A cyclist speeding at 18 ; km/h on a level ro | vedclass

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(A) On an unbanked road,the frictional force provides the necessary centripetal force for circular motion. The condition for the cyclist not to slip i

Vedclass · Accepted Answer

Yes,the cyclist will slip.

Vedclass · Answer

(A) On an unbanked road,the frictional force provides the necessary centripetal force for circular motion. The condition for the cyclist not to slip is $v^{2} \leq \mu_{s} R g$.Given: $v = 18 \; km/h = 18 	imes \frac{5}{18} = 5 \; m/s$,$R = 3 \; m$,$\mu_{s} = 0.1$,and $g = 9.8 \; m/s^{2}$.Calculating the maximum allowed speed squared: $\mu_{s} R g = 0.1 	imes 3 	imes 9.8 = 2.94 \; m^{2}/s^{2}$.Calculating the actual speed squared: $v^{2} = (5)^{2} = 25 \; m^{2}/s^{2}$.Since $v^{2} > \mu_{s} R g$ $(25 > 2.94)$,the frictional force is insufficient to provide the required centripetal force. Therefore,the cyclist will slip.

$A$ cyclist speeding at $18 \; km/h$ on a level road takes a sharp circular turn of radius $3 \; m$ without reducing the speed. The coefficient of static friction between the tyres and the road is $0.1$. Will the cyclist slip while taking the turn?

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