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Rotational Kinematics - Angular velocity

  1. Jan 18, 2010 #1
    1. The problem statement, all variables and given/known data
    http://online.physics.uiuc.edu/cgi/courses/shell/common/showme.pl?courses/phys211/oldexams/exam3/fa07/fig19.gif [Broken]
    A dumbbell consists of a slender rod of negligible mass and length L = 1 m and small steel balls attached to each end with mass 1 kg and 2 kg, respectively. It is pivoted at its center about a horizontal frictionless axle and initially held in place horizontally. The dumbbell is then released. What is the angular velocity of the system when the rod is vertical?

    2. Relevant equations



    3. The attempt at a solution

    I have figured out the angular acceleration upon release is 6.5rad/s^2 and the angular acceleration is zero at vertical (because of torque = 0 at vertical)
    But from there, i got stuck and i don't know how to link this acceleration into the velocity..


    Please could someone help me out here?
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Jan 18, 2010 #2

    ehild

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    It is easier to apply conservation of energy. When the rod is vertical, the centre of gravity is lower than in the horizontal position, the difference is equal to the kinetic energy.

    ehild
     
  4. Jan 20, 2010 #3
    I have tried to use the hint you have given..

    Iw^2/2 = mgh

    h = (2/3 - 1/2) ; length of center of gravity from pivot when it's at vertical

    And i took moment of inertia I as 3*(2/3-1/2)^2

    But when i calculated this, it was wrong...

    I think RHS of the eqn is correct...

    Is something wrong in LHS of the equation?

    Please help me out....:(

    I would really appreciate it .
     
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