Past exam of the mathematics course of the University of Oxford / 2021 / Prelims / Mathematics IV / 5 / b

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- Solution b

Solution

For a smooth surface the friction force vanishes. The

e_{θ}

component of Newton's second law then gives

r^{2} \dot{θ} = h

, constant. Along the constraint

z = H (r)

, the chain rule gives

\overset{z}{˙} = H^{'} \overset{r}{˙}, \overset{z}{¨} = H^{'} \overset{r}{¨} + H^{''} \overset{r}{˙}^{2} .

The acceleration in cylindrical coordinates is

\ddot{r} = (\overset{r}{¨} - r \dot{θ}^{2}) e_{r} + (r \ddot{θ} + 2 \overset{r}{˙} \dot{θ}) e_{θ} + \overset{z}{¨} e_{z} .

Take the dot product of Newton's second law with the tangent vector

e_{r} + H^{'} e_{z}

. The normal force disappears, while the gravity term is

- m g H^{'}

. Dividing by

m

gives

\overset{r}{¨} - r \dot{θ}^{2} + H^{'} (H^{'} \overset{r}{¨} + H^{''} \overset{r}{˙}^{2}) = - g H^{'} .

Using

\dot{θ} = h / r^{2}

yields

[1 + (H^{'})^{2}] \overset{r}{¨} + H^{'} H^{''} \overset{r}{˙}^{2} - \frac{h ^{2}}{r ^{3}} = - g H^{'} .

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