Question

(a) A fluid is rotating at constant angular velocity w about the central vertical axis of a cylindrical container. Show that the variation of pressure in the radial direction is given by
(b) Take p = pc at the axis of rotation (r = 0) and show that the pressure p at any point r is
(c) Show that the liquid surface is of paraboloidal form (Fig. 15-26); that is, a vertical cross section of the surface is the curve y = ω2r2/2g. (d) Show that the variation of pressure with depth is p = pgh.

Answer 1

Explanation:

Part a)

Let the pressure variation in radial direction is given as

$\frac{dP}{dr} = \frac{dF}{dV}$

here we used that the pressure at small element is given as

$\frac{dF}{dA} = dP$

now we also know that system is rotated with constant angular speed

so we have

$dF = (dm)\omega^2 r$

now we have

$\frac{dP}{dr} = \frac{dm}{dV} \omega^2 r$

so we have

$\frac{dP}{dr} = \rho \omega^2 r$

here we know that

$\rho = \frac{dm}{dV}$ = density of the liquid

Part B)

As we know that the pressure at the central axis is given as

$P = P_c$

now from above relation

$\int dP = \int \rho \omega^2 r dr$

$P - P_c = \frac{\rho \omega^2 r^2}{2}$

so we will have

$P = P_c + \frac{\rho \omega^2 r^2}{2}$

Part C)

As we know that y is the height of the liquid from the base

so here pressure with radial variation is given as

$P = P_c + \frac{\rho \omega^2 r^2}{2}$

now gauge pressure variation with height is given as

$P = -\rho g y$

so we have

$P = P_c - \rho g y + \frac{\rho \omega^2 r^2}{2}$

as we reach the top layer we have

$P = P_c = P_o$

$0 = \frac{\rho \omega^2 r^2}{2} - \rho g y$

so we have

$y = \frac{\omega^2 r^2}{2g}$

Part D)

As we know that pressure at depth y1 and y2 from the top is given as

$P_1 = P_o - \rho g y_1$

$P_2 = P_o - \rho g y_2$

now variation with depth is given as

$\Delta P = \rho g (y_2 - y_1)$

$\Delta P = \rho g h$

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Topic : Pascal's law

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