 Solution

ID: past-exam-of-the-short-options-paper-shared-by-prelims-and-parts-a-and-b-of-the-physics-course-of-the-university-of-oxford/2018/s7/2/b/solution

At the initially vertical configuration, the two conserved momenta coincide:

p_{ϕ} (t) = p_{ψ} (t) = I_{3} ω, H (t) = \frac{I _{1} ω _{0}^{2}}{2} + \frac{I _{3} ω ^{2}}{2} + m g a .

The apparent

0/0

angular term at

θ = 0

is interpreted by its regular limit with these equal momenta. Substitution into energy conservation, followed by

1 - cos θ = 2 sin^{2} (θ /2)

, gives

ω_{0}^{2} = \dot{θ}^{2} + \frac{I _{3}^{2} ω ^{2}}{I _{1}^{2}} tan^{2} \frac{θ}{2} - \frac{4 m g a}{I _{1}} sin^{2} \frac{θ}{2} .

C = I_{3} ω / I_{1}

yields

V (θ) = C^{2} sin^{2} \frac{θ}{2} [sec^{2} \frac{θ}{2} - α], α = \frac{4 m g a I _{1}}{I _{3}^{2} ω ^{2}} .

The zero of the potential at the vertical configuration includes the initially specified transverse kinetic energy separately as

ω_{0}^{2}

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