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Acceleration of a falling pinned stick

  1. Jul 15, 2007 #1
    1. The problem statement, all variables and given/known data
    A thin, uniform stick of length 3.1 m and mass 6.2 kg is pinned through one end and is free to rotate. The stick is initially hanging vertically and at rest. You then rotate the stick so that you are holding it horizontally. You release the stick from that horizontal position. Remember that the moment of inertia for a stick of mass m and length L about its end is (1/3)m L2.

    Note: Unless other wise specified, the following questions refer to the situation when the stick has traveled 57.2 degrees (the stick makes an angle of 57.2 degrees with the horizontal).

    Also, using conservation of energy, it can be shown that the square of the angular speed as a function of angle is given by:

    w2 = 3 g sin(q)/L
    with θ the angle measured clockwise from horizontal and L the length of the stick.

    I have the accelerations, (tangential, centripetal and angular)

    2. Relevant equations

    I need to find what the x component of the acceleration is, with the positive x direction defined as pointing horizontally towards the center of the stick.

    3. The attempt at a solution

    My attempt was to take my tangential and centripetal acceleration and multiple them by the cosine of my angle theta, then add them together, i.e.;
    A(tangential)*cos(θ) + A(centripetal)*cos(θ) = A(x).
    This doesnt give me the right answer, any help would be appreciated. Thanks! :)
     
  2. jcsd
  3. Jul 16, 2007 #2

    Doc Al

    User Avatar

    Staff: Mentor

    Since the tangential and centripetal accelerations are perpendicular to each other, you can't find their horizontal components by multiplying both by cos(θ).
     
    Last edited: Jul 16, 2007
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