If a cyclist moving with a speed of 4.9 , m/s | vedclass

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(C) For a cyclist taking a circular turn on a level road,the necessary centripetal force is provided by the static friction between the tyres and the

Vedclass · Accepted Answer

$0.61$

Vedclass · Answer

(C) For a cyclist taking a circular turn on a level road,the necessary centripetal force is provided by the static friction between the tyres and the road.The condition for safe turning is given by the formula: $v^2 \leq \mu rg$,where $v$ is the speed,$\mu$ is the coefficient of friction,$r$ is the radius,and $g$ is the acceleration due to gravity.To find the minimum coefficient of friction,we use: $\mu = \frac{v^2}{rg}$.Given: $v = 4.9 \, m/s$,$r = 4 \, m$,and $g = 9.8 \, m/s^2$.Substituting the values: $\mu = \frac{(4.9)^2}{4 	imes 9.8} = \frac{24.01}{39.2} = 0.6125$.Rounding to two decimal places,we get $\mu = 0.61$.

If a cyclist moving with a speed of $4.9 \, m/s$ on a level road can take a sharp circular turn of radius $4 \, m$,then the coefficient of friction between the cycle tyres and the road is:

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