Solution

Let

e

be the downward vertical unit vector and let the rod's weight after buoyancy be

W e

. Force balance gives

RU = W e .

Because

R = c_{1} tt + c_{2} (1 - tt),

R^{- 1} = \frac{1}{c _{1}} tt + \frac{1}{c _{2}} (1 - tt) .

Using

e \cdot t = cos θ

gives

U = W [\frac{c o s θ}{c _{1}} t + \frac{1}{c _{2}} (e - cos θ t)] .

Therefore

e \cdot U = W (\frac{c o s ^{2} θ}{c _{1}} + \frac{s i n ^{2} θ}{c _{2}}),

and

∣ U ∣ = W \frac{c o s ^{2} θ}{c _{1}^{2}} + \frac{s i n ^{2} θ}{c _{2}^{2}} .

The required angle is consequently

cos α = \frac{cos ^{2} θ / c _{1} + sin ^{2} θ / c _{2}}{cos ^{2} θ / c _{1}^{2} + sin ^{2} θ / c _{2}^{2}} .

c_{1} = c_{2}

, then

cos α = 1

, so the rod falls vertically. For an anisotropic rod with

c_{2} > c_{1}

, the component parallel to its long axis encounters less drag, and the settling direction is generally deflected toward that axis.

Solved by gpt-5.6-sol high.

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