Chapter 5: Problem 37
If a car is to travel with a speed \(v\) along the frictionless banked circular track of radius \(r\), the required angle of banking so that the car does skid is [J\&K 2010] (a) \(\theta=\tan ^{-1}\left(\frac{v^{2}}{r g}\right)\) (b) \(\theta=\tan ^{-1}\left(\frac{v}{r g}\right)\) (c) \(\theta=\tan ^{-1}\left(\frac{r^{2}}{v g}\right)\) (d) \(\theta<\tan ^{-1}\left(\frac{\partial^{2}}{r g}\right)\)
Short Answer
Step by step solution
Understanding the Problem
Identifying Forces Acting
Using Newton's Second Law
Equating Forces
Solving for Angle of Banking
Selecting the Correct Option
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Centripetal Force
- The centripetal force is calculated using the formula \( F_c = \frac{mv^2}{r} \), where \( m \) is mass, \( v \) is velocity, and \( r \) is the radius of the circle.
- This force acts toward the center of the circle, altering the direction of the car's velocity and thus keeping it in a circular orbit.
- In scenarios with no friction, like on a perfectly smooth banked track, the angle and the speed of the car must be just right to generate this force naturally via bank angle design.
Angle of Banking
- The road is tilted at an angle \( \theta \), which means a component of the normal force will act as the centripetal force.
- This angle ensures that even in the absence of friction, vehicles can maintain their motion along the desired circular path.
- Mathematically, it is determined by the formula \( \tan(\theta) = \frac{v^2}{rg} \), where \( \theta \) is the angle of banking, \( v \) is the velocity, \( r \) is the radius, and \( g \) is the acceleration due to gravity.
Newton's Second Law
- Vertical Component: The gravitational force pulling the car down is balanced by the vertical component of the normal force \( N \sin(\theta) \), keeping the car from falling straight downward.
- Horizontal Component: The centripetal force required for circular motion \( N \cos(\theta) \) is derived from the same normal force but acts horizontally towards the track center, enabling the car to turn.