# Lagrange-Euler equations for a bead moving on a ring

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.

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$$as the angle around the z-axis, starting from the x-axis. The z-axis passes through a diameter of the ring, and the ring rotates about the z-axis. \omega = \mathrm{d}\phi/\mathrm{d}t; the \omega^2\sin^2 \theta term comes from the rotation of the hoop, and the \dot\theta^2 term comes from the motion of the particle along the hoop. After you plug in the definitions of (r, \theta, \phi) in terms of (x, y, z) and apply the restriction that a^2 = x^2 + y^2 + z^2, the rest is algebra. Most of the terms cancel and/or simplify down to those two terms. I'm not sure where that factor of sin^2 \theta is coming from in the kinetic energy. If \omega is the angular velocity of the ring then it should just be:$$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$$• Where$\theta$is the position of the bead on the ring, and the coordinate system is fixed in its center? Do we need the extra summand$\omega t$in order to account for the movement of the circle? – superAnnoyingUser Feb 21 '13 at 22:25 The$\sin(\theta)$comes in by taking account the rotational kinetic energy of the mass. So, taking$\theta$as originating from the bottom of the loop, if the mass hangs there, it will have zero rotational kinetic energy. At$\theta=\pi\$, the mass will have the maximum rotational kinetic energy possible.