A point mass on top of a frictionless hemispherical dome is given an initial velocity v. Loss of contact?

Question

At what point does the point lose contact with the dome? Thorough explaination please. I remember doing this back in classical mechanics and saw it in a trade magazine about a year ago. Its been driving me nuts ever since.

Answer 1

This question seems incomplete, or relies on a diagram with additional information. If the point mass is given a horizontal velocity v, the mass either loses contact with the hemisphere immediately, or it doesn't lose contact with it at all. Assuming gravity is present, the path of the mass would be a parabola if the dome weren't present. If that parabola is outside of the semicircular cross-section of the hemisphere, then the mass loses contact with the dome immediately and follows the parabola. If the parabola is inside the semicircle, the mass doesn't lose contact at all; it slides along the dome instead of following a parabolic path. If the parabola is initially outside of the semicircle but then intersects it again, the point mass initially loses contact but then hit the dome and follows it.

We can define the path of the point parametrically. The horizontal velocity, v, does not change, so x = vt. The vertical velocity is governed by gravitational acceleration, so y = r - 0.5gt^2, where r is the radius of the dome, and thus the point mass's original vertical position. You will need to know r to solve the problem. Rewriting t as x/v, we can say y = r - 0.5g(x/v)^2 = r - (0.5g/v^2)x^2. The curve of the semicircular cross-section is part of the circle x^2 + y^2 = r^2, or y = sqrt(r - x^2). So if r - (0.5g/v^2)x^2 is greater than sqrt(r - x^2), the path of the point will be beyond the surface of the dome, and contact will be lost.