Angular Momentum

What is Angular Momentum?

Angular momentum is the moment of linear momentum of a body with respect to an axis of rotation.

Angular momentum of a particle about O is defined as ${\bf{L}} = {\bf{r}} \times {\bf{P}}$

Where P is the linear momentum and r is the position vector of the particle from the given point O. The angular momentum of a system of particles is the vector sum of the angular momenta of the particles of the system. So,

${\bf{L}} = \sum\limits_{i = 1}^n {{{\bf{r}}_i} \times {{\bf{P}}_i}}$

$\sin \theta = \frac{{OA}}{{OP}}\quad \Rightarrow \quad OA = OP\sin \theta$

Let a particle P of mass m moves at velocity v. Its angular momentum about a point O can be written as

$\ell = {\bf{OP}} \times (m{\bf{v}})$

$\Rightarrow \quad \ell = (OP)\,(mv)\sin \theta \,\hat n$

$\Rightarrow \quad \ell = (OA)\,(mv)\,\hat n$

$\Rightarrow \quad \ell = {{\bf{r}}_ \bot } \times m{\bf{v}} = {\bf{r}} \times m{\bf{v}}$

Where $r = OA = (OP)\sin \theta$ is the perpendicular distance of the line of motion from O.

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Proof of L=I*ω

Suppose a particle is going in a circle of radius r, and at some instant, the particle’s speed is v.

The origin may be chosen anywhere on the axis. We choose it at the center of the circle.

Now, angular momentum of particle about centre,

$\ell = {\bf{r}} \times m{\bf{v}} = r\,\hat i \times mv\,\hat j = rmv\,\hat k$

Also, ${\bf{r}} \times {\bf{P}}$ is perpendicular to r and P and hence is along the axis. So, component of ${\bf{r}} \times {\bf{P}}$ along the axis is mvr itself.

Now, consider a rigid body rotating about an axis AB.

Let the angular velocity of the rigid body be $\omega$ .

Consider the i^th particle going in a circle of radius ${r_i}$ with its plane perpendicular to AB. The linear velocity of this particle at this constant is ${v_i} = {r_i}\omega$ .

Now, the angular momentum about AB (for considered particle)

$\ell = {{\bf{r}}_i} \times {m_i}{{\bf{v}}_i}\,\,;\,\,{{\bf{v}}_i} = {\bf{\omega }} \times {{\bf{r}}_i}$

$\ell = {m_i}r_1^2{\bf{\omega }}$

$\sum \ell = \sum {m_i}r_1^2{\bf{\omega }}$

${\bf{L}} = {I_{sy}},\,\,axis\,of\,\,{\bf{\omega }}\,\,rotation$

${\bf{L}} = {I_{sy,\,AB}}\,{\bf{\omega }}$

Where I is the moment of inertia of the rigid body about axis AB.

Problems based on angular momentum

Example 1: Angular momentum of a Rotating disk

The diameter of a disc is 1 m. It has a mass of 20 kg. It is rotating about its axis with a speed of 120 rotations in one minute. Its angular momentum is $kg - {m^2}{\rm{/}}s$ is __________.

Ans.- Here, the body rotates about an axis, so we will use $L = I\omega$ .

Here, $r = \frac{1}{2}m$ ; mass = m = 20 kg ;

$\omega = \frac{{120 \times 2\pi r}}{{60 \times r}}rad{\rm{/}}s\quad \Rightarrow \quad \omega = 4\pi \,rad{\rm{/}}s$

$\left| {\bf{L}} \right| = \frac{1}{2}m{r^2}\omega = \frac{1}{2} \times 20 \times {\left( {\frac{1}{2}} \right)^2} \times 4\pi kg - {m^2}{\rm{/}}s$

$\Rightarrow \quad L = 10\pi = 31.4\,kg - {m^2}{\rm{/}}s$

Example 2: Particle moving in a circle

A particle of mass m is moving along a circle of radius r with a time period T. Its angular momentum is _____________.

Ans.- Since the particle is moving along a circle

$\left| {{{\bf{L}}_C}} \right| = I\omega = m{r^2}\omega = m{r^2}\frac{{2\pi }}{T}$

$\Rightarrow \quad {L_C} = \frac{{2\pi m{r^2}}}{T}$

Example 3: Angular momentum of a projectile

A particle is projected at time t = 0 from a point O with a speed u at an angle ${45^o}$ to horizontal. Find the angular momentum of the particle at time $\frac{u}{g}$ .

Ans.- Velocity of particle at time t is ${\bf{v}} = {v_x}\hat i + {v_y}\hat j$ .

Position vector of particle at time t is ${\bf{r}} = x\,\hat i + y\,\hat j$ .

Hence, angular momentum of the particle about the origin,

${{\bf{L}}_0} = {\bf{r}} \times m{\bf{v}} = m({\bf{r}} \times {\bf{v}}) = m(x\,\hat i + y\,\hat j) \times ({v_x}\hat i + {v_y}\hat j)$

$\Rightarrow \quad {{\bf{L}}_0} = m(x{v_y} - y{v_x})\,\hat k$ …(i)

Here, ${v_x} = \frac{u}{{\sqrt 2 }}\,;$

$x = {u_x}t + \frac{1}{2}{a_x}{t^2} = \frac{u}{{\sqrt 2 }} \times \frac{u}{g} + 0 = \frac{{{u^2}}}{{g\sqrt 2 }}$

${v_y} = \frac{u}{{\sqrt 2 }} - g\frac{u}{g} = \left( {\frac{{1 - \sqrt 2 }}{{\sqrt 2 }}} \right)u$

& $y = u\sin {45^o}t - \frac{1}{2}g{t^2} = \frac{u}{{\sqrt 2 }} \times \frac{u}{g} - \frac{1}{2}g \times \frac{{{u^2}}}{{{g^2}}}$

Now putting values of $x,\,{v_y},\,y$ and ${v_x}$ into equation (i),

${{\bf{L}}_0} = \frac{{m{u^3}}}{{2\sqrt 2 g}}( - \hat k)$ .

Rajesh Kumar

20 years of experience teaching high school and college physics to students across the globe.

When not teaching or mentoring, I write informative articles in physics and related subjects. So far, I have written more than 200 articles on different topics in physics. Apart from physics, I am proficient in engineering statics, dynamics, and calculus. I love spending time with my kids and listening to old Hindi songs.

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