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
$$\sum$$F = 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).

 

[rwisz]

I can understand your confusion about the forces at play in this scenario. In circular motion, there are two main forces at play: the centripetal force, which is directed towards the center of the circle and keeps the object moving in a circular path, and the centrifugal force, which is directed away from the center and is a result of the object's inertia trying to keep it moving in a straight line. In this case, the bowl is providing the centripetal force to keep the ball moving in a circular path. Even without friction, the ball will still experience a force from the bowl, as the bowl is constantly changing the direction of the ball's motion. This force is known as the normal force, and it is perpendicular to the surface of the bowl.

In order to calculate the height the ball will rise within the bowl, we can use the equation you provided, Fr = m*w^2/r. This equation relates the force of the normal force (N) to the mass of the object (m), the angular velocity (w), and the radius of the circle (r). By rearranging this equation, we can solve for the height (h) as h = (N/m)*r.

In this scenario, we can assume that the ball is not sliding or slipping on the surface of the bowl, so there is no friction present. This means that the normal force is equal to the weight of the ball (N = mg). Plugging in the given values, we can calculate the height as h = (0.05 kg*9.8 m/s^2*0.1 m)/(0.05 kg) = 0.098 m, or 9.8 cm.

I hope this explanation helps to clarify the forces at play in circular motion, and how they affect the motion of the ball in the bowl.