 Solution

ID: past-exam-of-the-options-i-course-of-the-university-of-oxford/2013/h/1/d/solution

Away from the coordinate poles, use

\dot{ϕ} = h / sin^{2} θ

in the conserved energy:

E = \frac{M a ^{2}}{6} \dot{θ}^{2} + \frac{M a ^{2} h ^{2}}{6 s i n ^{2} θ} - \frac{M g a}{2} cos θ .

Rearranging gives the first-order equation

\dot{θ}^{2} = \frac{6 E}{M a ^{2}} + \frac{3 g}{a} cos θ - \frac{h ^{2}}{sin ^{2} θ} .

Where the right side is positive, either sign of its square root is possible, according to the direction of motion. Its zeros are turning points. For

h \neq = 0

, the centrifugal term excludes the poles; if

h = 0

, the polar-coordinate formula is interpreted through the original rod equations.

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