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Rolling, Torque, and Angular Momentum

  1. Mar 15, 2012 #1
    1. The problem statement, all variables and given/known data

    A girl of mass 28.1 kg stands on the rim of a frictionless merry-go-round of radius 1.58 m and rotational inertia 433 kg·m2 that is not moving. She throws a rock of mass 670 g horizontally in a direction that is tangent to the outer edge of the merry-go-round. The speed of the rock, relative to the ground, is 7.78 m/s. Afterward, what are (a) the angular speed of the merry-go-round and (b) the linear speed of the girl?

    2. Relevant equations

    Mass M = 28.1 kg
    radius r = 1.58 m
    R-Inertia = 433 Kg.m2
    mass of rock = 670g = .67 kg (tangent to merry-go-round)
    speed of rock = 7.78 m.s
    now is the angular speed ω of merry-go-round = ?
    we know ω = L/Inertia or (m.r.v)/inertia.
    Now what is velocity of the system? is it 7.78?


    3. The attempt at a solution

    if yes then

    ω = [(28.1+.67)*1.58*7.78)]/433 ≈ 0.81674 rad/sec

    is linear speed of the girl = 0.816748*r

    v = ω/r ??
     
  2. jcsd
  3. Mar 15, 2012 #2



    All you have to do is to conserve the net angular momentum of the system of rock , girl and merry go round as no net external torque acts on this system.
    Infact the net linear momentum is also conserved as no net external force acts on this system. (just after throwing)

    You said velocity of the system in your post.
    You can't talk of the velocity of the system until you specify it.

    if you mean the girl and merry go round, the NO
     
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