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Rigid body dynamics (kinematics of a point)

  1. May 17, 2012 #1
    1. The problem statement, all variables and given/known data
    See attachment "problem"


    2. Relevant equations



    3. The attempt at a solution
    See attachment "work" and "answer"

    I got 0.92, but the book gives 1.73. Something went wrong, but I can't find it.
     

    Attached Files:

  2. jcsd
  3. May 17, 2012 #2

    tiny-tim

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    hi xzibition8612! :smile:

    why have you made xA negative? :confused:
     
  4. May 17, 2012 #3
    xA is negative because if you look at the attachment "question", I set point C as the origin. Left of C is in the negative direction. Thanks for responding.
     
  5. May 17, 2012 #4
    I think you have made some sign errors. Check your quadratic equation, I don't think q is correct.
     
  6. May 17, 2012 #5
    You made Xa positive the first time you used the law of cosines. Move the axis to A and see if you get a better result.
     
  7. May 18, 2012 #6
    My error, after deciding to do the math, your length of q is 13.9 m but I agree with tiny-tim, I think Xa= 10 m. The curious thing is that dKb/dt is -1.73. Right magnitude, wrong direction
     
  8. May 18, 2012 #7
    goddamn this problem is driving me insane, i set origin at A and the answer turns out to be negative.
     
  9. May 18, 2012 #8
    I think the negative sign means that point B is decelerating. It is does not mean that Vb is pointing toward c.

    There is another way to do this problem just using vectors and the kinematics of rigid bodies. The plane motion of the rigid bar can replaced by a translation defined by the motion of an arbitrary reference point and by a rotation about that point.

    The vector equation is VB= VB/A + VA

    You know the magnitude and direction of VA, the direction of VB. VB/A is the vector perpendicular to the rigid bar at Pt. B causing the rigid bar to rotate CCW. You can determine the angle of VB/A from the horizontal by geometry. Through vector addition you can determine the magnitude of VB. You will see that it's magnitude is 1.73 m/s away from B. So forget the negative sign, it is not a direction sign.
     
  10. May 18, 2012 #9
    alright man thanks a lot
     
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