## Angular Momentum:

For a particle having momentum , the angular momentum about the point O is

$ \displaystyle \vec{L} = \vec{r}\times \vec{p} $

Here $\vec{r}$ = Position vector of the particle with respect to the given point

and $\vec{P}$ = Linear momentum of the particle.

The direction of $\vec{L}$ can be determined by using the rule for vector products.

The magnitude of the angular momentum of a particle about a fixed point is equal to the product of its linear momentum and the length of the perpendicular to the linear momentum from the fixed point O.

Exercise : Find the angular momentum of a particle of mass m describing a circle of radius r with angular speed ω

Illustration : A particle of mass m is projected with a velocity v at an angle θ with the horizontal. Find its angular momentum about the point of projection when it is at the highest point of its trajectory

Solution :

At the highest point it has only horizontal velocity v_{x} = v cosθ

Length of the perpendicular to the horizontal velocity from ‘ O ‘ is the maximum height, where

$ \displaystyle H = \frac{v^2 sin^2\theta}{2 g} $

Angular momentum = (mvcosθ)×H

⇒ Angular momentum

$ \displaystyle L = \frac{m v^3 sin^2\theta cos\theta}{2 g} $

### Relation between Angular momentum & Moment of Inertia :

Suppose that a rigid body describes pure rotational motion. All its constituent particles also describe circular motion about the same axis of rotation.

The angular momentum of the particles about their corresponding centres are given by

L_{1} = m_{1} r_{1}^{2}ω , L_{2} = m_{2} r_{2}^{2}ω and similarly L_{n} = m_{n} r_{n}^{2}ω

where ω is the angular speed of the body.

Therefore angular momentum of the whole body is given by

L = L_{1} + L_{2} + ……….. + L_{n}

= (m_{1} r_{1}^{2} + m_{2} r_{2}^{2} + m_{3} r_{3}^{2} + …………m_{n} r_{n}^{2})ω

as the angular momentum of all the particles are in the same direction i.e. along the axis of rotation.

Hence, L = Iω

__Relation b/w torque & angular momentum :__

$ \displaystyle \vec{L} = \vec{r}\times \vec{p} $

$ \displaystyle \frac{d\vec{L}}{dt} = \frac{d\vec{r}}{dt} \times \vec{p} + \vec{r}\times \frac{d\vec{p}}{dt}$

$ \displaystyle = \vec{v}\times \vec{mv} + \vec{r}\times \vec{F}$

$ \displaystyle = 0 + \vec{r}\times \vec{F}$

$ \displaystyle \frac{d\vec{L}}{dt} = \vec{\tau_{ext}} $

### Conservation of Angular Momentum:

When there is no net external torque acting on a particle,

$ \displaystyle \frac{d\vec{L}}{dt} = 0 $

$\vec{L} = constant $

Therefore, the angular momentum of the particle remains invariant in the absence of any net external torque.

Condition under which Torque acting on a body is Zero

$\tau = r F sin\theta $

τ = 0 , when r = 0, that means that the force passes through the axis of rotation.

τ = 0 , when F = 0 (or negligibly small) that means no external force acts on the particle.

τ = 0 when θ = 0° or 180° the force is parallel to the radius vector.

(Radial forces cannot impart any torque on the particle i.e. centripetal force causes no torque).

When any or all of the above conditions are satisfied, τ = 0 and thus the angular momentum of the particle is conserved.

That means the particle will keep on rotating with a uniform or constant angular velocity.

Exercise : What is the moment of the gravitational force of the sun on earth about the axis of its rotation about the sun?