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.]

  • $\begingroup$ Pretty good first post, but the things you should probably add is your reasoning for choosing B (and the reasoning others might use for choosing C). $\endgroup$ – Kyle Oman Oct 2 '13 at 13:21
  • $\begingroup$ OK - my argument is that there is a lot of friction in this bowl - hence the ball barely makes it back to half the original height. I argue that before B it is essentially in free-fall and derive a speed based on the distance it falls/slips before it starts rolling. I show that this is higher than the implied speed at C given where it ends up (equating KE to PE, ignoring friction, so it is a minimum speed). I also note that the ball starts rolling after B, and this takes up energy and thus the speed along the path is reduced. Those who choose C say it is the lowest point so... $\endgroup$ – JustYield Oct 2 '13 at 13:40
  • $\begingroup$ Always assume pure rolling if not given. $\endgroup$ – evil999man Mar 14 '14 at 17:57
  • $\begingroup$ The given question is partially wrong because you cannot define absolute potential at a point. So asking that in which point ...the potential energy of the ball is highest without specifying a reference point is absurd! $\endgroup$ – Shivansh J Apr 27 '19 at 9:09

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.

  • $\begingroup$ If it is in pure roll along the bottom of the curve, at C, then what accounts for it coming to rest at D [end of arrow]? To me this must be due to friction acting against the rolling (friction now in the direction of the ball's motion) and this friction has been acting on the ball since B. Consequent question: the ball spins up for some duration, and friction gives the required torque. Once possibility of slipping is past (i.e. it is no longer moving vertically down) the direction of friction switches (dissipating the rolling energy). Where is the friction greatest? $\endgroup$ – JustYield Oct 2 '13 at 22:47
  • $\begingroup$ The diagram does not show it coming to rest at D, but continuing to move. Some of the height was lost at the initial slipping, but from then on the system is conservative. Also the coefficient of rolling friction is tiny in general. I often use 0.002-0.005 for steel on steel. For slipping it can be 0.08-1.00 depending on the material. So the rolling friction effect is negligible and C is still the point you are looking for. $\endgroup$ – ja72 Oct 3 '13 at 12:52
  • $\begingroup$ You can also view it as, max speed is point where acceleration is zero, and that is at C even with rolling. If it was only slipping then that point would be when the incline is $\mu = \tan \theta$. $\endgroup$ – ja72 Oct 3 '13 at 12:54
  • $\begingroup$ Many thanks for your additional comments. Can I please ask you to consider the case where the ball DOES come to rest at D [end of arrow] on its first pass? I appreciate you don't believe the diagram indicates that (and my colleagues agree with you), but if the material of the ball and bowl were such that there was damping (as mentioned below by @user1800 - under-damped harmonic motion) the greatest KE is attained by the ball before C (I realise that does not prove it is greater at B than C). $\endgroup$ – JustYield Oct 5 '13 at 2:43
  • $\begingroup$ @JustYield you are adding a hypothetical to a hypothetical. So now the ball is rolling in a bath of oil in order to make it rest at D? That is a different question that has no general answer. $\endgroup$ – ja72 Oct 6 '13 at 0:00

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.

  • $\begingroup$ I take your point that KE here includes both translational and rotational. (My focus on proving the ball was likely translating slower at C than at B clouded my thinking a little.) I think you've identified the area of uncertainties. If you had to put some numbers on it could you make some heroic assumptions e.g. full height H, B at H/4, C at 0 and D at H/2 etc? If we assume the ball comes to rest at D [end of arrow], does that help determine where the most losses to friction occurred? $\endgroup$ – JustYield Oct 2 '13 at 22:58

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.


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.


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