Angular Momentum of Rotation About a Fixed Axis: Definition, Derivation, Relation to Torque, Conservation of Angular Momentum, and Others

Angular momentum is a measure of the rotational inertia and angular velocity of a rotating object or system. In simple terms, the angular momentum $\vec{L}$ about a point or axis is defined as the equivalent of its linear momentum.

Motion of a body can be categorized into two types:

• Translational Motion (one which moves in a straight line)
• Rotational Motion (one which rotates around a fixed axis)

In the case of rotational dynamics, angular momentum depends on the angular velocity of the body rotating around the axis as well as on how the centre of mass is distributed thorughout the surface area of the body. If the mass is more inclined towards the centre point or axis, the body will rotate with comparatively higher speed. Similarly, if the mass is accumulated away from the axis, the body will rotate slower. This happens because it will start to oppose the angulr acceleration which leads to less angular velocity of the object rotating.

For a single particle of mass $m$ moving with velocity $\vec{v}$ at position $\vec{r}$ from the axis, the angular momentum is $\vec{L}=\vec{r}\times m\vec{v}$

For a rigid body rotating about a fixed axis with angular velocity $\vec{\omega }$ , the angular momentum about the axis will be:

$L=I\omega$

Where

$I$ = moment of inertia about the axis,

$\omega$  = magnitude of the angular velocity along the axis.

Angular Momentum is a vector quantity directed along the axis of rotation (determined by the right-hand rule) and it's SI unit is $\mathrm{k}\mathrm{g}\cdot {\mathrm{m}}^2/\mathrm{s}$ .

For a single particle with linear momentum $\vec{p}=m\vec{v}$  at position $\vec{r}$ relative to the axis, the angular momentum is:

$\vec{L}=\vec{r}\times \vec{p}=\vec{r}\times m\vec{v}$

Where,

r = position of the particle from the point of axis

p(mv) = linear momentum

 

Consider a particle of mass m which is moving in a circular orbit of

radius r

angular velocity ω.

So, the linear momentum will be: 

p = mv

But, speed: v =rω

Therefore, p = mrω

Now, considering the equation

$\stackrel{\to }{L}=\stackrel{\to }{r}\times \stackrel{\to }{p}$

On substituting the values, we get the new formula:

$L=mvr=m{r}^2\omega$

Now, we will dive into the derivation of angular momentum for a rigid body. A rigid body comprises of all the particles combined together. So, Total angular momentum = sum of all the particles.

We obtain the formula: 

$L=\underset{i}{\sum }m\_i{r}^i²\omega$

Now, taking the equation of Moment of Inertia into account:

$I=\underset{i}{\sum }{m}_i{r_i}^2$

We obtain the equation:

$L=I\omega$

Hence, Proved!

Now, let us dive into the relation of angular momentum with torque. First of all, torque is defined as the rotational equivalent of force in rotational dynamics. Just as we use force (f=ma) in case of linear motion, rotational motion onvolves using the term torque. It's general formula is denoted by:

$\stackrel{\to }{\tau }=\stackrel{\to }{r}\times \stackrel{\to }{F}$

Where,

r = radius of the circle

f = force applied

According to Newton's Second law of Motion,

$\stackrel{\to }{F}=\frac{d\stackrel{\to }{p}}{dt}$

Substitute this equation into torque. We obtain:

$\stackrel{\to }{\tau }=\stackrel{\to }{r}\times \frac{d\stackrel{\to }{p}}{dt}$

Next, differentiate the equation of angular momentum to get a new relation:

$\frac{d\stackrel{\to }{L}}{dt}=\frac{d}{dt}(\stackrel{\to }{r}\times \stackrel{\to }{p})$

Apply the product rule.

$\frac{d\stackrel{\to }{L}}{dt}=\frac{d\stackrel{\to }{r}}{dt}\times \stackrel{\to }{p}+\stackrel{\to }{r}\times \frac{d\stackrel{\to }{p}}{dt}$

On solving this equation, we arrive at the final answer:

$\stackrel{\to }{\tau }=\frac{d\stackrel{\to }{L}}{dt}$

So, these are the steps involved in the derivation of torque relation.

The law of conservation of angular momentum states that if the net external torque acting on a system about a fixed axis is zero, the total angular momentum about that axis will remain constant i.e.:

$\frac{d\stackrel{\to }{L}}{dt}=0⟹\stackrel{\to }{L}=\mathrm{constant}$

Therefore, L=Iω is a constant.

Hence, we conclude that if no force is acting on the object:

Linitial​=Lfinal​

I1​ω1​=I2​ω2

Relevant Links: NCERT Chapter 6: Systems of Particles and Rotational Motion