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Chapter 3: Problem 79

A particle of mass \(m\) is projected with a velocity \(v\) making an angle of \(45^{\circ}\) with the horizontal. The magnitude of angular momentum of projectile about the point of projection when the particle is at its maximum height \(h\) is : (a) zero (b) \(\frac{m v h}{\sqrt{2}}\) (c) \(\frac{m v h^{2}}{\sqrt{2}}\) (d) none of these

Short Answer

Expert verified
Option (b), \( \frac{m v h}{\sqrt{2}} \), is correct.

Step by step solution

01

Understand Angular Momentum at Maximum Height

The angular momentum of the particle about the point of projection at maximum height can be calculated using the formula for angular momentum, \( L = r \cdot mv \sin(\theta) \), where \( r \) is the perpendicular distance from the point of projection to the velocity vector at maximum height. At maximum height, the vertical component of velocity is zero, and only horizontal motion is present. Thus, the perpendicular distance \( r \) is the maximum height , \( h \), itself.
02

Identify Horizontal Velocity

Since the launch angle \( \theta = 45^{\circ} \), the horizontal and vertical components of the initial velocity are equal. Therefore, the horizontal velocity \( v_x \) at all points in the projectile motion, including at maximum height, is \( v \cos(45^{\circ}) = \frac{v}{\sqrt{2}} \).
03

Calculate Angular Momentum

The angular momentum \( L \) of the particle when it is at its maximum height \( h \) is given by the formula:\[ L = m \cdot (\frac{v}{\sqrt{2}}) \cdot h \].After simplifying, we find:\[ L = \frac{mvh}{\sqrt{2}} \].
04

Determine Correct Option

From the calculation, the magnitude of the angular momentum at maximum height is \( \frac{m v h}{\sqrt{2}} \), which matches option (b). Therefore, option (b) is the correct answer.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Understanding Projection Motion
Projection motion refers to the movement of an object thrown or projected into the air, subject to only the acceleration of gravity. It's the basis for analyzing many real-world phenomena, such as the flight of a kicked soccer ball or a dart thrown at a board. In our exercise, the particle is projected at a specific angle, which means we need to consider both horizontal and vertical components of the motion.
  • **Angle of Projection**: The particle in our discussion is projected at a 45掳 angle. This angle is special because it maximizes the horizontal range for a given velocity.
  • **Horizontal and Vertical Components**: Using trigonometry, the initial velocity vector can be broken down into horizontal and vertical components. These components help us determine the particle's path trajectory.
  • **Gravity's Role**: Gravity only affects the vertical component of the particle's velocity, decreasing it as it rises and increasing it as it falls.
Understanding projection motion allows us to predict the path and behavior of the particle in the air, which is crucial for further calculations like maximum height and angular momentum.
Maximum Height in Projectile Motion
Maximum height is a critical point in projectile motion where the vertical component of velocity becomes zero. At this point, the particle stops going up and starts descending.
  • **Vertical Component Zero**: As the particle reaches its maximum height, gravity has slowed down its vertical motion until it momentarily stops before descending. This is why the velocity component vertically is zero at the maximum height.
  • **Calculating Maximum Height**: The maximum height can be calculated using the formula \[ h = \frac{v^2 \sin^2(\theta)}{2g} \], where \( v \) is the initial velocity, \( \theta \) is the launch angle, and \( g \) is the acceleration due to gravity.
  • **Importance in Angular Momentum**: Knowing the maximum height is vital for calculating angular momentum about the point of projection, as it affects the distance in calculations.
Understanding this helps in predicting and analyzing the particle's motion precisely at various stages of its flight.
The Role of Horizontal Velocity
Horizontal velocity in projection motion describes how fast the particle moves along the horizontal plane. Unlike vertical velocity, it remains constant if air resistance is negligible.
  • **Constant Value**: Since there's no horizontal acceleration in idealized projectile motion, the horizontal velocity remains constant throughout the flight of the particle.
  • **Calculation**: For our particle projected at 45掳, the horizontal component of velocity is calculated using \[ v_x = v \cos(45^\circ) = \frac{v}{\sqrt{2}} \].
  • **Relevance to Angular Momentum**: Horizontal velocity is crucial when determining the angular momentum at the maximum height, as it is the only velocity component not equal to zero.
By maintaining focus on horizontal velocity, we understand how far the particle travels horizontally and analyze its motion in terms of both distance and time.

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Most popular questions from this chapter

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