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Rotational Motion of a long pole

  1. Oct 14, 2004 #1
    A 3.3m long pole is balanced vertically on its tip. It is given a tiny push. What will be the speed of the upper end of the pole just before it hits the ground? Assume the lower end does not slip.

    mgh=.5mv^2 + .5Iw^2 Initial PE = final KE translation + final KE rotational
    v=rw, I=.33mr^2

    Combine these equations and I got
    v=(1.5gh)^(1/2)

    Now I think this is the speed of the CM, located 1.65m above the ground (h=1.65). So to find the speed of the tip I covert v into radians per second and then multiply that by 3.3 to find speed of the tip.

    Is this method correct? The book contains no answer.
     
  2. jcsd
  3. Oct 14, 2004 #2

    Doc Al

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    Staff: Mentor

    Your basic idea of using energy conservation is correct, but the details are not. The total KE of the rod as it hits the ground can be viewed as a combination of the translational KE of the CM plus the rotational KE about the CM. ([itex]I_{cm} = 1/12 M L^2[/itex])

    A simpler approach is to realize that the stick is in pure rotational about the pivot point, thus the KE of the rod is just the rotational KE about the pivot point. ([itex]I_{end} = 1/3 M L^2[/itex])
     
  4. Oct 14, 2004 #3
    Thanks for your help.
     
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