/*! 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 34 (1) What are the \(x, y,\) and \... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

(1) What are the \(x, y,\) and \(z\) components of the angular momentum of a particle located at \(\vec{\mathbf{r}}=x \hat{\mathbf{i}}+y \hat{\mathbf{j}}+z \hat{\mathbf{k}}\) which has momentum \(\mathbf{p}=p_{x} \mathbf{i}+p_{y} \mathbf{j}+p_{z} \mathbf{k} ?\)

Short Answer

Expert verified
The components are \( L_x = y p_z - z p_y \), \( L_y = z p_x - x p_z \), and \( L_z = x p_y - y p_x \).

Step by step solution

01

Understanding Angular Momentum

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

Express Vectors in Component Form

Given \( \vec{r} = x \hat{i} + y \hat{j} + z \hat{k} \) and \( \vec{p} = p_x \hat{i} + p_y \hat{j} + p_z \hat{k} \), we need to find each component of the angular momentum using the cross product formula.
03

Cross Product Formula

The formula for the cross product \( \vec{A} = \vec{B} \times \vec{C} \) is:\[\vec{A} = \begin{vmatrix}\hat{i} & \hat{j} & \hat{k} \x_1 & y_1 & z_1 \x_2 & y_2 & z_2 \end{vmatrix}\]which translates to \( \vec{A} = ((y_1 z_2 - z_1 y_2) \hat{i}, (z_1 x_2 - x_1 z_2) \hat{j}, (x_1 y_2 - y_1 x_2) \hat{k}) \).
04

Calculate the x-component

Using the formula above, the \( x \)-component of the angular momentum, \( L_x \), is given by:\[L_x = y p_z - z p_y\]
05

Calculate the y-component

The \( y \)-component of the angular momentum, \( L_y \), is given by:\[L_y = z p_x - x p_z\]
06

Calculate the z-component

The \( z \)-component of the angular momentum, \( L_z \), is given by:\[L_z = x p_y - y p_x\]
07

Final Components of Angular Momentum

The components of the angular momentum vector \( \vec{L} \) are:- \( L_x = y p_z - z p_y \)- \( L_y = z p_x - x p_z \)- \( L_z = x p_y - y p_x \)

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Cross Product
The cross product is a crucial operation when dealing with vectors, especially in physics. It combines two vectors to create a third vector, which is perpendicular to the plane of the first two.

For example, when finding angular momentum, we use the cross product formula, \( \vec{L} = \vec{r} \times \vec{p} \). This calculates the angular momentum \( \vec{L} \), by taking the positional vector \( \vec{r} \) and the momentum vector \( \vec{p} \) of a particle.

The cross product is not like the typical multiplication of numbers. Instead, it involves the sine of the angle between the vectors and the magnitude of the resulting vector depends on this angle. If \( \vec{A} \) and \( \vec{B} \) are two vectors, then \( \vec{A} \times \vec{B} \) produces a vector perpendicular to both. You can compute it using the determinant of a matrix formed by the components of the vectors.

Thus, understanding the cross product is essential when dealing with rotational systems and is fundamental for calculating vector-related physical quantities like torque and moment of inertia, apart from angular momentum.
Position Vector
The position vector \( \vec{r} \) represents the location of an object in a coordinate system. It originates from the origin of the system and points to the location of the object.

In a 3-dimensional space, the position vector can be expressed in its component form as \( \vec{r} = x \hat{i} + y \hat{j} + z \hat{k} \), where:
  • \( x, y, z \) are the respective coordinates of the object.
  • \( \hat{i}, \hat{j}, \hat{k} \) are the unit vectors in the direction of the x, y, and z-axes respectively.

The position vector is essential in determining the physical location of particles and points in space. Knowing \( \vec{r} \) allows us to calculate other vector quantities such as displacement, velocity, and of course, angular momentum when combined with the momentum vector.
Momentum Vector
The momentum vector of a particle, denoted as \( \vec{p} \), is a physical quantity that represents the motion of that particle. It is calculated by the mass of the particle multiplied by its velocity. In a vector form, for three dimensions, it is expressed as:
  • \( \vec{p} = p_x \hat{i} + p_y \hat{j} + p_z \hat{k} \)
  • \( p_x, p_y, p_z \) are the momentum components along the x, y, and z directions.
  • \( \hat{i}, \hat{j}, \hat{k} \) are the unit vectors along their respective axes.

The momentum vector is crucial in the study of dynamics as it holds both magnitude and direction, giving us insights into how a particle moves through space. In conjunction with the position vector, it helps us determine the angular momentum and understand rotational motion practically.
Vector Components
Vector components are the individual parts of a vector that show how much of the vector lies along each axis of a coordinate system. In a three-dimensional space, any vector can be broken down into its x, y, and z components.

When computing angular momentum, understanding vector components is key. The expression for the components of angular momentum, for instance, involves the separate components of both the position and momentum vectors:
  • \( L_x = y p_z - z p_y \)
  • \( L_y = z p_x - x p_z \)
  • \( L_z = x p_y - y p_x \)

This reveals how each component of a vector contributes to the total vector sum. The breakup into components makes it easier to work with vectors using basic algebraic operations separately on each axis, thus simplifying complex vector calculations. Understanding vector components is foundational for tackling a wide array of physics and engineering problems.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

(II) A potter's whecl is rotating around a vertical axis through its center at a frequency of 1.5 \(\mathrm{rev} / \mathrm{s}\) . The whecl can be considered a uniform disk of mass 5.0 \(\mathrm{kg}\) and diameter 0.40 \(\mathrm{m}\) . The potter then throws a 2.6 -kg chunk of clay, approximately shaped as a flat disk of radius \(8.0 \mathrm{cm},\) onto the center of the rotating whecl. What is the frequency of the whecl after the clay sticks to it?

(II) A person stands on a platform, initially at rest, that can rotate frecly without friction. The moment of inertia of the person plus the platform is \(I_{\mathrm{P}}\) . The person holds a spinning bicycle whecl with its axis horizontal. The whecl has moment of inertia $$I_{\mathrm{w}} \text { and angular velocity } \omega_{\mathrm{W}} $$ What will be the angular velocity $$ \omega_{\mathbf{Y}} $$ of the platform if bthe person moves the axis of the wheel so that it points (a) vertically upward, (b) at a \(60^{\circ}\) angle to the vertical, \((c)\) vertically downward? (d) What will \(\omega_{\mathrm{P}}\) be if the person reaches up and stops the whecl in part \((a) ?\)

(II) A uniform disk turns at 3.7 rev/s around a frictionless spindle. A nonrotating rod, of the same mass as the disk and length equal to the disk's diameter, is dropped onto the freely spinning disk, Fig. 31. They then turn together around the spindle with their centers superposed. What is the angular frequency in \(\mathrm{rev} / \mathrm{s}\) of the combination?

(III) Let \(\vec{\mathbf{A}}, \vec{\mathbf{B}},\) and \(\vec{\mathbf{C}}\) be three vectors, which for generality we assume do not all lie in the same plane. Show that $$ \vec{\mathbf{A}} \cdot(\vec{\mathbf{B}} \times \vec{\mathbf{C}})=\vec{\mathbf{B}} \cdot(\vec{\mathbf{C}} \times \vec{\mathbf{A}})=\vec{\mathbf{C}} \cdot(\vec{\mathbf{A}} \times \vec{\mathbf{B}}) $$

A spherical asteroid with radius \(r=123 \mathrm{m}\) and mass \(M=2.25 \times 10^{10} \mathrm{kg}\) rotates about an axis at four revolutions per day. A "tug" spaceship attaches itself to the pole (as defined by the axis of rotation) and fires its engine, applying a force \(F \quad\) tangentially to the asteroid's surface as shown in Fig. \(44 .\) If \(F=265 \mathrm{N},\) how long will it take the tug to rotate the asteroid's axis of rotation through an angle of \(10.0^{a}\) by this method?

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.