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Mathematics course of the University of Cambridge

Past exam of the mathematics course of the University of Cambridge

2020

Paper 2

Solving the two momentum equations gives
$$
\dot\phi=\frac{L_z-L_3\cos\theta}{I_1\sin^2\theta},
\qquad
\dot\psi+\dot\phi\cos\theta=\frac{L_3}{I_3}.
$$
Substitution into the conserved energy reduces the <heavy symmetric top> to one-dimensional <effective potential> motion:
$$
\boxed{\frac12I_1\dot\theta^2+V_{\mathrm{eff}}(\theta)=E'},
$$
where
$$
\boxed{V_{\mathrm{eff}}(\theta)
=\frac{(L_z-L_3\cos\theta)^2}{2I_1\sin^2\theta}
+Mgl\cos\theta},
\qquad
\boxed{E'=E-\frac{L_3^2}{2I_3}}.
$$
This is the <heavy symmetric top reduction>.

Solution

Past exam of the mathematics course of the University of Cambridge / 2020 / ii / Paper 2 / 14B / iii / Solution

Past exam of the mathematics course of the University of Cambridge / 2020 / ii / Paper 2 / 14B / iii / Solution

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Past exam of the mathematics course of the University of Cambridge / 2020 / ii / Paper 2 / 14B / iii / Solution

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