A cylindrical container of water with a radius of 6.0 cm is placed on a phonograph turntable so that its outer edge just touches the outer edge of the turntable. The radius of the turntable is 14.5 cm, and the turntable rotates at 33 and a 1/3 revolutions per minute. How much is the water depressed in the centre compared to the outer edge of its container?[/B]
gz − 1/2 ω2 (x2 + y2 ) = constant[/B]
The Attempt at a Solution
I know that x2+y2 is equal to the radius squared. I think from my notes it's equal to the radius squared of the water in the container. So plugging this in, and omega I calculated to be 3.49 rad/s, I get a height of 0.0022 m. I'm not sure where to go from here though.[/B]
So there are two forces acting on the surface, gravity and centripetal force. The acceleration from gravity is g, the acceleration from centripetal force is a.
a is defined as a = rw^2.
The infinitesimal elements of the surface are dr (radial distance), dz (vertical distance).
The ratio between the forces and the elements are equal such that:
dz/dr = rw^2/g
Now integrate to get z(r). Then keep in mind the volume is still the same as when it was standing still to find the depression.