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A cheerleader is practicing her baton twirling. The baton has two .6kg masses

  1. Nov 12, 2012 #1
    Hi, everyone! Thank you for using your time to try to help me! You are kind! :)

    1. A cheerleader is practicing her baton twirling. The baton has two .6kg masses attached to opposite ends of a massless rod 70cm long. If the baton is twirled about a point 30cm from one end at 1.6 revolutions per second, what is the baton's kinetic energy?

    2. KE(total) = KE(1) + KE(2)
    v=ωr
    # of revolutions = # of radians / 2pi
    I for a baton with a massless rod in between: I = mr^2

    3. Okay, so I don't know if I'm doing this right...

    *Changing revolutions/s to radians/s to know what the angular velocity (ω) is:
    1.6 revolutions/s x #rad/2pi = 10.05rad/s

    So now we know the angular velocity (ω) = 10.05rad/s

    *KE(total) = KE(1) + KE(2)
    KE(total) = 1/2(I)(ω^2) + 1/2(I)(ω^2)
    KE(total) = 1/2(.6)(.30^2)(10.05^2) + 1/2(.6)(.40^2)(10.05^2)
    KE(total) = 7.58J

    ...I have no idea if that's right... I'm not sure if I use the same thing for the angular velocity... I think both .6kg masses move at the same speed... not sure though. :frown:

    Thank you for helping me to figure this out! :)
     
  2. jcsd
  3. Nov 12, 2012 #2

    haruspex

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    looks right to me.
     
  4. Nov 13, 2012 #3
    Oh, really? Yay! Thanks, Haruspex! :)
     
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