Past exam of the mathematics course of the University of Oxford / 2017 / Prelims / Mathematics IV / 6 / a

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- Solution a

Solution

In cylindrical coordinates,

θ = ω t

and

z = r^{2} / (2 a)

. Thus

\overset{z}{˙} = \frac{r}{a} \overset{r}{˙}, \overset{z}{¨} = \frac{r r ¨ + r ˙ ^{2}}{a} .

The acceleration is

A = (\overset{r}{¨} - r ω^{2}) e_{r} + 2 ω \overset{r}{˙} e_{θ} + \frac{r r ¨ + r ˙ ^{2}}{a} e_{z} .

A tangent vector along the wire is

t = e_{r} + (r / a) e_{z}

. Because the wire is smooth, its constraint force is orthogonal to

t

. Taking the dot product of Newton's second law with

t

eliminates that force; gravitational weight contributes

- m g r / a

. Hence

m [\overset{r}{¨} - r ω^{2} + \frac{r}{a ^{2}} (r \overset{r}{¨} + \overset{r}{˙}^{2})] = - \frac{m g r}{a} .

Multiplying by

a^{2} / m

gives

(a^{2} + r^{2}) \overset{r}{¨} + r \overset{r}{˙}^{2} - a^{2} ω^{2} r = - a g r .

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