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Stick thrown in air

  1. Nov 17, 2008 #1
    1. The problem statement, all variables and given/known data

    A student throws a stick of length L up in the air. At the moment the stick leaves his hand, the speed of the stick's end is zero. The stick completes N turns as its caught by the student at the initial release point. Show that the height to which the centre of mass of the stick rose is [tex]\pi NL/4[/tex]



    3. The attempt at a solution
    All that I can make out of the problem is that t if the velocity of the centre of mass of the stick was v then by energy conservation the centre of mass of the stick would have risen by [tex]v^{2}/2g[/tex]

    I cannot make out what can I do with that N and all. Help me please.
     
  2. jcsd
  3. Nov 17, 2008 #2

    tiny-tim

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    Hi ritwik06! :smile:

    The way to approach a question like this is to start:

    "Assume the velocity of the two ends are v and 0, vertically …",

    and then calculate both the maximum height and the number of turns. :wink:
     
  4. Nov 17, 2008 #3
    I tried this.
    Assuming the velocity of the two ends are v and 0
    Using conservation of angular momentum:
    mvL/2=I [tex]\omega[/tex]

    then 0.5 I [tex]\omega^{2}[/tex]=mgh
    ????
     
  5. Nov 17, 2008 #4

    Redbelly98

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    I think you'll also need the time taken to reach the maximum height. And from that figure out the time taken to come back down as well.
     
  6. Nov 18, 2008 #5

    tiny-tim

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    exactly what it says on the box! …

    Hi ritwik06! :smile:

    (have an omega: ω and a pi: π and a squared: ² :wink:)

    uh-uh … "conservation of angular momentum" means exactly what it says on the box!

    the angular momentum will carry on the same, no matter where the stick goes (unless there's a torque, which there isn't … the weight acts through the c.o.m) …

    so this is two separate problems …

    for the height you can ignore the angular momentum, and for the angular momentum you can ignore the height. :biggrin:
     
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