Question

A curved in a level road has a radius 75 m. The maximum speed of a car turning this curved road can be 30 m/s without skidding. If radius of curved road is changed to 48 m and the coefficient of friction between the tyres and the road remains same, then maximum allowed speed would be ______ m/s.

Answer 1

Correct Answer: 24

To solve the problem of determining the maximum allowed speed of a car on a curved road, we use the relationship between centripetal force and frictional force. For a car moving in a circle, the centripetal force \( F_c \) required to keep it moving in a circle of radius \( r \) at velocity \( v \) is given by \( F_c = \frac{mv^2}{r} \), where \( m \) is the car's mass. The frictional force \( F_f \) that prevents skidding is given by \( F_f = \mu mg \), where \( \mu \) is the coefficient of friction and \( g \) is the acceleration due to gravity. The maximum speed \( v_{\text{max}} \) is when the frictional force equals the centripetal force: \( \frac{mv_{\text{max}}^2}{r} = \mu mg \). Simplifying, \( v_{\text{max}}^2 = \mu gr \) and \( v_{\text{max}} = \sqrt{\mu gr} \).
For the initial condition with \( r = 75 \, \text{m} \) and \( v_{\text{max}} = 30 \, \text{m/s} \), we calculate the coefficient of friction: \( \mu = \frac{v_{\text{max}}^2}{gr} = \frac{30^2}{75g} = \frac{900}{75g} = \frac{12}{g} \).
Keeping the same \( \mu \) for the new radius, \( r = 48 \, \text{m} \), the new \( v_{\text{max}} \) is \( v_{\text{max}} = \sqrt{\mu g \times 48} = \sqrt{\left(\frac{12}{g}\right) \times g \times 48} = \sqrt{12 \times 48} = \sqrt{576} = 24 \, \text{m/s} \).
This falls within the specified range of 24—24 m/s, confirming the solution is correct.