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Two Skaters Attached by Pole

  1. Mar 25, 2014 #1
    1. The problem statement, all variables and given/known data
    In the figure below, two skaters, each of mass 60 kg, approach each other along parallel paths separated by 3.4 m. They have opposite velocities of 2.1 m/s each. One skater carries one end of a long pole with negligible mass, and the second skater grabs the other end of it as she passes. The skaters then rotate around the center of the pole. Assume that the friction between skates and ice is negligible.
    (a) What is the radius of the circle they now skate in? Answer=1.7 m
    (b) What is the angular speed of the skaters?
    (c) What is the kinetic energy of the two-skater system?
    (d) Next, the skaters pull along the pole until they are separated by 0.8 m. What is their angular speed then?
    (e) Calculate the kinetic energy of the system now.
    Hint:The angular momentum of the two-skater system cannot change because there is no net external torque to change it. The angular momentum of a particle is the product of the particle's momentum (mv) and the perpendicular distance from its path to the center about which we calculate angular momentum (here the center of the pole). How is rotational kinetic energy related to rotational inertia and angular speed?

    2. Relevant equations
    L(initial)=L(final)
    L=mvr=Iw
    I=Sum(mr^2)
    w=v/r


    3. The attempt at a solution
    I'm stuck at part b. I have tried w=v/r (w being angular speed) but the answer is incorrect. I have tried using mvr=Iw and the answer is not correct. The answer is not 2.1, 1.2, 1.1, or 1.05. I don't know how else to find w, and I don't understand why my answer is incorrect.
     
  2. jcsd
  3. Mar 25, 2014 #2

    gneill

    User Avatar

    Staff: Mentor

    Is the problem perhaps related to the units or significant figures you're specifying in your answer? ω = v/r is a good approach.
     
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