3.5 Angular Momentum

3.5 Angular Momentum

We have already seen that linear momentum plays an important role in translational motion of bodies. Similarly, another quantity known as angular momentum has important role in the study of rotational motion.

A particle is said to possess an angular momentum about a reference axis if it so moves that its angular position changes relative to that reference axis.

The angular momentum $\mathbf{L}$ of a particle of mass $m$ moving with velocity $\mathbf{v}$ and momentum $\mathbf{p}$ (Fig. 3.17) relative to the origin $O$ is defined as:

where $\mathbf{r}$ is the position vector of the particle at that instant relative to the origin $O$. Angular momentum is a vector quantity. Its magnitude is:

where $\theta$ is the angle between $\mathbf{r}$ and $\mathbf{p}$. The direction of $\mathbf{L}$ is perpendicular to the plane formed by $\mathbf{r}$ and $\mathbf{p}$ and its sense is given by the right hand rule of vector product. SI unit of angular momentum is $\mathrm{kg\,m^{2}\,s^{-1}}$ or $\mathrm{J\,s}$.

If the particle is moving in a circle of radius $r$ with uniform angular velocity $\omega$, then angle between $\mathbf{r}$ and tangential velocity is $90^{\circ}$. Hence:

But $v = r\omega$. Hence $L = mr^{2}\omega$

Now consider a symmetric rigid body rotating about a fixed axis through the centre of mass as shown in Fig. 3.18. Each particle of the rigid body rotates about the same axis in a circle with an angular velocity $\omega$. The magnitude of the angular momentum of the particle of mass $m_{i}$ is $m_{i}v_{i}r_{i}$ about the origin $O$. The direction of $\mathbf{L}$ is the same as that of $\omega$. Since $v_{i} = r_{i}\omega$, the angular momentum of the $i$th particle is $m_{i}r_{i}^{2}\omega$. Summing this over all particles gives the total angular momentum of the rigid body:

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

Example 3.5

The mass of Earth is $6.00 \times 10^{24}\,\mathrm{kg}$. The distance $r$ from Earth to the Sun is $1.50 \times 10^{11}\,\mathrm{m}$. As seen from the direction of the North Star, the Earth revolves counter-clockwise around the Sun. Determine the orbital angular momentum of the Earth about the Sun, assuming that it traverses a circular orbit about the Sun once a year ($3.16 \times 10^{7}\,\mathrm{s}$).

Solution

To find the Earth's orbital angular momentum, we must first know its orbital speed from the given data. When the Earth moves around a circle of radius $r$, it travels a distance of $2\pi r$ in one year. Its orbital speed $v_{0}$ is thus:

Orbital angular momentum of the Earth $= L_{0} = mv_{0}r$

The sign is positive because the revolution is counterclockwise.