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 m . If the coefficient of friction between the tyres and the road is 0.6, the frictional force required by the rider to

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 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

Smart Achievers · Accepted Answer

(2)

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.

Solution

Q.
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 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

1 300 N

2 600 N

3 1200 N

4 150 N

Ans.
(2)

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.

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