Kvant Physics Problem 27
A rigid hemispherical bell of radius $R$ rests on a horizontal table with its rim in tight contact with the table, preventing fluid flow under the rim until lift-off.
Verified: no
Verdicts: SKIP + SKIP
Solve time: 1m02s
Source on kvant.digital
Problem
In a hemispherical bell, tightly resting on a table, water is poured through an opening at the top (Fig. 3). When the water reaches the opening, it lifts the bell and begins to flow out from below. Find the mass of the bell if its radius is $R$, and the density of water is $\rho$.
Fig. 3
Setup and Assumptions
A rigid hemispherical bell of radius $R$ rests on a horizontal table with its rim in tight contact with the table, preventing fluid flow under the rim until lift-off. Water of density $\rho$ is poured through a small opening at the top of the hemisphere (at the pole of the spherical surface). The bell has mass $m$, which is unknown.
The water level rises inside the bell until it reaches the top opening, after which any additional water flows out and the internal free surface remains fixed at the opening level. At the instant of lift-off, the bell loses contact with the table everywhere along its rim.
The flow is assumed quasi-static, so at every stage the fluid is in hydrostatic equilibrium. The water is incompressible, viscosity is neglected, and surface tension effects at the rim and opening are ignored. The air pressure outside and at the opening is $p_0$ and is taken as constant. Gravitational acceleration is $g$.
The goal is to determine the mass $m$ of the bell at the instant when the net vertical force on it becomes zero and it begins to lift.
Physical Principles
The hydrostatic pressure field in a fluid at rest satisfies
$p = p_0 + \rho g h,$
where $h$ is the depth measured downward from the free surface.
The resultant vertical force exerted by a static fluid on a curved surface equals the weight of the imaginary column of fluid bounded by the surface and the free surface above it, independent of the detailed pressure distribution.
Newton’s second law at the threshold of lift-off requires that the total vertical force on the bell satisfies
$F_{\uparrow} - F_{\downarrow} = 0.$
The vertical force from the fluid on the bell is obtained from the hydrostatic pressure integrated over the inner surface, while the weight of the bell is $mg$ acting downward.
Derivation
When the water inside the bell reaches the top opening, the free surface is located at the apex of the hemisphere. The fluid occupies the entire hemispherical volume beneath this point. Because the opening is at atmospheric pressure, the pressure field inside the water is purely hydrostatic relative to this reference level.
The vertical resultant force exerted by the water on the hemispherical inner surface is equal to the weight of a fictitious column of water occupying the volume geometrically enclosed by the hemisphere up to the free surface. This standard hydrostatic result yields a net upward force on the bell equal to the weight of the water contained in the hemispherical region,
$F_{\text{fluid}} = \rho g V_{\text{hemi}}.$
The volume of a hemisphere of radius $R$ is
$V_{\text{hemi}} = \frac{2}{3}\pi R^3.$
Therefore the upward force due to the water is
$F_{\text{fluid}} = \rho g \frac{2}{3}\pi R^3.$
At the threshold of lift-off, this upward force balances the weight of the bell,
$\rho g \frac{2}{3}\pi R^3 = mg.$
Cancelling $g$ from both sides gives the required mass of the bell,
$m = \frac{2}{3}\pi \rho R^3.$
Result
The mass of the bell is
$m = \frac{2}{3}\pi \rho R^3.$
Substituting the given quantities in SI units, with $\rho$ in $\mathrm{kg,m^{-3}}$ and $R$ in $\mathrm{m}$,
$m = \frac{2}{3}\pi \rho R^3 \ \mathrm{kg}.$
The final numerical expression is
$\boxed{m = \frac{2}{3}\pi \rho R^3 \ \mathrm{kg}}.$
Sanity Checks
Dimensional consistency follows from $\rho R^3$, since $\rho$ has units $\mathrm{kg,m^{-3}}$ and $R^3$ has units $\mathrm{m^3}$, giving $\mathrm{kg}$ for $m$.
In the limit $R \to 0$, the mass tends to zero proportionally to the volume of the hemisphere, matching the expectation that a smaller fluid column generates a smaller hydrostatic force.
The result depends only on the fluid density and geometric volume, which is consistent with hydrostatics where pressure forces integrate to a weight-like quantity. Any sign error in the pressure force direction would invert the balance condition and incorrectly predict a negative or unphysical bell mass.