18A (Fluid Dynamics)

Solution

u = \nabla ϕ

\nabla \cdot u = 0

, hence

Δ ϕ = 0

. In the laboratory frame, the no-penetration condition at the translating sphere is

\partial_{r} ϕ (a, θ) = U cos θ

For separation of variables, take the angular factor

g (θ) = cos θ

, which satisfies

\frac{1}{s i n θ} (sin θ g^{'})^{'} = - 2 g

. The radial equation is

(r^{2} f^{'})^{'} - 2 f = 0,

with solutions

r

and

r^{- 2}

. Rest at infinity excludes the

r

ϕ = - \frac{U a ^{3} cos θ}{2 r ^{2}}, u = \frac{U a ^{3}}{r ^{3}} cos θ e_{r} + \frac{U a ^{3}}{2 r ^{3}} sin θ e_{θ} .

There is no azimuthal velocity. With fluid density

ρ_{f}

K_{f} = \frac{ρ _{f} U ^{2} a ^{6}}{2} \int_{a}^{\infty} r^{- 4} d r \int_{0}^{2 π} \int_{0}^{π} (cos^{2} θ + \frac{1}{4} sin^{2} θ) sin θ d θ d φ .

The angular integral is

2 π

, and the radial integral is

1/ (3 a^{3})

, so

K_{f} = ρ_{f} π a^{3} U^{2} /3

. Since

M = (4 π a^{3} /3) ρ_{s}

and

k = ρ_{f} / ρ_{s}

, $K_{f} = k M U^{2} /4$ . Thus the added mass of a sphere is

k M /2

During a fall through

h

M g h

- k M g h

. Starting from rest, conservation of energy gives

M (1 - k) g h = \frac{1}{2} M U^{2} + \frac{k M U ^{2}}{4} .

Consequently $U = 4 (1 - k) g h / (2 + k)$ . The mass cancels: the reduced speed comes from both buoyancy and the fluid's added mass.

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