1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Throwing a stick in the air

  1. Jan 31, 2007 #1
    1. The problem statement, all variables and given/known data

    A girl throws a stick of length .27 m and mass .18 kg into the air so that the center of mass rises vertically. At the moment it leaves her hand, the stick is horizontal and the speed of the end of the stick nearest to her is zero. When the center of mass reaches its highest point, the stick has made 31 complete revolutions. How long did it take the center of mass to reach its highest point?

    2. Relevant equations

    I let moment of inertia = (1/3)ML^2 (rod with rotational axis through end). I assume the problem uses conservation of energy combined with basic kinematics.

    3. The attempt at a solution

    At start, the rod has KE .5mv^2, v being the upwards velocity of the center of mass, and rotational KE .5Iw^2. At its highest point, it has potential energy mgh and rotational KE .5Iw^2. I assume the rotational kinetic energy does not change as it goes up? So the translational kinetic energy must equal mgh. I just don't know how to use the other information to get the height or velocity of the center of mass.
     
  2. jcsd
  3. Jan 31, 2007 #2

    Doc Al

    User Avatar

    Staff: Mentor

    You can solve this purely as a kinematics problem. Hint: How does the initial speed of the the center of mass relate to the angular speed?
     
  4. Jan 31, 2007 #3
    v=rw, so velocity at center of mass would equal .135w? I also figured that the time of one rotation would be 2pi(.27)/w, so for all 31 rotations it would equal 62pi(.27)/w. Am I on the right track?
     
  5. Feb 1, 2007 #4

    Doc Al

    User Avatar

    Staff: Mentor

    Good.
    How did you figure this? (It's not quite right.)

    Correct your mistake and you are on the right track. Hint: Relate v and total time.
     
  6. Feb 1, 2007 #5
    Why is v=ωr? Is it because the end of the stick is initially ascending with the same velocity as the center of mass?
     
  7. Feb 1, 2007 #6

    Doc Al

    User Avatar

    Staff: Mentor

    No, it's because you are told that "the stick is horizontal and the speed of the end of the stick nearest to her is zero". The initial speed of the center is v (by definition), so the only way that the end near her can have speed zero is if it rotates such that v=ωr.
     
  8. Feb 1, 2007 #7
    OK heres what I did.

    v = wr

    t (of one rotation) = (2pi)/w

    t (total) = (62pi)/w

    I also used Vf = Vi + at ... so 0 = v - 9.8t, t = v/9.8

    Set those ts equal (62pi)/w = v/9.8

    substitute w = v/r and cross multiply

    (9.8)(62pi) = (v^2)/r

    Now would r be the length of the stick or half of it? I said the entire length since one end serves as the axis of rotation.

    (9.8)(62pi)(.27) = v^2
    v = 22.7

    0 = 22.7 - 9.8t
    t = 2.3165

    Where'd I go wrong?
     
  9. Feb 1, 2007 #8

    Doc Al

    User Avatar

    Staff: Mentor

    That's OK, but you made a tactical error: By setting the ts equal, you eliminate them from the equations. But t is what you are trying to find. :wink: (So you are forced to waste time calculating v, when you don't need too. No big deal.)
    No, the stick spins about its center. (The stick's motion is the sum of the motion of its center of mass plus its rotation about its center of mass.) That's your real error.

    Also, the v in v=ωr is the initial speed of the center of mass. (See my response to Gyroscope.)
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Throwing a stick in the air
  1. Stick thrown in air (Replies: 4)

Loading...