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.
Best Answer
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.