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How much time goes by before the merry-go-round is turning 0.5 revolutions per second

  1. Nov 22, 2011 #1
    Torque = 2m * 50N
    M.o.I = (400kg * 2m^2)/2
    a = 1/8 rad/s^2 = 0.019894368 revs/s^2

    I don't even know which formula to use

    But I tried this one

    ω= ω0 + αt

    0.5rev/s = 0 + 0.019894368rev/s^2 * t

    t = 23.132

    That's seems like a lot of time... not sure if it's right.
     
  2. jcsd
  3. Nov 22, 2011 #2
    Re: How much time goes by before the merry-go-round is turning 0.5 revolutions per se

    You're nearly there. ω is the angular velocity, you have the revolutions per second. Now, if I remember correctly,

    ω=2∏η
    where ω is angular velocity
    η is revolutions per second
     
  4. Nov 22, 2011 #3
    Re: How much time goes by before the merry-go-round is turning 0.5 revolutions per se

    Ooops, I mean 25.132, that was a typo. Is that the correct answer?
     
  5. Nov 22, 2011 #4
    Re: How much time goes by before the merry-go-round is turning 0.5 revolutions per se

    Yes, that is the answer I get, just got there in a different way.
     
  6. Nov 22, 2011 #5
    Re: How much time goes by before the merry-go-round is turning 0.5 revolutions per se

    Good then

    I figured it out, my mistake was thinking that the 50N push was just one push and then release, so I was confused as to why the merry-go-round would accelerate.

    But it's a constant 50 N force, which is why it's accelerating, did not realize that at first.

    Thanks
     
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