Question

1. 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​

Answer 1

Answer:

He will

Explanation:

Since the road is a circle hence the bicyclist must overcome the centripetal force to remain in the circle and the velocity that he is travelling with is 10 m/s with the friction between the tyres and the road is 0.6.

So, the velocity with which he must cross the turn will be v^2=μrg where μ is the coefficient of friction, r is the radius of the turn and g is the acceleration due to gravity. So, v^2=0.6*100*10=600 or v=√600 =10√6 m/s. Which is less than the velocity of the cyclist. Hence, he will be able to cross the turn.

Answer 2

the frictional force required by the rider to cross the turn is 600N