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Homework Help: Angular Momentum - Train Placed on a Wheel

  1. May 11, 2010 #1
    I've been trying this problem for a while now and can't seem to get the given answer. I know I'm probably making an elementary mistake somewhere but I can't find it.

    1. The problem statement, all variables and given/known data
    A track is mounted on a large wheel that is free to turn with negligible friction about a vertical axis. A toy train of mass m is placed on the track and, with the system initially at rest, the train's electrical power is turned on. The train reaches a steady speed of 0.176 m/s with respect to the track. What is the angular speed of the wheel if its mass is 1.97m and its radius is 0.584 m? (Treat the wheel as a hoop, and neglect the mass of the spokes and hub.) (Answer 0.101 rad/s)

    Ok so:
    Vt= .176 m/s
    mt= M
    m2= 1.97M
    r= 0.584m
    I2=m2r2
    Xf= ?
    (X is angular velocity, subscript t for train and 2 for wheel)

    2. Relevant equations

    L=IX=rmv
    Li=Lf
    Lt+L2=0



    3. The attempt at a solution

    So I set it up with the initial equation
    Lt = -L2
    rmtvt = IX

    Then expanded and substituted for corresponding masses
    rMv = 1.97Mr2X

    M cancels and so does an r, so
    vt = 1.97rX

    Then substitute the values
    .176 = 1.97(.584)(X)

    I get a final answer of
    X = .153 rad/s

    The answer is supposed to be .101 rad/s which I can't seem to get.


    Also I used X for angular speed instead of omega because the text formatting wasn't working, and about 4 different symbols showed up instead of omega. Not sure why.
     
    Last edited: May 11, 2010
  2. jcsd
  3. May 11, 2010 #2

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    Welcome to PF!

    Hi Crusher8576! Welcome to PF! :smile:

    (have an omega: ω :wink:)
    hmm … I got the same answer as you until I read the question more carefully. :redface:

    To find L, you need the actual speed of the train, which isn't 0.176, is it? :smile:
     
  4. May 11, 2010 #3
    Aha I've got it. Thanks for the help.

    If anyone was wondering how to do it:

    The wheel is spinning in the opposite direction of the train, so the train may have a speed of .176, but that's in relation to the track. I had to find the actual speed of the train in relation to the ground.

    So in the part where I have vt = 1.97rX, replace the Vt with what I called Va (for V actual), and also created Vw for (V wheel).

    And Va = Vt - Vw = Vt - Xr

    So then we're left with Vt - Xr = 1.97rX to solve for X.
     
    Last edited: May 11, 2010
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