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## Homework Statement

A bar of soap of mass m is at rest on a frictionless square plate of side-length D rests on a horizontal table. At t=0 I start to raise one edge of the plate the plate pivots about the opposite edge. The angular velocity about the pivot is a constant ω Show that the equation of motion for the soap is of the form [itex]\ddot{x} - \omega ^{2} x = - g sin( \omega t )[/itex]. Solve this for x(t) given that . In this equation x is the distance from the pivot to the soap (i.e. measured up the slope).

## Homework Equations

## The Attempt at a Solution

We will find that the potential energy is [itex]V = mgx sin( \theta )[/itex], and the kinetic energy is just [itex]T = \frac{1}{2} m \ddot{x}^{2}[/itex]. Then our Lagrangian is just [itex]L = \frac{1}{2} m \ddot{x}^{2} - mgx sin( \theta )[/itex] and since we know omega is constant, then θ = ωt, so our Lagrangian becomes [itex]L = \frac{1}{2} m \ddot{x}^{2} - mgx sin( \omega t)[/itex].

But using the Euler-Lagrange equation yields [itex]\ddot{x} = -g sin( \theta )[/itex].

Where did the [itex] - \omega ^{2} x[/itex] come from? i want to say that it has something to do with the fact that the angle is changing over time...but i can't see it right now. any help would be great. thanks in advance.