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 if the coefficient of friction between the tyres and the road is 0.6 will the rider be able to cross the turn ?​

Answer 1

Answer:

The centripetal force acting on the bicycle is given by,

F

c

=

r

mv

2

=

100

100×10×10

=100N

Required frictional force to cross the turn,

F

f

=μmg

=0.6×100×10=600N

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

Answer 2

Answer:

The centripetal force acting on the bicycle is given by,

F

c

=

r

mv

2

=

100

100×10×10

=100N

Required frictional force to cross the turn,

F

f

=μmg

=0.6×100×10=600N

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