

Angular Momentum
Angular momentum is the vector product of the angular velocity of a particle and its moment of inertia. If a particle of mass m has linear momentum (p) and 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. Mathematically
l = r × p
Derivation of Equation for Angular Momentum
Since 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 = (dp/dt)× r + (dr/dt )×p
Since velocity is the change in position at some time interval, thence, dr/dt = v and p = mv,
Thus, dr/dt × p = v × mv
Now since both the vectors are parallel, their products shall be a zero (o). Now let’s take dp/dt × r,
Since F = dp/dt
thus ,(dp/dt ) × r = F × r = τ
This means that, d (r×p)/ dt = τ. Since l= r×p, therefore,
dl / dt = τ
Angular Momentum of the Rigid Body Rotating about a Fixed Axis
The angular momentum what we studied above is on a particle about any point which states that the rate of change of total angular momentum wrt time about a point equals the total net external torque acting on the system about the same point. Thus, the Angular momentum remains conserved when the total external torque is zero.
But in order to calculate the net rate of change of angular momentum of a rotating object about a fixed axis, we will be learning about the concept of angular momentum of a particle undergoing the rotational motion about a fixed axis.
Now, we shall deal with angular momentum about a fixed axis .
Thus in order to study rotational momentum in reference to a rigid body, we consider it as a vector acting on a system of particles. Since during rotational motion every particle behaves differently, hence we can calculate the angular momentum for a system of many particles.
The angular momentum of any particle rotating about a fixed axis depends on the net external torque acting on that body.
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Let us Consider an object rotating about a fixed axis, as shown in the figure. Now consider a particle P in the body that rotates about the axis as shown above. Thus the total angular momentum for this system is given by,
L = \[\sum_{i=1}^{N}\] ri X pi
Where, P is the momentum and is equal to mv and r is the distance of the particle from the axis of rotation.
The contribution of individual particle to the total angular momentum is given as, l= r×p
Using vector law of addition OP = OC + CP.
So we can write, l = (OC +CP) ×p = (OC×p) +(CP×p)
v = rpw where rp is the perpendicular distance of point P from axis of rotation.
Also, the tangential velocity v at the point p is perpendicular to the vector rp.
Using the right-hand thumb rule, the direction of product CP×v is parallel to the axis of rotation.
Similarly, the product of the vectors OC×V is perpendicular to the axis of rotation.
So, we can write it as: l = OC × mv + lz
The component of angular momentum parallel to the fixed axis of rotation, which is along the z-axis is Iz
L = ∑ l = ∑ (lp + lz )
Here Lp is the perpendicular component of momentum can be given as,
Lp = ∑ OC i × mivi
And the parallel component of the momentum is ,
Lz = (∑ miri2 ) ωk’
Lz =Iz ωk’
Each and every particle possessing a velocity vi has a corresponding particle possessing velocity –vi located diametrically opposite on the circle since the object under consideration is generally symmetric about the axis of rotation thus at a particular perpendicular distance rp,, the total angular momentum due to these particles cancel each other.
For such symmetrical objects, the total momentum of the object is given by,
L = Lz =Iz ωk’
Where, ω is the angular velocity of the body and gives the direction of the total angular momentum.
and ,I is the moment of inertia of the body.
FAQs on Angular Momentum About Fixed Axis
1. What is angular momentum for a rigid body rotating about a fixed axis?
For a rigid body rotating about a fixed axis, angular momentum is the rotational equivalent of linear momentum. It is defined as the product of the body's moment of inertia (I) about the axis of rotation and its angular velocity (ω). It quantifies the amount of rotational motion a body possesses.
2. What is the formula for calculating the angular momentum of a rigid body about a fixed axis?
The formula for the angular momentum (L) of a rigid body rotating about a fixed axis is given by: L = Iω. In this equation, 'I' represents the moment of inertia of the body about the fixed axis, and 'ω' is its uniform angular velocity. The direction of the angular momentum vector lies along the axis of rotation.
3. How is angular momentum related to torque for a body in rotational motion?
The relationship between angular momentum (L) and torque (τ) is the rotational analogue of Newton's second law. It states that the time rate of change of the angular momentum of a rigid body is equal to the net external torque acting on it. This is expressed mathematically as τ = dL/dt.
4. What are some real-world examples of angular momentum about a fixed axis?
Common real-world examples of angular momentum in action include:
- The rotation of the Earth on its axis, which helps maintain its stable orientation in space.
- A spinning top, which uses its high angular momentum to resist falling over.
- The wheels of a moving bicycle, where their angular momentum contributes to the bike's stability.
- A gyroscope in a navigation system, which relies on the conservation of angular momentum to maintain a fixed direction.
5. Why is angular momentum considered a vector quantity, and how is its direction determined?
Angular momentum is a vector quantity because it is defined by the cross product of two vectors: the position vector (r) and the linear momentum vector (p). For a rigid body rotating about a fixed axis, its direction is always along the axis of rotation and can be found using the right-hand thumb rule. If you curl the fingers of your right hand in the direction of the body's rotation, your extended thumb will point in the direction of the angular momentum vector.
6. How does changing the distribution of mass in a rotating body affect its angular momentum?
The distribution of mass affects angular momentum through the body's moment of inertia (I). If mass is distributed far from the axis of rotation, the moment of inertia is large. If it is concentrated close to the axis, the moment of inertia is small. For a constant angular velocity (ω), a body with a larger moment of inertia will possess a greater angular momentum, as per the formula L = Iω.
7. What is the principle of conservation of angular momentum, as seen in a spinning ice skater?
The principle of conservation of angular momentum states that if the net external torque acting on a system is zero, its total angular momentum remains constant. An ice skater is a classic example:
- When her arms are stretched out, her mass is distributed farther from her axis of rotation, resulting in a high moment of inertia (I) and a slower spin.
- When she pulls her arms in, she concentrates her mass closer to the axis, decreasing her moment of inertia.
- Since angular momentum (L = Iω) must be conserved (as there is no external torque), a decrease in 'I' causes her angular velocity (ω) to increase, making her spin much faster.
8. What is the fundamental difference between the angular momentum of a single particle and that of a rigid body?
The primary difference lies in their definition and calculation:
- Single Particle: Its angular momentum (l) is calculated about a point (origin) using the formula l = r × p, where 'r' is the position vector and 'p' is the linear momentum. Its direction depends on the plane formed by 'r' and 'p'.
- Rigid Body: Its total angular momentum (L) about a fixed axis is the sum of the angular momenta of all its particles. For a symmetrical body, this simplifies to L = Iω. The vector 'L' is always directed along the fixed axis of rotation.

















