The angle which the bicycle and its rider must make with the vertical when going round a curve of $8.1 m$ radius at $9 m s^{- 1}$ is (Take $g = 10 m / s^{2}$ )

20^{\circ}

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45^{\circ}

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30^{\circ}

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60^{\circ}

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Solution

The correct option is B $45^{\circ}$

The components of Normal reaction for cyclist need to satisfy the equation of dynamics i.e. supporting the weight & centripetal force.

So, we have
$N sin θ = \frac{m v^{2}}{r}$ .......(i)
$N cos θ = m g$ ..........(ii)
On dividing (i) by (ii), we get
$tan θ = \frac{v^{2}}{r g}$
Here, $r = 8.1 m, v = 9 m s^{- 1}$
Putting values,
$tan θ = \frac{v^{2}}{r g} = \frac{9 \times 9}{8.1 \times 10} = 1$
$θ = {tan}^{- 1} (1) = 45^{\circ}$

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The angle which the bicycle and its rider must make with the vertical when going round a curve of 8.1 m radius at 9 ms−1 is (Take g=10 m/s2)

The angle which the bicycle and its rider must make with the vertical when going round a curve of $8.1 m$ radius at $9 m s^{- 1}$ is (Take $g = 10 m / s^{2}$ )