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Angular Velocity after addition of mass?

  1. Dec 3, 2006 #1
    1. The problem statement, all variables and given/known data

    A merry-go-round is a common piece of playground equipment. A 4 m diameter merry-go-round with a mass of 220 kg is spinning at 16 rpm. John runs tangent to the merry-go-round at 5.0 m/s, in the same direction that it is turning, and jumps onto the outer edge. John's mass is 31 kg. What is the merry-go-round's angular velocity, in rpm, after John jumps on?


    2. Relevant equations

    Li = Lf
    KE = 1/2mv^2
    KE = 1/2Iw^2


    3. The attempt at a solution

    I tried doing the following:

    Li = Lf
    (.5)(220)(2^2)(16) = (.5(220) + 31)(2^2)wf
    wf = 12.482 rpm

    That appears to be wrong.


    I also tried using:

    1/2Iw^2 (KE of John and merry-go-round) = 1/2mv^2 (KE of John) + 1/2Iw^2 (KE of merry-go-round) and I also do not get the right answer.

    Does something have to be converted such as 5 m/s to rpm?



    Thank you!
     
  2. jcsd
  3. Dec 3, 2006 #2
    1) ANSWER : conservation of angular momentum :

    2) the initial system is the merry go round with no guy on it + the guy that is "about" to jump onto the merry go round. You have been given the intial angular mometum of the merry go round Iw and the runner gives you an angular momentum equal to : mvr (you forgot this !!!)

    3) the final system is the merry go round + guy on it. The rotational inertia of the guy is I_runner = mr^2 (we treat him as a point particle) with r equal to the radius because the guy is sitting at the outer edge. The angular momentum of the merrry go round is Iw' and it is this w' that you need

    4) Solve Iw + mvr = (I + I_runner)w' for w'

    5) make sure you use THE CORRECT UNITS for w and w' (ie rad/s)

    6) good luck

    7) greets marlon
     
    Last edited: Dec 3, 2006
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