# Lagrangian of particle on a horizontal, square rotating hoop.

The problem of interest is worded as follows:

A horizontal square wire frame with vertices $ABCD$ and side length $2a$ rotates with constant angular frequency $\omega$ about a vertical axis through $A$. A bead of mass $m$ is threaded on $BC$ and moves without friction. The bead is connected to $B$ and $C$ by two identical light springs of force constant $k$ and equilibrium length $a$.

By considering a small displacement of the particle away from the midpoint of $BC$, find the Lagrangian.

I'm struggling to find a good coordinate system to work in. I considered taking A to be the origin, and using a polar angle measured from the line produced by the initial position of $AM$ where $M$ is the midpoint of $BC$, but the expression for the angle that the particle makes with the initial line seems far too complicated to be the best way.

Any hints on how to proceed?

• Is the position of the bead (a+r, wt) in polar coordinates where r is the distance from the equilibrium position and will satify a DE of the form r'' = -cr for some constant c (' is derivative wrt time t).
– Paul
Nov 14 '14 at 11:52
• @Paul, Where are you taking the origin? Also, surely the angular component varies differently to linearly in time, given that the springs will be moving the particle so that it doesn't trace out a circle as it goes around? Nov 14 '14 at 12:16
• @Paul, it's more likely that I've misinterpreted your hint. Nov 14 '14 at 12:17
• The angular position will just be wt as it rotates round at constant angular velocity. The radial position will vary harmonically either side of a and satisfy $mr'' = -2kr$. your 2 variables will be r and $\theta = wt$
– Paul
Nov 14 '14 at 14:35

## 1 Answer

You know the explicit polar coordinates of $B$ and $C$ and you know the bead is constrained to move between them, i.e. its position is always of the form $${\bf b}=\lambda {\bf B} + (1-\lambda){\bf C} \text{ for } \lambda \in \left[0,1\right].$$ I would try using $\lambda$ (or $\mu = \lambda-\frac{1}{2}$, the displacement from the centre of $BC$) as the coordinate of the bead. You can calculate the kinetic and potential energy in terms of this.