To tackle this question, we need to delve into the concept of angular momentum and how it can be expressed in different forms. Angular momentum is a vector quantity that represents the rotational motion of a particle about a point. It can be calculated using the position vector \( \mathbf{r} \) and the linear momentum \( \mathbf{p} \). Let's break this down step by step.
Understanding Angular Momentum
The angular momentum \( \mathbf{L} \) of a particle is defined as:
\( \mathbf{L} = \mathbf{r} \times \mathbf{p} \)
Where \( \mathbf{r} \) is the position vector from the point of rotation to the particle, and \( \mathbf{p} \) is the linear momentum given by \( \mathbf{p} = m \mathbf{v} \), with \( m \) being the mass of the particle and \( \mathbf{v} \) its velocity vector.
Components of Angular Momentum
When we express \( \mathbf{r} \) and \( \mathbf{v} \) in terms of their components, we have:
- Position vector: \( \mathbf{r} = (x, y, z) \)
- Velocity vector: \( \mathbf{v} = (v_x, v_y, v_z) \)
Substituting these into the angular momentum formula, we can calculate the components of \( \mathbf{L} \) along the x, y, and z axes.
Calculating the Components
The cross product \( \mathbf{r} \times \mathbf{p} \) can be expanded using the determinant of a matrix:
\( \mathbf{L} = \begin{vmatrix}
\mathbf{i} & \mathbf{j} & \mathbf{k} \\
x & y & z \\
mv_x & mv_y & mv_z
\end{vmatrix} \)
Calculating this determinant gives us the components:
- For \( L_x \):
\( L_x = m(y v_z - z v_y) \)
- For \( L_y \):
\( L_y = m(z v_x - x v_z) \)
- For \( L_z \):
\( L_z = m(x v_y - y v_x) \)
This confirms the expressions for the components of angular momentum along the x, y, and z axes as stated in your question.
Angular Momentum in the XY Plane
Now, let's consider the scenario where the particle moves only in the xy plane. In this case, the z-component of the velocity \( v_z \) is zero. Thus, we can simplify our expressions:
- Since \( v_z = 0 \), the equations for \( L_x \) and \( L_y \) become:
\( L_x = m(y \cdot 0 - z v_y) = -m z v_y \)
\( L_y = m(z v_x - x \cdot 0) = m z v_x \)
- For \( L_z \):
\( L_z = m(x v_y - y v_x) \)
In this case, \( L_x \) and \( L_y \) will depend on the z-coordinate of the particle, but if the particle is constrained to the xy plane, the z-component of angular momentum \( L_z \) is the only non-zero component. Thus, the resultant angular momentum vector will point in the z-direction.
Summary
In summary, we derived the components of angular momentum from the position and momentum vectors and demonstrated that when a particle moves solely in the xy plane, its angular momentum vector is directed along the z-axis. This highlights the importance of understanding the spatial constraints on motion when analyzing angular momentum.
