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Year 1 of the physics course of the University of Oxford

Physics 1 paper of the University of Oxford

Past exam of the physics 1 paper of the University of Oxford

2015 resit

The $\theta$ <Euler-Lagrange equation> reads
$$
\frac d{dt}(ml^2\dot\theta)-\left(ml^2\sin\theta\cos\theta\dot\phi^2-mgl\sin\theta\right)=0.
$$
Hence
$$
\boxed{\ddot\theta=\sin\theta\cos\theta\dot\phi^2-\frac gl\sin\theta.}
$$
The first term reflects the coupling to azimuthal motion; the second is the gravitational restoring contribution for an <angle> measured from the downward vertical.

Solution

Past exam of the physics 1 paper of the University of Oxford / 2015 resit / 10 / c / Solution

Past exam of the physics 1 paper of the University of Oxford / 2015 resit / 10 / c / Solution

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Past exam of the physics 1 paper of the University of Oxford / 2015 resit / 10 / c / Solution

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