1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Angular Velocity of Ice Skater Around a Pole

  1. Nov 21, 2013 #1
    1. The problem statement, all variables and given/known data
    An Ice skater of mass m = 70kg is initially traveling at a speed v = 4 m/s along a straight path that brings his center of mass to within a distance b = .6m of a vertical pole fixed in the ice. He reaches out with his hand as he passes the pole and hangs on so that he pulls himself into a circular path of radius r = .8m (measured from the axis of rotation to his center of mass) around the pole. If the skater's moment of inertia about his center of mass is I = 1.40 kgm2 , what is his final angular velocity ω about the pole, assuming that no torques act during this maneuver?


    2. Relevant equations
    Li = Lf
    Iiωi = Ifωf
    or maybe...
    L = mvr = Iω
    or possibly...
    mviri = mvfrf

    or if energy is conserved...
    .5mv2 = .5Iω2

    I'm mostly confused about which formula to use.

    3. The attempt at a solution

    mvri = Iω
    (70)(4)(.6) = (1.4)ω
    ω = 120 rad/s
    But I thought that this was a relatively unreasonable speed for the skater to be going, so I tried this...

    .5(70)(4)2 = .5(1.4)ω2
    ω = 28.3 rad/s

    But I don't think energy is conserved, so I also tried this...

    mvr = mvr
    (70)(4)(.6) = (70)(v)(.8)
    v = 2.14 m/s
    v=ωr
    3 = ω(.8)
    ω = 3.75 rad/s

    This seemed reasonable to me, but I didn't use the moment of inertia, and I have nothing to check my work against. How do I know which formula is the correct one to use in this situation?
     

    Attached Files:

    • 18.jpg
      18.jpg
      File size:
      16.9 KB
      Views:
      64
  2. jcsd
  3. Nov 22, 2013 #2

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    That I is the MoI about the skater's centre of mass, but the rotation is about the pole. How do you correct for that?
     
  4. Nov 22, 2013 #3
    I = Icm + Md2
    I = (1.4) + (70)(.8)2
    I = 46.2 kgm2

    mvr = Iω

    (70)(4)(.6) = (46.2)ω
    3.64 rad/s = ω

    Would this be the correct answer?
     
  5. Nov 22, 2013 #4

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    Looks right to me.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Angular Velocity of Ice Skater Around a Pole
Loading...