How High Does the Ball Rise in a Rotating Bowl?

[Isiudor]

Homework Statement

a ball with a mass of 50 grams is placed inside a circular bowl with a radius of 10cm.
the bowl rotates at a rate of 5 hertz

what is the height in cms the ball will rise within the bowl?

Homework Equations

Fr = m*w^2/r
\sumF = ma

The Attempt at a Solution

well the solution is pretty obvious to me;

since we are talking about max height then Fnety = 0 hence N*cosa = mg and N*sina = m * w^2/r.

the problem is I don't understand why the ball would move at all, what force is the bowl exerting on the ball? there is no mention of friction and the ball initially has no inertia of it's own, how does the spinning bowl effect it without friction?

 

[arkofnoah]

The bowl has a curvature, draw a tangent to it and find the normal force.

http://img14.imageshack.us/img14/2571/screenshot20100803at142.png

The horizontal component of this normal force is simply the centripetal force exerted on the ball. The vertical component of this normal force moves the ball upwards.

 

[Isiudor]

that is when we assume the ball is already rotating in a certain height within the bowl at the same rate as the bowl.

what I am asking is why does the bowl exert any force on the ball at all

say, we place the ball in the dead center of the bowl and then begin rotating the bowl, there's no friction, why doesn't the ball simply stay in the bottom? there's no force acting on it.

 

[arkofnoah]

oh. in that case you are quite right. the ball would presumably just remain at the bottom of the bowl without spinning at all if there's no friction. if there's no friction at all the ball wouldn't even undergo circular motion at the same rate as the ball, so i think for this question you have to assume friction and that the ball started off slightly "off-center" from the bottom of the bowl (else there will be no horizontal vector accounting for the centripetal force).