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Stuck with 2D kinetics problem

  1. Jul 21, 2016 #1
    1. The problem statement, all variables and given/known data
    Hi, I've been trying to solve this exercise for the last four hours and I'm totally stuck. The problems goes like this:

    Given the mechanism in the image, located in the vertical plane, the OB cord is cut when t=0. Find the reactions Rx, Ry and N in {Kgf} for the instants when t=0s, t=0.2s and t=0.4s. The initial values are θ(0)=60° and θ'(0)=0. Both rods have mass m=25kg and lenght L=3m
    (i uploaded a image showing the mechanism)


    2. Relevant equations
    The first thing I realized was that the three reactions were time/angle dependent, so I inmediately thought of solving this problem with energy considerations, instead of newton laws. To do this I choose my system to be both rods so the external forces are the gravity and the three reactions. Also, those reactions don't do any work to the system so in terms of equation I got:
    (initial kinetic energy) + (initial potential energy) = (final kinetic energy) + (final potential energy)

    3. The attempt at a solution
    The problem I'm facing here is this, I get a real nasty looking differential equation with no analytic solution. This is preventing me to find θ(t). Also, i not sure if this is really the approach i should be following, is there any easier way im not taking into account?
     

    Attached Files:

  2. jcsd
  3. Jul 22, 2016 #2
    I decided to use the lagrangian to get the differential equation in the generalized coordinate θ to obtain its differential equation. I've got this
    (I0+Icr)θ'' + (mgLcosθ)θ' = 0
    Icr is the moment of inertia of the second rod with respect to its instantaneus center of rotation.
    Ok so I solved this diff. eq. using wolpram alpha and the solution looks really complicated. Im sure this is not the approach this exercise requires, but I cant seem to figure out any other way, because as i said, the three reactions are not constant. Any idea on this? I would really appreciate any help
     
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