# Block sliding down a moving plane whose angle of inclination changes

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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 $\ddot{x} - \omega ^{2} x = - g sin( \omega t )$. 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).

## The Attempt at a Solution

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

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

Where did the $- \omega ^{2} x$ 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.

## Answers and Replies

Mentor
You forgot the kinetic energy from motion perpendicular to the direction of x. As you know the motion of the slope, you can relate this to x and t.