Kvant Physics Problem 27

Verified: yes
Verdicts: PASS + PASS
Solve time: 12m36s
Source on kvant.digital

Problem

Water is poured through the top opening into a hemispherical bell that sits snugly on a table (Fig. 3). When the water reaches the opening, it lifts the bell and begins to flow out from below. Determine the mass of the bell if its radius is $R$ and the density of water is $\rho$.

Fig. 3

Setup and Assumptions

The physical system consists of a rigid hemispherical bell of radius $R$ resting on a horizontal table. Water is poured through the opening at the top until the water level reaches that opening. At that instant the free surface of the water is at height $R$ above the table, and the bell is just about to lift. The unknown quantity is the mass of the bell, denoted by $M$. The density of water is $\rho$, and the acceleration due to gravity is $g$.

The water is treated as an incompressible fluid in hydrostatic equilibrium. Dynamic effects associated with the pouring process are neglected. The bell remains in contact with the table until the lifting threshold is reached. The condition for the onset of lifting is that the resultant upward force exerted by the water equals the weight of the bell.

Physical Principles

The pressure in a fluid at rest satisfies

$$p=\rho g h,$$

where $h$ is the depth below the free surface.

The force exerted by a fluid on a surface is obtained by integrating the pressure over that surface:

$$F=\int p,dA.$$

Since the water under the bell occupies a hemispherical cavity, the pressure acting on the horizontal plane of the table is not uniform. The depth below the free surface depends on the distance from the center. The lifting force must therefore be obtained by integrating the hydrostatic pressure over the circular base.

At the instant the bell begins to rise,

$$F_{\mathrm{up}}=Mg.$$

Derivation

Choose cylindrical coordinates with origin at the center of the circular base. Let $r$ be the distance from the axis of symmetry.

The hemispherical cavity is described by

$$r^2+z^2=R^2,$$

with the table at $z=0$. Thus the height of the cavity above the table at radius $r$ is

$$z(r)=\sqrt{R^2-r^2}.$$

The free surface of the water is at the top opening, at height $R$ above the table. Hence the depth of the point $(r,0)$ below the free surface is

$$h(r)=R-z(r) =R-\sqrt{R^2-r^2}.$$

The pressure acting on the underside of the bell over an annular element of area

$$dA=2\pi r,dr$$

is

$$p(r)=\rho g,h(r) =\rho g!\left(R-\sqrt{R^2-r^2}\right).$$

The total upward force is

$$F_{\mathrm{up}} = \int_0^R p(r),dA = 2\pi\rho g \int_0^R \left(R-\sqrt{R^2-r^2}\right)r,dr.$$

Evaluating the first integral,

$$\int_0^R Rr,dr = R\left[\frac{r^2}{2}\right]_0^R = \frac{R^3}{2}.$$

For the second integral, let

$$u=R^2-r^2, \qquad du=-2r,dr.$$

Then

$$\int_0^R r\sqrt{R^2-r^2},dr = \frac12\int_0^{R^2}u^{1/2},du = \frac12\left[\frac{2}{3}u^{3/2}\right]_0^{R^2} = \frac{R^3}{3}.$$

Substituting these results,

$$F_{\mathrm{up}} = 2\pi\rho g \left( \frac{R^3}{2} - \frac{R^3}{3} \right) = 2\pi\rho g \cdot \frac{R^3}{6} = \frac13\pi\rho gR^3.$$

The bell starts to lift when

$$Mg=F_{\mathrm{up}}.$$

Hence

$$Mg=\frac13\pi\rho gR^3,$$

and after dividing by $g$,

$$M=\frac13\pi\rho R^3.$$

Result

The mass of the bell is

$$\boxed{M=\frac13\pi\rho R^3}.$$

For $R=0.2,\mathrm{m}$ and $\rho=1000,\mathrm{kg/m^3}$,

$$M = \frac13\pi(1000)(0.2)^3 = \frac13\pi(8) \approx 8.38,\mathrm{kg}.$$

Thus

$$\boxed{M\approx 8.4,\mathrm{kg}}.$$

Sanity Checks

The dimensions are correct because

$$[\rho R^3] = \frac{\mathrm{kg}}{\mathrm{m}^3},\mathrm{m}^3 = \mathrm{kg}.$$

The result can also be written as

$$M=\rho V_{\text{hemisphere}},$$

since

$$V_{\text{hemisphere}} = \frac{2}{3}\pi R^3$$

and

$$\frac12\rho V_{\text{hemisphere}} = \frac12\left(\frac{2}{3}\pi\rho R^3\right) = \frac13\pi\rho R^3.$$

The original derivation treated the pressure as uniformly equal to $\rho gR$ over the entire base. The depth below the free surface actually varies from $0$ at the center to $R$ at the rim, so the pressure must be integrated over the circular area. Carrying out that integration produces the factor $1/3$ in the final answer.