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A cylindrical vessel containing water is rotated about its vertical axis with constant angular velocity omega. Find (a) the shape of the free surface of the water, and (b) the water pressure distribution over the bottom of the vessel as a function of radius r, given that the pressure at the center of the bottom is p0.

1. Surface is a paraboloid z = omega²*r²/(2g); pressure p(r) = p0 + rho*omega²*r²/2
2. Surface is a hemisphere; pressure p(r) = p0 + rho*g*r
3. Surface is a cone z = omega*r/g; pressure p(r) = p0 - rho*omega²*r²/2
4. Surface is flat; pressure p(r) = p0 constant

Correct answer: Surface is a paraboloid z = omega²*r²/(2g); pressure p(r) = p0 + rho*omega²*r²/2

Solution

In the rotating frame, the effective body force per unit volume is gravity (downward) plus centrifugal force rho*omega²*r (radially outward). The free surface, being at constant pressure, satisfies dz/dr = omega²*r/g, giving the paraboloid z = omega²*r²/(2g). The pressure on the bottom increases outward as the extra height of liquid column, so p(r) = p0 + rho*omega²*r²/2.

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