# Applying Lagrange's equations ignoring normal reaction

A small bead is sliding on a smooth vertical circular hoop of radius $a$, which is constrained to rotate with constant angular velocity $\omega$ about its vertical diameter.
$\theta$ is an angle between the downward vertical and the radius to the bead.

I've calculated that:

Lagrangian for this motion is $L=\frac{1}{2}m(a^2\dot\theta^2+a^2\omega^2\sin^2\theta)+mga\cos\theta$
There are 4 positions of equilibrium when $g<a\omega^2$: $\theta_1=0$, $\theta_2=\pi$, $\theta_{3,4}=\ arc\cos(\frac{g}{a\omega^2})$ and 2 positions of equilibrium ($\theta_1=0$,$\theta_2=\pi$) when $g>a\omega^2$.

For what values of $\omega$ is the equilibrium position at the lowest point of the hoop stable?

Why can we apply Lagrange's equations to this problem , thus ignoring the normal reaction between the hoop and the bead?

For question 1, try writing $\theta=\theta_0+\epsilon$ where $\theta_0$ is a point where the bead can sit stably on the hoop, and $\epsilon$ is a small perturbation. Try making the lagrangian for $\epsilon$ look like the lagrangian for a harmonic oscillator,

\begin{equation} \frac{1}{2} m \dot{\epsilon}^2 - \frac{1}{2} k \epsilon^2 \end{equation}

Then you should be able to compute the frequency in terms of $m$ and $k$ using standard formulas.

For question 2, what direction does the normal force point in?

• Ok, I will try to do that for the first question. About the second one... is it directed from a bead to the center of the hoop?
– gov
Aug 1 '13 at 11:07
• It is directed towards the center. Think about what that means. In polar coordinates $r$ and $\theta$, are 0, 1, or 2 of the coordinates affected by the normal force? Aug 1 '13 at 16:47
• Well, if it is directed towards the center, than in lays along $\vec e_r$. And it's perpendicular to $\vec e_\theta$. So, one coordinate, $r$, is affected by normal force?
– gov
Aug 1 '13 at 17:55
• You are * this * close to answering part b. You have $\theta$ in your action and you are neglecting $r$ because of the constraint. The normal force doesn't affect $\theta$, it affects $r$. So... why can you neglect the normal force? Aug 2 '13 at 11:51
• I don't know really :D I only know now that if I multiply Newton's second law by $\vec e_\theta$, normal force will be canceled.
– gov
Aug 2 '13 at 12:54