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Angular Velocity of a System

  1. Oct 25, 2008 #1
    1. The problem statement, all variables and given/known data
    A runner of mass 51.0 kg runs around the edge of a horizontal turntable mounted on a vertical, frictionless axis through its center. The runner's velocity relative to the earth has magnitude 3.60 m/s. The turntable is rotating in the opposite direction with an angular velocity of magnitude 0.160 rad/s relative to the earth. The radius of the turntable is 3.20m , and its moment of inertia about the axis of rotation is 79.0 kg*m^2.


    2. Relevant equations
    w=v/r
    I1w1=I2w2


    3. The attempt at a solution
    okay first we are given the veocity of the runner, to determine his angular velocity its just
    w1=vrunner/rtable=3.6/3.2=1.125 rad/s

    I1(runner)=mr^2=51.0*3.2^2=522.24 kg*m^2

    now using the conversation of angular momentum

    I1w1+I2w2=(I1+I2)w2'

    the w and I provided in the problem statement can be used as w2 & I2 respectivly and we can now solve for w2' which is

    I1w1+I2w2/(I1+I2)=w2'

    subbing in all the values

    (522.24*1.125)+(79.0*0.160)/(522.24+79.0)=600.16/601.24=0.9982=0.988(sig figs)

    Now the problem I'm having with this is that I still receive an error with this answer and I've been through my work half a dozen times and I dont belive I made any rounding answers so if any one could tell me what Im doing wrong I would be very thankful.
     
  2. jcsd
  3. Oct 25, 2008 #2

    alphysicist

    User Avatar
    Homework Helper

    Hi anubis01,

    I have not checked all of your numbers, but remember that the runner and platform are rotating in opposite direction. What has to be changed here?
     
  4. Oct 26, 2008 #3
    Oh so then since the table is rotating in the opposite direction of the runner w2=-0.160 rad/s

    I1w1+I2w2/(I1+I2)=w2'

    subbing in the values

    (522.24*1.125)+(79*-0.160)/(522.24+79)=574.88/601.24
    w2'=0.95615=0.956(sig figs)

    Thanks for the help, I would have never have caught that error by myself.
     
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