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Finding angular speed & KE using moment of inertia?

  1. Apr 6, 2013 #1
    In the figure, two skaters, each of mass 48.1 kg, approach each other along parallel paths separated by 2.90 m. They have opposite velocities of 1.61 m/s each. One skater carries one end of a long pole of negligible mass, and the other skater grabs the other end as she passes. The skaters then rotate about the center of the pole. Assume that the friction between skates and ice is negligible. What are (a) the radius of the circle, (b) the angular speed of the skaters, and (c) the kinetic energy of the two-skater system? Next, the skaters each pull along the pole until they are separated by 1.82 m. What then are (d) their angular speed and (e) the kinetic energy of the system?

    ω=v/r
    I=Ʃmr2
    K=1/2Iω2

    a) I found the radius obviously by doing 2.90/2=1.45m
    b) ω =1.61m/s / 1.45m = 1.11rad/s
    c) I = 2(48.1kg)(1.45m)2=202.3 kgm2
    K = (1/2)(202.3)(1.11)2=125 J

    a, b and c were all the correct answers but then when I tried to solve d and e the exact same way the answers are incorrect. Isn't the only thing you have to change the fact that the radius is now 1.82/2=.91m?

    d) ω=1.61m/s / .91m = 1.77 rad/s
    e) I = 2(48.1kg)(.91m)2=79.66kgm2
    K = (1/2)(79.66)(1.77)2=125 J

    d and e were incorrect. What am I doing wrong?
     

    Attached Files:

  2. jcsd
  3. Apr 6, 2013 #2

    SammyS

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    It's not the same problem. How can it be solved the same way?

    Why would the skaters speed remain the same as before?
    Ask yourself, 'what is conserved as the skaters get closer together?'
     
  4. Apr 6, 2013 #3
    Angular momentum. Thanks, got it!
     
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