Angular Momentum | Physics Class 11 - NEET PDF Download

We know the concept of torque as a turning effect of force. Torque is regarded as a rotational analog of force. According to Newton's second law of motion, force is equal to the rate of change of linear momentum. Therefore, if we can represent torque as the rate of change of some quantity, that quantity would be the rotational analog of linear momentum, and it can be called angular momentum, which we will study in detail in this document. 

Angular momentum of a particle about a point

If p is the linear momentum of a particle in a given reference frame, then the angular momentum of the particle about an origin O in this reference frame is defined as

  = position vector of the particle with respect to the point about which angular momentum is to be calculated.

θ = angle between vectors r and p

 = perpendicular distance of the line of motion of particle from point O.

 = perpendicular component of momentum.

SI unit of angular momentum is kg m² s^-1.

Relation between Torque and Angular Momentum

As we know that,

Differentiating with respect to the time we get,

Q1. A particle of mass m is moving along the line y = b, z = 0 with constant speed v. State whether the angular momentum of the particle about the origin is increasing, decreasing, or constant. 

Solution:

 constant as m, v, and b all are constants.

Direction of  also remains the same. Therefore, the angular momentum of the particle about the origin remains constant with due course of time.

Note: In this problem  is increasing, q is decreasing but r sin q, i.e., b remains constant. Hence, the angular momentum remains constant.

Q2. A particle of mass m is projected with velocity v at an angle q 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 cos q. The length of the perpendicular to the horizontal velocity from 'O' is the maximum height, where

⇒ Angular momentum L =              

Angular Momentum of a rigid body rotating about a fixed axis

Suppose a particle P of mass m is going in a circle of radius r and at some instant the speed of the particle is v. For finding the angular momentum of the particle about the axis of rotation, the origin may be chosen anywhere on the axis. We choose it at the center of the circle. In this case   are perpendicular to each other and are along the axis. Thus, the component along the axis is mvr itself. The angular momentum of the whole rigid body about AB is the sum of components of all particles, i.e.,

L =

Here, I is the moment of inertia of the rigid body about AB.

Note: Angular momentum about the axis is the component of  along the axis. In most of the cases, angular momentum about axis is Iω.

Q3. Two small balls A and B, each of mass m, are attached rigidly to the ends of a light rod of length d. The structure rotates about the perpendicular bisector of the rod at an angular speed w. Calculate the angular momentum of the individual balls and of the system about the axis of rotation.

Solution:

Consider the situation shown in the figure. The velocity of the ball A with respect to the center O is 

v = .

The angular momentum of the ball with respect to the axis is

L₁ = mvr =  = mwd². The same is the angular momentum L₂ of the second ball. The angular momentum of the system is equal to the sum of these two angular momenta i.e., L = 1/2 mwd².

Conservation of Angular momentum

The time rate of change of angular momentum of a particle about some reference point in an inertial frame of reference is equal to the net torques acting on it.

or  Now, suppose that , then , so that  = constant.

When the resultant external torque acting on a system is zero, the total vector angular momentum of the system remains constant. This is the principle of the conservation of angular momentum.

For a rigid body rotating about an axis (the z-axis, say) that is fixed in an inertial reference frame, we have

L_z = Iw

It is possible for the moment of inertia I of a rotating body to change by rearrangement of its parts. If no net external torque acts, then L_z must remain constant and if I do change, there must be a compensating change in w. The principle of conservation of angular momentum in this case is expressed.

Iw = constant.

Q4. A wheel of the moment of inertia I and radius R is rotating about its axis at an angular speed of w₀. It picks up a stationary particle of mass m at its edge. Find the new angular speed of the wheel. 

Solution: The net external torque on the system is zero. Therefore, angular momentum will remain conserved. Thus,

Note: 

Comments on Linear Momentum :

In case I: Linear momentum is not conserved just before and just after collision because during collision hinge force acts as an external force.

In case II: Linear momentum is conserved just before and just after collision because no external force on the string.

Comments on Angular Momentum :

In case I: Hinge force acts at an external force during collision but except point A all the other reference points given . So angular momentum is conserved only for point A.

In case II: Angular momentum is conserved at all points in the world.

Q5. A uniform rod of mass m and length l can rotate freely on a smooth horizontal plane about a vertical axis hinged at point H. A point mass having the same mass m coming with an initial speed u perpendicular to the rod, strikes the rod in-elastically at its free end. Find out the angular velocity of the rod just after collision.

Solution: 

Angular momentum is conserved about H because no external force is present in the horizontal plane which is producing torque about H.

Q6. A uniform rod of mass m and length l can rotate freely on a smooth horizontal plane about a vertical axis hinged at point H. A point mass having the same mass m coming with an initial speed u perpendicular to the rod, strikes the rod and sticks to it at a distance of 3l/4 from the hinge point. Find out the angular velocity of the rod just after collision.

Solution:

From angular momentum conservation about H, initial angular momentum = final angular momentum

Q7. A uniform rod AB of mass m and length 5a is free to rotate on a smooth horizontal table about a pivot through P, a point on AB such that AP = a. A particle of mass 2m moving on the table strikes AB perpendicularly at the point 2a from P with speed v, the rod being at rest. If the coefficient of restitution between them is , find their speeds immediately after impact. 

Solution: 

Let the point of impact be Q so that

PQ = 2a

Let P be the point of pivot that AP = a

Let the velocities of point, Q, and the particle after impact be v_q and v_p respectively then from momentum conservation about point P.

L_i = L_f