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Hibbler dynamics problem 13-97

  1. Feb 20, 2007 #1
    1. The problem statement, all variables and given/known data
    Smooth particle with mass of 80 grams is attached by an elastic cord from O to P. O is at the base of a circle with radius .4m. Angle theta is 60deg from horizontal. The cords unstretched length is .25m. K of the cord is 30N/m
    The path the particle makes is r=0.8sin(theta)m. The constant angular velocity is 5 rad/s. An arm moves this particle.


    2. Relevant equations
    What I am looking for is the force that the guide has on the particle. This particular one is in the horizontal plane so gravity is out. How do you do this?


    3. The attempt at a solution
    Force that the cord produces is (0.69m-.25m)30N/m or 6.64 newtons
    The listed answer is 7.67N. I know this has something to do with the tangential and normal forces due to the particle itself.
     
  2. jcsd
  3. Feb 20, 2007 #2
    Can you post a diagram of the problem>?
     
    Last edited: Feb 20, 2007
  4. Feb 20, 2007 #3
    Drawing for Hibbler 13-97

    I attached a word drawing of the dynamics problem. Hopefully it will make the problem clearer.
    Thanks
     

    Attached Files:

  5. Feb 21, 2007 #4
    you should check your math on the force that the cord produces


    Go back to the basics, and draw a FBD of the particle with all the forces acting on it,

    establish a coordinate system for the radial and transverse components

    (hint: the elastic cord's force is collinear with the radial component, and the guide's force is collinear with the transverse component)

    once you have this FBD, it should be easy to write down your equations of motion to solve for the unknowns
     
    Last edited: Feb 21, 2007
  6. Feb 21, 2007 #5
    Solution. Woo hoo

    Npsin(θ ) − 30N/m(r − .25m) = .08kg(r'' − rθ'2)
    F−Npcos(θ ) = .08kg(rθ'' + 2r'θ')

    If you work this all the way out it finally does work out to 7.67N
     
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