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Pivoting Stick (Rotational Dynamics)

  1. Jun 12, 2010 #1
    I misunderstood the problem, and the answer is the correct one! sorry, and someone can close this post if necessary

    1. The problem statement, all variables and given/known data
    A stick of uniform density with mass M = 7.9 kg and length L = 1 m is pivoted about an axle which is perpendicular to its length and located 0.15 m from one end. Ignore any friction between the stick and the axle.

    The stick is held horizontal and then released. What is its angular speed as it passes through the vertical.

    [PLAIN]https://wug-s.physics.uiuc.edu/cgi/courses/shell/common/showme.pl?cc/DuPage/phys2111/summer/homework/Ch-10-Rotation/pivoting_stick/5.gif [Broken]

    3. The attempt at a solution
    1) First I calculated the moment of inertia about the axle using the Parallel Axis Thm:
    Itotal = Icm + Md2
    d = 0.50 - 0.15 => d = 0.35
    Itotal = (1/12)ML2 + Md2 => Itotal = 1.626

    2) I calculated the potential energy when the center of mass is at its highest point:
    U=Mgh => U=Mgd => U=27.13 J

    2) When this potential energy is 0, it's kinetic energy is maximum, so:
    (1/2)Itotalw2=U => w=sqrt(2U/Itotal) => w=5.776 rad/seg

    I will appreciate any help!!!
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Jun 12, 2010 #2


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    Why is the potential energy Mgd? Where is your zero of potential energy?
  4. Jun 12, 2010 #3
    Hi santi_h87,
    Your answers are correct. But d symbol is not the same on your diagram and the calculation d.
    Last edited by a moderator: May 4, 2017
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