A car of mass 1500 kg is moving with a speed | vedclass

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Question

(D) For a car moving on a level circular road,the necessary centripetal force is provided by the static friction between the tires and the road.Let $m

Vedclass · Accepted Answer

$0.8$

Vedclass · Answer

(D) For a car moving on a level circular road,the necessary centripetal force is provided by the static friction between the tires and the road.Let $m$ be the mass of the car,$v$ be the speed,$r$ be the radius of the circular path,and $\mu$ be the coefficient of friction.The centripetal force required is $F_c = \frac{mv^2}{r}$.The maximum frictional force available is $f_{max} = \mu N = \mu mg$.To prevent slipping,we must have $F_c \leq f_{max}$,which implies $\frac{mv^2}{r} \leq \mu mg$.Thus,the minimum coefficient of friction required is $\mu = \frac{v^2}{rg}$.Substituting the given values: $v = 12.5 \ m/s$,$r = 20 \ m$,and $g = 9.8 \ m/s^2$:$\mu = \frac{12.5 	imes 12.5}{20 	imes 9.8} = \frac{156.25}{196} \approx 0.797$.Rounding to the nearest provided option,we get $\mu = 0.8$.

$A$ car of mass $1500 \ kg$ is moving with a speed of $12.5 \ m/s$ on a circular path of radius $20 \ m$ on a level road. What should be the coefficient of friction between the car and the road,so that the car does not slip?

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