(I would certainly appreciate comments by those people considering this question to be too unclear. Or am I to guess what is unclear? The How to Ask section was of no help in this regard.)

This was given as an exercise in one of the beginner physics courses at our university this year

A particle starts at rest at the top of a frictionless sphere of radius $R$ and slides down on the sphere under the force of gravity. How far below its starting point does it get before flying off the sphere?

It can also be found in (and was propably taken from) the Feynman Lectures. The solution given to the students was the same one that can be found on the Fey. Lec. Website (I will refer to it as the FL solution), and you can find similar solutions on Phys.SE, since questions about similar or identical exercises have been asked before.

I have some questions about the exercise, the solution and beyond

  • In the exercise: What is meant by "flying off the sphere"?

  • In the FL solution: Why can the particle not fly off before $N=0$ ?

  • What would the trajectory of the particle look like with the constraint $\boldsymbol{r}^2(t) \geq R^2$ and some starting velocity $\neq 0$ ?

  • Are there some good books dealing with nonholonomic constraints (mainly inequalities) in the lagrangian formalism?

Background on the questions

About the first question: In my eyes "flying off the sphere at time $t_0$" has (at least) 2 very natural possible meanings; The right side showing how I would formulate it mathematically.

  • $(M1)$ "loosing contact at $t_0$" $ ~:\Leftrightarrow ~~t_0 = \inf\{ ~t \mid \boldsymbol{r}(t)^2 > R^2 \} $

(yes; the above means it could touch the sphere again after $t_0$)

  • $(M2)$ "never being on the sphere again after $t_0$" $ ~:\Leftrightarrow ~~ t_0 = \inf\{ ~t \mid \forall s > t : ~ \boldsymbol{r}(s)^2 > R^2 \} $

But which one is meant in the exercise? Or is it meant in a completely different way?

And as a follow-up question: If either $M1$ or $M2$ is meant; Where does the FL solution show - since it does not calculate $\boldsymbol{r}(t)$ - that one of them is fullfilled?

About the second question: The FL solution treats the problem in the following way: Aussume circle path up to $h$ where $N=0$; calculate $h$. So they have to assume $$ N > 0 \Rightarrow \text{circle path op to }h \Leftrightarrow \text{no flying off before }h $$ But this does not seem obvious to me, especially if "flying off" means either $M1$ or $M2$.

About the third question: The FL solution assumes the particle to be in touch with the sphere at all times before reaching some height $h$. This is partly due to choosing the starting velocity to be 0 (or at least very very small) . But how does the trajectory look for different starting velocities, and assuming the constraint $\boldsymbol{r}^2(t) \geq R^2$ for all $t$? It can not be the same as before. For Example: Giving the particle a big starting velocity will send it off the sphere into free fall; no sliding at all.

About the fourth question: Normally I would try and derive the equations of motion by using the lagrange formalism. But in the case of the $\boldsymbol{r}^2(t) \geq R^2$ constrain, I don't know how to do this. Nonholonomic contrains were never dealt with (at most mentioned) in any lectures I went to or books I have looked into.

  • 1
    $\begingroup$ -1 Not clear what you are asking. You have over-explained your problem to such an extent that it is confusing, and not clear what the issue is. The mathematical formalism does not help. Are you asking a semantic question about what "flies off" means? Are you asking if the particle, having first left contact with the sphere, can make contact again (for some values of initial speed)? Moreover, have you done any calculations to check your suspicion that the particle bounces on the sphere? It is easily within your ability to do so. $\endgroup$ Dec 30, 2017 at 13:56
  • $\begingroup$ @sammygerbil Indeed; The first question is pretty much a semantic one, but nonetheless one you have to answer: What does "flying off" mean? I presented two possible meanings, already showing that it is not a trivial question. Have I done some calculations about the bouncing? Yes. And they lead me to that suspicion. At the beginnig I wanted to include them, but then there would have been even more text ^^' (apparently it's still too much). $\endgroup$
    – user122721
    Dec 30, 2017 at 18:11
  • $\begingroup$ The question is better, but what does Q2 mean? ... In your comment, if you have done the calculations, why is the the bouncing still a suspicion? Why haven't your calculations confirmed it? $\endgroup$ Dec 30, 2017 at 19:17
  • $\begingroup$ @sammygerbil I guess I will try to reformulate this one as well ;) Q2 pretty much asks: Why is there some centripetal Force acting on the particle?. I find it intuitive that the following two forces are acting on it: normal force and gravity. One coming from the sphere and the other due to the earth. But where does the centripetal force come from? $\endgroup$
    – user122721
    Dec 30, 2017 at 19:25
  • $\begingroup$ @sammygerbil The calulations showed that: no matter how small the tangential starting velocity, it will first go into some free fall trajectory, and be it only for a very short time. Then it will hit the sphere again. At this point I don't know how it will continue. It depends on what you assume will happen on the impact: Does the particle get reflected by the surface of the sphere? Does the sphere absorb some part of the kinetic energy? Some other possibility? $\endgroup$
    – user122721
    Dec 30, 2017 at 19:31

2 Answers 2

  1. "Flying off" means leaving contact with the sphere. Practically this happens when the normal reaction between the particle and sphere becomes zero.

  2. Not clear what you mean by this. While the particle is moving on the sphere it is following a circular path, so it is accelerating radially. The net force in the radial direction must be $ma_c=mv^2/R$. There is also a tangential component of force, causing tangential acceleration, but this does not affect $a_c$. Only the tangential speed, not tangential acceleration, affects the value of $a_c$.

  3. The constraint $r \ge R$ (I don't understand why you write $r^2 \ge R^2$) means that the particle cannot penetrate the surface of the sphere. The problem does not imply that the particle is constrained to stay on the surface of the sphere, otherwise it could never leave and the problem would be meaningless. The trajectory is obviously in 2 parts : (i) the arc of a circle until condition 1 is met and the particle leaves contact with the sphere; (ii) thereafter, the arc of a parabola.

If, instead of sliding from rest, the particle is given an initial speed $v_0$, then it will lose contact with the sphere at a smaller angle $\theta$. Having left contact with the sphere it follows a parabolic trajectory which does not intersect the sphere again. This is because the sphere curves faster than the parabola. The radius of curvature of the parabola decreases on either side of the vertex, whereas the curvature of the sphere is constant. If the sphere and parabola touch on one side of the vertex they cannot intersect or touch again.

The particle does not make contact with the sphere again for any initial speed $v_0$. So there is no bouncing.

The centripetal force is not an extra force. It is the component of other forces which accounts for centripetal (radially inward) acceleration, ie circular motion. Likewise tangential force is the component of forces which account for tangential acceleration.

The 2 forces you mentioned together provide the centripetal force : the normal force $N$ and the radial component of gravity $mg\cos\theta$. While the particle remains in contact with the sphere there is a radial inward acceleration $a_c$. Applying $F=ma$ along the inward radial direction :

As the partice descends (ie as $\theta$ increases) its speed $v$ increases, accelerated by the tangential component of gravity $mg\sin\theta$. This means its centripetal acceleration $a_c=v^2/R$ also increases. However, the component $mg\cos\theta$ decreases. $N$ adjusts to whatever value makes the above equation hold. However, the constraint means that it cannot fall below zero, ie it cannot be inward. So the particle leaves contact when $N=0$, because there is not enough centripetal force to keep it moving in a circle of radius $R$.

If the particle were a bead constrained to move on a circular hoop then the hoop could provide an inward normal reaction force $N$ (ie +ve in the above equation), so that there is enough force for centripetal acceleration with fixed radius $R$ and increasing speed $v$.


It flies off like M1. You are overthinking this.

As you slide down the sphere, you move slowly at first. If the sphere suddenly vanished, you would follow a parabola. If you are moving slowly in a place where the sphere was nearly horizontal, you would fall through the floor - The parabola would pass through where the sphere had been.

As you slide farther, you move faster and the surface gets steeper. There is a place where the parabola you would follow becomes tangent to the sphere. Beyond this point, you follow a parabola that leaves the surface of the sphere behind.

Yes, it is possible for a ball to start bouncing on the upper part of the sphere and bounce its way down the sphere until it falls off and never returns. But in this problem, it slides smoothly.

The constraint $r^2 > R^2$ means nothing more than the particle is always outside the sphere. The distance of the particle from the center of the sphere, r, is always larger than the distance of the surface of the sphere, R.

So if the particle flew toward the sphere, it would stick, or slide along the surface, or bounce off, or something similar. But it would not penetrate the sphere.

In this problem, the nonholonomic constraint just means the particle cannot penetrate into the sphere. There are many elementary physics problems with this kind of constraint. You shouldn't need a textbook on the subject.

When you stand on the floor, gravity pushes you down and the floor holds you up. The force the floor exerts is just big enough to keep you from penetrating the floor.

If you carry a weight, the combined force of gravity from you and the weight increase. The floor pushes upward with a force just big enough to keep you and the weight from penetrating the floor.

If gravity turns off, the floor exerts no force.

If a helicopter lifts you, the floor exerts no force to prevent it.