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Rolling, Torque, and Angular Momentum

  1. Mar 14, 2012 #1
    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?

    please help me solve this question.

    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?

    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
     
  2. jcsd
  3. Mar 14, 2012 #2

    NascentOxygen

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    Staff: Mentor

    Hi ranjaxt2012! http://img96.imageshack.us/img96/5725/red5e5etimes5e5e45e5e25.gif [Broken]

    This sounds like a homework question, so should really be posted in the homework sub-forum.
     
    Last edited by a moderator: May 5, 2017
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