Question

The mass of a bicycle rider along with the bicycle is 100 kg. He wants to cross over a circular turn of radius 100m 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 _______
A) 330N
B) 600N
C) 1200N
D) 150N

Answer 1

We need to understand the relation between the mass of the bicycle, the velocity of the system, the radius of the circular turn with the frictional force provided by the road for a safe journey without slipping on taking the turn on the circular path.

Complete step by step solution:
We are given the information about the circular turn on the path of a bicycle rider. We know that for a body to take a safe turn about a circular turn the velocity of the body should be such that the force offered by the centripetal force will be always lesser than the frictional force offered by the road on contact with the body.

Now, we can consider the situation given to us in which, the bicycle and the rider of mass 100 kg together is moving at velocity of 10 $m{{s}^{-1}}$taking a turn around a circular path of radius 100m. From this information, we can find the centripetal force provided by the circular motion for the bicycle as –
${{F}_{C}}=\dfrac{m{{v}^{2}}}{r}$
Where, m is the mass of the system,
v is the velocity of the system,
r is the radius of the circular path.
The centripetal force is given as –

& {{F}_{C}}=\dfrac{m{{v}^{2}}}{r} \\\ & \Rightarrow {{F}_{C}}=\dfrac{100kg\times {{(10m{{s}^{-1}})}^{2}}}{100} \\\ & \therefore {{F}_{C}}=100N \\\ \end{aligned}$$ The centripetal force on the bicycle is 100N. Now, let us find the frictional force acting on the bicycle as – $$\begin{aligned} & {{F}_{f}}=\mu mg \\\ & \Rightarrow {{F}_{f}}=0.6\times 100\times 10 \\\ & \therefore {{F}_{f}}=600N \\\ \end{aligned}$$ The road is providing a frictional force of 600N on the tyres of the bicycle. From the two forces, we understand that the centripetal force acting on the bicycle is lesser than the frictional force. The rider can safely turn around the curve without slipping with the given speed. The required frictional force is, therefore, 600N **The correct answer is option B.** **Note:** Usually, a road is banked to provide a greater balance between the centripetal force and the frictional force. This enables the rider to take the curve with a greater velocity than we have used in the present situation as more stability is attained by this.