 Solution

ID: past-exam-of-the-theoretical-mechanics-course-of-the-university-of-oxford/2008/a/3/i/solution

u_{θ} = 0

. Continuity gives

\partial_{r} (r u_{r}) = 0

, hence

u_{r} = g (θ) / r

. Substituting into radial and angular momentum gives

p_{r} = \frac{ρ}{r ^{3}} (g^{2} + ν g^{''}), p_{θ} = \frac{2 ρ ν}{r ^{2}} g^{'} .

For example

\nabla^{2} u_{r} - u_{r} / r^{2} = g^{''} / r^{3}

, and the radial acceleration is

- g^{2} / r^{3}

. Equality of the mixed pressure derivatives now requires

ρ (2 g g^{'} + ν g^{(3)}) / r^{3} = - 4 ρ ν g^{'} / r^{3} .

ν (g^{(3)} + 4 g^{'}) + 2 g g^{'} = 0.

It can also be integrated once as

ν (g^{''} + 4 g) + g^{2} = constant

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