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Rotational inertia, angular speed, and kinetic energy

  1. Apr 7, 2010 #1
    1. The problem statement, all variables and given/known data

    Two skaters, each of mass 50 kg, approach each other along parallel paths separated by 4.0 m. They have opposite velocities of 1.7 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?
    (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 0.7 m. What is their angular speed then?
    (e) Calculate the kinetic energy of the system now.


    2. Relevant equations

    w= v/r
    ke= 0.5*I*w^2


    3. The attempt at a solution

    I have figured out parts (a), (b), and (c). For part (d), I used the same process as in part (b). In (d), I assumed that the new radius was .35 m, and the velocity was still 1.7. So, 1.7/.35 should equal the new angular speed, but that answer does not work. In part (e), I still used what I think is the new radius, .35 m, and put the values into the Kinetic energy equation: 0.5*(50*(.35)^2+50*(.35)^2)*(1.7/.35)^2 = 0.5*I*w^2. This answer does not work either, but I have to be missing something in part (d) to begin with, and I have to get that first anyway to solve (e). Does anyone see what I'm forgetting here? Thanks.
     
  2. jcsd
  3. Apr 7, 2010 #2

    collinsmark

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    Gold Member

    So far so good. o:)

    I think that's where you went wrong. :frown:

    Use conservation of angular momentum for part (d). The only things acting on the system are the skaters themselves, thus there are no external torques, which means that conservation of angular momentum applies. Set the angular momentum, before and after the skaters pull themselves closer together, equal to each other. Solve for the new [tex] \omega [/tex].

    Why does the kinetic energy change? The skaters have to do work to pull themselves inwards. The skaters are adding energy to the system in the form of work. But change in kinetic energy or not, angular momentum is still conserved.
     
  4. Apr 9, 2010 #3
    Thanks. I got it.
     
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