/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 6 Find the components along the \(... [FREE SOLUTION] | 91Ó°ÊÓ

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Find the components along the \(x, y, z\) axes of the angular momentum 1 of a particle, whose position vector is \(\mathbf{r}\) with components \(x, y, z\) and momentum is \(\mathbf{p}\) with components \(p_{x} \cdot p_{y}\) and \(p_{z} .\) Show that if the particle moves only in the \(x-y\) plane the angular momentum has only a z-component.

Short Answer

Expert verified
The particle's angular momentum has a z-component only when it moves in the x-y plane.

Step by step solution

01

Understand angular momentum

The angular momentum \( \mathbf{L} \) of a particle is given by the cross product of its position vector \( \mathbf{r} \) and its momentum \( \mathbf{p} \). Mathematically, it is expressed as \( \mathbf{L} = \mathbf{r} \times \mathbf{p} \).
02

Write the position vector and momentum vector

For a particle with position vector \( \mathbf{r} = (x, y, z) \) and momentum vector \( \mathbf{p} = (p_x, p_y, p_z) \), we have the complete description of the necessary components to calculate angular momentum.
03

Calculate the cross product \( \mathbf{r} \times \mathbf{p} \)

The components of the cross product \( \mathbf{L} = \mathbf{r} \times \mathbf{p} \) can be calculated using the determinant:\[\mathbf{L} = \begin{vmatrix}\mathbf{i} & \mathbf{j} & \mathbf{k} \x & y & z \p_x & p_y & p_z\end{vmatrix}\]This results in:\[\mathbf{L} = (yp_z - zp_y)\mathbf{i} - (xp_z - zp_x)\mathbf{j} + (xp_y - yp_x)\mathbf{k}\]
04

Find components of angular momentum

The components of \( \mathbf{L} \) are:- \( L_x = yp_z - zp_y \)- \( L_y = zp_x - xp_z \)- \( L_z = xp_y - yp_x \)
05

Analyze if motion is in the x-y plane

If the particle moves only in the \(x-y\) plane, then \(z = 0\) and the momentum component \(p_z = 0\). Substitute these into the equations for \(L_x, L_y,\) and \(L_z\).
06

Simplify for motion in the x-y plane

Substitute \(z = 0\) and \(p_z = 0\):- \( L_x = y \cdot 0 - 0 \cdot p_y = 0 \)- \( L_y = 0 \cdot p_x - x \cdot 0 = 0 \)- \( L_z = xp_y - yp_x \)Thus, the angular momentum has no \(x\)- or \(y\)-components, only a \(z\)-component.

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Key Concepts

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

Cross Product
The cross product is a fundamental operation in vector mathematics, particularly useful when dealing with three-dimensional space. It takes two vectors and produces a third vector that is perpendicular to both of the original vectors.
The magnitude of this resulting vector is given by the product of the magnitudes of the original vectors and the sine of the angle between them. This basis of calculation is represented as:
  • \( \| \mathbf{L} \| = \| \mathbf{r} \| \| \mathbf{p} \| \sin(\theta) \ \)
Where \( \theta \) is the angle between \( \mathbf{r} \) and \( \mathbf{p} \).
The direction of the vector resulting from a cross product is determined by the right-hand rule. In the context of angular momentum, this results in a vector that points in a direction that describes the axis of rotation, with the right-hand rule offering a quick way to determine this direction.
Position Vector
In the study of physics, particularly when discussing motion and forces, the position vector \( \mathbf{r} \) is a key entity. It essentially provides a way to describe the location of a point in space, usually relative to some origin. The position vector can be written in terms of its component form as \( \mathbf{r} = (x, y, z) \).
Each component \( x, y, \) and \( z \) represents the position of the particle along each respective axis.
The utility of a position vector becomes clear when determining the angular momentum of a particle. Its ability to offer spatial orientation aids in calculating the cross product with the momentum vector, hence allowing us to find the angular momentum.
Momentum Vector
A momentum vector \( \mathbf{p} \) is fundamental in describing the motion of particles. It combines the mass of an object and its velocity into a single vectorial representation: \( \mathbf{p} = m \cdot \mathbf{v} \). For a three-dimensional space, it is broken down into components as \( \mathbf{p} = (p_x, p_y, p_z) \).

Each component aligns with one of the axes in a Cartesian coordinate system and represents the momentum in that specific direction. Essentially, it indicates how much motion the particle has along the given directions.
  • \( p_x = m \cdot v_x \)
  • \( p_y = m \cdot v_y \)
  • \( p_z = m \cdot v_z \)
Understanding the momentum vector is crucial for determining how a particle moves and interacts with forces, which includes its contribution to the angular momentum when calculated with its position vector.
Components of Angular Momentum
The angular momentum \( \mathbf{L} \) of a particle describes its rotational motion about an origin. It is computed as the cross product of the position vector \( \mathbf{r} \) and the momentum vector \( \mathbf{p} \), given as \( \mathbf{L} = \mathbf{r} \times \mathbf{p} \).
When expanded, this yields components:
  • \( L_x = yp_z - zp_y \)
  • \( L_y = zp_x - xp_z \)
  • \( L_z = xp_y - yp_x \)
These components correspond to the rotation about the \( x, y, \) and \( z \) axes respectively.
When analyzing specific conditions, like a particle moving only in the \( x-y \) plane, some simplifications occur. With \( z = 0 \) and \( p_z = 0 \), the angular momentum simplifies to only having a \( z \)-component, demonstrating the unique scenarios that can occur depending on motion parameters.

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