Question

The minimum velocity (in \[m{s^{ - 1}}\]) with which a car driver must traverse a flat curve of radius 150 m and coefficient of friction 0.6 to avoid skidding is
A. \[60\,m{s^{ - 1}}\]
B. \[30\,m{s^{ - 1}}\]
C. \[15\,m{s^{ - 1}}\]
D. \[25\,m{s^{ - 1}}\]

Answer 1

Hint:Before we start addressing the problem, we need to know about centripetal force. Whenever a body passes through the curve centripetal force acts on the body and the direction of the force is at the body and is always acting away towards the center of the curve. The centripetal force is necessary to act on the body as the curve arises to balance the forces on the body.

Formula Used:
To find the centripetal force the formula is,
\[F = \dfrac{{m{v^2}}}{r}\]
Where, m is the mass of the body, v is the speed of the body and r is radius.

Complete step by step solution:
Here, when a car driver is moving in a flat curve, he must possess a centripetal force. Centripetal force is not a fundamental force which means some other force behaves like centripetal force. Here, the static friction acts as centripetal force. The expression for the centripetal force is,
\[\dfrac{{m{v^2}}}{r} = F\]
\[\Rightarrow \dfrac{{m{v^2}}}{r} = \mu N\]
Here, the force is nothing but the frictional force i.e., \[\mu N\] and N is the normal force given by, \[N = mg\].

Then, the above equation becomes,
\[\dfrac{{m{v^2}}}{r} = \mu \left( {mg} \right)\]
\[\Rightarrow {v^2} = \mu rg\]
\[\Rightarrow v = \sqrt {\mu rg} \]
Substitute the value of \[\mu = 0.6\], \[r = 150\,m\] and \[g = 9.8\,m{s^{ - 2}}\]
\[v = \sqrt {0.6 \times 150 \times 9.8} \]
\[\Rightarrow v = 29.69\,m{s^{ - 1}}\]
\[\therefore v = 30\,m{s^{ - 1}}\]
Therefore, the minimum velocity with which a car driver must traverse a flat curve is \[30\,m{s^{ - 1}}\].

Hence, option B is the correct answer.

Note: In this problem it is important to remember that the equation for the centripetal force that is, \[F = \dfrac{{m{v^2}}}{r}\]. Here, the force is nothing but the frictional force between the car and the road.