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 co~efficient of static-friction between the tyres and the road is 0.1. Will the cyclist slip while taking the turn?

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Solution

Speed of the cyclist $v = 18 k m / h = 18 \times \frac{5}{18} = 5 m / s$
Given : Radius $r = 3 m$ $μ = 0.1$
Safe limit of velocity $v_{s} = \sqrt{μ r g}$
So, $v_{s} = \sqrt{0.1 \times 10 \times 3} = 1.732 m / s$
Since, cyclist rides at a faster speed than safe limit. So, the cyclist slips.

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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 co~efficient of static-friction between the tyres and the road is 0.1. Will the cyclist slip while taking the turn?

Speed of the cyclist v=18 km/h=18×518=5 m/sGiven : Radius r=3 m μ=0.1Safe limit of velocity vs=√μrgSo, vs=√0.1×10×3=1.732 m/sSince, cyclist rides at a faster speed than safe limit. So, the cyclist slips.

Speed of the cyclist $v = 18 k m / h = 18 \times \frac{5}{18} = 5 m / s$
Given : Radius $r = 3 m$ $μ = 0.1$
Safe limit of velocity $v_{s} = \sqrt{μ r g}$
So, $v_{s} = \sqrt{0.1 \times 10 \times 3} = 1.732 m / s$
Since, cyclist rides at a faster speed than safe limit. So, the cyclist slips.