1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Mass on a stick model.

  1. Feb 5, 2008 #1

    Integral

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    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

    Integral

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    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.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Mass on a stick model.
  1. Meter Stick (Replies: 6)

  2. The owie stick (Replies: 8)

Loading...