5 ms·
No. What all this shows is that both cases (hemispherical dome or the other one in OP) of a ball in stable equilibrium are outside the domain of applicability o
by abdullahkhalids 2y ago
No. What all this shows is that both cases (hemispherical dome or the other one in OP) of a ball in stable equilibrium are outside the domain of applicability of the theory. You should simply not use the theory as is to make predictions about when and how the ball will roll off.
- tsimionescu 2y agoActually, there is nothing mysterious here. The theory, applied correctly, predicts the expected thing: a particle that starts at rest at the apex will never leave it. The other "solution" is not actually a particle at rest (it has a non-0 6th derivative of position, that is the rate of change of the rate of change of its acceleration is 1/6). So it's actually describing a particle whose acceleration will increase at an increasing rate forever, but that just happens to be 0 at time T.
- abdullahkhalids 2y agoYes. The situation is correctly described by a modified version of standard Newton's theory, in which you start caring about higher derivatives.
- foxglacier 2y agoJust in case I'm completely wrong about reality, both cases are unstable equilibrium, aren't they? Equilibrium in that forces balance and zero acceleration but unstable in that any disturbance will lead to it rolling off rather than returning to where it was.