Angular Momentum for Continuous Mass Distribution

Angular momentum is a fundamental concept in physics, particularly in the study of rotational motion. It is the rotational equivalent of linear momentum and plays a crucial role in understanding the dynamics of systems with continuous mass distribution, such as solid bodies, fluids, or gases.

Definition of Angular Momentum

For a single particle, the angular momentum $\vec{L}$ about a point is defined as the cross product of the position vector $\vec{r}$ (from the point to the particle) and the linear momentum $\vec{p}$ of the particle:

$$ \vec{L} = \vec{r} \times \vec{p} $$

For a particle of mass $m$ moving with velocity $\vec{v}$, the linear momentum is $\vec{p} = m\vec{v}$, so the angular momentum becomes:

$$ \vec{L} = \vec{r} \times m\vec{v} $$

Angular Momentum for Continuous Mass Distribution

When dealing with a continuous mass distribution, such as a solid object, we must integrate over the entire volume of the object to find the total angular momentum. The angular momentum of an element of mass $\delta m$ is:

$$ \delta \vec{L} = \vec{r} \times \delta \vec{p} $$

where $\delta \vec{p} = \delta m \vec{v}$ is the linear momentum of the mass element. To find the total angular momentum $\vec{L}$, we integrate over the entire volume $V$ of the mass distribution:

$$ \vec{L} = \int_V \vec{r} \times \delta m \vec{v} $$

If the mass distribution has a density $\rho(\vec{r})$, then $\delta m = \rho(\vec{r}) dV$, where $dV$ is the differential volume element. The integral becomes:

$$ \vec{L} = \int_V \vec{r} \times \rho(\vec{r}) \vec{v} dV $$

Moment of Inertia

The moment of inertia $I$ plays a significant role in the angular momentum of continuous mass distributions. It is a measure of an object's resistance to changes in its rotational motion and depends on the mass distribution relative to the axis of rotation.

For a continuous mass distribution, the moment of inertia about an axis is defined as:

$$ I = \int_V r_\perp^2 \rho(\vec{r}) dV $$

where $r_\perp$ is the perpendicular distance from the axis of rotation to the element of mass.

Angular Momentum for Rigid Body Rotation

For a rigid body rotating about a fixed axis with angular velocity $\vec{\omega}$, every point in the body has the same angular velocity. The angular momentum of the body is then given by:

$$ \vec{L} = I \vec{\omega} $$

where $I$ is the moment of inertia of the body about the axis of rotation.

Table: Key Differences and Important Points

Aspect	Point Mass	Continuous Mass Distribution
Definition	$\vec{L} = \vec{r} \times m\vec{v}$	$\vec{L} = \int_V \vec{r} \times \rho(\vec{r}) \vec{v} dV$
Moment of Inertia	Not applicable	$I = \int_V r_\perp^2 \rho(\vec{r}) dV$
Angular Velocity	Same as the particle	Same for all points in a rigid body ($\vec{\omega}$)
Calculation	Direct product	Integration over volume

Examples

Example 1: Angular Momentum of a Uniform Rod

Consider a uniform rod of length $L$ and mass $M$ rotating about one end with angular velocity $\omega$. The rod has a linear mass density $\lambda = \frac{M}{L}$. The moment of inertia of the rod about the end is:

$$ I = \int_0^L x^2 \lambda dx = \frac{1}{3}ML^2 $$

The angular momentum of the rod is:

$$ \vec{L} = I \vec{\omega} = \frac{1}{3}ML^2 \vec{\omega} $$

Example 2: Angular Momentum of a Solid Sphere

A solid sphere of radius $R$, mass $M$, and uniform density rotating about an axis through its center with angular velocity $\omega$ has a moment of inertia:

$$ I = \frac{2}{5}MR^2 $$

The angular momentum of the sphere is:

$$ \vec{L} = I \vec{\omega} = \frac{2}{5}MR^2 \vec{\omega} $$

Understanding the angular momentum for continuous mass distributions is essential for analyzing the rotational motion of various physical systems, from simple mechanical devices to celestial bodies. It is a key concept in classical mechanics and has applications in many areas of physics and engineering.