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Physics · LAWS OF MOTION
हिन्दीA car negotiates a level, unbanked circular turn of radius \( R = 40\text{ m} \). The coefficient of static friction between the tyres and the road is \( \mu_s = 0.25 \). Taking \( g = 10\text{ m/s}^2 \), calculate the maximum safe speed \( v_{\max} \) to avoid slipping. If the speed of the car is increased to \( 15\text{ m/s} \), determine the new minimum coefficient of static friction \( \mu_s' \) required to prevent the car from slipping.
Correct answer · Option 4
\( v_{\max} = 10\text{ m/s} \) and \( \mu_s' = 0.5625 \)
Explanation
The maximum safe speed on a level road is \( v_{\max} = \sqrt{\mu_s R g} = \sqrt{0.25 \times 40 \times 10} = 10\text{ m/s} \). For a new speed \( v' = 15\text{ m/s} \), the required static friction coefficient is \( \mu_s' = \frac{v'^2}{R g} = \frac{225}{400} = 0.5625 \).
