I have derived a model for what happens when one rolls a gear down a slope, and I have performed an experiment, however the experiment doesn't fit the model. Where did I go wrong?
The model
When a gear rolls down a slope, I model it as constantly switching between rotating around the end of a spike, and a spike colliding with the surface.
Moment of inertia around spike
Let the moment of inertia aroud the center of the gear be $I_{CM} = \alpha mr^2$. Then the Moment of inertia around a spike is $I = I_{CM} + mr^2 = (1+\alpha)mr^2$.
What happens to the angular velocity when a collision happens?
The image below is an illustration of a gear just before a collision. The velocity $v$ is perpendicular to the left line, since the gear is rotating around that spike. The angle $\theta$ and the distance $\delta$ is the values shown in the image.
I'm going to assume there is conservation of angular momentum in this collision. Let's find the angular momentum around the collision point before and after the collision: \begin{align} L_{\mathrm{before}} &= I_{CM}\omega_1 + mv_1r\sin(90^\circ-2\theta) = I_{CM}\omega_1 + mv_1r\cos(2\theta) \\ L_{\mathrm{after}} &= I\omega_2 \end{align} where $v_1,\omega_1$ is the velocity and angular velocity before the collision, and $\omega_2$ is the angular velocity after the collision.
Because of the conservation, we can write $I_{CM}\omega_1 + mv_1r\cos(2\theta) = I\omega_2$.
I'd like to find a formula for $\omega_2 / \omega_1$.
\begin{align*} \frac{\omega_2}{\omega_1} &= \frac{I_{CM} + mr^2\cos(2\theta)}{I_{CM}+mr^2} \\ &= \frac{I_{CM} + mr^2 - mr^2 + mr^2\cos(2\theta)}{I_{CM}+mr^2} \\ &= \frac{(1+\alpha)mr^2 - mr^2 + mr^2\cos(2\theta)}{I_{CM}+mr^2} \\ &= \frac{mr^2\alpha + mr^2\cos(2\theta)}{I_{CM}+mr^2} \\ &= \frac{mr^2\alpha + mr^2\cos(2\theta)}{(1+\alpha)mr^2} \\ &= \frac{\alpha + \cos(2\theta)}{1+\alpha} \\ &= 1 - \frac{1 - \cos(2\theta)}{1+\alpha} \\ &= 1 - \frac{2(\sin\theta)^2}{1+\alpha} \end{align*} Finally using right angle trigonometry $\sin \theta = \frac \delta r$, and we get: $$\frac{\omega_2}{\omega_1} = 1 - \frac{2\delta^2}{(1+\alpha)r^2}$$
What happens to the angular velocity when the gear rotates?
The image below is an illustration of a gear on a slope just before a collision. The angle $\varphi$ is the angle of the slope.
In order to compute the change in angular velocity, I'm going to set the energies just after a collision and just before the next collision equal to one another. Let $h$ be the change in height, then:
$$\frac12I\omega_2^2 = \frac12I\omega_3^2 + mgh = \frac12I\omega_3^2 + 2\delta mg\sin\varphi$$ where $\omega_3$ is the angular velocity just before the next collision.
I'm going to define $\beta$ as: $$\beta = \frac{4\delta mg\sin\varphi}{I} = \frac{4\delta g\sin\varphi}{(\alpha+1)r^2}$$ This allows us to write the following relationship: $$\omega_2^2 = \omega_3^2 + \beta$$
The equation for $\omega_2 / \omega_1$ allows us to write: \begin{align*} \omega_1^2 &= \left(1 - \frac{2\delta^2}{(\alpha+1)r^2}\right)^2\omega_2^2 \\ \omega_1^2 &= \left(1 - \frac{2\delta^2}{(\alpha+1)r^2}\right)^2(\omega_3^2 + \beta) \end{align*}
Putting it all together
Since $\omega_1$ and $\omega_3$ both are angular velocities just before a collision, we now have a recursion formula, that allows you to compute the angular velocity just before any collision. It turns out this recursion has a fixed point, and it converges to that fixed point. Let $\omega_\infty$ be the fixed point. \begin{equation*} \omega_\infty^2 = \left(1 - \frac{2\delta^2}{(\alpha+1)r^2}\right)^2(\omega_\infty^2 + \beta) \label{eq:inf} \end{equation*} Now $2\delta^2$ is pretty small, so using $(1+x)^n \approx 1+nx$ allows us to simplify to \begin{equation*} \omega_\infty^2 = \left(1 - \frac{4\delta^2}{(\alpha+1)r^2}\right)\omega_\infty^2 + \left(1 - \frac{4\delta^2}{(\alpha+1)r^2}\right)\beta \end{equation*} \begin{equation*} \omega_\infty^2 = \omega_\infty^2 - \frac{4\delta^2}{(\alpha+1)r^2}\omega_\infty^2 + \left(1 - \frac{4\delta^2}{(\alpha+1)r^2}\right)\beta \end{equation*} \begin{equation*} \frac{4\delta^2}{(\alpha+1)r^2}\omega_\infty^2 = \left(1 - \frac{4\delta^2}{(\alpha+1)r^2}\right)\beta \end{equation*} No $\frac{4\delta^2}{(\alpha+1)r^2}$ is pretty small as well, so lets just say it's zero: \begin{equation*} \frac{4\delta^2}{(\alpha+1)r^2}\omega_\infty^2 = \beta \end{equation*} \begin{equation*} \omega_\infty^2 = \frac{(\alpha+1)r^2}{4\delta^2}\cdot\frac{4\delta mg\sin\varphi}{I} = \frac{g\sin\varphi}\delta \end{equation*} Allowing us to conclude that the fixed point is: \begin{align*} \omega_\infty &= \sqrt{\frac{g\sin\varphi}\delta} \\ v_\infty &= r\sqrt{\frac{g\sin\varphi}\delta} \end{align*}
Now, I've made an experiment with $\delta = 1.2\mathrm{cm}, r = 4.8\mathrm{cm}$ and 24 spikes on the gear. The experiment was made on two different slopes by filming the gear rolling down the slopes and saving the x and y coordinates in each frame.
The results in the experiment are as follows: (data files here and here)
It's pretty clear that the fact that the prediction that it converges to some fixed point is correct, but the line with the final velocity is not parallel to the $v_\infty \cdot t$ line at all. The actual terminal speeds appear to be $52.19 \mathrm{cm s^{-1}}$ and $72.39 \mathrm{cm s^{-1}}$.
Now the question is: What did I do wrong? I have some ideas to what could be wrong, here are my thoughts:
- There is not actually conservation of moment of inertia, since the gears do not have spikey spikes, but the spikes are actually flat on the end, like the ones in the pictures are. I can't really reason whether this should give a larger or smaller velocity either.
- I'm pretty sure there is conservation of energy.
- There doesn't seem to be large errors in the measurements.
- Maybe it's not perfectly rotating around the spike, it could be sliding a bit too, which is consistent with the real velocity being larger.
- I made some approximations during the derivation, I have attempted to have a computer solve the fixed point equation without adding those approximations, but I didn't get anything useful. The two approximations I did shift the expected velocity in different directions, so it's also not clear if this would give a larger or smaller velocity.
