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Impulse on a rod.

  1. Feb 23, 2012 #1
    1. The problem statement, all variables and given/known data
    Suppose there is a rod on a friction less horizontal plane and an impulse "J" is given to it at a distance "x" from its center.

    Using this data I have to find out the instantaneous center of rotation.

    3. The attempt at a solution

    By impulse-momentum theorem,
    J = M*Vcm
    (Vcm= Velocity of center of mass)

    and by angular impulse momentum theorem,
    J*x = I*ω

    What do I do next?
     
  2. jcsd
  3. Feb 23, 2012 #2
    Is the length l and mass m ?
     
  4. Feb 23, 2012 #3
    Well, the source of the question isn't very reliable, but no that data is not given.

    But if it cant be done without knowing that, go on ahead and use l and m.
     
  5. Feb 23, 2012 #4
    See I think ( I am not sure :( )

    J(x+r)= ( Ml^2 /12 + mr^2 )( v/(x+r))
     
  6. Feb 23, 2012 #5
    Here I have assumed instantaneous axis of rotation is r units below centre of mass
     
  7. Feb 23, 2012 #6
    v here is Vcm right?

    And I just rechecked the question, the length is given as L but the mass isn't given.

    The answer is given as (L^2)12x
     
  8. Feb 23, 2012 #7
    L is given ? and j is not coming in the answer?
     
  9. Feb 23, 2012 #8
    Yes, L is given and nope, no J.

    In fact, when we solve this equation : J(x+r)= ( Ml^2 /12 + Mr^2 )( v/(x+r))

    we get a quadratic: (on substituting Vcm as J/M by linear momentum impulse theorem)

    x^2 + 2xr - (L^2)/12 = 0

    Which unfortunately doesn't give the correct answer, but I think we're close.
     
  10. Feb 23, 2012 #9
    ohk let me try once again
     
  11. Feb 23, 2012 #10
    add conservation of moentum also to find v
     
  12. Feb 23, 2012 #11
    On further speculation, had it been v/r instead of v/(x+r) we'd have got the correct answer, without a quadratic forming.
     
  13. Feb 23, 2012 #12
    yea i guess it should be v/r , i am not still sure
     
  14. Feb 23, 2012 #13
    Why should it be v/r ?
     
  15. Feb 23, 2012 #14
    ω = v/(perpendicular distance) = v/(x+r) as you earlier said.
     
  16. Feb 23, 2012 #15
    even if we put v/r answer is not coming
     
  17. Feb 23, 2012 #16
    J(x+r)= ( Ml^2 /12 + Mr^2 )( v/r)

    Now J = M*v

    J(x+r)= ( Ml^2 /12 + Mr^2 )*J/(Mr)

    r(x+r)= l^2 /12 + r^2

    xr = l^2 /12

    Whoops! I realized I made an error typing the correct answer in my earlier post! I'm very sorry!
     
  18. Feb 23, 2012 #17
    i have some figures :bugeye:
     

    Attached Files:

  19. Feb 23, 2012 #18
    And the last one :redface:
     

    Attached Files:

  20. Feb 23, 2012 #19
    Oh yes x.x distance from the com! Thanks a lot!
     
  21. Feb 23, 2012 #20
    your welcome :D
    keep coming back to PF
     
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