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Determine the sign of angular momentum

  1. Nov 21, 2009 #1
    1. The problem statement, all variables and given/known data

    A 2 kg disk traveling at 3 m/s strikes a 1 kg stick of length 4 m that is lying flat on nearly frictionless ice .Assume the collision is elastic and the disk does not deviate from its original line of motion. Find the translational speed of the disk, the translational speed of the stick, and the angular speed of the stick after collision. The moment of inertia of the stick about center of mass is 1.33 kg m2

    Attempt:

    conservation. of linear momentum

    mdvdi = mdvdf + msvs
    (2 kg) (3 m/s)=(2 kg) vdf + (1 kg)vs
    6 kg m/s-(2 kg) vdf = (1 kg)vs----------------------(1)

    conservation. of angular momentum
    -rmdvdi = -rmdvdf + I*omega
    -12kgm2/s=(- 4kg m) vdf + (1.33 kg m2)*omega----------------------(2)
    -9rad/s+(3rad/m) vdf = omega

    elastic collision
    1/2 mdvdi2 = 1/2mdvdf2 +1/2 I*omega2 + 1/2msvs2
    18 m2 /s2 =2mdvdf2 +(1.33 m2 )*omega2 + vs2--------------------------(3)

    solve .(1), (2), (3) >>> vdf =2.3 m/s, omega = -2rad/s , vs =1.3 m/s



    For my solution above, I have no idea why the rmdvdi and rmdvdf are negative, but I*omega is positive. I tried to figure it with right hand rule, but i still find three of them should be in the same direction.
     
  2. jcsd
  3. Nov 22, 2009 #2
    Hi, could anyone help me to figure out the reason?
    I still don't get it =(
     
  4. Nov 22, 2009 #3
    anyone knows? perhaps i attach a picture to explain the motion better.

    http://img97.imageshack.us/img97/4315/rotation.jpg [Broken]
     
    Last edited by a moderator: May 4, 2017
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