Question

a vessel of height H, is filled uto height 3H/4 by a liquid, is rotated with constant angular velocity w about the acis pasing through the middle. The radius ofcylinder is R. Find the value of w for which the base of the container is just exposed

Answer 1

$\sqrt{\frac{3Hg}{R^2}}$

Answer 2

The initial volume of the liquid in the cylindrical vessel is $V_0 = \pi R^2 (3H/4)$.

When the vessel is rotated with angular velocity $\omega$ about its central vertical axis, the free surface of the liquid forms a paraboloid of revolution. The equation of the free surface relative to the base of the cylinder (at $z=0$) is given by $z(r) = h_c + \frac{\omega^2 r^2}{2g}$, where $h_c$ is the height of the liquid at the center ($r=0$).

The volume of the liquid in the rotating vessel is conserved. The volume can be calculated by integrating the height of the liquid surface over the base area. $V = \int_0^R z(r) 2\pi r dr = \int_0^R (h_c + \frac{\omega^2 r^2}{2g}) 2\pi r dr$ $V = 2\pi \int_0^R (h_c r + \frac{\omega^2 r^3}{2g}) dr = 2\pi \left[ h_c \frac{r^2}{2} + \frac{\omega^2 r^4}{8g} \right]_0^R$ $V = 2\pi \left( h_c \frac{R^2}{2} + \frac{\omega^2 R^4}{8g} \right) = \pi R^2 h_c + \frac{\pi \omega^2 R^4}{4g}$.

By conservation of volume, $V = V_0$: $\pi R^2 (3H/4) = \pi R^2 h_c + \frac{\pi \omega^2 R^4}{4g}$. Dividing by $\pi R^2$: $3H/4 = h_c + \frac{\omega^2 R^2}{4g}$.

The condition "the base of the container is just exposed" means that the liquid level at the center of the base is zero. So, $h_c = 0$. Substituting $h_c = 0$ into the volume conservation equation: $3H/4 = 0 + \frac{\omega^2 R^2}{4g}$. $\frac{3H}{4} = \frac{\omega^2 R^2}{4g}$. $3H = \frac{\omega^2 R^2}{g}$. $\omega^2 = \frac{3Hg}{R^2}$. $\omega = \sqrt{\frac{3Hg}{R^2}}$.

At this angular velocity, the height of the liquid at the edge ($r=R$) would be $h_e = h_c + \frac{\omega^2 R^2}{2g} = 0 + \frac{(3Hg/R^2) R^2}{2g} = \frac{3Hg}{2g} = \frac{3H}{2}$. Since the height of the vessel is $H$, and $3H/2 > H$, the liquid will spill out of the vessel when the base is just exposed. This is a common scenario in such problems; the calculated $\omega$ is the value required to achieve the condition $h_c=0$, assuming the vessel is tall enough to contain the liquid profile up to $r=R$.

The value of $\omega$ for which the base of the container is just exposed is $\sqrt{\frac{3Hg}{R^2}}$.