A polygon is rolling down a hill

[etotheipi]

Homework Statement

A polygon is rolling down a hill, find the angular velocity after the $N^{\text{th}}$ impact with the hill (assume polygon has slight concavity)

Relevant Equations

N/A

An $n$-sided regular polygon is rolling down a frictional ramp at angle $\theta$ to the horizontal. I define $\beta := \frac{2\pi}{n}$ as the angle at the top of each of the $n$ isosceles triangles that make up the polygon. Let $\omega_{k, 1}$ be the angular velocity just after the $k^{\text{th}}$ impact with the ramp, and $\omega_{k, 2}$ be the angular velocity just before the $(k+1)^{\text{th}}$ impact with the ramp. Finally, let the moment of inertia about the centre of the polygon be $I_{cm} := pmr^2$ and the moment of inertia about one of the vertices be $I_{v} := qmr^2$, where $r$ is the distance from the centre to a vertex. The side length of the polygon is $l = \sqrt{2r^2(1-\cos{\beta})}$.

At the $(k+1)^\text{th}$ impact, we conserve angular momentum about the point on the ground with which the next vertex is just about to come into contact,$$I_{cm} \omega_{k,2} + mr^2 \omega_{k,2} \cos{\beta} = I_v \omega_{k+1, 1}$$ $$(p+\cos{\beta})\omega_{k,2} = q\omega_{1, k+1}$$Between impacts, the polygon gains kinetic energy $mgl\sin{\theta}$ due to the decrease in height of its centre of mass, so we can also say that$$\frac{1}{2}I_v \omega_{k, 1}^2 + mgl\sin{\theta} = \frac{1}{2}I_v \omega_{k,2}^2$$ $$\omega_{k,2}^2 - \omega_{k, 1}^2 = \frac{2g\sin{\theta}}{qr}\sqrt{2(1-\cos{\beta})}$$I put this into the first equation to obtain$$\omega_{k+1,1} = \frac{p +\cos{\beta}}{q} \sqrt{\omega_{k,1}^2 + \frac{2g\sin{\theta}}{qr}\sqrt{2(1-\cos{\beta})}}$$I wondered if anyone can help me to solve this recurrence relation? Thanks

 

[haruspex]

obtain$$\omega_{k+1,1} = \frac{p +\cos{\beta}}{q} \sqrt{\omega_{k,1}^2 + \frac{2g\sin{\theta}}{qr}\sqrt{2(1-\cos{\beta})}}$$I wondered if anyone can help me to solve this recurrence relation? Thanks

It would have been easier algebra to have worked in terms of the half angle.
Isn't your last equation of the form $x_{k+1}=Ax_k+B$?

Of course, if it is accelerating then at some point it will lose contact before the next impact.

 

[etotheipi]

It would have been easier algebra to have worked in terms of the half angle.
Isn't your last equation of the form $x_{k+1}=Ax_k+B$?

Ah you're right, I could just let $u_k := \omega_k^2$ and then it is just a linear recurrence relation. Also a good idea about the half angle, I think the nested root would become $2\sin{\frac{1}{2}\beta}$, for one thing.

Of course, if it is accelerating then at some point it will lose contact before the next impact.

I hadn't thought of that! Loosely, I imagine should just be a case of writing the Newton II equation for circular motion of the centre of mass of the polygon about the point of contact (in terms of the normal force and the weight), and letting $N$ be zero to see when weight can no longer provide the necessary centripetal acceleration. Is that also what you had in mind?

I'll try to tie off these loose ends tomorrow, because it's getting a little late. Thanks for the help!

 

[jbriggs444]

It would have been easier algebra to have worked in terms of the half angle.
Isn't your last equation of the form $x_{k+1}=Ax_k+B$?

Oh, right. Square both sides. Don't work with velocity. Work with energy.

Edit: I see that I am late to the party.