Ball rolling into a bowl - where is its maximum KE (speed)... given there is friction. See diagram 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.]
 A: 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.
A: with no concrete values about the height AB and BC and the coefficient of static friction, angle of slope, it is all guesswork and assumptions. 
for sphere at C to have lower total KE than at B, this requires that there is deceleration after B--ie, friction MUST equal or exceed the gravitational force along that slope (this is why angle of slope is key). if this condition is not met, the ball will continue to accelerate either by sliding or rolling due to height BC being nonzero--even a miniscule amount of acceleration definitively increases the sphere's total KE over the amount at point B.
Disregarding rotational KE.
in a rolling sphere, 40% of the total KE is rotational (always true)--using the following equations:
Moment of inertia, $I_{sphere}$ = $\frac{2}{3}mr^2$
Rotational KE = $\frac{1}{2}I\omega^2$ = $\frac{1}{2}\frac{2}{3}mr^2\frac{v^2}{r^2}$, where v is linear velocity at the circumference
= $\frac{1}{3} mv^2$
Translational KE = $\frac{1}{2}mv^2$
if we assume the ball slides from A to B and only starts to roll after B, and we take the question to mean only translational KE (not reasonable IMHO) this accounts only for a 40% loss due to technicality. the gain in KE from B to C is reduced by 40%, plus additional compounded percentage due to parasitic losses. if we assume the percentage parasitic losses are the same through all points, then $\frac{BC}{AC}<0.4.$ fulfills the condition that the additional drop from B to C does not add enough to the KE of the ball to account from loss to rotation, ie, ball at C has lower KE than at B.
A: For my answer I am making the following assumptions:
1. The bowl/canyon spherical and is very shallow
2. Friction force is independent of rotation.
Equating all the forces thus:
$$ ma = - cv - mgsin(\theta) $$
where $\theta$ is the angle between the vertical  axis and the line joining the center of the spherical bowl and the ball. a = acceleration, v = velocity, c = velocity dependent friction coefficient. If $l$ is the radius of the spherical bowl, the equation reduces to :
$$ ml (\theta'') = - cl(\theta') - mgsin(\theta) $$
and thus 
$$ m (\theta'') = - c(\theta') - (mg/l)sin(\theta) $$
The above diff equation is unsolvable , so make an approximation and take the taylor series of the  sine term. Since the bowl is shallow, only the first order of the taylor series will do. (The Differential equation is still unsolvable for higher orders). The D.E. thus becomes:
$$ m (\theta'') = - c(\theta') - (mg/l)\theta $$
Which is nothing but the damped harmonic oscillator with spring constant $$ k = mg/l $$
Since the ball crosses the bottom-most point it probably losses energy; returns back to the bottom-most point; oscillates about it and eventually comes to rest. And hence it is safe to assume it's an under damped oscillator. Any plot for x for a DHO will tell you the maximum velocity is attained between time = 0 and the moment the oscillator reaches the equilibrium point (the point about which it oscillates)for the first time. Since in this case the equilibrium point is the bottom most point, the maximum velocity (KE) is attained before that point. The answer (with these assumptions) is thus B.    
A: From A to B  maximum change in potential energy takes place, this energy converts to kinetic energy (both rotational and translational) however some energy is also lost as heat due to friction, but the slope is high in magnitude it is safe to assume that the normal force will be very small and hence the dissipated energy would not be very large.
Now, from B to C, there is not very much change in height, so definitiely not much KE is gained during this journey, however due to its very small slope it has a normal force of high magnitude this would result in high dissipation of energy.

$ \Delta PE = mgh $
     $ W_{frictional} = mgcos(\theta)d$

Assuming that the journey from B to C is pretty much rectilinear, now since $\theta$ is very small $cos(\theta)$ would be near 1 hence $cos(\theta)d$ would be greater than $h$  because clearly d is much more thab h. Since more energy is wasted as heat due to friction than is gained by change in $PE$ we assert that the ** point B** haw maximum KE.
