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Baton twirler problem

  1. Nov 17, 2008 #1
    1. The problem statement, all variables and given/known data
    A baton twirler throws a spinning baton directly upward. As it goes up and returns to the twirler's hand, the baton turns through four revolutions. Ignoring air resistance and assuming that the average angular speed of the baton is 1.80 rev/s, determine the height to which the center of the baton travels above the point of release.

    2. Relevant equations
    w=[tex]\theta[/tex]/Delta t
    *v=vo+at ~ w=wo+[tex]\omega[/tex]t
    *v2=vo2+2ax ~ w2=wo2+2[tex]\omega[/tex][tex]\theta[/tex]
    *x=vot+(1/2)at2 ~ [tex]\theta[/tex]=wot+(1/2)at2

    the ones with * means i changed it to what the problem is about. means the same thing just different letters so you wont get confused i guess.

    3. The attempt at a solution
    you find the height which is the same as distance but how would you plug that in? would the rev/s be the velocity part? its not in m/s are you supposed to convert it or something? i know it makes 4 revolutions but i still have no idea what to do just get the answer and multiply it by 4? please help!

    thank you!
    sweedel joseph
  2. jcsd
  3. Nov 17, 2008 #2
    Well if the baton goes through 4 revolutions in the air, and it makes 1.8 revolutions per second, then it should have stayed 4 rev /(1.8 rev/sec) = 2.22222222 seconds in the air.

    So now I think you can consider the center of the baton as a particle which was shot straight up in the air and came back down in 2.2222222 seconds. So then you can find the maximum height of the particle.
  4. Nov 17, 2008 #3
    wait how did you find time? just divide that by 60? then that answer divide it by the total you had.

    since the answer is 2.222222 just plug that into the equation to find the distance? would it be the same does it matter if its height or length? just the negative stuff right but you dont need to worry about it because its going up.
  5. Nov 17, 2008 #4
    I think the crux of the problem is in finding time... if you assume that the twirler holds it in the middle, spinning it, then throws it up and catches it in the same spot at the same height, then you shouldn't need to worry about how long it is or anything.

    But think about it, if the baton will make 1.8 revolutions or spins every second, then 1.8 times the seconds should equal 4, since it made 4 revolutions. Solving for the seconds there i got 2.22222.

    Then you could say that it went up in half the time, so that it came down in half the time, 1.1111111 seconds. Since it starts from rest at its highest point, consider it a particle that was dropped from rest and fell for 1.111111111111 seconds. The distance it fell should equal the distance from the maximum height to the twirler's hand, or what you are looking for.
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