 Solution

ID: past-exam-of-the-probability-statistics-and-options-ii-course-of-the-university-of-oxford/2008/h/2/b/i/solution

The centre-of-mass coordinates are

a = h sin θ cos ϕ, b = h sin θ sin ϕ, c = z + h cos θ .

Differentiation and cancellation of cross terms gives

\overset{a}{˙}^{2} + \dot{b}^{2} + \overset{c}{˙}^{2} = \overset{z}{˙}^{2} + h^{2} (\dot{θ}^{2} + \dot{ϕ}^{2} sin^{2} θ) - 2 h sin θ \overset{z}{˙} \dot{θ} .

Insert this into the supplied kinetic energy and subtract the gravitational potential

m g c

. Thus

L = \frac{A + m h ^{2}}{2} (\dot{θ}^{2} + \dot{ϕ}^{2} sin^{2} θ) + \frac{C}{2} (\dot{ψ} + \dot{ϕ} cos θ)^{2} + \frac{m}{2} (\overset{z}{˙}^{2} - 2 h sin θ \overset{z}{˙} \dot{θ}) - m g (z + h cos θ) .

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