# Banked Tracks for Turning 17.6

1. How does rotating the bowl make the marble leave the bowl - surely if the marble is in the centre of the bowl then there would be no movement and it would just stay in the centre.
2. when it says "A spherically shaped bowl of inner radius 16cm is 8.0cm deep at its centre" does that mean that the bowl looks like the diagram I've drawn (below)
Working:
Original post by mosaurlodon

1. How does rotating the bowl make the marble leave the bowl - surely if the marble is in the centre of the bowl then there would be no movement and it would just stay in the centre.
2. when it says "A spherically shaped bowl of inner radius 16cm is 8.0cm deep at its centre" does that mean that the bowl looks like the diagram I've drawn (below)

I would answer q2 first. No.
You are drawing a bowl with a thickness of 8.0 cm NOT 8.0 cm deep.
To think of this "A spherically shaped bowl of inner radius 16cm is 8.0cm deep at its centre", first draw a sphere of radius 16 cm or in a 2D, a circle of radius 16 cm, then measure a distance of 8.0 cm from the bottom of the circle upward. The circular arc that “covers” this 8.0 cm from the bottom of the circle is the shape of the bowl.
Original post by mosaurlodon

1. How does rotating the bowl make the marble leave the bowl - surely if the marble is in the centre of the bowl then there would be no movement and it would just stay in the centre.
2. when it says "A spherically shaped bowl of inner radius 16cm is 8.0cm deep at its centre" does that mean that the bowl looks like the diagram I've drawn (below)

For q1, the physics behind the problem of a penny placed at some distance from the centre of a rotating turntable is very similar to the question that you are asking.
There is a video that explains and demonstrates the penny problem very well. You can watch the video to see if you can come out with an explanation for your question.
Ok thanks for the help with the video - the explanation Ive got is that lets call the distance from the centre, R, and the bigger R gets, the less centripetal force it has, so its easier to leave the bowl - I think when R is 0, ie the marble is on the centre, then the centripetal force is infinite - in other words it can never leave the bowl, no matter how much the bowl rotates.

I've drawn a diagram, with an angle theta to resolve forces.
Diagram of forces. So N-mgcos(theta) = Fcsin(theta) and mgsin(theta)=Fccos(theta)
I know youre meant to cancel R somehow but it doesnt seem to quite work out
Original post by mosaurlodon
Ok thanks for the help with the video - the explanation Ive got is that lets call the distance from the centre, R, and the bigger R gets, the less centripetal force it has, so its easier to leave the bowl - I think when R is 0, ie the marble is on the centre, then the centripetal force is infinite - in other words it can never leave the bowl, no matter how much the bowl rotates.

I've drawn a diagram, with an angle theta to resolve forces.

Its an (inverted/blue) spherical cap
https://en.wikipedia.org/wiki/Spherical_cap
so using their notation r=16 and h=r-h=8 so "a" and theta at the (your) top of the cap are easy to work out/well known triangle.

If its on the point of leaving, the ratio of the two forces (horizontal and vertical) is known (trig), and you dont have to calculate the actual value of normal force.
(edited 1 month ago)
Oh so this diagram.
Ok I've equated forces and now I've got the answer, thanks to both of you.
Looks like my previous method was definitely wrong.
Might just be a typo, but the 16 radius should be the vertical height (as marked, so 8+8) as well as the hypotenuse marked N, not the horizonntal radius which is 8sqrt(3) as its a 30-60-90. From a quick calc, it gives the same answer though.
Oh yeah sorry I drew my diagram wrong - its what I meant to draw but flipped those 2 around.