What Is the Minimum Speed Needed for a Particle to Stay on a Rippled Surface?

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To keep a particle on a rippled surface described by h(x) = d Cos[k x], the minimum speed required is v ≤ Sqrt[g/(k^2 d]. The particle will lose contact with the surface at the point of maximum curvature, which occurs at the crest of the wave. At this crest, the centripetal acceleration needed to maintain contact is greatest. Although the particle could potentially leave the surface at other points, it must exceed the crest's maximum speed to do so. Understanding these dynamics is crucial for analyzing motion on non-linear surfaces.
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Consider a particle moving without friction on a rippled surface, given by the equation h(x) = d Cos[k x]. Gravity acts down in the negative h direction. If the particle starts at x=0 with a speed in the x direction, for what value of v will the particle stay on the surface at all times?

The answer is if v<= Sqrt[g/(k^2 d)].

I found a general formula that involves x but is too complicated to solve. There must be another way to do this without using a calculator. Thanks!

Ying
 
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Look at the boundary conditions.

The point where the the particle will leave the surface is the point with the largest curvature. You need to solve for the required speed at the crest of the wave, and it will hold for all other points on the wave.
 
How do you know that the particle will leave the surface at the crest of the wave?
 
At the crest, the centripetal acceleration needed to maintain contact is the highest.

It can leave at another point, but it has to be going faster than the crest's max speed to do so.
 
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