# Mass on a stick model.

1. Feb 5, 2008

### Integral

Staff Emeritus
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 :

$$\tau = I \ddot {\theta}$$

$$I = m r^2$$

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 $\phi$ the angle between the cylinder and the pivot arm. So I have :

$$\tau = r F sin( \phi)$$

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 $\theta$

By the law of sines I get

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

I get x in terms of $\theta$ from the law of cosines

$$x^2 = L^2 + r^2 -2lr cos(\theta)$$

The differential equation is:

$$\ddot{ \theta } = \frac { \tau } {I}$$

See the attachment for a diagram.

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2. Feb 5, 2008

### Integral

Staff Emeritus
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.