Lagrange-Euler equations for a bead moving on a ring

Question

A bead with mass $m$ is free to glide on a ring that rotates about an axis with constant angular velocity. Form the Lagrange-Euler equations for the movement of the bead.

Solution: Let us introduce the generalized coordinates $\theta$ to determine the beads position. $$\Rightarrow L=L(\theta, \dot\theta)=K-P$$ $$K={mv^2\over2}=\frac m2a^2(\omega^2\sin^2\theta+\dot\theta^2)$$ Here $a$ is the radius of the circle, and $\omega$ I believe to be the angular velocity. Now I know that $v = (\dot x,\dot y)$ and $v^2 = \dot x^2+\dot y^2$.

Then $\dot x=a\omega\sin\theta\land\dot y=a\dot\theta$. Then $x=-a\omega\cos\theta\land y=a\theta$. But I believe that is what they used to get $v^2$.

Am I wrong? Is there a law, giving $x$ and $y$? How did they arrive at $v = (\omega^2a\sin\theta, a\dot\theta)$?

The rest of the solution I understand.

Answer 1

I think you have your geometry wrong. You need to set up the speed in three dimensions: $\dot{\bf x} = (\dot x,\dot y,\dot z)$. Then $\dot{\bf x}^2 = \dot x^2 + \dot y^2 + \dot z^2$. Convert that into spherical coordinates $(r, \theta, \phi)$, with $\theta$ as the angle down from the z-axis towards the xy-plane, and $\phi$$ as the angle around the z-axis, starting from the x-axis.

The z-axis passes through a diameter of the ring, and the ring rotates about the z-axis. $\omega = \mathrm{d}\phi/\mathrm{d}t$; the $\omega^2\sin^2 \theta$ term comes from the rotation of the hoop, and the $\dot\theta^2$ term comes from the motion of the particle along the hoop. After you plug in the definitions of $(r, \theta, \phi)$ in terms of $(x, y, z)$ and apply the restriction that $a^2 = x^2 + y^2 + z^2$, the rest is algebra. Most of the terms cancel and/or simplify down to those two terms.

Answer 2

I'm not sure where that factor of $sin^2 \theta$ is coming from in the kinetic energy. If $\omega$ is the angular velocity of the ring then it should just be:

$$K=\frac{1}{2} ma^2(\omega + \dot{\theta})^2$$

This is because $x=a cos(\theta +\omega t)$ and $y=a sin(\theta +\omega t)$, which can be found using some simple geometry. When you take the time derivative you get:

$$\dot{x}=-a(\dot{\theta}+\omega ) sin(\theta +\omega t)$$

$$\dot{y}=a(\dot{\theta}+\omega ) cos(\theta +\omega t)$$

So:

$$\dot{x}^2+\dot{y}^2=a^2(\dot{\theta}+\omega )^2$$