Problem 44
Question
A motorcycle moving with a velocity of \(72 \mathrm{kmh}^{-1}\) on a flat road takes a turn on the road at a point where the radius of curvature of the road is \(20 \mathrm{~m}\). The acceleration due to gravity is \(10 \mathrm{~ms}^{-2}\). In order to avoid skidding, he must not bent with respect to the vertical plane by an angle greater than [BVP Engg. 2006] (a) \(\theta=\tan ^{-1}(2)\) (b) \(\theta=\tan ^{-1}(6)\) (c) \(\theta=\tan ^{-1}(4)\) (d) \(\theta=\tan ^{-1}(25.92)\)
Step-by-Step Solution
Verified Answer
Option (a): \(\theta = \tan^{-1}(2)\).
1Step 1: Convert velocity to meters per second
The motorcycle's velocity is given as \(72 \text{ kmh}^{-1}\). First, we need to convert this velocity into meters per second (\(\text{ms}^{-1}\)). We know 1 km/h is equal to \(\frac{5}{18} \, \text{ms}^{-1}\). So, \[72 \, \text{kmh}^{-1} = 72 \times \frac{5}{18} = 20 \, \text{ms}^{-1}.\]
2Step 2: Understand the formula for banking angle
The banking angle \(\theta\) is given by the formula \(\tan \theta = \frac{v^2}{rg}\), where \(v\) is the velocity, \(r\) is the radius of curvature, and \(g\) is the acceleration due to gravity.
3Step 3: Plug values into the formula
We know that \(v = 20 \, \text{ms}^{-1}\), \(r = 20 \, \text{m}\), and \(g = 10 \, \text{ms}^{-2}\). Substituting these values into the formula, we get \[\tan \theta = \frac{(20)^2}{20 \times 10} = \frac{400}{200} = 2.\] Therefore, \(\theta = \tan^{-1}(2)\).
4Step 4: Identify the correct option
From the calculation, we found that \(\theta = \tan^{-1}(2)\). Comparing with the given options, we identify that the correct answer is option (a): \(\theta = \tan^{-1}(2)\).
Key Concepts
Banking AngleCircular MotionTrigonometric Functions
Banking Angle
When driving around curves, especially at high speeds, vehicles often tilt inward, angling themselves to make better contact with the road. This tilt is known as the banking angle. Understanding this concept is crucial for both drivers and designers of vehicles and roads. A banking angle helps prevent skidding by using the normal force and friction to counteract the centrifugal force that pulls the vehicle outward.
The banking angle can be calculated using the formula \( \tan \theta = \frac{v^2}{rg} \), where \( v \) is the velocity, \( r \) is the radius of the curve, and \( g \) is the gravitational acceleration. This formula illustrates how the required angle increases with speed and decreases for larger curve radii. In practical terms:
The banking angle can be calculated using the formula \( \tan \theta = \frac{v^2}{rg} \), where \( v \) is the velocity, \( r \) is the radius of the curve, and \( g \) is the gravitational acceleration. This formula illustrates how the required angle increases with speed and decreases for larger curve radii. In practical terms:
- A higher speed requires a steeper angle to maintain stability.
- A larger curve radius lessens the required angle for the same speed.
Circular Motion
Circular motion occurs when an object travels along a circular path. In this situation, the object experiences a centripetal force directed towards the center of the circle that keeps it in motion. This force is crucial for maintaining motion on a curved path.
For an object moving in a circle with radius \( r \) at speed \( v \), the centripetal force \( F_c \) is given by \( F_c = \frac{mv^2}{r} \), where \( m \) is the mass of the object. This force results in acceleration that is also directed toward the center of the circle, known as centripetal acceleration, \( a_c = \frac{v^2}{r} \).
For an object moving in a circle with radius \( r \) at speed \( v \), the centripetal force \( F_c \) is given by \( F_c = \frac{mv^2}{r} \), where \( m \) is the mass of the object. This force results in acceleration that is also directed toward the center of the circle, known as centripetal acceleration, \( a_c = \frac{v^2}{r} \).
- Effect of Speed: Faster speeds increase the centripetal force needed to maintain the circular path.
- Effect of Radius: A larger radius reduces the required force for a given speed.
Trigonometric Functions
Trigonometric functions are mathematical tools used to relate the angles and sides of triangles, especially right triangles. In physics, these functions, such as sine, cosine, and tangent, play a significant role in analyzing forces and motion.
The tangent function \( \tan(\theta) \) is particularly useful because it relates the angle \( \theta \) to the opposite and adjacent sides of a right triangle. In the context of banking angles, the relationship \( \tan \theta = \frac{v^2}{rg} \) uses the tangent function to help calculate the angle required to prevent skidding on a curve.
The tangent function \( \tan(\theta) \) is particularly useful because it relates the angle \( \theta \) to the opposite and adjacent sides of a right triangle. In the context of banking angles, the relationship \( \tan \theta = \frac{v^2}{rg} \) uses the tangent function to help calculate the angle required to prevent skidding on a curve.
- Understanding Tangent: The tangent of an angle in a right triangle equals the ratio of the opposite side to the adjacent side.
- Inverse Tangent: The inverse function \( \tan^{-1}(x) \) provides the angle when given its tangent.
Other exercises in this chapter
Problem 43
A car of mass \(1000 \mathrm{~kg}\) negotitates a banked curve of radius \(90 \mathrm{~m}\) on a frictionless road. If the banking angle is \(45^{\circ}\), the
View solution Problem 43
A curved road of \(50 \mathrm{~m}\) radius is banked at correct angle for a given speed. If the speed is to be doubled keeping the same banking angle, the radiu
View solution Problem 45
A particle is moving in a circle of radius \(R\) in such a way that at any instant the normal and tangential components of its acceleration are equal. If its sp
View solution Problem 42
A particle moves in circular path of radius \(R\). If centripetal force \(F\) is kept constant but the angular velocity is double, the new radius of the path wi
View solution