Question

The coefficient of friction between the road and the tyres of a cyclist is 0.1. The maximum speed with which he can take a circular turn of radius 2 m without skidding is ( \( g = 10 \) m/s\( ^2 \) )

• A. \( \sqrt{2} \, {ms}^{-1} \)
• B. \( \sqrt{3} \, {ms}^{-1} \)
• C. \( \sqrt{5} \, {ms}^{-1} \)
• D. \( 2 \, {ms}^{-1} \)
• E. \( 3 \, {ms}^{-1} \)

Hint:

When solving problems involving friction and circular motion, use the formula for maximum velocity in circular motion: \( v_{{max}} = \sqrt{r \cdot g \cdot \mu} \).

Answer 1

The Correct Option is A

The maximum speed \( v_{{max}} \) at which the cyclist can take the turn without skidding is given by the formula: \[ v_{{max}} = \sqrt{r \cdot g \cdot \mu} \] where: 
- \( r = 2 \, {m} \) (radius), 
- \( g = 10 \, {m/s}^{2} \) (acceleration due to gravity), 
- \( \mu = 0.1 \) (coefficient of friction). 
Substitute the values: \[ v_{{max}} = \sqrt{2 \times 10 \times 0.1} = \sqrt{2} \, {ms}^{-1} \] 
Hence, the correct answer is (A).