2.1 For better understanding angular momentum is classified as following four types,

  Rotational Dynamics

    1.0 Introduction
    2.0 Angular momentum or moment of a momentum
    3.0 Relation between torque and angular momentum
    4.0 Combined translational and rotational motion of a rigid body
    5.0 Rotational kinetic energy
    6.0 Uniform pure rolling
    7.0 Accelerated pure rolling
    8.0 Instantaneous axis of rotation
    9.0 Toppling

---

2.1 For better understanding angular momentum is classified as following four types,

• Angular momentum of a particle about a given point
• Angular momentum of a particle describing circular motion
• Angular momentum of a rigid body rotating about an axis of rotation
• Angular momentum of a combined translational and rotational motion of a rigid body

---

2.1.1 Angular momentum of a particle about a given point

Consider a particle $P$ of mass $m$ is moving with velocity $v$

Linear momentum $\left( {\overrightarrow p } \right)$ of the particle is, $$\overrightarrow p = m\overrightarrow v $$

So, its angular momentum $\overrightarrow L $ about point $O$ is given by, $$\begin{equation} \begin{aligned} \overrightarrow L = \overrightarrow r \times \overrightarrow p \\ \overrightarrow L = \overrightarrow r \times \left( {m\overrightarrow v } \right) \\ \overrightarrow L = m\left( {\overrightarrow r \times \overrightarrow v } \right) \\\end{aligned} \end{equation} $$

where,

$\overrightarrow r $: is the position vector of particle $P$ from point $O$ at any time $t$

Magnitude of angular momentum is, $$L = mvr\sin \theta $$ or $$L = mv{r_ \bot }$$

where, ${r_ \bot }$ is the perpendicular of the velocity vector $\left( {\overrightarrow v } \right)$ from point $O$

---

2.1.2 Angular momentum of a particle describing a circular motion

Consider a particle $P$ of mass $m$ moving in a circle of radius $r$ about point $O$ as shown in the figure.

So, angular momentum of the particle $P$ about $O$ is, $$\begin{equation} \begin{aligned} \overrightarrow L = \overrightarrow r \times \overrightarrow p \\ \overrightarrow L = \overrightarrow r \times \left( {m\overrightarrow v } \right) \\ \overrightarrow L = m\left( {\overrightarrow r \times \overrightarrow v } \right) \\\end{aligned} \end{equation} $$ As, $$\overrightarrow v = \overrightarrow \omega \times \overrightarrow r $$ So, $$\overrightarrow L = m\left( {\overrightarrow r \times \left( {\overrightarrow \omega \times \overrightarrow r } \right)} \right)$$

Vector triple product is given by, $\overrightarrow a \times \left( {\overrightarrow b \times \overrightarrow c } \right) = \left( {\overrightarrow a .\overrightarrow c } \right)\overrightarrow b - \left( {\overrightarrow a .\overrightarrow b } \right)\overrightarrow c $ So, $$\overrightarrow L = m\left[ {\left( {\overrightarrow r .\overrightarrow r } \right)\overrightarrow \omega - \left( {\overrightarrow r .\overrightarrow \omega } \right)\overrightarrow r } \right]$$ As, $$\overrightarrow r \bot \overrightarrow \omega $$ So, $$\overrightarrow r .\overrightarrow \omega = r\omega \cos 90^\circ = 0$$ Therefore, $$\overrightarrow L = m{r^2}\overrightarrow \omega $$

As $I = m{r^2}$, $$\overrightarrow L = I\overrightarrow \omega $$

So, the direction of angular momentum is same as the direction of rotation about an axis of rotation.

---

2.1.3 Angular momentum of a rigid body rotating about an axis of rotation

Consider a rigid body rotating about an axis of rotation with an angular velocity $\omega $.

Now consider a $i^{th}$ particle of mass $m_i$ is moving in a circle of radius $r_i$ about point $O$ with velocity $v_i$ as shown in the figure. $$\overrightarrow v = \overrightarrow \omega \times {\overrightarrow r _i}$$

${\overrightarrow r _i}$: is the position vector from point $O$

So, the angular momentum of the particle $P$ is, $$\begin{equation} \begin{aligned} {{\vec L}_i} = {{\vec r}_i} \times {{\vec p}_i} \\ {{\vec L}_i} = {m_i}\left( {{{\vec r}_i} \times {{\vec v}_i}} \right) \\ {{\vec L}_i} = {m_i}\left[ {{{\vec r}_i} \times \left( {\vec \omega \times {{\vec r}_i}} \right)} \right] \\ {{\vec L}_i} = {m_i}\left[ {\left( {{{\vec r}_i}.{{\vec r}_i}} \right)\vec \omega - \left( {{{\vec r}_i}.\vec \omega } \right){{\vec r}_i}} \right] \\ {{\vec L}_i} = {m_i}\left[ {r_i^2\vec \omega - 0} \right]\quad \left( {As,\;{{\vec r}_i} \bot \vec \omega } \right) \\ {{\vec L}_i} = {m_i}r_i^2\vec \omega \\\end{aligned} \end{equation} $$

Taking summation of the above equation for the whole rigid body we get, $$\begin{equation} \begin{aligned} \sum {{{\overrightarrow L }_i} = \sum {{m_i}r_i^2} \overrightarrow \omega } \\ \overrightarrow L = I\overrightarrow \omega \\\end{aligned} \end{equation} $$

As, $\sum {mr_i^2 = I} $

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

---

2.1.4 Angular momentum of a combined translation and rotational motion of a rigid body

Consider a rigid body undergoing a rotational motion with an angular velocity $\left( {\overrightarrow \omega } \right)$ and translational motion with velocity $\left( {\overrightarrow v } \right)$ in the direction as shown in the figure.

Let $O$ be a fixed point in an inertial frame of reference.

Now, let us consider an $i^{th}$ particle of mass $m_i$ at a distance $\left( {{{\overrightarrow r }_{i,\;com}}} \right)$ moving with a velocity $\left( {{{\overrightarrow v }_{i,\;com}}} \right)$.

where,

${\overrightarrow r _{i,\;com}}$: Position vector of the $i^{th}$ particle with respect to center of mass

${\overrightarrow v _{i,\;com}}$: Velocity of the $i^{th}$ particle with respect to center of mass

As we know, $$\begin{equation} \begin{aligned} {\overrightarrow v _{i,\;com}} = {\overrightarrow v _i} - {\overrightarrow v _{com}} \\ {\overrightarrow v _i} = {\overrightarrow v _{i,\;com}} + {\overrightarrow v _o}\quad \left( {As,\,{{\overrightarrow v }_{com}} = {{\overrightarrow v }_o}} \right) \\\end{aligned} \end{equation} $$

where,

${\overrightarrow v _i}$: Velocity of the $i^{th}$ particle

Also, $$\begin{equation} \begin{aligned} {\overrightarrow r _{i,\;com}} = {\overrightarrow r _i} - {\overrightarrow r _{com}} \\ {\overrightarrow r _i} = {\overrightarrow r _{i,\;com}} + {\overrightarrow r _o}\quad \left( {As,\,{{\overrightarrow r }_{com}} = {{\overrightarrow r }_o}} \right) \\\end{aligned} \end{equation} $$

So, the angular momentum of the $i^{th}$ particle about point $O$ is, $$\begin{equation} \begin{aligned} {\overrightarrow L _i} = {\overrightarrow r _i} \times {\overrightarrow p _i} \\ {\overrightarrow L _i} = {\overrightarrow r _i} \times \left( {{m_i}{{\overrightarrow v }_i}} \right) \\ {\overrightarrow L _i} = {m_i}\left[ {{{\overrightarrow r }_i} \times {{\overrightarrow v }_i}} \right] \\ {\overrightarrow L _i} = {m_i}\left[ {\left( {{{\overrightarrow r }_{i,\;com}} + {{\overrightarrow r }_o}} \right) \times \left( {{{\overrightarrow v }_{i,\;com}} + {{\overrightarrow v }_o}} \right)} \right] \\ {\overrightarrow L _i} = {m_i}\left[ {\left( {{{\overrightarrow r }_{i,\;com}} \times {{\overrightarrow v }_{i,\;com}}} \right) + \left( {{{\overrightarrow r }_{i,\;com}} \times {{\overrightarrow v }_o}} \right) + \left( {{{\overrightarrow r }_o} \times {{\overrightarrow v }_{i,\;com}}} \right) + \left( {{{\overrightarrow r }_o} \times {{\overrightarrow v }_o}} \right)} \right] \\ \\\end{aligned} \end{equation} $$

Taking summation of the above equation for calculating angular momentum of the whole rigid body, $$\begin{equation} \begin{aligned} \sum {{{\overrightarrow L }_i}} = \sum {{m_i}} \left[ {\left( {{{\overrightarrow r }_{i,\;com}} \times {{\overrightarrow v }_{i,\;com}}} \right) + \left( {{{\overrightarrow r }_{i,\;com}} \times {{\overrightarrow v }_o}} \right) + \left( {{{\overrightarrow r }_o} \times {{\overrightarrow v }_{i,\;com}}} \right) + \left( {{{\overrightarrow r }_o} \times {{\overrightarrow v }_o}} \right)} \right] \\ \sum {{{\overrightarrow L }_i}} = \sum {{m_i}} \left( {{{\overrightarrow r }_{i,\;com}} \times {{\overrightarrow v }_{i,\;com}}} \right) + \left( {\sum {{m_i}} {{\overrightarrow r }_{i,\;com}} \times {{\overrightarrow v }_o}} \right) + \left[ {{{\overrightarrow r }_o} \times \left( {\sum {{m_i}} {{\overrightarrow v }_{i,\;com}}} \right)} \right] + \sum {{m_i}} \left( {{{\overrightarrow r }_o} \times {{\overrightarrow v }_o}} \right) \\\end{aligned} \end{equation} $$

As we know,

$\begin{equation} \begin{aligned} \sum {{{\overrightarrow L }_i}} = L \\ {\sum m _i} = M \\ \sum {{m_i}{r_{i,\;com}} = M{R_{com,\;com}}} = 0 \\ \sum {{m_i}{v_{i,\;com}}} = M{v_{com,\;com}} = 0 \\\end{aligned} \end{equation} $

So, the above equation becomes, $$\begin{equation} \begin{aligned} \overrightarrow L = \sum {{m_i}} \left( {{{\overrightarrow r }_{i,\;com}} \times {{\overrightarrow v }_{i,\;com}}} \right) + M\left( {{{\overrightarrow r }_o} \times {{\overrightarrow v }_o}} \right) \\ \overrightarrow L = \sum {{m_i}} \left( {{{\overrightarrow r }_{i,\;com}} \times \left( {\overrightarrow \omega \times {{\overrightarrow r }_{i,\;com}}} \right)} \right) + M\left( {{{\overrightarrow r }_o} \times {{\overrightarrow v }_o}} \right) \\ \overrightarrow L = \sum {{m_i}} \left( {r_i^2\overrightarrow \omega } \right) + M\left( {{{\overrightarrow r }_o} \times {{\overrightarrow v }_o}} \right) \\ \overrightarrow L = \left( {\sum {{m_i}} r_i^2} \right)\overrightarrow \omega + \left( {{{\overrightarrow r }_o} \times M{{\overrightarrow v }_o}} \right) \\ \overrightarrow L = {I_{com}}\overrightarrow \omega + \left( {{{\overrightarrow r }_o} \times {{\overrightarrow p }_{com}}} \right) \\ \overrightarrow L = {\overrightarrow L _{com}} + \left( {{{\overrightarrow r }_o} \times {{\overrightarrow p }_{com}}} \right) \\\end{aligned} \end{equation} $$

where,

${\overrightarrow p _{com}}$: Linear momentum of the rigid body about the center of mass

$\left( {{{\overrightarrow r }_o} \times {{\overrightarrow p }_{com}}} \right)$: angular momentum of the center of mass about point $O$