Modeling the damping of a physical rigid pendulum

  • #1
Bossmand Per
1
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TL;DR Summary
This question is about modeling the damping of a physical rigid pendulum. With more than 26,000 data points it will be quite easy to test the different hypotheses. However, the data seems to be structures in an unexpected way.
I'm engaged in a research project focused on pendulums.
I'm trying to model a rigid pendulum's motion with a second order differential equation (where time is the independent variable) describing the relationship between θ (theta) , ω (omega) and α (alpha) where:
- θ is the angle of the pendulum measured in radians. θ is zero when the center of mass is directly below the axis of rotation.
- ω is the derivative of θ also known as the angular velocity.
- α is the derivative of ω also known as the angular acceleration.
I need your help to find the best model for the pendulum.
Using a precise measuring technique, I have gathered approximately 26,000 data points (θ, ω, α) detailing the movement of my rigid physical pendulum. This should be sufficient to verify whatever model you would come up with. This GIF gives an impression of the data:



Honestly, I have no idea why my data looks like this. I just can't explain away the clear but unexpected structures in the data. I expected the data to fit to this differential equation:
α=-μ*ω-mgd/I*sin(θ), where:
- μ (mu) is a friction constant
- m is the mass of the pendulum
- g is the gravitational acceleration
- d is the distance between the center of mass and the axis of rotation
- I is the moment of inertia
Here I'm taking advantage of the formula for net torque (τ) given by τ=α*I (only valid for rigid objects). The way I see it the torque consists of two things:
1) The pendulum's weight Fg = mg. The torque generated by the weight is the cross product of the weight vector and the vector from the pivot point to the center of mass:
τweight = Fg x r
Therefore the size of the torque generated by the weight is mgd sinθ.
2) The damping. In the quite simple model provided earlier, the damping as a torque is modeled as -μωI (since we divide by I to get from τ to α). I'm open to that it could be more complicated than that. Since I'm using a bike wheel as my pendulum I think it would make sense to subdivide the damping into at least two: a) the air resistance which probably only depends on ω and b) the friction in the hub which probably depends on ω as well as on θ. The reason why it might depend on θ is because of the centrifugal force which plays a role in the pressure on the ball bearings I guess. For example if the wheel is spinning and θ=π (meaning that the center of mass is directly above the pivot point) the weight (which I assume to be constant) will be counteracted by the centrifugal force, probably resulting in less friction. I've worked out that the magnitude of net force on the hub (the sum of the weight vector and the centrifugal force vector) will be:
Fhub = m√(d2ω4+2dgω2cosθ+g2)

* ω was obtained by measuring the angular velocity with a cellphone attached to the pendulum. θ was obtained by integrating ω (with respect to time). α was found by differentiating ω (with respect to time). The experiment was repeated five times with varying initial conditions (angle and angular velocity) in order to cover more of the (θ, ω) plane. The data covers the 3D space [-πrad, πrad]x[-12.05rad/s , 13.82rad/s]x[-6.56rad/s/s 6.49rad/s/s].

** The pendulum that I am studying consists of a ½kg weight attached to a bike wheel so that it spins unhindered around its axis of rotation. Here’s a sketch of the setup:
unavngivet.png

The thing with my data that confuses me the most is best explained by looking at this section curve:
1715806135875.png
1715806154928.png


This section curve shows the angular acceleration (α or alpha) for θ=-π/2rad. What I want you to notice is that α seems to be largest in magnitude when ω=0rad/s. I don't see how to include this behavior in a model, and what is worse: I don't think that this is intuitive: Why doesn't α "care" which way the pendulum is rotating when the angle is -π/2rad? It almost looks like the damping should be ω2 but I can't find a way to justify this. I understand that if the pendulum were to move downwards from this position (meaning that ω>0rad/s), the pendulum would experience a damping torque in the opposite direction (upwards) resulting in a smaller α which is what we observe. But if the pendulum were to move upwards from this position (meaning that ω<0rad/s), the pendulum should experience a damping torque in the opposite direction (downwards) resulting in an even greater α which is the exact opposite of what we observe! In other words: I really didn't see this parabola shape coming. I had expected a monovariant curve. Why isn't the sign of the damping torque opposite of the sign of ω?
To make this even more clear I have actually found a way to isolate the damping torque (actually not the torque but τdamping*I) from the torque provided by the weight (still not the torque but τweight*I you know). This can be done because the constant mgd/I corresponds to the the partial derivative of the plot (θ, ω, α) with respect to θ evaluated at (θ, ω)=(0rad, 0rad/s):
1715809279504.png
=
1715809265914.png
-
1715809320226.png


So just to summarize my question: If the simple model provided earlier was sufficient, the dampning would be a plane instead of this curvy structure (figure named "Damping alpha). But what can explain this structure? Why is the bike wheel pendulum damped like that?
 

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  • #2
Bossmand Per said:
This section curve shows the angular acceleration (α or alpha) for θ=-π/2rad. What I want you to notice is that α seems to be largest in magnitude when ω=0rad/s.
The free body diagram has a vertical force ##Mg##, a force ##Ma## in the direction of motion, and a force perpendicular to the direction of motion. The only forces that affect the angular acceleration are ##Mg## and ##Ma##, and those forces are not affected by the angular velocity. Your plot of ##\omega## vs ##\alpha## is not consistent with the theory.
Bossmand Per said:
ω was obtained by measuring the angular velocity with a cellphone attached to the pendulum.
When things don't make sense, it's time to look very hard at the source of your numbers. I would start by plotting ##\theta## against time, then compare the period of oscillation to the period measured by hand with a stopwatch. Also, exactly how are you measuring angular velocity with a cellphone? Please explain in far more detail than you think necessary, including a photo or diagram.
 

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