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Instantanoeus axis of rotation

  1. Jul 7, 2012 #1
    1. The problem statement, all variables and given/known data
    A uniform rod of length l is given an impulse at right angles to its length as shown. Find the distance of instantaneous centre of rotation from the centre of rod.
    2dltzkm.png

    2. Relevant equations



    3. The attempt at a solution
    Impulse=change in momentum
    or Impulse=mvCM (Am i right here?)

    I am not sure about this but i think i need to conserve angular momentum. But how should i make the equation for initial angular momentum. Should it be mvCMx or mvCM(x+d), d is the distance of instantaneous centre of rotation of rod from the CM.

    Any help is appreciated!
     
  2. jcsd
  3. Jul 7, 2012 #2

    ehild

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    Impulse is FΔt,force multiplied by its time of action. And it is equal to the change of momentum. So the equation mvCM=Impulse is right.
    The applied impulse also means torque for time Δt: τΔt=Impulse*(x+d) The applied torque changes the angular momentum. You can calculate both the torque and angular momentum with respect to the instantaneous axis, considering the motion of the rod a pure rotation. Use the relation between torque and change of angular momentum:τ =I(Δω)/(Δt).You know ω from the initial speed of the CM: The CM will move around the instantaneous axis with angular speed ω=vCM/d.


    ehild
     
  4. Jul 7, 2012 #3
    Thanks ehild for your reply! :smile:

    I have got the answer using the relations you posted but i still don't understand what do you mean by "pure rotation"?
     
  5. Jul 8, 2012 #4

    ehild

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    How is the instantaneous axis defined?

    See picture. It shows a rod moved from position 1 to position 2 and rotated by 90°. You can do it by translating the rod from 1 to 2 and then rotating about the CM, or by rotating the whole rod about the instantaneous axis O.

    ehild
     

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    Last edited: Jul 8, 2012
  6. Jul 8, 2012 #5
    Thank you ehild for the explanation!
     
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