very little idea about this problem . The contact force has to vanish for particle B to loose contact .Not sure how to relate the velocity of upper particle with the normal contact force of the lower particle .

    I don't think any conservation laws would work.

    I would be grateful if somebody could help me with the problem.

    Attached Files:

    • rod.GIF
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  2. jcsd
  3. May 30, 2015 #2
    Assuming that the rope have constant length and the B particle can move only horizontally, this problem must ask for the limit velocity of A just it is over B particle. If A move to right, the distance go to be larger that impossible, so B start to move with some acceleration. Think about what can be this acceleration.

    Or, change frame, take origin to A particle so the problem is the same that to give horizontal velocity ##-v## to B particle. The rope is constant (again).
  4. May 30, 2015 #3


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    If B loses contact, there is no force between B and the ground any more. What happens to the center of mass? Where is B relative to the center of mass, where is it relative to the ground?
  5. May 30, 2015 #4
    It is a limit process ##t\to0##.
  6. May 30, 2015 #5
    The CM moves with speed v/2 towards right . The only external force acting on the system would be gravitational force .

    B moves with speed v/2 towards left relative to CM .

    Since it just looses contact , it is effectively at rest relative to the ground .
  7. May 30, 2015 #6


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    I haven't solved the problem, but would start as follows.

    Remember, motion of a rigid body can be considered as motion of the CM with acceleration determined by the external forces and rotation about the CM according to the external torques.
    At t=0, the system has some momentum and angular momentum.
    You can write the motion of ball B in terms of the motion of the CM and the rotation about it. B should stay on the ground or lift from it: its vertical acceleration can not be negative even without a normal force.
  8. May 30, 2015 #7
    Sorry , I can't see how to implement the above strategy to deal with this problem.
  9. May 30, 2015 #8


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    If B were to stay in contact with the floor, what would be the height, ##y##, of the CM of the system as a function of angle of rotation of the system?

    Can you use this relation to find an expression for ##\ddot{y}##?
  10. May 31, 2015 #9
    Taking upward positive and ##\theta## to be angle made by rod with the vertical .

    ##N - 2mg =2m\ddot{y}##

    ##\ddot{y} = -\frac{l}{2}[cos\theta \dot{\theta}^2+sin\theta\ddot{\theta}]##
  11. May 31, 2015 #10


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    What doe the second equation reduce at the time of interest?
  12. May 31, 2015 #11
    Are you getting v = 2m/s ?
  13. May 31, 2015 #12


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    My answer is in terms of the length of the rod and g. I haven't yet put in numbers. Lets see....

    Yes, I get 2 m/s.
  14. May 31, 2015 #13
    ##v=\sqrt{2gl}## . Right ?
  15. May 31, 2015 #14


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  16. May 31, 2015 #15
    I have been thinking about this problem for last two days and here you come and make this problem look so ordinary .

    How do you think so clearly ?
  17. May 31, 2015 #16


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    I don't know. It always starts out kind of foggy and then the fog lifts (sometimes :smile:).

    You got the answer very quickly after I gave a hint. So, I think I might have given too much away. It's always difficult to find the right nudge.
  18. May 31, 2015 #17
    Thank you very very much .

    Good night :oldsmile: .
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