The mass of a bicycle rider along with the bicycle is 100 kg. He wants to cross over a circular turn of radius 100 m with a speed of 10 ms-1. If the coefficient of friction between the tyres and the road is 0.6, the frictional force required by the rider to cross the turn, is

Question

The mass of a bicycle rider along with the bicycle is 100 kg. He wants to cross over a circular turn of radius 100 m with a speed of 10 ms-1. If the coefficient of friction between the tyres and the road is 0.6, the frictional force required by the rider to cross the turn, is (A) 300 N (B) 600 N (C) 1200 N (D) 150 N

Tardigrade Admin · Accepted Answer

Centripetal force =(mv2/r) =(100×10×10/100) =100 N Required frictional force to cross the turn, =μ mg=0.6×100×10 =600 N As the frictional force is greater than the centripetal force, so the rider will be able to cross the turn.

Q. The mass of a bicycle rider along with the bicycle is $100 k g$ . He wants to cross over a circular turn of radius $100 m$ with a speed of $10 m$ $s^{- 1}$ . If the coefficient of friction between the tyres and the road is $0.6$ , the frictional force required by the rider to cross the turn, is

$300 N$

$600 N$

$1200 N$

$150 N$

Centripetal force
$= \frac{m v ^{2}}{r}$ $= \frac{100 \times 10 \times 10}{100}$ $= 100 N$
Required frictional force to cross the turn,
$= μ m g = 0.6 \times 100 \times 10$ $= 600 N$
As the frictional force is greater than the centripetal force, so the rider will be able to cross the turn.

Q. The mass of a bicycle rider along with the bicycle is 100kg. He wants to cross over a circular turn of radius 100m with a speed of 10ms−1. If the coefficient of friction between the tyres and the road is 0.6, the frictional force required by the rider to cross the turn, is

300N

600N

1200N

150N