Question

A train runs along an unbanked circular track of radius 30 m at a speed of $54\dfrac{{km}}{{hr}}$. The mass of the train is ${10^6}kg$. What provides the centripetal force required for this purpose-the engine or the rails? What is the angle of banking required to prevent wearing out of the rail?

Answer 1

When a car undergoes circular motion the centripetal force comes into play. Using Newton’s law, frictional force will help cars to avoid skidding on banked roads. This friction helps the cars to nullify the centrifugal force and acts as centripetal force. We will use $\tan \theta = \dfrac{{{{\text{v}}^{\text{2}}}}}{{{\text{rg}}}}$ to find the angle of the banked road.

Complete step by step answer:
When a car makes a turn on a flat road, the centrifugal force starts to act away from the imaginary center of the turning. The static frictional force of the car’s tyre acts against the slipping. This friction is used as the centripetal force.
Moving on to the next part:
Radius of circular train track = 30 m
Speed of train,
$= 54\dfrac{{km}}{{hr}}$
$= 54 \times \dfrac{{1000}}{{3600}}\dfrac{m}{{\sec }}$
=$= 15\dfrac{m}{{\sec }}$
Mass of the train $= {10^6}kg$
Lateral thrust provides centripetal force to the rails. Newton’s third law is responsible for repulsive force which acts on the rail. This law states that to every action, there is equal and opposite reaction. Wear and tear are prevented from repulsive force which wheels exert on the rail.
Banked roads:
When the outermost portion of curved roads is raised to certain heights than inner portions then it is termed as banked roads. This arrangement is used to increase the normal reaction so that the friction acting against the centrifugal force is increased.
Let us consider, banked road with radius ‘r’, a train of mass ‘m’ is moving with velocity ‘v’, then ‘$\theta$’ be the angle of banked road,
Using formula,
$\tan \theta = \dfrac{{{{\text{v}}^{\text{2}}}}}{{{\text{rg}}}}$
g = acceleration due to gravity $\approx 10\dfrac{m}{{{{\sec }^2}}}$
$\tan \theta = \dfrac{{{{15}^2}}}{{30 \times 10}}$
$\Rightarrow \tan \theta = \dfrac{{225}}{{300}}$
$\Rightarrow \tan \theta = 0.75$
$\Rightarrow \theta = {\tan ^{ - 1}}\left( {0.75} \right)$
$\Rightarrow \theta = 36.87^\circ$
$36.87^\circ$ is the angle of banking required to prevent wearing out of the rail.

Note:
If we take the value of g as $9.8\dfrac{m}{{\sec }}$ then we will get a slightly different solution. Conversion of $\dfrac{{km}}{{hr}}$ into $\dfrac{m}{{\sec }}$ is must because value of g and radius everything is mention in CGS system.