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Mass on a stick model.

  1. Feb 5, 2008 #1


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    I have need of a mathematical model of a mass at the end of a stick, pivoting about the opposite end. What I have done so far is not giving realistic results, would appreciate it if some one could point out my error.

    Basic model :

    [tex] \tau = I \ddot {\theta} [/tex]

    [tex] I = m r^2 [/tex]

    The system is being pushed by an air cylinder (force F) with the fixed end mounted a distance L directly below the pivot, the moving end of the cylinder is mounted a distance r along the pivot arm from the pivot point. With [itex] \phi [/itex] the angle between the cylinder and the pivot arm. So I have :

    [tex] \tau = r F sin( \phi) [/tex]

    My variable of interest will be the angle between the pivot arm and the line defined by the pivot and the fixed end of the cylinder, call this angle [itex] \theta [/itex]

    By the law of sines I get

    [tex] \frac {Sin(\phi)} {L} = \frac {sin(\theta)} {x} [/tex]
    where x is the length of the cylinder.

    I get x in terms of [itex] \theta [/itex] from the law of cosines

    [tex] x^2 = L^2 + r^2 -2lr cos(\theta) [/tex]

    The differential equation is:

    [tex] \ddot{ \theta } = \frac { \tau } {I} [/tex]

    See the attachment for a diagram.

    Attached Files:

  2. jcsd
  3. Feb 5, 2008 #2


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    Ok, The above model is correct. For the last week I have been running this through an excel spreadsheet using a Runga Kutta method, the results were not correct, the system simply did not respond the way I KNEW it should.

    My error? using the fricking English measurement system. I was using lbs where I needed slugs. So I was throwing 32 times the mass I thought I was. That sort of slowed things down. I finally made the simulation work by changing everything to metric, Newtons for force, kg for mass and meters everywhere, It worked great. It was then that I seriously began to figure out the lbs mass lbs force issue.
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