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Rotating rod returning to its initial position

  1. Apr 1, 2017 #1
    1. The problem statement, all variables and given/known data
    A uniform rod (mass 'm', length 'l') held horizontally above the ground. It is released from its supports and struck vertically upwards at one end. With what impulse should the rod be struck if it is to return to exactly its starting position and orientation?

    2. Relevant equations

    I know the moment of inertia of a rod about its centre is ml^2/12,
    Torque = Moment of Inertia x Angular acceleration
    and the rod moves in the vertical plane around the CM in spirals, but apart from that I'm really stuck.

    I can't seem to get the concept of how the rod returns to its starting position.

    3. The attempt at a solution

    I tried to conserve energy, mgh=0.5 Iω^2 + KE but I'm not sure where that gets me.

    Thanks for any help or pointers :)
  2. jcsd
  3. Apr 1, 2017 #2


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    This is not an energy conservation situation. Note that the torque about any point on the vertical path of the CM is zero. Also note that in the time the CM goes up and comes back down, the rod undergoes an integer number of revolutions about the CM. Start by figuring out expressions for the initial speed of the CM and the initial angular speed about the CM.
  4. Apr 1, 2017 #3
    Ahh thank you
    So I just found the change in velocity as ωL/6 (from change of angular momentum=radius*linear momentum).
    Also that the time taken for it to reach the same place is (2πLn/3g)^0.5 where n is the number of revolutions.
    Now how do I get the new required impulse to return it to its original orientation? o_O
    How do I know that it actually needs a counter impulse?
  5. Apr 1, 2017 #4


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    This doesn't look right. The time depends on the impulse J.
    What new required impulse? The CM is kicked up by the initial impulse J and is pulled back down by the force of gravity.
  6. Apr 1, 2017 #5
    can't believe I was so silly, I thought there was a second impulse we had to give the rod doh!
    Thanks for your patience, got it sorted now
  7. Apr 1, 2017 #6


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    OK, what's your answer?
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