Question

If the banking of the tracks is${{\tan }^{-1}}\left( 2 \right)$, the maximum velocity with which a train can be moved on a circular track of radius $20m$will be,

& A.1.0m/\sec \\\ & B.4.5m/\sec \\\ & C.8.3m/\sec \\\ & D.20m/\sec \\\ \end{aligned}$$

Answer 1

Hint : This question is related to the topic circular motion. In this question, first we need to know about centripetal force and circular motion. We have to find maximum velocity. By using the expression for the velocity of the vehicle on the curved banked road and trigonometric concept we can find maximum velocity.

Complete step-by-step solution:
This question is related to the topic circular motion. Circular motion is described as a movement of an object while rotating along a circular path. There are two types of circular motion such as Uniform circular motion and non-uniform circular motion. There are basically various types of motion such as rotary motion, oscillating motion, linear motion, rectilinear motion, circular motion, periodic motion, reciprocating motion, rotational motion etc.
In the special case of Earth circular motion around the sun or any satellite or circular motion around any celestial body, then CENTRIPETAL FORCE causing the motion is the result of gravitational attraction between them. Centripetal forces are always directed toward the centre of the circular path.
When the track is banked, then the inclination of the surface of the road with the horizontal is known as the angle of banking. As the speed of the vehicle increases, the centripetal force needed for the circular motion of the vehicle also increases. Without proper friction, a vehicle will not be able to move on the curved road with a large speed. To avoid this we may increase the force of friction making the road rough. Also, the force of friction is not always reliable because it changes when roads are oily or wet due to rains etc; to eliminate this difficulty the curved roads are generally banked. Due to banking the necessary centripetal force is provided by the component of the normal reaction.
The expression for the velocity of a vehicle on a curved banked road is:
$v=\sqrt{rg\tan \theta }.......Equation\left( 1 \right)$
Where,

& v=velocity \\\ & r=radius \\\ & g=gravitational force \\\ & \theta =angle of banking \\\ \end{aligned}$$ Here we have given values as:- $$radius=20m$$ $$\theta =angle of banking={{\tan }^{-1}}\left( 2 \right)$$ $$g=10m/{{s}^{2}}$$ Now, using $$Equation\left( 1 \right)$$ we will put the values and on simplifying we will get the value of velocity. $$v=\sqrt{20\times 10\times \tan \left( {{\tan }^{-1}}2 \right)}......Equation\left( 2 \right)$$ From trigonometric concept we know that, $${{\tan }^{-1}}\left( \tan \theta \right)=\theta $$ Where, $$\dfrac{-\pi }{2}\angle \theta \langle \dfrac{\pi }{2}$$ Using trigonometric concept, we will simplify $$Equation\left( 2 \right)$$as: $$v=\sqrt{20\times 10\times 2}$$ $$v=\sqrt{400}$$ Therefore, we will find out the value of$$\sqrt{400}$$, The SI unit of velocity is meter per second, $$v=20\dfrac{m}{s}$$ Final answer is$$20\dfrac{m}{s}$$. Therefore, the maximum velocity with which a train can be moved on a circular track of radius $$20m$$will be$$20\dfrac{m}{s}$$. **Note:** When a vehicle moves along a curved path with very high speed, then there is a chance of overturning of the vehicle. Inner wheel leaves the ground first .A cycling race track is called Velodrome which has a saucer-shaped track. The rider has to take position on a track as per his/her speed.