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There is a fairly common problem in elementary physics class that goes something like this:

A small block slides from rest from the top of a frictionless sphere. How far below the top does it lose contact with the sphere? The sphere does not move.

—paraphrased from problem 4.6 of An Introduction to Mechanics by Kleppner/Kolenkow

It’s fairly easy to derive the solution to this problem. However, a slightly more difficult problem would be to explicitly calculate the trajectory of the block as a function of time. Here we will do so.

First, with T and U denoting kinetic and potential energy, respectively, we have conservation of mechanical energy:

Taking potential energy to be zero at y = 0 and applying the fact that the top of the sphere is an equilibrium point:

Solving for the speed v:

Since the point mass is constrained to lie on the circle the sphere makes in the xy-plane, we know that the velocity vector must be tangent to the circle. Letting α be the angle the point mass makes with the vertical:

Expressing everything in terms of y and simplifying, we get:

Separating variables:

and integrating, we get an equation expressing t in terms of y:

The antiderivative can be found in closed form, as one can verify by inputting the integral expression to The Integrator. To solve the integral by hand, it suffices to do u-substitution on √(r + y′) and recognize the transformed integral as being amenable to a hyperbolic trigonometric substitution.

Now we can assume that t₀ is zero, but since y = r is an equilibrium point, we have to take y₀ to be very close but not exactly equal to r; if we don’t, we end up with a singularity when we evaluate the antiderivative at y₀.

Solving the integral and simplifying, we get:

For simplicity, we make the substitution:

We can think of ε as a dimensionless parameter which controls how close y₀ is to r.

Our function then becomes:

and inverting it gets us:

with x(t) easily derivable from the constraint:

This form is quite amenable to quantitative analysis. First of all, one can quickly verify that the equation makes dimensional sense, as the argument to tanh is indeed dimensionless and the total expression has dimensions of length. tanh is an odd function which approaches ±1 as its argument approaches ±∞, so we can conclude that y(t) stays within [−r, r). The argument to tanh is just a line with a large y-intercept so the result of tanh starts at just under +1 and decreases slowly for a while, then when its argument intersects the x-axis, it briefly shoots down to 0 and shoots back up slowly approaching +1. That means that y(t) starts at just under +r and decreases slowly for a while until it falls off. If there were another constraint force acting on the point mass, say if it were a bead on a loop, it would not shoot off but quickly shoot down to −r and then just as quickly shoot back up to approach +r from the left. This seems unphysical as we would expect the bead to rebound, but we’re neglecting the frictional forces that would cause that effect in practice.