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Hinge Reaction

  1. Nov 23, 2013 #1
    1. The problem statement, all variables and given/known data


    a uniform rod of mass m and length L is free to rotate in the vertical plane about a horizontal axis passing through its end. the rod is released from rest in the position shown by slightly displacing it clockwise. Find the hinge reaction at the axis of rotation at the instant the rod turns through
    (a) 90* (b) 180*

    2. Relevant equations
    i am not getting the answer by the following attempt, any help will be highly appreciated....!!!!


    3. The attempt at a solution
    i have conserved energy as shown in the following equation:-
    mgL/2 = mv2/2 + Iω2/2

    and also ω=2v/L
    so i have further got v2=3gL/7
     

    Attached Files:

  2. jcsd
  3. Nov 23, 2013 #2

    TSny

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    If I denotes the moment of inertia about the fixed axis (that is, about the end of the rod) then the total kinetic energy is Iω2/2 instead of Mvc2/2 + Iω2/2.
     
  4. Nov 23, 2013 #3
    okay will try it once again
     
  5. Nov 23, 2013 #4
    by your method i have got v=3g/2 which does not matches the ans..
     
  6. Nov 23, 2013 #5
    please help me....!!!
     
  7. Nov 23, 2013 #6

    TSny

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    What velocity have you calculated? Does v represent the velocity of the center of mass or the velocity of the far end of the rod or the velocity of some other point?

    Note that v = 3g/2 is inconsistent in terms of units. You have velocity on the left and acceleration on the right.

    Can you show more detail of how you got this result?
     
    Last edited: Nov 23, 2013
  8. Nov 24, 2013 #7
    v represents the velocity of centre of mass
     
  9. Nov 24, 2013 #8

    TSny

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    OK. Since v = 3g/2 cannot be correct (because the units don't check), you must have made a mistake. But there is no way for us to know where you made the mistake unless you show your work.
     
  10. Nov 24, 2013 #9
    mgl/2=Iω2/2
    mgl/2= 1/2x1/3ml2x(2v/l)2
    m,1/2 and also l2 gets cancelled
    then
    gl=1/3x4v2
    v=(3gl/4)1/2
     
  11. Nov 24, 2013 #10

    TSny

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    That's correct.
     
  12. Nov 24, 2013 #11
    but i have not got the hinge reaction
     
  13. Nov 24, 2013 #12
    i have got v
     
  14. Nov 24, 2013 #13

    TSny

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    Yes, that's right. What concepts are you going to use to get the hinge reaction?
     
  15. Nov 24, 2013 #14
    H= hinge reaction
    Hx=2mv2/l
    and for y direction
    Hy=mg
     
  16. Nov 24, 2013 #15

    TSny

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    OK, but since I don't know which direction you chose for the positive x direction, I don't know whether Hx is toward the left or toward the right.

    This is not correct. ƩFy = May.
    Does the center of mass have any acceleration in the y direction?
     
  17. Nov 24, 2013 #16
     
  18. Nov 24, 2013 #17
    whats the problem with this??
     
  19. Nov 24, 2013 #18

    TSny

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    In what direction is the x component of the acceleration of the center of mass when the rod is horizontal?
     
  20. Nov 24, 2013 #19
    right
     
  21. Nov 24, 2013 #20

    TSny

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    In the equation ƩFy = May, what are the forces that appear on the left side and what are the signs for these forces?
     
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