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

Short Answer

Expert verified
\(L_x = y p_z - z p_y\) ; \(L_y = z p_x - x p_z\) ; \(L_z = x p_y - y p_x\)

Step by step solution

01

Identify the cross product formula

Angular momentum \( \overrightarrow{L} \) can be calculated using the cross product of position vector \( \overrightarrow{r} = x \hat{\mathbf{i}} + y \hat{\mathbf{j}} + z \hat{\mathbf{k}} \) and momentum vector \( \overrightarrow{p} = p_x \hat{\mathbf{i}} + p_y \hat{\mathbf{j}} + p_z \hat{\mathbf{k}} \). The formula is \( \overrightarrow{L} = \overrightarrow{r} \times \overrightarrow{p} \).
02

Compute the x-component of angular momentum

Using the determinant method for computing cross products, the x-component \( L_x \) is given by eliminating the first row and first column of the determinant, resulting in:\[L_x = (y \cdot p_z - z \cdot p_y)\]
03

Compute the y-component of angular momentum

The y-component \( L_y \) is obtained by eliminating the first row and second column of the determinant, resulting in:\[L_y = (z \cdot p_x - x \cdot p_z)\]
04

Compute the z-component of angular momentum

The z-component \( L_z \) is found by eliminating the first row and third column of the determinant, which gives:\[L_z = (x \cdot p_y - y \cdot p_x)\]
05

Compile the full angular momentum vector

The resulting angular momentum vector \( \overrightarrow{L} \) is:\[\overrightarrow{L} = (y \cdot p_z - z \cdot p_y) \hat{\mathbf{i}} + (z \cdot p_x - x \cdot p_z) \hat{\mathbf{j}} + (x \cdot p_y - y \cdot p_x) \hat{\mathbf{k}}\]

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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 mathematical operation used to find a vector that is perpendicular to two given vectors. In three-dimensional space, the cross product of vectors \( \overrightarrow{a} \) and \( \overrightarrow{b} \) is denoted as \( \overrightarrow{a} \times \overrightarrow{b} \).
This operation is essential in calculating quantities like torque and angular momentum.

In the context of angular momentum, given a position vector \( \overrightarrow{r} \) and a momentum vector \( \overrightarrow{p} \), the angular momentum \( \overrightarrow{L} \) can be determined using the cross product. The direction of the resulting angular momentum vector \( \overrightarrow{L} \) will be perpendicular to both \( \overrightarrow{r} \) and \( \overrightarrow{p} \).

The formula for the cross product in vector components is as follows:
  • \( \overrightarrow{L} = \overrightarrow{r} \times \overrightarrow{p} \)
This results in a new vector that represents angular momentum when applied to the context of particle dynamics.
Determinant Method
The determinant method is a systematic way to calculate the cross product of two vectors.
This method uses a 3x3 matrix to organize the unit vectors and the components of the vectors involved.

For the position vector \( \overrightarrow{r} = x \hat{\mathbf{i}} + y \hat{\mathbf{j}} + z \hat{\mathbf{k}} \) and momentum vector \( \overrightarrow{p} = p_x \hat{\mathbf{i}} + p_y \hat{\mathbf{j}} + p_z \hat{\mathbf{k}} \), the determinant form is set up as follows:

\[\begin{vmatrix} \hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \x & y & z \p_x & p_y & p_z\end{vmatrix}\]
This matrix is expanded to calculate each component of the resulting cross product. The structure involves:
  • Ignoring one of the unit vector columns while calculating the minor determinant for each.
  • The first column omitted results in the \( x \)-component;
  • The second column omitted gives the \( y \)-component;
  • The third results in the \( z \)-component.
This method simplifies the process of finding the angular momentum components in physics.
Momentum Vector
A momentum vector represents the motion of a particle in terms of its mass and velocity.
Defined as \( \overrightarrow{p} = p_x \hat{\mathbf{i}} + p_y \hat{\mathbf{j}} + p_z \hat{\mathbf{k}} \), it indicates how much motion a particle has and in which direction.

The components of the momentum vector \( (p_x, p_y, p_z) \) signify:
  • \( p_x \): motion across the x-axis
  • \( p_y \): motion across the y-axis
  • \( p_z \): motion across the z-axis
When calculating the angular momentum, the values from the momentum vector interact with those of the position vector in the determinant method.
This interaction through the cross product ultimately describes the rotation or angular motion of a particle in space.
Position Vector
The position vector is essential in determining the location of a particle in three-dimensional space.
It is expressed as \( \overrightarrow{r} = x \hat{\mathbf{i}} + y \hat{\mathbf{j}} + z \hat{\mathbf{k}} \), where \( x \), \( y \), and \( z \) describe the particle's coordinates.

Each component represents a specific axis:
  • \( x \): horizontal position along the x-axis
  • \( y \): vertical position along the y-axis
  • \( z \): depth position along the z-axis
The position vector acts as one part of the cross product calculation with the momentum vector.
The interaction between the position vector and the momentum vector through the cross product determines the complete angular momentum vector of a particle.

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Most popular questions from this chapter

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

(1) If vector \(\vec{\mathbf{A}}\) points along the negative \(x\) axis and vector \(\vec{\mathbf{B}}\) along the positive \(z\) axis, what is the direction of \((a) \vec{\mathbf{A}} \times \vec{\mathbf{B}}\) and \((b) \mathbf{B} \times \vec{\mathbf{A}} ?(c)\) What is the magnitude of \(\overline{\mathbf{A}} \times \vec{\mathbf{B}}\) and \(\vec{\mathbf{B}} \times \vec{\mathbf{A}} ?\)

(II) \(\mathrm{A} 220\) -g top spinning at 15 \(\mathrm{rev} / \mathrm{s}\) makes an angle of \(25^{\circ}\) to the vertical and precesses at a rate of 1.00 rev per 6.5 \(\mathrm{s}\) . If its \(\mathrm{CM}\) is 3.5 \(\mathrm{cm}\) from its tip along its symmetry axis, what is the moment of inertia of the top?

A toy gyroscope consists of a 170 -g disk with a radius of \(5.5 \mathrm{~cm}\) mounted at the center of a thin axle \(21 \mathrm{~cm}\) long (Fig. 11-41). The gyroscope spins at \(45 \mathrm{rev} / \mathrm{s}\). One end of its axle rests on a stand and the other end precesses horizontally about the stand. (a) How long does it take the gyroscope to precess once around? \((b)\) If all the dimensions of the gyroscope were doubled (radius \(=11 \mathrm{~cm},\) axle \(=42 \mathrm{~cm})\) how long would it take to precess once?

If there were a great migration of people toward the Earth's equator, would the length of the day \((a)\) get longer because of conservation of angular momentum; (b) get shorter because of conservation of angular momentum; (c) get shorter because of conservation of energy; \((d)\) get longer because of conservation of energy; or (e) remain unaffected?

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