1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Minimum velocity for particle to lose contact

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

    ?temp_hash=836da03d58b2dee809e9158633a204bb.gif

    2. Relevant equations


    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
      rod.GIF
      File size:
      24.5 KB
      Views:
      194
  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

    mfb

    User Avatar
    2016 Award

    Staff: Mentor

    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

    ehild

    User Avatar
    Homework Helper
    Gold Member

    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

    TSny

    User Avatar
    Homework Helper
    Gold Member

    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

    TSny

    User Avatar
    Homework Helper
    Gold Member

    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

    TSny

    User Avatar
    Homework Helper
    Gold Member

    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

    TSny

    User Avatar
    Homework Helper
    Gold Member

    Yep.
     
  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

    TSny

    User Avatar
    Homework Helper
    Gold Member

    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: .
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Minimum velocity for particle to lose contact
  1. Minimum velocity (Replies: 1)

Loading...