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Rotational mechanics sphere collision problem

  1. May 24, 2012 #1
    1. The problem statement, all variables and given/known data

    Hello everyone,
    Please help me with this problem, as I am getting stuck solving this problem.

    A solid sphere of Mass M and Radius R rolling (pure) encounters a protrusion on the surface of height R/4. Its in the form of a step. What can be the maximum velocity V of sphere so that it doesn't leave contact from surface and final velocity?

    Attached below is a figure:



    2. Relevant equations
    ζ=Iα
    I=MR.R.2 / 5 for sphere
    MV=P.dt where P is impulse

    3. The attempt at a solution

    Assume the impule duration is dt
    Impulse force in in X and Y direction is X,Y

    I used linear conservation equations.
    Then conserve angular momentum about the corner point.
    Also The torque about the same point which is only due to weight.
    The normal force from ground and friction will be 0 just after it lifts off.
    The velocity component would be perpendicular to line joining corner to center of sphere for it to not loose contact.
     
  2. jcsd
  3. May 24, 2012 #2

    tiny-tim

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    welcome to pf!

    hi shreyashebbar! welcome to pf! :smile:
    (your attachment has disappeared, it is https://www.physicsforums.com/attachment.php?attachmentid=47577&d=1337847790)

    you need to find the centripetal acceleration

    from that, find the reaction force (from the step) :wink:
     
  4. May 24, 2012 #3
    Hi Well thats a hint I wrote the following equations.
    X any Y are reaction forces on the sphere
    and assuming the impulse for time delta T
    Ac is centripetal acceleration
    Vx1 and Vy1 are velocities just after the impact and the ball tries to lift off.

    Let me know if my equations are right. An can make use of energy conservation:
    Kinetic energy(Linear+Rotational) initially = Kinetic energy(Rotation about the corner + Velocity) + Potential energy for raising it till r/4?

    Or do I use the torque equation which would be just because of the own weight?

    Thanking you for all the help.
     

    Attached Files:

  5. May 24, 2012 #4

    tiny-tim

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    no!

    this is a collision

    energy is not usually conserved in collisions!

    you can only use conservation of momentum and conservation of angular momentum

    start again :smile:
     
  6. May 24, 2012 #5
    Hi,
    Well I have written down the 7 equations by conserving angular momentum about the corner of step and linear momentum along x and y axes just after the collision. Do let me know if it is correct :)
     

    Attached Files:

  7. May 24, 2012 #6

    tiny-tim

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    (on second thoughts, we don't know the direction of the impulse, so conservation of linear momentum won't help :redface:)

    i don't follow your conservation of angular momentum equation :confused:

    you need the angular momentum before, and the angular momentum after

    (and no force or torque: the weight, mg, is not impulsive, so it doesn't contribute in the infinitesimal time between "before" and "after")
     
  8. May 24, 2012 #7
    Hi I messed up the previous solution I think.
    Here are the new equations. Hope this one is the correct approach ... :x

    If not correct kindly let me know how do I go about this.

    Thanking you in anticipation
     

    Attached Files:

  9. May 24, 2012 #8

    tiny-tim

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    fine down to v0 = 28/23 v1 :smile:

    after that, i'm confused :confused:

    what does "at" mean?

    you simply need an equation relating the centripetal acceleration to the weight and the reaction force

    (and remember you know what cosθ equals)
     
  10. May 24, 2012 #9
    Hi,
    At i have assumed as the tangential acceleration of the centre of sphere about P.
    Is that assumption wrong as well?

    Regards
     
  11. May 24, 2012 #10

    tiny-tim

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    you don't need tangential acceleration, you're not interested in tangential motion
     
  12. May 24, 2012 #11
    oh, was such a silly mistake :(
    Well i hope this one should be the right answer.. .
     

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    Last edited: May 24, 2012
  13. May 24, 2012 #12

    tiny-tim

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    all looks fine now! :biggrin:
     
  14. May 24, 2012 #13
    :biggrin::smile:
    Thank you so much for the replies and enjoyed the way I learnt :)
     
  15. May 27, 2012 #14

    haruspex

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    Shreyashebbar, please explain why you started a fresh thread for the same problem. Was the answer wrong previously?
    I must say, I do not see any answers on this thread that look the same as what I get:
    v2 = 5Rg cos θ / (2 + 5 cos θ)
     
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