sophiecentaur said:
I think I have. What is the difference with the path of the CM of the pencil and the CM of the puck? Are they not both longitudinal lines on a sphere? Doesn't the PE vary in the same way?
The shape of the potential hill is different. This means that the differential equation of motion is different. Naturally, the solution(s) to the differential equation are different. Not just quantitatively different. Qualitatively different.
This has been explained up-thread. However, I will explain again.
In the case of an approximately hemispherical hill we have a horizontal force law that is directly proportional to the distance from the center of the hilltop. Without explicitly solving the associated differential equation (second order, linear, homogenous, easy to solve) the primary solutions are two exponentials, one decaying as time increases toward ##+\infty## and the other decaying as time decreases toward ##-\infty##. Any linear combination of these solutions will also be a solution.
As is usual in the case of differential equations, we use the boundary conditions for position and velocity at ##t=0## to determine which linear combination fits our scenario.
The resulting motion can be characterized as the ball or puck approaching the hill and rebounding away, approaching the hill and passing over the top, approaching the top of the hill forever, receding from the top forever or starting at the top and staying there forever.
Those are the only solutions. There is no solution where the ball or puck starts at rest at the top and then comes down.
That is for an approximately hemispherical hill.
If the hill is shaped differently, we can encounter a different differential equation where the boundary conditions (e.g. ##x=0## and ##v=0## at ##t=0##) are insufficient to uniquely pick out a solution. The laws of Newtonian mechanics are not predictive in this case.
The idea that Newtons laws fail to make a prediction can be a jarring realization.