It was stated in the previous module that angular momentum is defined such that its first time derivative gives torque on the particle. This condition was specified keeping in mind about Newton's second law for translation. Following the logic, let us consider angular momentum of a moving particle moving with respect to a point and find whether first derivative actually yields torque as expected or not?

Angular momentum of the particle about the origin of the coordinate system is :

The "r" and "v" vectors represent position and velocity vectors respectively as shown in the figure. Taking differentiation of the terms with respect to time, we have :

Angular momentum of particle in motion |
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For constant mass,

By definition, the first time derivative of velocity is the acceleration and first time derivative of position vector is the velocity of the particle. Putting the appropriate terms for these quantities,

But, the vector product of a vector with itself is equal to zero as sinθ = sin0° = 0. Hence,

Rearranging, we have :

Indeed the first derivative of angular momentum equals the toque on the particle as expected. Since force on the particle is external force, we can qualify the above relation that first time derivative of angular momentum equals external torque applied on the particle. Next, we should think about a situation when more than one force acts on the particle. According to Newton's second law in translation, we have :

Substituting the same in the equation above, we have :

We note here that all forces are acting at the same point (occupied by the particle). As such, position vectors for all forces are same. Thus, we can enclose the vector product inside the summation sign,

In words, Newton's second law of motion for a particle in angular form is read as :

- Definition 1: Newton's second law of motion in angular form
- The net torque on a particle is equal to the time rate of change of its angular momentum.

### Example 1

Problem : Consider projectile motion of a particle of mass "m" and show that time derivative of angular moment about point of projection is equal to torque on the particle about that point.

Solution : For this question, we express various vector quantities in terms of unit vectors, because two directions involved in projectile motion i.e. horizontal and vertical are mutually perpendicular. This makes it easy to represent vectors in the unit vector form.

For the problem in hand, we consider horizontal and vertical directions as "x" and "y" reference directions respectively. Now, the torque, τ, on the particle about the point of projection "O" is :

Projectile motion |
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The position vector of the position of the projectile at a given time "t" is :

We note here that the external force on the projectile is force due to gravity. Hence,

Putting in the equation of the torque, we have :

Negative sign shows that torque is clockwise i.e. into the plane of the projection "xy" plane. Next, we evaluate torque about "O" in terms of angular momentum. Here, angular momentum is given as :

Differentiating with respect to time "t", we have :

But

Putting this value in the equation,

The quantities torque and angular momentum are both measured or calculated about a common point "O". If it is not so, then the relationship is not valid. This aspect should always be kept in mind while applying the law.

We have emphasized that the angular form of Newton's second law is defined about a point - not an axis and as such not limited to rotational motion. Hence, we can employ this law even in situation where motion is strictly along a straight line. We shall, as a matter of fact, work out an example to illustrate this.

### Example 2

Problem : A bunjee jumper of mass 60 kg jumps from the middle of a bridge 200 m in length. Find the torque as time derivative of angular momentum about a point "O" as shown in the figure, before the rope attached to the jumper becomes taught.

Bunjee jumping |
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Solution : Initially, the jumper is at a horizontal distance of 100 m from the end of the bridge. We must observe here that the jumper is not tied to the bridge. The rope is loose during the fall. He or she falls through a vertical straight path. It is a linear motion for which we are required to calculate torque. It is also evident that we can treat the jumper as point mass as all body parts are moving with same motion.

Now, angular momentum of the jumper with respect to the point, O, is :

Here, r sinθ = L/2, where "L" is the length of the brifge.

However, we do not know the speed of the jumper. We note here that the force on the jumper is force due to gravity. He/she falls with a constant acceleration. Thus, applying equation of motion, speed of the jumper after time "t" is :

Substituting in the equation above :

Differentiating with respect to time,

In a similar situation in the earlier module titled Angular momentum , we had found that a particle moving in a straight line parallel to one of the axes of coordinate system has constant angular momentum and zero torque. The difference has arisen as velocity was constant in that case, whereas velocity is not constant in this case. We need force (hence torque) to cause acceleration or deceleration. The result, therefore, is consistent with the basic percept of Newton's law.

Further, we can also calculate torque using relation