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Homework Help: Rotational Kinetic Energy of ride

  1. Mar 10, 2007 #1
    1. The problem statement, all variables and given/known data

    A horizontal 800.0 N merry-go-round with a radius
    of 1.5 m is started from rest by a constant horizontal
    force of 50.0 N applied tangentially to the merry-go-
    round. Find the kinetic energy of the merry-go-round
    after 3.0 s. Assume it is a solid cylinder.


    2. Relevant equations

    KE = (1/2)(Iw^2)


    3. The attempt at a solution

    Answer in the back is 280 J

    My answer, however, turns out to be 40ish.

    Here's how I did it:

    I found moment of inertia I of combined objects (I treated the forces as objects here):

    I = I_(mgr) + I_(object_at_edge)
    I = (.5MR^2) + (MR^2)
    I = (.5*80*1.5^2) + (5*1.5^2)
    I = 101.25 kg.m^2

    Then, I chose the 50 N constant force as the force that maintains circular motion, so

    50 = (mv^2)/r

    50 = ((I/r^2)*v^2)/r

    50 = ((101.25/1.5^2)*v^2)/1.5

    v^2 = 1.667
    v = 1.29

    Now, KE = (.5)(Iw^2)
    = (.5)(101.25*(0.86)^2)
    = 43.5 J

    Haha, awfully messy.
     
    Last edited: Mar 10, 2007
  2. jcsd
  3. Mar 10, 2007 #2

    mjsd

    User Avatar
    Homework Helper

    note you have not used the fact that it is asking you what happen after 3.0s. it said a constant force is applied tangentially so can't equate 50 = mv^2/r either... my guess from the wording of the question is that the merry go round is under constant acceleration.. you should work out the angular acceleration and torque... [tex]\tau= I\alpha[/tex] where, I, here is moment of inertial of object.
     
  4. Mar 10, 2007 #3
    Why do you state "combined objects?"
    The 50 N is applied tangentially.
     
  5. Mar 11, 2007 #4
    Thanks for the help. I have made a fundamental error here and I should have known better. I have reached the solution by using the net torque identity, which equals to I*angular acceleration.

    t=Ia

    Then since t=Fd

    Fd = Ia
    (50)(1.5) = (.5*80*1.5^2)(a)

    a = 0.83 rad/s^2

    Since a = (w_f - w_i)/t

    0.83 = w_f/3

    w_f = 2.49

    Now we can easily conclude KE_f:

    KE_f = (.5)(90)(2.49)^2
    = 280 J
     
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