# Ball in bowl dispute

Imagine a solid ball rolling around in a hemispherical bowl without slipping. The bowl has radius L and the ball has radius R. I need to write down the constraint relating the angular velocity of the ball, call it omega , to the angular velocity the center of the ball is sweeping out in the bowl. Theta is the angle between the line from the center of the bowl to the center of the ball and the vertical. So I need the relationship between omega and the time derivative of theta.

My professor is being extremely stubborn and claiming that the constraint equation is:

$$\omega = L\theta/R$$

But it actually should be this:
$$\omega = (L-R)\theta/R$$

Is there anything short and to the point I can say for him to listen to me? I've sent a long explanation to him but he simply dismisses me and tells me to keep thinking about the problem.

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A.T.
Is there anything short and to the point I can say for him to listen to me?
Try this puzzle. It's very similar, but instead of rolling inside, you roll outside:
http://www.cut-the-knot.org/pythagoras/two_coins.shtml

Or you can adapt it to your case, and roll a coin within some ring. Choose a simple radius ratio, like L/R=2. Then see which formula gives the right number of coin revolutions, after one full round along the ring.

Try this puzzle. It's very similar, but instead of rolling inside, you roll outside:
http://www.cut-the-knot.org/pythagoras/two_coins.shtml

Or you can adapt it to your case, and roll a coin within some ring. Choose a simple radius ratio, like L/R=2. Then see which formula gives the right number of coin revolutions, after one full round along the ring.
Thanks. I've actually showed him a gif of circles rolling within another circle, like this one: http://www.math.ucr.edu/home/baez/mathematical/astroid_animation.gif

His response was, I kid you not, was that I "simply am missing the point of the constraint condition."

Then perhaps you could formulate the constraint condition and show that the prof's equation is incompatible with it?

Then perhaps you could formulate the constraint condition and show that the prof's equation is incompatible with it?
I did precisely that. Either he's too arrogant to actually listen to anything I'm saying, or he doesn't actually understand the problem at hand.

I did precisely that. Either he's too arrogant to actually listen to anything I'm saying, or he doesn't actually understand the problem at hand.
The worst part is that I and many other students failed an entire problem on his midterm because of this disagreement

A.T.
Thanks. I've actually showed him a gif of circles rolling within another circle, like this one: http://www.math.ucr.edu/home/baez/mathematical/astroid_animation.gif
That shows it quite clearly.

##\omega = 3 \dot{\theta}## for ##L = 4 R##

which is consistent with your formula. You professor probably makes the mistake of unrolling the circumference to get

##\omega = 4 \dot{\theta}##

and forgetting that when you roll it into a circle, you remove one revolution of the ball.

The worst part is that I and many other students failed an entire problem on his midterm because of this disagreement
Well; In don't know what possibilities you have on your school in such a case. Is there some kind of mediator? He should talk to your professor, and eventually consult with other math/physics professors.

Another thing to try is the limiting case ##R \to L##.

A.T.
Another thing to try is the limiting case ##R \to L##.
Yep, makes it very obvious.

Hm, well, maybe my own confusion can be of use. It seems to me like the professor is correct. As I seem to be very much in the minority, I will assume I'm making the same mistake the professor is. So, I will explain my reasoning and hopefully convincing me of my error will furnish the way to convince the professor of his.

That shows it quite clearly.

ω=3θ˙\omega = 3 \dot{\theta} for L=4RL = 4 R
Curious, because to me this "clearly" shows that ##\omega = 4\dot{\theta}##! The animation shows that in the time that the centre of the ball makes one full revolution around, the ball itself has rotated four times. Thus, the angular velocity of the ball with respect to its centre is four times greater than the angular velocity of the centre of the ball with respect to the centre of the bowl.

You professor probably makes the mistake of unrolling the circumference to get

ω=4θ˙\omega = 4 \dot{\theta}

and forgetting that when you roll it into a circle, you remove one revolution of the ball.
I'm not sure what you mean by "removing one revolution of the ball". Suppose the ball rolls at a constant rate ##\omega##. Then after rolling for time ##t##, the distance between the intial and final contact points between the ball and and bowl is ##R \omega t##. Certainly, the centre of the ball doesn't travel as far but we're only interested in the overall angle traversed which is the same for both (simply draw a line from the centre of the bowl through the centre of the ball to the contact point). Those contact points are radial distance ##L## from the centre of the bowl, and so ##\theta = R \omega t/L##. Differentiating with respect to time again gives the professor's constraint.

Another thing to try is the limiting case RL.
Well, the R=L case can't be reconciled with the no slip condition but it certainly makes sense to me in the limit of the ball approaching the size of the bowl, the centre should move in smaller and smaller circles at a rate approaching that of the ball's rotation. On the other hand, if the constraint is indeed ##\omega = (L-R)\dot{\theta}/R## then solving for ##\dot{\theta}## would imply that as R approaches L the centre of the ball revolves around the bowl infinitely fast! That makes even less sense to me.

So, I'll accept on the basis of majority vote that I'm probably making a mistake, but it seems pretty convincing to me so I guess I can understand the professor's stubbornness!

Aha, I see my mistake. At least with the animation. Well, no need to convince me!

You are overthinking this now. In the limiting case, your formula gives zero angular velocity for the ball, yet arbitrary angular velocity for its centre of mass. Which is compatible with the no-slip condition.

A.T.
Aha, I see my mistake.
Then maybe you can explain it, in a way that would have helped you.

Then maybe you can explain it, in a way that would have helped you.
Well, my mistake with the animation was not realizing at first that the same point on the ball successively coming into contact with the bowl does not constitute a full rotation, it's a bit less. In other words, I was watching how the marker on the bowl moved relative to the bowl, not relative a fixed coordinate system. Similarly with the other argument, I was imagining a ball being unwrapped along the inside of the bowl, forgetting that dividing the distance the contact point travels by the bowl's radius does not actually correspond to a rotation angle but something a little more. Pretty much exactly what you said you thought was the problem. So I suppose what would have made me see it more immediately with the animation was explicitly pointing out the difference between a full rotation and the contact point travelling a full circumference and then telling me to watch how many rotations the ball makes.

So, maybe the professor will see it if it's explained that way. But possibly he's dug himself into a deep enough hole at this point that he'll need a colleague or two to tell him he's wrong.

haruspex
Homework Helper
Gold Member
Seems to me that no-one is reading this correctly:
Theta is the angle between the line from the center of the bowl to the center of the ball and the vertical.
That makes theta a constant. The circular path taken by the centre of the ball has radius ##(L-R)\sin(\theta)##.
We need another angle, ##\phi## say, to represent the progress of the ball around the axis of the bowl.
By considering the linear velocity of the centre of the ball at some instant, we have:
##\omega R = (L-R)\sin(\theta) \dot \phi##.

A.T.
Seems to me that no-one is reading this correctly:
I was initially also wondering, about the "rolling around" formulation, and why spheres, not cylinders, are used if it's only rolling within a vertical plane. But since both proposed solutions (prof & dwieker) seemed to interpret it that way, I went along with it. It would be good to have the exact formulation as written in the exam.

That makes theta a constant.
I agree that two degrees of freedom are required to formalize the problem as initially stated. But theta needn't be constant; in fact, the description is quite explicit about its non-constancy. Like A.T., I inferred from the original description that it was the other degree of freedom that was fixed. @dwieker may want to clarify this.

haruspex