How Does a Rotating Pail Affect Water Surface Shape?

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SUMMARY

The discussion focuses on determining the shape of the water surface in a rotating pail, specifically when the pail has an angular velocity of w = 2π rad/sec and a diameter of 0.45 m. The surface slope is defined by dy/dx = tan(theta), which requires resolving forces to express tan(theta) as a function of x. The goal is to calculate the height of the water surface at the outer edge of the pail, which is positioned at a radius of 0.225 m from the center.

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Homework Statement


A rotating pail of water with an angular velocity w has a surface curving up from the centre to the outer edge. Find the shape of the surface. The slope of the surface is dy/dx = tan(theta) at position x is y (= function of x) is the shape of the surface. You must resolve forces to find tan(theta) as a function of x, and then solve for the height, y, as a function of x.

If the pail rotates once per second, and has a diameter of 0.45 m, how high above the centre is the outer edge?



Homework Equations





The Attempt at a Solution


All I see is that w = 2pi rad/sec and r = 0.225 m.
 
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Method I:
Ideal Fluid, by definition, can not tolerate any shear force. Use this fact.
Assume the centre of the water surface to be origin. Let P(x,y) be a point on the water surface. Consider forces on it. Impose the condition that tangential force will be zero.
I hope you can proceed now.
 

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