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Merry-go-round physics question

  1. Mar 20, 2009 #1
    The problem statement, all variables and given/known data

    A playground merry-go-round has radius 2.00m and moment of inertia 3000kgm² about a vertical axle through its center, and it turns with negligible friction.
    A child applies an 18.0N force tangentially to the edge of the merry-go-round for 25.0s . If the merry-go-round is initially at rest, what is its angular speed after this 25.0s interval?


    The attempt at a solution

    I have no idea what to apply actually. I only thinking of treating the child as a particle.
    Moment of inertia of child = mr²

    Can apply consv of angular momentum? But is the child considered to be acting external force?

    Can someone help me pls? Thanks!!
     
  2. jcsd
  3. Mar 20, 2009 #2
    Re: Merry-go-round

    Based on what you wrote, it sounds like the child isn't actually on the merry-go-round (especially since the child's weight is not given) so I would assume the child's push is simply an external force.

    I'd think you could use torque/angular acceleration or angular impulse/momentum to solve
     
  4. Mar 20, 2009 #3

    Doc Al

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    Staff: Mentor

    Re: Merry-go-round

    The child isn't riding the merry-go-round, she's just exerting an external force on it. Angular momentum is not conserved. What torque does she exert? Apply Newton's 2nd law for rotation.
     
  5. Mar 20, 2009 #4

    Delphi51

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    Homework Helper

    Re: Merry-go-round

    The child is not on the merry-go-round so it does not make sense to calcuate his/her moment of inertia. This question is analogous to applying a force to a mass and asking how fast it would go after a given time. You would use F = ma to find the acceleration and then use v = at to find the velocity. In the merry-go-round problem, use the analogous circular motion formulas.
    F = ma ---> [tex]Torque = I\omega [/tex]
    v = vi + at ----> [tex]\omega = \omega i + \alpha t[/tex]
     
  6. Mar 20, 2009 #5
    Re: Merry-go-round

    Okay, i got it. thanks! angular speed = 0.3rad/s

    The next part is actually:

    How much work did the child do on the merry-go-round?



    I tried working out.
    W = Fs = Frθ = Fr(integration of ω) = 18*2*0.3*25 = 270J

    But its wrong..
     
  7. Mar 21, 2009 #6

    Doc Al

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    Staff: Mentor

    Re: Merry-go-round

    It's wrong because the angular speed is not constant, so θ ≠ ωt. Use your kinematic formulas to find the angle.
     
  8. Mar 21, 2009 #7
    Re: Merry-go-round

    Write the torque equation and find out angular acceleration.
    Use it to find the angular velocity after a certain period of time. (Use kinematic relationship)
     
  9. Mar 21, 2009 #8
    Re: Merry-go-round

    The angle through which it has rotated can be found out by kinematics again. :wink:
    The angle won't be of the form [tex]\theta = \omega t[/tex] cause there is acceleration.
     
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