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I'm currently stumped by the following situation. Say we've got a rectangular physical pendulum (think ruler with a hole-punch at one end).

It's trivial to analyze the motion of the pendulum with the pivot point as the chosen "base." There's no translational motion, and gravity simply creates a torque. There's only angular motion to consider.

However, I'm currently writing a physics engine as a personal project, and I've been doing fine until I ran into this problem (which is purely physics).

I need, at each step of the simulation, provide every entity (such as the pendulum bar) with it's net force at the center of mass, and net torque about the center of mass.

In attempting to analyze a pendulum as such to find these qualities, it's much more complicated:

Pendulum Situtation

Now, considering the center of mass as the point of reference, The forces in black are those acting on the bar: they are (1) gravity and (2) some concoction of contact forces from the pin that it's swinging on.

The quantities in red are the things that the forces SHOULD produce. I should find a net translational force, and a net torque.

What I've ever learned about physics suggests that I should be able to consider an arbitrary point on the object, and the torque calculations will add up to the same amount. So there should be an equally valid perspective on a pendulum where treat it as if it's both moving and rotating from the center of mass's point of view.

My question is, what is the nature of these contact forces? What insight am I missing about this situation?

In other words, What are the forces in this perspective that give rise to correct pendulum motion?

Thanks for any insight on this, - Chase

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In this case, the reaction forces depend upon the velocity of the system as well. –  Manishearth Mar 12 '12 at 4:51
That's fine if the velocities appear in the equation, but I'd be fantastic if you could elaborate on that relationship –  Chase Meadors Mar 12 '12 at 4:53
If your system has an angular velocity $\omega$, then it must have a resultant acceleration of $\omega^2 r_{cm}$ directed towards the point of contact. Applying Newton's laws in the radial direction gets you one equation. Now, equating torques (torque due to $mg$ about hinge = torque due to reaction forces about com), we get another equation. We can simultaneously solve these to get the components of the reactin force. Using these, you can solve for everything else, including the tangential acceleration of com. –  Manishearth Mar 12 '12 at 5:14
Note that I am still solving it in the ground frame. Calculations in the com frame involve psuedoforces. Equipped with your value for the reaction force, you should be able to proceed –  Manishearth Mar 12 '12 at 5:18
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