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Main Question or Discussion Point
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.
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.
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