Question
To what angle a cyclist has to bend to negotiate a curved path? Derive an expression.
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Answer
As the cyclist leans by an angle $\theta$ with the vertical, the normal can be acting as shown. If v is the speed of the cyclist, the necessary centripetal force is provided by the component $\text{N}\sin\theta$ of normal reaction and $\text{N}\cos\theta$ provides balancing for weight. 
$\therefore\ \text{N}\sin\theta=\frac{\text{mv}^2}{\text{r}},$ $\text{N}\cos\theta=\text{mg}$ Dividing, $\tan\theta=\frac{\text{v}^2}{\text{rg}}$ $\Rightarrow\ \theta=\tan^{-1}\Big(\frac{\text{v}^2}{\text{rg}}\Big)$
$\therefore\ \text{N}\sin\theta=\frac{\text{mv}^2}{\text{r}},$ $\text{N}\cos\theta=\text{mg}$ Dividing, $\tan\theta=\frac{\text{v}^2}{\text{rg}}$ $\Rightarrow\ \theta=\tan^{-1}\Big(\frac{\text{v}^2}{\text{rg}}\Big)$Need a full question paper?
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