Originally Posted bymaxhk

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A cylinder partly filled with water is rotated abour its axis with constant angular velocity (\omega).

As the rotation proceeds, the water level rises along the wall and sinks in the center to form a concave surface.

(A) show that this surface has the shape of a surface formed by revolving the parabola y = (omega^2)/2*g + h;

where g is acceleration due to gravity and h distance from the vertex of the parabola to the bottom of the cylinder.

(B) If V0 is the volume of water, express h as a function of \omega

(C) If the cylinder rotates faster and faster then either the bottom will be exposed or water begin to spill out the top. If the cylinder is originally half full, which happens first ?

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Done part (B)

h = V0/(pi*r^2) - (r^2/(4*g))*omega

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