# Forces and torques about the CENTER OF MASS of a physical pendulum

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:

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

• 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 – cemulate 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
• I'm not sure this qualifies as a answer, so commenting here. I think the difficulties arise because you are modelling them as rigid bodies. When rigid bodies are in contact, the only way to know the forces is by working backwards from the resulting accelerations, i.e. future motion; so in a physics engine, you want the forces to predict future motion, but you need to know future motion to calculate the forces, and oops. – Carl Lei Aug 8 '17 at 9:09

## 1 Answer

To make it easier to analyze, let's decompose this 'reaction' force on the pivot in directions perpendicular $(x')$ and parallel $(y')$ to the bar (instead of the indicated $x$ and $y$ you indicated). Let's call these components $F_x'$ and $F_y'$.

Equations for translational motion:

$$F_y' - P_y' = 0$$

$$F_x' - P_x' = m*a$$

Now, let's calculate the torque about the center of mass, assuming the pendulum's length is L:

$$T(cm) = -F_x'*L/2 = I(cm)*b$$

Where $b$ is the angular acceleration of the pendulum about the center of mass. $I(cm)$ for a pendulum is equal to $mL²/12$. Hence:

$$F_x' = -mL*b/6$$

The final problem is determining the relationship between $b$ and $a$. That is easily done considering the center of mass rotates in a circular motion about the pivot, hence:

$$a = b*L/2$$

$L/2$ being the radius of the circumference described by the pendulum's center of mass. Making the substitution:

$$-mLb/6 - P_x' = mLb/2$$

Remember that $P_x' = mg*sin(\theta), \theta = angle$ between the pendulum and the vertical direction $y$.

• $mgsin(\theta) = -2mLb/3$

Which is the same differential equation you would obtain considering the torque about the pivot? Don't forget $b$ is the second time derivative of $\theta$.

I hope this clarifies your doubts a bit. As for the nature of these forces, they are of electromagnetic origin (just as any normal force) caused by the bond strength of whichever materials are in contact at the pivot. And yes, I am considering there is no attrition and hence energy dissipation.

• Except there is a component of acceleration parallel to the pendulum, the centripetal acceleration. If there wasn't, it wouldn't move in a circle. – Johnathan Gross Sep 17 '17 at 5:25