Question

A bus turns a slippery road having coefficient of friction $ 0.5 $ with a speed of $ 10{m \mathord{\left/
 {\vphantom {m s}} \right.} s} $ . The minimum radius of the arc in metre in which the bus turns is? [Take $ g = 10{m \mathord{\left/
 {\vphantom {m {{s^2}}}} \right.} {{s^2}}} $ ]
(A) $ {\text{4m}} $
(B) $ {\text{10m}} $
(C) $ {\text{15m}} $
(D) $ {\text{20m}} $

Answer 1

Hint : Frictional force resists motion, and it is less on a slippery road. The centripetal force is balanced by the force of friction acting on the tyres, this keeps the balance intact.

Formula Used: The formulae used in the solution are given here.
Force of friction $ f = \mu N $ where $ \mu $ is the coefficient of friction and $ N $ is the normal force.
Centripetal Force $ {F_c} = \dfrac{{m{v^2}}}{r} $ .

Complete step by step answer
Friction is a force between two surfaces that are sliding, or trying to slide, across each other.Friction is the force resisting the relative motion of solid surfaces, fluid layers, and material elements sliding against each other.
Mathematically, $ f = \mu N $ where $ \mu $ is the coefficient of friction and $ N $ is the normal force.
Now we know that the normal force is equal to the weight of a body. Mathematically, $ N = mg $ .
Thus, the force of friction can be expressed as, $ f = \mu N = \mu mg $ .
It has been given, a car turns a corner on a slippery road at a constant speed of $ 10{m \mathord{\left/
 {\vphantom {m s}} \right.} s} $ and the coefficient of friction is $ 0.5 $ .
In this case, a centripetal force acts on the body. The magnitude of this force is given by, $ {F_c} = \dfrac{{m{v^2}}}{r} $ where $ m $ is the mass, $ v $ is the velocity and $ r $ is the radius of the circle, the turn is a part of.
Thus, balancing the forces, we can write,
 $ \mu mg = \dfrac{{m{v^2}}}{r} $
Since the mass of the body remains the same, we can write,
 $ \mu g = \dfrac{{{v^2}}}{r} $
 $ \Rightarrow r = \dfrac{{{v^2}}}{{\mu g}} $
It has been given that, the coefficient of friction $ \mu = 0.5 $ , acceleration due to gravity $ g = 10{m \mathord{\left/
 {\vphantom {m {{s^2}}}} \right.} {{s^2}}} $ and velocity $ v = 10{m \mathord{\left/
 {\vphantom {m s}} \right.} s} $ .
The minimum radius of the arc in metre in which the car turns is given by assigning these values to the equation above.
 $ {\text{r = }}\dfrac{{{\text{1}}{{\text{0}}^{\text{2}}}}}{{{\text{0}}{{.5 \times 10}}}}{\text{ = }}\dfrac{{100}}{5}{\text{ = 20m}} $ .
The minimum radius of the arc is $ 20m $ .
Hence, the correct answer is Option D.

Note
The causes of friction are molecular adhesion, surface roughness, and deformations.
Adhesion is the molecular force resulting when two materials are brought into close contact with each other. Trying to slide objects against each other requires breaking these adhesive bonds. For years, scientists thought that friction was caused only by surface roughness, but recent studies have shown that it is actually a result of adhesive forces between the materials. But surface roughness is a factor when the materials are rough enough to cause serious abrasion. This is called the sandpaper effect.
When one or both of the materials is relatively soft, much of the resistance to movement is caused by deformations of the objects or by a ploughing effect.