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A cylindrical bucket of liquid (density rho) is rotated about its symmetry axis, which is vertical. If the angular velocity is ω, show that the pressure at a distance r from the rotation axis is P=P 0+ 21rhoω 2r 2where P 0 is the pressure at r=0,

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Answer:

To derive the formula for the pressure at a distance r from the rotation axis, you can use the principle of hydrostatic equilibrium. This states that the pressure at a point in a fluid is equal in all directions and the net force on a fluid element is equal to zero.

Consider a small cylindrical element of fluid at a distance r from the rotation axis, as shown in the diagram below. The dimensions of the element are dr in the radial direction, 2πr in the circumferential direction, and h in the vertical direction. The weight of the element is equal to the product of its volume, density, and the acceleration due to gravity: W = (2πr)(dr)(h)(ρ)(g). The pressure at the top and bottom faces of the element is P + dP, where P is the pressure at the point and dP is a small change in pressure. The forces acting on the element are the weight of the element, the pressure force at the top face, and the pressure force at the bottom face.

[asy]

unitsize(2cm);

pair P1, P2, P3, P4;

P1 = (0,0);

P2 = (1,0);

P3 = (1,1);

P4 = (0,1);

draw((-0.5,0)--(1.5,0));

draw((0,-0.5)--(0,1.5));

draw(P1--P2--P3--P4--cycle);

draw((0.5,0)--(0.5,1));

draw((0.25,0)--(0.25,1));

draw((0.75,0)--(0.75,1));

label("$r$", (0.5,1.5), red);

label("$

User Jez D
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