It is the rotational analogue of linear momentum, it is denoted by l, and angular momentum of a particle in rotational motion is defined as:

\(l\) = \(r~×~p\) |

This is a cross product of r i.e. the radius of the circle formed by the body in rotational motion, and p i.e. the linear momentum of the body, the magnitude of a cross product of two vectors is always the product of their magnitude multiplied with the sine of the angle between them, therefore in the case of angular momentum the magnitude is given by,

\(l\) = \(r~p~sinθ\) |

Relationship between torque and angular momentum can found as follows,

\(\overrightarrow{l}\) = \(\overrightarrow{r}~×~\overrightarrow{p}\)

Differentiating LHS and RHS,

\(\frac{d\overrightarrow{l}}{dt}\) = \(\frac{d}{dt}(\overrightarrow{r}~×~\overrightarrow{p})\)

By the property of differentiation on cross products the above expression can be written as follows,

\(\frac{d\overrightarrow{l}}{dt}\) = \(\frac{dr}{dt}~\times~\overrightarrow{p}~+~r~\times~\frac{d\overrightarrow{p}}{dt}\)

\(\frac{d\overrightarrow{r}}{dt}\) is change in displacement with time, therefore it is linear velocity \(\overrightarrow{v}\)

\(\frac{d\overrightarrow{l}}{dt}\) = \(\overrightarrow{v}~\times~\overrightarrow{p}+~r~\times~\frac{d\overrightarrow{p}}{dt}\)

\(\overrightarrow{p}\) is linear momentum i.e. mass times velocity,

\(\frac{d\overrightarrow{l}}{dt}\) = \(\overrightarrow{v}~\times~m\overrightarrow{v}~+~r~\times~\frac{d\overrightarrow{p}}{dt}\)

Now notice the first term, there is \(\overrightarrow{v}~\times~\overrightarrow{v}\) magnitude of cross product is given by

\(\overrightarrow{v}~\times~\overrightarrow{v}sinθ\) where the angle is 0 hence the whole term becomes 0.

From newton’s 2nd law we know that \(\frac{d\overrightarrow{p}}{dt}\) is force,

\(\frac{d\overrightarrow{l}}{dt}\) = \(\overrightarrow{r}~\times~\overrightarrow{F}\)

We know that \(r~\times~f\) is torque, hence

\(\frac{d\overrightarrow{l}}{dt}\) = \(\overrightarrow{τ}\),torque

So rate of change of angular momentum is torque.

Angular momentum of a system is conserved as long as there is no net external torque acting on the system, the earth has been rotating on its axis form the time the solar system was formed due to the law of conservation of angular momentum,

There are two ways to calculate the angular momentum of any object, if it is a point object in a rotation, then our angular momentum is equal to the radius times the linear momentum of the object,

\(\overrightarrow{l}\) = \(\overrightarrow{r}~\times~\overrightarrow{p}\)

If we have an extended object, like our earth for example, the angular momentum is given by moment of inertia i.e. how much mass is in motions in the object and how far it is from the centre, times the angular velocity,

\(\overrightarrow{l}\) = \(\overrightarrow{I}~\times~\overrightarrow{\omega}\)

But in both case as long as there is no net force acting on it, the angular momentum before is equal to angular momentum after some given time, imagine rotating a ball tied to a long string, the angular momentum would be given by,

\(\overrightarrow{l}\) = \(\overrightarrow{r}~\times~\overrightarrow{p}\) = \(\overrightarrow{r}~\times~m\overrightarrow{v}\)

Now when we somehow decrease the radius of the ball by shortening the string while it is in rotation, the r will reduce, now according to the law of conservation of angular momentum L should remain the same, there is no way for mass to change, therefore \(\overrightarrow{v}\) should increase, to keep the angular momentum constant, this is the proof for conservation of angular momentum.

Law of conservation of angular momentum has many applications, including:

- Electric generators,
- Aircraft engines, etc.