Hi!
I've just come across the following problem, and I quote(translate): A hollow hemisphere is held at rest, vertically, concave side facing upwards. A small ball placed inside against the surface, at a distance "h" from the bottom, is shot horizontally with velocity "vo". Find the value of "vo", as a function of "h", so that ball reaches the hemisphere's rim and remains turning around without leaving the hemisphere.
First, intuition says (to me, at least) that the ball will try to gain in altitude as it rotates. And this is in accordance with the fact that there is a torque provided by appropriate components of the force of gravity and the centrifugal force.
What I wished to ask you is: a) is it true that for each "vo", the ball either reaches an altitude at which it remains turning with constant angular speed or it leaves the hemisphere, and consequently there is a certain value of "vo" at which the ball just reaches the rim and remains there?; is my approach correct?
Thanks in advance for your help.

You need to consider inertia here. Each ring around the inside at a constant height would have an associated velocity that would maintain it upward by centrifugal pressing outward against the slope of the hemisphere versus gravity.

But two things to note are that for a lossless system inertia will be preserved so if you start low but emit the ball at a high velocity it will rise and "overshoot" the steady state height, and then it will drop below it again, though possibly not as low as it began ... I guess it's not impossible there's an ever decaying value that "damps" this overshoot and undershoot, though if started at the extreme of launching the ball from the center it would never stabilize at a constant circular height, as it would have no angular velocity. But it could be that the distances travelled during an overshoot versus an undershoot are different and this might cause a gradual phase shift that cancels out the oscillation.

For the "vo" at the rim, it would be infinite, because centrifugal force would only push outward and the slope of the hemisphere would be vertical at that point and allow for no upward force. Also it would be likely impossible to reach the exact rim stably as a ball rising toward it would overshoot and fly out of the hemisphere, unless you emitted the ball at the rim with infinite velocity, but realistically you could only get very close to the rim with high velocities.

Thanks StevenA, many thanks! You have given me some material to think about this problem. Taking into account your observations I intend to keep working on it and if I come to an answer that looks reasonable I will post it. Tnaks again.

My pleasure and thank you too. Have fun. (That's what science is suppose to be about anyway )

Hi StevenA!
Here I am, back with the same problem. I have spent many an hour on it and all I have to say for that is that the more I think and work on it the more confused I get. And what worries me is that the problem appears in a physics text book for engineering students, with a hint for its solution: to use the conservation laws. Where I am stuck in is: the ball is shot when standing still at a given height "h" over the reference, which I take to be the bottom of the hemisphere, with a given initial horizontal velocity "vo"; associated with this velocity there is an angular momentum "Lo", which is that makes the ball gain in altitude. Depending on the value of "vo", the ball will -as you pointed- either overshoot or reach a maximum height and then begin to fall back (NOT, keep on rotating with a constant angular speed as I said in my first post, shame on me!). So the initial energy has a component due to the velocity "vo" (1/2*m*vo^2), a component due to its height over the reference surface(m*g*h), AND a component due to the angular momentum "Lo". "Lo" is the cross product of : <radius X po> (all vectors). "po" is <m*vo>. This component should have the form of <1/2*I*w^2>, where "I" is the ball's moment of inertia (simply m*R^2). But what must I take as the "radius"? And what as "w"? When the ball reaches its maximum altitude there is no more angular momentum. I can write the equations for the torque provided by the force of gravity (and again, shame on me: the centripetal force doesn't enter into the picture), and equate this torque with the change in angular momentum (dT/dt = dL/dt). This torque is what causes the angular momentum to eventually diminish to zero. I am sorry to bother you with something that presumably should be not that complicated. (Maybe it is my 75 years of age). Thanks for your patience.

Well, trying to estimate the vertical inertia and how the ball might seek a specific height is a bit too complex for me to try here, but we should be able to at least get a formula for the stable height at which the ball can roll without either rising or falling:

Let's first change coordinates a bit to make the math a bit easier and assume the top of the hemisphere is aligned with a flat table and the ball effectively falls down to negative heights. I'll ignore mass, though if they give a mass for the ball then that will add more to the math (you'd subtract out the kinetic energy lost rising over that height).

If we call the radius of the hemisphere rh, and we have a vertical distance down, we'll call h, then the ball would be rotating along a path with radius r and:

r^2 = rh^2-h^2.

......r
...______
...|....../
h.|..../ rh
...|../
...|/

(It didn't let me put spaces so I had to use periods ... ignore all the periods ).

The centrifugal acceleration for the ball orbiting around radius r, would be:

a=v^2/r

But this accel is only outward. To calculate the mechanical gain upwards, we scale this by the ratio h/r, so:

upward accel = v^2/r * h/r = h*v^2/r^2

This needs to match the gravitational constant so we have:

g=h*v^2/r^2

We want to solve for h, so let's first expand r and get:

g=h*v^2/(rh^2-h^2)

We should be able to use the pythagorean theorem here:

g*(rh^2-h^2)=h*v^2

-g*h^2 - v^2*h + g*rh^2 = 0

So you can try plugging in A=-g B=-v^2 and C=g*rh^2

At the end remember to calculate the height as rh-h, because I used the top of the hemisphere as a reference for 0 height.

... hopefully I didn't make any mistakes in these. You should go back and double check everything.