Section9.6Angular Momentum

Angular momentum is a fundamental property of a rotating body. In this section we will develop the concept systematically.

Subsection9.6.1Angular Momentum of a Particle

Let $\vec r$ be the position of a particle of mass $m$ and velocity $\vec v\text{.}$ The momentum of the particle is $\vec p = m\vec v$ and the angular momentum, to be denoted by $\vec l \text{,}$ is defined by the cross product of $\vec r$ and momnetum $\vec p \text{.}$

$$\vec l = \vec r \times \vec p.\tag{9.6.1}$$

From this definition, magnitude of angular momentum will be

$$l = r\,p\,\sin\,\phi,\tag{9.6.2}$$

where $\phi$ is the angle between vectors $\vec r$ and $\vec p\text{.}$

The direction of angular momentum is obtained by applying the right hand rule of cross product on vectors $\vec r$ and $\vec p\text{.}$ First to note is that vector $\vec l$ will be perpendicular to the plane formed by vectors $\vec r$ and $\vec p\text{.}$ To use right hand rule, place $\vec r$ along the thumb of your right hand and $\vec p$ along any of the other fingers. Then, $\vec l$ will be coming out of the palm.

Subsection9.6.2Angular Momentum of a Particle in a Circular Motion

Consider a particle of mass $m$ moving in a circle of radius $R$ as shown in Figure 9.6.3. For a particle moving in a circle, the velocity is in the direction of the tangent to the circle.

Since the tangent of a circle is perpendicular to the radial line from the center, the angle between vectors $\vec r$ and $\vec p$ is $90^{\circ}\text{.}$

\begin{equation*} \phi = 90^{\circ}. \end{equation*}

This gives the magnitude of angular momentum to be

$$l = m v R.\label{eq-angular-momentum-magnitude-mvR}\tag{9.6.3}$$

Subsection9.6.3A Particle in Circular Motion as Rotation

We can also think of the circular motion in Figure 9.6.3 as rotation about $z$-axis by quantifying the rotation by angle $\theta$ subtended by the radial line to the particle with positive $x$-axis. The distance $s$ covered on the arc of the circle will be related with angle covered $\theta$ by arc-angle relation.

\begin{equation*} s = R \theta. \end{equation*}

The rate of change in arc distance $s$ is the speed $v$ and rate of change of $\theta$ is the angular speed $\omega\text{.}$ Therefore,

\begin{equation*} v = R \omega. \end{equation*}

Substituting in Eq. (9.6.3) we get another useful expression for angular momentum.

$$l = m R^2 \omega.\tag{9.6.4}$$

The quantity $m R^2$ in this formula is called moment of inertia of the particle about the axis. It is denoted by letter $I \text{.}$

$$I\, (\text{of a particle at a distance }R \text{ from axis}) = m R^2.\tag{9.6.5}$$

In terms of moment of inertia and the angular velocity, the formula for the magnitude of the angular momentum for a particle moving in a circle takes a really simple form

$$l = I\,\omega.\tag{9.6.6}$$

If you want to be more complete in your description, you could include information about the axis of rotation. For $z$-axis, this equation would be

\begin{equation*} l_z = I_{zz}\,\omega_z\text{.} \end{equation*}

The two subscripts for moment of inertia will become clear when we discuss moments of inertia of more complicated systems.

A ball of mass $200\text{ grams}$ is moving in a circle of radius $25\text{ cm}$ with a uniform speed of $10\text{ m/s}\text{.}$ When observed from above the motion appears counterclockwise. Find the angular momentum about the center of the circle treating the ball as a point particle. Give both the magnitude and the direction of the angular momentum.

Hint

Use definition

$0.5\text{ kg.m}^2\text{/s}\text{,}$ up.

Solution

The magnitude is

\begin{align*} L \amp = m v R\\ \amp = 0.2\text{ kg}\times 10\text{ m/s} \times 0.25\text{ m} \\ \amp = 0.5\text{ kg.m}^2\text{/s}. \end{align*}

Using the right hand rule, we find that the direction of angular monentum vector is pointed up. The direction is shown in the figure.

The magnitude of the angular momentum of a steel ball of mass $400\text{ grams}$ moving in a circle in a horizontal plane of radius $50\text{ cm}$ is $3\text{ kg. m}^2\text{/s}$ about the center and the direction of the angular momentum vector is towards the ground.

Find the speed of the ball and the clockwise or counterclockwise sense of its motion in the circle as observed from above the circle. Treat the ball as a point particle.

Hint

Use definition

$15\text{ m/s}\text{,}$ clockwise.

Solution

Let $v$ be the speed of the ball. From the definition, we get $m v R = L\text{,}$ which gives

\begin{equation*} v = \dfrac{L}{m R } = \dfrac{3\text{ kg. m}^2\text{/s}}{ 0.4\text{ kg} \times 0.5\text{ m} } = 15\text{ m/s}. \end{equation*}

By the right-hand rule, we get the direction of the motion is clockwise since the direction of the angular momentum is down.

Subsection9.6.4Angular Momentum of a Collection of Particles

Since, angular momentum is a vector, to get the momentum of the collection, we just need to work out the momenta of each particle and then add them up vectorially. Say, we have $N$ particles in the collection, with positions $\vec r_1\text{,}$ $\vec r_2\text{,}$ $\cdots\text{,}$ and similarlry for their momenta $\vec p_1\text{,}$ $\vec p_2\text{,}$ $\cdots\text{.}$ Then, their individual momenta will be

\begin{align*} \amp \vec l_1 = \vec r_1 \times \vec p_1,\\ \amp \vec l_2 = \vec r_2 \times \vec p_2,\\ \amp \cdots, \end{align*}

and the total angular momentum, to be denoted by $\vec L \text{,}$ will be

$$\vec L = \vec l_1 + \vec l_2 + \cdots + \vec l_N.\tag{9.6.7}$$

Subsection9.6.5Angular Momentum of a Rotating Rigid Body of Point Masses

Suppose we have $N$ balls tied together by light rigid bars. We can ignore the masses of the bars and focus on the angular momenta of the balls only. Now, this is the same problem as the last subsection --- with one big simplification --- all of the balls have the same angular velocity $\omega\text{.}$

The angular momentum of each of the ball will simply be its moment of inertia times the angular velociy, and the total angular momentum will just be their sum.

\begin{align*} \amp l_1 = m_1 R_1^2 \omega,\\ \amp l_2 = m_2 R_2^2 \omega,\\ \amp l_3 = m_3 R_3^2 \omega,\\ \amp \cdots. \end{align*}

Adding them all, we get the magnitude of the total angular momentum will be

$$L = \left( m_1 R_1^2 + m_2 R_2^2 + \cdots + m_N R_N^2 \right)\ \omega,\tag{9.6.8}$$

which we can again write as moment of inertia times angular velocity.

$$L = I\, \omega,\tag{9.6.9}$$

with $I$ for the $N\text{-}$particle system to be

$$I = m_1 R_1^2 + m_2 R_2^2 + \cdots + m_N R_N^2.\tag{9.6.10}$$

Subsection9.6.6Angular Momentum of a Rotating Rigid Body

The angular momentum of a rotating body depends on angular velocity and moment of inertia, which takes into account how the mass is distributed in the body. In our treatment of angular momentum above, we have noted that if an object is rotating at angular velocity $\omega$ about an axis about which its moment of inertia is $I\text{,}$ then, we could expect its angular momentum to be

$$L = I\,\omega.\label{eq-angular-momentum-equal-I-omega}\tag{9.6.11}$$

This is true for rotation about special axes, called principal axes. Figure 9.6.8 and Figure 9.6.9 illustrate two types of principal axes for a disk.

For instance, for a disk of mass $M$ and radius $R\text{,}$ the moment of inertia $I$ for rotation about the axis through the center of the axis and perpendicular to the disk is

$$I_\text{disk}^\perp = \frac{1}{2}M\,R^2.\label{eq-I-disk-perp}\tag{9.6.12}$$

This wouldn't be the case if you were rotating the disk about an axis through the center but lying in the plane of the disk. Then, we will have

$$I_{\text{disk}}^\parallel = \dfrac{1}{4}M\,R^2.\label{eq-I-disk-parallel}\tag{9.6.13}$$

For a fixed-axis rotation about axis in Figure 9.6.8, angular momentum will be in the same direction as angular velocity with magnitude

\begin{equation*} L = \frac{1}{2}M\,R^2\,\omega. \end{equation*}

For a fixed-rotation about axis in Figure 9.6.9, angular momentum will be in the same direction as angular velocity with magnitude

\begin{equation*} L = \frac{1}{4}M\,R^2\,\omega. \end{equation*}

In general, our axis of rotation, may be in some arbitrary direction. What happens then? Figure 9.6.10 shows rotation of a disk about an arbitrary axis. We choose Cartesian axes along principal axis of Figure 9.6.9 and Figure 9.6.8. Angular momentum vector will be

\begin{equation*} \vec \omega = \omega_x\hat i + \omega_y\hat j+ \omega_z\hat k. \end{equation*}

Now, angular momentum will have the following components

\begin{gather*} L_x = I_{\text{disk}}^\parallel\, \omega_x.\\ L_y = I_{\text{disk}}^\parallel\, \omega_y.\\ L_z = I_\text{disk}^\perp\, \omega_z. \end{gather*}

Clearly, $\vec L$ is not in the same direction as angular velocity $\vec \omega\text{.}$ This is very different than what happened in the case of velocity and momentum - they are always in the same direction.