Originally Posted by **GrahamJamesK** A ball rolls under gravity inside a smooth bowl with circular cross section and inner radius *a*. Show that the ball can roll in a **horizontal** circle at a height *h* above the bottom of the bowl provided:
v^2=g*h(2**a*-*h*/*a*-*h*)
OK, well I'm not certain at all how to begin this problem. The only one of the values I know is g=10 m/s^2.
I suspect this has something to do with a conical pendulum?
I know there is no vertical motion so i can ignore that for a start. The only force I can reason to acting on the ball is gravity
The ball has to be moving at some angle theta to the vertical and that at the given height *h* the radius is *a*
Any suggestions where to start? |

I would first like to make a comment. It is going to be next to impossible to lead you through without essentially "giving away" the solution. I urge you to carefully study this and make sure you can derive every equation below. Else you will gain nothing.

This is going to be a pain in the neck without a decent diagram, but I'm afraid I'm not up to sketching it. (And my scanner is down.)

So. Let's assume our spherical bowl has its minimum at the origin (with the usual +x and +y directions). I am going to arbitrarily pick the problem to be in the xy plane...any cross-section of the problem will do. Our segment of the sphere in the xy plane is given by x^2 + (y - a)^2 = a^2. I am choosing to work in the lower half of that, so our equation for the cross-section becomes y = a - sqrt(a^2 - x^2).

I am considering the ball to be (momentarily) at the point (x, h) and I'm going to do a FBD about this point. So I'm going to need two things before analyzing the FBD. I need to find x, which is the radius of the circle the ball is making, and I need to know at what angle the normal force acts at this point.

The radius is easy...x = r and y = h, so from the circle equation:

y = a - sqrt(a^2 - x^2)

h = a - sqrt(a^2 - r^2)

I get r = sqrt(2ah - h^2).

Now to the hard part. First of all, going back to your Calculus I class given a function y(x) we know that y' = tan(alpha), where alpha is the angle between the tangent line to the curve and the +x axis. In this case, that means

y = a - sqrt(a^2 - x^2)

y' = -x/sqrt(a^2 - x^2) = -x/(y - a)

Putting in the point (x, y) = (r, h) gives

y' = -r/(h - a) = r/(a - h) = tan(alpha)

(We'll need the tangent in a moment.)

Now that we have the angle we can move onto the FBD. I have a weight acting downward (-y direction) and a normal force acting perpendicular to the tangent line at the point (x, y) = (r, h). Notice that the ball moves in a circle, so the net force in the x direction is equal to the centripetal force. So we get

(sum)Fx = -N*sin(alpha) = Fc = -mv^2 / r

In the y direction we get

(sum)Fy = N*cos(alpha) - mg = 0

N = mg/cos(alpha)

Putting this N into the (sum)Fx equation:

mv^2 / r = mg/cos(alpha) * sin(alpha) = mg*tan(alpha)

v^2 = gr*sqrt(2ah - h^2)/(a - h)

And since r = sqrt(2ah - h^2)

v^2 = g*sqrt(2ah - h^2)*sqrt(2ah - h^2)/(a - h)

which you can finish simplifying yourself.

-Dan