Question

If we are given \(r, p\), and \(\theta\), we can calculate the angular momentum of a particle from Eq. 10-2. Sometimes, however, we are given the components \((x, y, z)\) of \(\overrightarrow{\mathbf{r}}\) and \(\left(v_{x}, v_{y}, v_{z}\right)\) of \(\overrightarrow{\mathbf{v}}\) instead. ( \(a\) ) Show that the components of \(\overrightarrow{\mathbf{1}}\) along the \(x, y\), and \(z\) axes are then given by $$ \begin{aligned} &l_{x}=m\left(y v_{z}-z v_{y}\right) \\ &l_{y}=m\left(z v_{x}-x v_{z}\right) \\ &l_{z}=m\left(x v_{y}-y v_{x}\right) \end{aligned} $$ (b) Show that if the particle moves only in the \(x y\) plane, the resultant angular momentum vector has only a \(z\) component.

Short answer

The components of the angular momentum for a particle along the \(x, y\), and \(z\) axes are given by \(l_x=m(yv_z-zv_y)\), \(l_y=m(zv_x-xv_z)\), \(l_z=m(xv_y-yv_x)\). If the particle moves only in the \(x - y\) plane, the angular momentum vector has only a \(z\) component.

Step-by-step solution

Formula for Angular Momentum

The angular momentum vector of a particle is defined by \( \overrightarrow{L} = m \overrightarrow{r} \times \overrightarrow{v} \), where \(m\) is the mass, \(\overrightarrow{r}\) is the position vector, and \(\overrightarrow{v}\) is the velocity vector.

Angular Momentum Components

We can calculate the vector components of angular momentum using the formula for the cross product: \[ \overrightarrow{L} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ x & y & z \ v_x & v_y & v_z \end{vmatrix} \] which extends to: \[ \begin{aligned} l_x=m(yv_z-zv_y),\ l_y=m(zv_x-xv_z),\ l_z=m(xv_y-yv_x). \end{aligned} \]

Particle Movement in the xy Plane

If the particle only moves in the xy plane, this implies \(z = 0\) and \(v_z = 0\). In this case, the \(x\) and \(y\) components of the angular momentum (\(l_x, l_y\)) become 0, and only the \(z\) component remains. This can be seen from the expressions for \(l_x, l_y, l_z\) derived in Step 2.

Key concepts

Cross Product in Physics

Understanding the cross product is essential when dealing with rotational motion and angular momentum in physics. The cross product, also known as the vector product, is a mathematical operation that applies to two vectors in three-dimensional space. The result of this operation is another vector, which is perpendicular to the plane formed by the original vectors. The magnitude of the resultant vector indicates the area of the parallelogram with sides equal to the magnitudes of the original vectors.

In physics, the cross product becomes highly relevant when calculating the angular momentum \( \overrightarrow{L} \) or the torque \( \overrightarrow{\tau} \) because both of these involve a vector perpendicular to the motion's plane. For angular momentum, we use the cross product of the position vector \( \overrightarrow{r} \) and the velocity vector \( \overrightarrow{v} \) with the relation \( \overrightarrow{L} = m \overrightarrow{r} \times \overrightarrow{v} \). Calculating this cross product as shown in the formulas gives us the components of the angular momentum vector.

Grasping the cross product enhances comprehension of how rotational effects arise, how the direction of rotation is determined, and why certain movements impact the system's dynamics in the way they do.

Angular Momentum Components

The components of the angular momentum vector are a significant aspect of its overall understanding. As previously described, the angular momentum \( \overrightarrow{L} \) is the cross product of the position vector \( \overrightarrow{r} \) and the velocity vector \( \overrightarrow{v} \) multiplied by the mass \(m\). In component form, this leads to three separate equations, one for each axis in Cartesian coordinates:

\[ l_x = m(yv_z - zv_y), \]
\[ l_y = m(zv_x - xv_z), \]
\[ l_z = m(xv_y - yv_x). \]
These equations describe how each component of angular momentum is affected by the respective positions and velocities along the other two axes. For example, the \(x\)-component \(l_x\) depends on the position and velocity in the \(y\) and \(z\) directions, but not on the \(x\) position or velocity.

Understanding the directional characteristics of these components helps visualize how a particle's motion in space contributes to its overall angular momentum and the distribution of that momentum along different axes in space.

Particle Motion in 2D Plane

When dealing with particle motion in a 2D plane, such as the \(xy\) plane, we make a simplifying assumption by considering only two dimensions. This simplification impacts the angular momentum of the particle as well.

If a particle moves solely within the \(xy\) plane, meaning there is no movement along the \(z\)-axis, the \(z\) component of the position \(r\) and the velocity \(v\) will both be zero. According to the components of angular momentum, this results in the \(x\) and \(y\) components of the angular momentum being nullified, as they both include terms with \(z\) or \(v_z\):

\[ l_x = m(yv_z - zv_y) = 0, \] because \(z\) and \(v_z\) are zero,
\[ l_y = m(zv_x - xv_z) = 0, \] for the same reason.

Therefore, only the \(z\)-component \(l_z = m(xv_y - yv_x)\) remains, giving the particle a single angular momentum vector that points along the \(z\)-axis, perpendicular to the \(xy\) plane of motion. This principle is crucial in 2D motion analysis and helps predict rotational outcomes and stability in planar systems such as a frisbee in flight or a ceiling fan's blades.