Angular Momentum Explained: Definition, Proof and Examples

By Pankaj Kumar|Last Updated: July 12, 2026|

What is Angular Momentum?
Proof of L=I*ω
Problems based on angular momentum

Key Takeaways

Angular momentum equals the moment of linear momentum about an axis of rotation.
For a rigid body, angular momentum L equals moment of inertia times angular velocity (L = Iω).
Angular momentum is a vector quantity derived from the cross product of position and linear momentum.
The perpendicular distance from the axis determines the magnitude of 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}} = sumlimits_{i = 1}^n {{{bf{r}}_i} times {{bf{P}}_i}} $

$sin theta = frac{{OA}}{{OP}}quad Rightarrow quad OA = OPsin 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 instant is ${v_i} = {r_i}omega $.

Now, the angular momentum about AB (for the 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.

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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 = Iomega $.

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

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

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

$ Rightarrow quad L = 10pi = 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:

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

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

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

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

${v_y} = frac{u}{{sqrt 2 }} – gfrac{u}{g} = left( {frac{{1 – sqrt 2 }}{{sqrt 2 }}} right)u$

$y = usin {45^o}t – frac{1}{2}g{t^2} = frac{u}{{sqrt 2 }} times frac{u}{g} – frac{1}{2}g times frac{{{u^2}}}{{{g^2}}}$

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

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

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