= Particle on a uniformly rotating inclined plane
For a smooth plane at inclination $\theta$ rotating about a vertical axis at constant <angular velocity> $\omega$, choose rotating coordinates $\xi$ horizontally and $\eta$ uphill, with their cross product the upward <normal vector>. A sliding particle under gravity satisfies
$$
\ddot\xi=\omega^2\xi+2\omega\dot\eta\cos\theta,\qquad\ddot\eta=\omega^2\eta\cos^2\theta-2\omega\dot\xi\cos\theta-g\sin\theta.
$$
The <normal reaction> is $N=m(g\cos\theta-2\omega\dot\xi\sin\theta+\omega^2\eta\sin\theta\cos\theta)$ while contact is maintained. The <Coriolis force> does no work in the rotating frame, giving the conserved specific <energy>
$$
\frac12(\dot\xi^2+\dot\eta^2)+g\eta\sin\theta-\frac{\omega^2}{2}(\xi^2+\eta^2\cos^2\theta).
$$
The last term is the <centrifugal potential>; this quantity need not equal the inertial <mechanical energy>.
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