Consider yourself rotating around the same axis, with hands stretched. The phenomenon seems simple, but did you ever wonder that with every rotation that you make the Angular momentum changes. Now if you bring your hands closer to the axis of your rotation, what will be the inertia in such case? Let’s learn more about system of particles and rotational motion.

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## Angular Momentum

Earth rotates around its own axis, a ballet dancer with his hands stretched rotates around his own axis! These are some of the most common examples of rotation around a fixed axis. The speed of the ballet dancer changes when his stretched hands are folded inwards. Why does this happen? Well, the reason is the changing angular momentum during the circular motion of the ballet dancer.

**Browse more Topics under System Of Particles And Rotational Dynamics**

- ntroduction to Rotational Dynamics
- Vector Product of Two Vectors
- Centre of Mass
- Motion of Centre of Mass
- Moment of Inertia
- Theorems of Parallel and Perpendicular Axis
- Rolling Motion
- Angular Velocity and Angular Acceleration
- Linear Momentum of System of Particles
- Torque and Angular Momentum
- Equilibrium of a Rigid Body
- Dynamics of Rotational Motion About a Fixed Axis
- Kinematics of Rotation Motion about a Fixed Axis

What exactly is angular momentum? Angular momentum can be defined as the vector product of the angular velocity of a particle and its moment of inertia. When a particle of mass m shows linear momentum (p) at a position (r) then the angular momentum with respect to its original point O is defined as the product of linear momentum and the change in position. Here the rotational momentum is taken as l, so,

l = r × p

The magnitude of l as a vector shall be,

l = rp_{1} sin θ

θ here is the angle between the relative position (r) and linear momentum (p) while p_{1 }is the magnitude of p. We can also write this as:

l= rp’ or r’p

r’ here is r sin θ and is a distance perpendicular from the point of origin. p’ is taken as p sin θ which is the component of p and is in a direction . to the relative position (r). Now at any point where r = 0 or when the linear momentum disappears, angular momentum also becomes a zero. Linear momentum (p) vanishes or becomes a zero when its line of direction passes through the origin or when θ is 0º or 180°.

When calculating the angular momentum for any particle we need to know the relation between the moment of a force and angular momentum. The relation between the two is same as that of force and linear momentum. From l=r × p, when we differentiate it with respect to time we get,

dl/ dt = d (r × p) / dt

Applying the product rule for differentiation,

d (r × p) / dt = dr/dt × p + dp/dt × r

Now, we already know that velocity is the change in position at some time interval, thence, dr/dt = v and p = mv,

dr/dt × p = v × mv

Now since both are parallel vectors, their products shall be a zero (o). Now let’s take dp/dt × r,

F = dp/dt

dp/dt × r = F × r = τ

This means that, d (r×p)/ dt = τ. Since l= r×p, therefore,

dl / dt = τ

From the above equation, we can say that the rate of change of angular momentum is directly related to the torque on it. Torque (τ) is the external force on the particle.

## Angular Momentum for Rotation About a Fixed Axis

From the above discussion we now know that for a system of particles, the time rate of total angular momentum about a point equals the total external torque acting on the system about the same point. Angular momentum is conserved when total external torque is zero. The angular momentum studied here is on a particle on any point.

Here we shall deal with angular momentum about a fixed axis. When we study rotational momentum in reference to a rigid body, we take it as a vector acting on a system of particles. During rotational motion every particle behaves differently, hence we calculate the angular momentum for a system of many particles.

The angular momentum of any particle rotating about a fixed axis depends on the external torque acting on that body. The angular momentum discussed here, is that of a rigid body rotating about a fixed axis. Before we start, let’s see the figure below:

## General Equation

For total angular momentum(L) we take the following general equation;

L = _{t=1}^{N}∑r_{t} × p_{t }

As already mentioned that when we calculate the angular momentum, we initially take it on an individual particle and then sum up the contributions of the individual particle. For any particle, l= r×p. From the figure above, r = OA. Using the right angle rule, we take OA = OB+BA. Substituting these values in r we get,

l = (OB +BA) ×p = (OB ×p) +(BA×p)

Since, p = mv, hence l= (OB×mv)+ (BA ×mv). The linear velocity (v) of the particle at point A is given by:

v = ω r_{1}

r_{1} is the length of BA which is the perpendicular distance of point A from the axis of rotation. v is tangential at A to the circular motion in which the particle moves. With the help of the right-hand rule, we know that BA × v, which is parallel to the fixed axis. The unit vector along the fixed axis is k’. From the above equation,

BA ×mv = r_{1 }×(mv) k’ = m r_{1}^{2} ω

Likewise, we can say that OB × v is perpendicular to the fixed axis. Denoting a part of l along fixed axis z as l_{z }we get,

l_{z }= BA ×mv =m r_{1}^{2} ω

l = l_{z } + OB×mv

We already know that l_{z }is parallel to the fixed axis while l is perpendicular. Generally, angular momentum l is not along the axis of rotation which means that for any particle l and ω are not impliedly parallel to one another, but for any particle p and v are parallel to each other. For a system of particles, total angular momentum,

L = l_{t} = l_{tz} + OB_{t} ×m_{t} v_{t}

We denote L_{1} and L_{z} as the components of L that are perpendicular to the z-axis and along the z-axis respectively; hence

L_{1 = }OB_{t} ×m_{t} v_{t}

Here, m_{t} and v_{t }are mass and velocity of a t^{th} particle and B_{t }is the centre of the circle of motion described by the particle t.

L_{z }= l _{tz} = _{t}∑ m_{t} r_{t }^{2 }ω k’

L_{z = }I ωk’

According to the definition of Moment of Inertia I,= m_{t} r_{t }^{2 }which is substituted in the above equation. As already said, in rotational motion we take angular momentum as the sum of individual angular momentums of various particles. Therefore,

L = L_{z }+ L_{1}

## Angular Momentum for Symmetrical and Asymmetrical Bodies

It has to be noted that, the rigid bodies that we are considering here are symmetric in nature. For symmetric bodies, the axis of rotation is in symmetry with one of their axes. For bodies like these, every particle experiences a velocity *v* which is opposed by another velocity *-v* that is located diametrically opposite in the circle covered during the rotation by the particle. Both the negative and positive gradients of *v* will give L_{1} as zero in symmetric bodies. Hence,

L=L_{z }= Iωk

For asymmetric bodies, however, L ≠ L_{z} and L hence do not lie along the rotational axis of the said particle.

## Differentiation of Angular Momentum

When we take the equation, L_{z }=Iωk, we get

d(L_{z})/dt = k d(Iω)/dt

k here is a constant vector, we already know that dL/dt = τ, therefore, the external torque of only those components need to be considered which are along the axis of rotation in a fixed axis. So now we take **τ =** τ_{1}k.Since the direction of L_{z} is fixed we assume the L= L_{z}+L_{1} therefore for a rotation along a fixed axis we get,

dL_{z} /dt= τ_{1}k and dL_{1}/dT = 0

The angular momentum which is perpendicular to the fixed axis is constant for the particle rotating along the same axis. Now since L_{z }= Iωk, therefore

d(Iω) /dt= τ_{1}

In a situation where the moment of inertia (I) does not change with time, the equation becomes:

d(Iω) /dt= I dω /dt = Iα

## Conservation of Angular Momentum

According to the principle of conservation of angular momentum, in the absence of any external torque, the angular momentum remains constant, no matter what the interaction and transformation within the system of rotation. This, therefore, means that if there is no external torque along the axis of rotation then Iω is constant. Here we neglect the friction acting during the mechanism of rotation.

With the help of Angular momentum in the rotation along the fixed axis, we can now understand how various acrobats and skaters manage their angular speed during their actions.

You must have noticed yourself that when on a rotating chair, the moment you stretch your hands the speed of rotation decreases while when your hands are brought back closer to your body, the same speed increases. This is due to the changing external torque and angular momentum. A principle that dictates the action of acrobats, skaters, and dancers during various performances, angular momentum helps them master the art with excellence!

## Solved Examples for You

- the moment of inertia
- a scalar vector
- directly proportional to the moment of inertia
- none of the above