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 $10 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ . 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:

• A. 300 N
• B. 600 N
• C. 1200 N
• D. 150 N

Answer 1

Answer: (B)

600 N

Answer 2

Centripetal force

$= \frac { m v ^ { 2 } } { r } = \frac { 100 \times 10 \times 10 } { 100 } = 100 \mathrm {~N}$

Required frictional force to cross the turn

$= \mu m g = 0.6 \times 100 \times 10 = 600 \mathrm {~N}$

As the frictional force is greater than the centripetal force, so the rider will be able to cross the turn.