# Effective potential of a particle sliding on an inclined rod

1. Apr 25, 2013

1. The problem statement, all variables and given/known data

Hey guys,

Here is the question:

A pointlike mass $m$ can slide along a rigid rod of length $l$ and negligible mass. One extremity of the rod is fixed at the origin $O$ of an inertial system $(x,y,z)$, and the rod forms a constant angle $α$ with the $z$-axis. The rod rotates about the $z$-axis with constant angular velocity $ω$. Gravity acts in the negative $z$-direction.

(i) How many degrees of freedom does the system have?

(ii) Write down the Lagrangian and the Lagrange equations.

(ii) Recast the system as a 1-dimensional motion in an effective potential. Find an expression for the effective potential and determine the equilibrium positions as well as their stability

2. Relevant equations
$T=1/2mv^{2}$
Lagrangian: $L = T - V$
Lagrange Equation: $\frac{d}{dt}\frac{\partial L}{\partial \dot{q}} = \frac{\partial L}{\partial q}$

3. The attempt at a solution
Here are my solutions:

(i) one degree of freedom - the distance from the origin to the point mass. call this distance $r$.

(ii) I wont put the calculations here, I'll just give my results:

Lagrangian $L = T - V = \frac{1}{2}m[\dot{r}^{2}+r^{2}ω^{2}sin^{2}α] - mgrcosα$

Lagrange equation: $\ddot{r}-rω^{2}sin^{2}α + gcosα = 0$

(iii)

Okay, so here is the issue. How do you get the effective potential from this? and do I just differentiate that to get the equilibrium positions and then differentiate again to determine stability?

2. Apr 26, 2013

### TSny

If you had a particle of mass m confined to 1-dimension of space (a straight line) and you let r be the distance from the origin, how would you express the kinetic energy T? How would you construct a potential energy function V(r) for this particle so that it would have the same Lagrangian L as your three-dimensional problem of the particle on the rotating wire?

Sounds good.