How Do You Derive the Angular Velocity of a Ball in a Hemispherical Bowl?

[mjolnir80]

Homework Statement

a small ball slides , with constant speed (a_\theta tangential acceleration = 0) , around a horizontal circle at height h inside a firctionless, hemispherical bowl of radius R.
derive a formula for the balls angular velocity \omega in terms of the radius of the radius of the bowl R, the height of the ball above the bottom of the bowl h and g

Homework Equations

The Attempt at a Solution

arent R and h constantly changing here (due to gravity the ball is sliding lower and lower in the bowl)
and if they are changing wouldn't the ball have a tangential acceleration because the angular acceleration is getting bigger and bigger?

 

[LowlyPion]

You might want to draw a force diagram for the ball.

Remember that it is frictionless so its motion is the only thing that can hold it up.

Where the ball is rolling is around a locus of points at height h within the bowl. That locus of points describes a circle that is at a radius r' that is a function of both the overall radius R and the height h that it's rolling at. You should be able to use normal geometry to determine that.

So as it is rolling it has a velocity v (ωr) and that v creates a centrifugal force outward that holds the ball against the angle of the bowl at that point. (Again use geometry to figure that angle. ) And at that angle there is a force of gravity component acting down the slope and the outward force acting up the slope. Since they must balance ...

 

[Dr.D]

I agree that you should draw a free body diagram (a force diagram), but remember that the only forces that act on this ball are (1) gravity, and (2) the normal contact force between the ball and the bowl. There are no other forces at all acting on the ball.

The concept of "centripetal force" is bogus because it refers to something that is not a force at all. It only serves to confuse people. Draw the actual forces, and then correctly write the acceleration of the ball. If you do, all will be well.