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Mechanics Question

  1. May 30, 2015 #1
    1. The problem statement, all variables and given/known data
    A mass m with velocity v0 makes an inelastic collision with second mass M that is suspended by a string of length L. The velocity v0 is perpendicular to the vertical string. After the collision the combined masses, m+M, rotate in a vertical plane around the point of suspension of the string. Find the minimum value v0 such that the string remains always under tension for a complete rotation.

    2. Relevant equations
    ∑Fr=mv2/r
    T=1/2mv2
    U=mgh

    3. The attempt at a solution
    The condition for the an inflexible string to remain under tension is that it never goes slack, i.e. length=L for the entire rotation.

    Since the collision is inelastic, momentum is conserved while kinetic energy is not. So,

    mv0=(m+M)v' where v' is the magnitude of the x-directed velocity of the combined masses (m+M) after the collision. so v'=m/(m+M)v0.

    At the bottom of the vertical circle (In Polar Coordinates),
    letting m'=(m+M)


    ∑Fr= Tb-m'g=m'v'2/L (*)

    At the top,

    ∑Fr=Tt+m'g=m'v''2/L (**)

    Using Energy,

    1/2m'v'2=1/2m'v''2+m'g(2L) (***)
    v'2=v''2+4gL

    v''2=v'2-4gL

    I am going to pause here to see if there is anything I am missing.
     
  2. jcsd
  3. May 30, 2015 #2

    Doc Al

    User Avatar

    Staff: Mentor

    You're on the right track. Find v'', then v', then v0.
     
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