Please examine this diagram and answer the apparently trivial questions. I am particularly interested in reasoned answers for part (a)(ii) – where is the maximum Kinetic energy?
I say it is at B (as does the answer key), but others are less convinced and prefer the lowest point, C. I also assume that the diagram implies the ball comes to rest at the end of the arrow marked D.
[Apologies if I've broken any forum etiquette, this is my first post.]
Best Answer
without friction it is obvious
Kinetic Energy + Potential Energy = Constant
so maximum KE is at lowest PE, or point C.
With friction, which induces rolling the total kinetic energy is still constant, because the ball is rolling. Only when slipping the energy is dissipated. The only time it is slipping would be initially (before point B). After that and near the bottom you are in pure rolling and therefore the answer is still point C.
With rolling on a circular bowl, the KE is $K=\frac{1}{2} I \omega^2 + \frac{1}{2} m \left(\omega r \right)^2 = m g (H-y) $, and the PE is $P=\frac{1}{2} m g y$ since a falling rolling ball has speed profile of
$$ \frac{1}{2} \omega^2 = \frac{g (H-y)}{\frac{I}{m}+r^2} $$
which is a result of the equations of motion
$$ \ddot{\theta} = \dot{\omega} = - \frac{g r \cos\left( \frac{r}{H} \theta \right)}{\frac{I}{m}+r^2}$$
and $ y = H + H \sin \left(\frac{r}{H} \theta \right) $ with $y$ the height of the ball and $\theta$ its rotation. Initially when $\theta=0$ the position is $y=H$ with $y=0$ at point C.