First slide

Banking of road

Question

A cyclist riding the bicycle at a speed of $14 \sqrt{3} {ms}^{- 1}$ takes a turn around a circular road of radius $20 \sqrt{3} m$ without skidding. Given g= 9.8 m/s², what is his inclination to the vertical.

Moderate

A

30°
B

90°
C

45°
D

60°

Solution

According to question, speed of bicycle,
$v = 14 \sqrt{3} {ms}^{- 1}$
Radius of circular road,
$r = 20 \sqrt{3} m and g = 9 .8 {ms}^{- 2} .$
Let the inclination of bicycle with normal be $θ$ .
For no slip condition
$θ = \tan^{- 1} (\frac{y^{2}}{rg}) = \tan^{- 1} (\frac{(14 \sqrt{3})^{2}}{20 \sqrt{3} \times 9 .8})$
$= \tan^{- 1} (\sqrt{3}) = 60^{\circ}$

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