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Chapter 11: Problem 37

(II) A particle is at the position \((x, y, z)=(1.0,20,3.0) \mathrm{m} .\) It is travcling with a vector velocity \((-5.0,+2.8,-3.1) \mathrm{m} / \mathrm{s}\) . Its mass is 3.8 \(\mathrm{kg}\) . What is its vector angular momentum about the origin?

Short Answer

Expert verified
The vector angular momentum is \((-267.52, -45.22, 390.64) \text{ kg m}^2/\text{s}\).

Step by step solution

01

Identify the given vectors

The position vector \( \mathbf{r} \) of the particle is \( (1.0, 20.0, 3.0) \) meters, and the velocity vector \( \mathbf{v} \) is \( (-5.0, 2.8, -3.1) \) meters per second.
02

Write the formula for angular momentum

The vector angular momentum \( \mathbf{L} \) of a particle about the origin is given by the cross product \( \mathbf{L} = \mathbf{r} \times \mathbf{p} \), where \( \mathbf{p} \) is the linear momentum vector \( \mathbf{p} = m \mathbf{v} \).
03

Calculate the momentum vector

The momentum vector \( \mathbf{p} \) can be calculated by multiplying the mass \( m = 3.8 \) kg with the velocity vector \( \mathbf{v} = (-5.0, 2.8, -3.1) \): \[ \mathbf{p} = (3.8 \times -5.0, 3.8 \times 2.8, 3.8 \times -3.1) = (-19.0, 10.64, -11.78) \].
04

Calculate the angular momentum cross product

The angular momentum \( \mathbf{L} = \mathbf{r} \times \mathbf{p} \) requires computing the cross product of \( \mathbf{r} = (1.0, 20.0, 3.0) \) and \( \mathbf{p} = (-19.0, 10.64, -11.78) \). The cross product is computed as: \[ \mathbf{L} = \left| \begin{matrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ 1.0 & 20.0 & 3.0 \ -19.0 & 10.64 & -11.78 \end{matrix} \right| \].
05

Solve the determinant

The determinant can be calculated as follows: \( \mathbf{L} = \mathbf{i}(20.0 \times -11.78 - 3.0 \times 10.64) - \mathbf{j}(1.0 \times -11.78 - 3.0 \times -19.0) + \mathbf{k}(1.0 \times 10.64 - 20.0 \times -19.0) \).
06

Compute the components and simplify

Compute each component: \(-235.6 - 31.92 = -267.52 \) for \( \mathbf{i} \), \( -11.78 + 57.0 = 45.22 \) for \( \mathbf{j} \), and \( 10.64 + 380.0 = 390.64 \) for \( \mathbf{k} \). Thus, \( \mathbf{L} = (-267.52, -45.22, 390.64) \).

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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 crucial mathematical operation when dealing with vectors. It is mainly used in physics to find quantities like torque and angular momentum. Given two vectors \( \mathbf{A} \) and \( \mathbf{B} \), the cross product (denoted by \( \mathbf{A} \times \mathbf{B} \)) results in a vector that is perpendicular to both \( \mathbf{A} \) and \( \mathbf{B} \).
  • This operation only works in three-dimensional space.
  • The resulting vector's direction is determined by the right-hand rule. Point your index finger in the direction of \( \mathbf{A} \) and your middle finger in the direction of \( \mathbf{B} \), and your thumb will point in the direction of \( \mathbf{A} \times \mathbf{B} \).
  • The magnitude of the cross product is equal to the area of the parallelogram that the vectors span.
For the given problem, the cross product \( \mathbf{r} \times \mathbf{p} \) was calculated using the determinant method. This involves placing the unit vectors \( \mathbf{i}, \mathbf{j}, \mathbf{k} \) along with the components of both vectors into a matrix, and solving for the determinant to find each component of the resulting vector.
Linear Momentum
Linear momentum is a vector quantity defined as the product of an object's mass and its velocity. Represented as \( \mathbf{p} = m \mathbf{v} \), where \( m \) stands for mass and \( \mathbf{v} \) stands for velocity.
  • It describes the quantity of motion an object possesses.
  • A larger mass or higher speed results in greater linear momentum.
  • Linear momentum is a fundamental concept in physics due to its conservation in isolated systems. This means that if there are no external forces acting on a system, the total momentum within remains constant.
In the exercise, the linear momentum of the particle was calculated by multiplying its mass (3.8 kg) and its velocity vector \((-5.0, 2.8, -3.1)\, \text{m/s} \), yielding the momentum vector \(( -19.0, 10.64, -11.78 )\). This momentum vector is essential for calculating the angular momentum using the cross product.
Vector Calculations
Vectors are mathematical entities with both a magnitude and direction, making them fundamental in physics when describing motion and force.
  • Vectors can be represented in component form, such as \((x, y, z)\), to simplify calculations in three-dimensional space.
  • Key operations with vectors include addition, subtraction, and multiplication (dot product and cross product).
  • Understanding vectors involves learning about unit vectors, which help ensure calculations are dimensionally accurate.
In this exercise, vector calculations were used to determine the particle's angular momentum. By knowing the position vector \( \mathbf{r} \) and the computed momentum vector \( \mathbf{p} \), the cross product \( \mathbf{L} = \mathbf{r} \times \mathbf{p} \) was carried out, resulting in the angular momentum vector \((-267.52, 45.22, 390.64)\). These calculations highlight the importance of mastering vector mathematics for solving complex physics problems.

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Most popular questions from this chapter

The time-dependent position of a point object which moves counterclockwise along the circumference of a circle (radius \(R\) ) in the \(x y\) plane with constant speed \(v\) is given by $$ \overrightarrow{\mathbf{r}}=\hat{\mathbf{i}} R \cos \omega t+\hat{\mathbf{j}} R \sin \omega t $$ where the constant \(\omega=v / R .\) Determine the velocity \(\overrightarrow{\mathbf{v}}\) and angular velocity \(\overrightarrow{\boldsymbol{\omega}}\) of this object and then show that these three vectors obey the relation \(\overrightarrow{\mathbf{v}}=\overrightarrow{\boldsymbol{\omega}} \times \overrightarrow{\mathbf{r}}\).

(II) (a) Show that the cross product of two vectors, $$ \vec{\mathbf{A}}=A_{x} \hat{\mathbf{i}}+A_{y} \hat{\mathbf{j}}+A_{z} \hat{\mathbf{k}}, \text { and } \vec{\mathbf{B}}=B_{x} \hat{\mathbf{i}}+B_{y} \hat{\mathbf{j}}+B_{z} \hat{\mathbf{k}} $$ $$ \begin{aligned} \vec{\mathbf{A}} \times \vec{\mathbf{B}}=&\left(A_{y} B_{z}-A_{z} B_{y}\right) \hat{\mathbf{i}}+\left(A_{z} B_{x}-A_{x} B_{z}\right) \hat{\mathbf{j}} \\ &+\left(A_{x} B_{y}-A_{y} B_{x}\right) \hat{\mathbf{k}} \end{aligned} $$ (b) Then show that the cross product can be written $$ \vec{\mathbf{A}} \times \vec{\mathbf{B}}=\left| \begin{array}{ccc}{\hat{\mathbf{i}}} & {\hat{\mathbf{j}}} & {\hat{\mathbf{k}}} \\\ {A_{x}} & {A_{y}} & {A_{z}} \\ {B_{x}} & {B_{y}} & {B_{z}}\end{array}\right| $$ where we use the rules for evaluating a determinant. (Note, however, that this is not really a determinant, but a memory aid.)

A particle of mass \(1.00 \mathrm{~kg}\) is moving with velocity \(\overrightarrow{\mathbf{v}}=(7.0 \hat{\mathbf{i}}+6.0 \hat{\mathbf{j}}) \mathrm{m} / \mathrm{s} . \quad(a)\) Find the angular momentum \(\overrightarrow{\mathbf{L}}\) relative to the origin when the particle is at \(\overrightarrow{\mathbf{r}}=(2.0 \hat{\mathbf{j}}+4.0 \hat{\mathbf{k}}) \mathrm{m} .(b)\) At position \(\overrightarrow{\mathbf{r}}\) a force of \(\overrightarrow{\mathbf{F}}=4.0 \mathrm{Ni} \hat{\mathbf{i}}\) is applied to the particle. Find the torque relative to the origin.

A thin rod of length \(\ell\) and mass \(M\) rotates about a vertical axis through its center with angular velocity \(\omega .\) The rod makes an angle \(\phi\) with the rotation axis. Determine the magnitude and direction of \(\overrightarrow{\mathbf{L}}\).

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?
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