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 :

ℓ = r x p = m ( r x v ) ℓ = r x p = m ( r x v )

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 :

Figure 2: The particle is moving with a velocity in 3-D reference system.

Angular momentum of particle in motion

ⅆ ℓ ⅆ t = ⅆ ⅆ t { m ( r x v ) } ⅆ ℓ ⅆ t = ⅆ ⅆ t { m ( r x v ) }

For constant mass,

⇒ ⅆ ℓ ⅆ t = m ⅆ ⅆ t ( r x v ) ⇒ ⅆ ℓ ⅆ t = m ⅆ ⅆ t ( r x v )

⇒ ⅆ ℓ ⅆ t = m ( r x ⅆ v ⅆ t + ⅆ r ⅆ t x v ) ⇒ ⅆ ℓ ⅆ t = m ( r x ⅆ v ⅆ t + ⅆ r ⅆ t x v )

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,

⇒ ⅆ ℓ ⅆ t = m ( r x a + v x v ) ⇒ ⅆ ℓ ⅆ t = m ( r x a + v x v )

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

⇒ ⅆ ℓ ⅆ t = m ( r x a ) ⇒ ⅆ ℓ ⅆ t = m ( r x a )

Rearranging, we have :

⇒ ⅆ ℓ ⅆ t = r x m a = r x F ⇒ ⅆ ℓ ⅆ t = r x m a = r x F

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 :

⇒ F net = m a ⇒ F net = m a

Substituting the same in the equation above, we have :

⇒ ⅆ ℓ ⅆ t = r x m a = r x F net = r x ∑ F i ⇒ ⅆ ℓ ⅆ t = r x m a = r x F net = r x ∑ F i

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,

⇒ ⅆ ℓ ⅆ t = ∑ r x F i = ∑ t i = t net ⇒ ⅆ ℓ ⅆ t = ∑ r x F i = ∑ t i = t net

⇒ t net = ⅆ ℓ ⅆ t ⇒ t net = ⅆ ℓ ⅆ t

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 :

Figure 3: Angular moment about point of projection.

Projectile motion

t = r x F t = r x F

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

r = x i + y j r = x i + y j

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

F = - mg j F = - mg j

Putting in the equation of the torque, we have :

t = ( x i + y j ) x - mg j t = ( x i + y j ) x - mg j

t = - x mg k t = - x mg k

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

ℓ = m ( r x v ) = m { ( x i + y j ) ) x v ℓ = m ( r x v ) = m { ( x i + y j ) ) x v

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

t = ⅆ ℓ ⅆ t = m { ( ⅆ x ⅆ t i + ⅆ x ⅆ t j ) ) x ⅆ v ⅆ t t = ⅆ ℓ ⅆ t = m { ( ⅆ x ⅆ t i + ⅆ x ⅆ t j ) ) x ⅆ v ⅆ t

But ⅆ v ⅆ t ⅆ v ⅆ t is equal to acceleration. In the case of projectile, acceleration due to gravity is the acceleration of the projectile,

a = ⅆ v ⅆ t = - g j a = ⅆ v ⅆ t = - g j

Putting this value in the equation,

t = m { ( ⅆ x ⅆ t i + ⅆ x ⅆ t j ) ) x - g j t = m { ( ⅆ x ⅆ t i + ⅆ x ⅆ t j ) ) x - g j

t = - x mg k t = - x mg k

We must note here that 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.

Figure 4: The bubjee jumper jumps from the middle of bridge.

Bunjee jumping

Solution : Initially, the jumper is at a horizontal distance of 100 m from the end of the bridge. We must note 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 :

ℓ = m r v sin θ ℓ = m r v sin θ

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

ℓ = m v L 2 ℓ = m v L 2

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 :

v = u + a t = 0 + g t = g t v = u + a t = 0 + g t = g t

Substituting in the equation above :

⇒ ℓ = m g t L 2 ⇒ ℓ = m g t L 2

Differnetiating with respect to time,

⇒ t = m g L 2 = 60 x 10 x 200 2 = 60000 N - m ⇒ t = m g L 2 = 60 x 10 x 200 2 = 60000 N - m

We note following four things about this problem :

(i) Applying right hand rule, we realize that torque is into the plane of the figure. If we choose the origin on the right side of the bridge, then torque is out of the plane of figure.

(ii) The torque is calculated on a particle like object even though the object is moving along straight line without any angular manifestations.

(iii) In a similar situation in the earlier module, we had noted that a particle moving in a straight line parallel to one of the axes of coordinate system has 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.

(iv) We can also calculate torque using relation t = r x F t = r x F , where F = mg. This relation should also give same result.

In rotation, a particle rotates about a fixed axis as shown in the figure. We consider haere a particle, which rotates about y-axis along a circular path in a plane parallel to "xz" plane. By the nature of the rotational motion, linear velocity,"v", and hence linear momentum, "p" are tangential to circular path and are perpendicular to the radius vector, "r".

Figure 5: Angular momentum of a particle rotating about an axis.

Angular momentum

The angle between radius vector "r" and velocity vector "v" is always 90°. This fact is not easily visualized. In order to visualize the same, we specifically consider a time instant when radius vector and moment arm, r ⊥ r ⊥ , are in "xy" plane. At this instant, the velocity vector, "v'", is tangential to the circle and is perpendicular to the "xz" plane. This figure clearly shows that radius vector is indeed perpendicular to velocity vector.

Figure 6: Angular momentum of a particle rotating about an axis.

Angular momentum

Thus, angular momentum of the particle in rotation is :

ℓ = m r v sin θ = m r v sin 90 0 = m r v ℓ = m r v sin θ = m r v sin 90 0 = m r v

The direction of angular momentum is perpendicular to the plane formed by radius and velocity vectors. For the specific situation as shown in the figure above, the direction of angular momentum is obtained by first shifting the velocity vector to the origin and then applying right hand rule. Importantly, the angular momentum vector makes an angle with extended z-axis as shown here.

Figure 7: Direction of angular momentum.

Angular momentum

The component of angular momentum in the direction along axis of rotation i.e. y-axis is :

ℓ y = m r v sin α ℓ y = m r v sin α

From geometry, we see that :

r sin α = r ⊥ r sin α = r ⊥

We should note the difference. It ( r ⊥ r ⊥ ) is the moment arm about the axis of rotation - not the moment arm about the point "O", which is equal to OO'.

ℓ y = m r ⊥ v = m r ⊥ x ω r ⊥ = m r ⊥ 2 ω ℓ y = m r ⊥ v = m r ⊥ x ω r ⊥ = m r ⊥ 2 ω

But, we know that m r ⊥ 2 = I m r ⊥ 2 = I . Hence,

ℓ y = I ω ℓ y = I ω

This has been the expected relation corresponding to p=mv for translational motion. The product of moment of inertia and angular velocity about a common axis is equal to component of angular momentum about the rotation axis.

If we define angular momentum of the particle for rotation as the product of linear momentum and moment arm about the axis of rotation (not radius vector as in the general case), then we can say that product of moment of inertia and angular velocity about a common axis is equal to the angular momentum about the rotation axis. Dropping the suffix referring the axis of rotation, we have :

ℓ = I ω ℓ = I ω

We must ensure, however, that all quantities in the equation above refer to the same axis of rotation. Also, we should also keep in mind that the definition of angular momentum about an axis has been equal to the component of linear momentum about the axis of rotation and is different to the one about a point.

Newton's law for a particle in rotation is established by taking first derivative of the angular momentum of a particle about an axis of rotation. As derived in previous section. it is given by :

ℓ = r ⊥ 2 p ℓ = r ⊥ 2 p

Differentiating with respect to time,

t net = ⅆ ℓ ⅆ t = ⅆ ( Iω ) ⅆ t t net = ⅆ ℓ ⅆ t = ⅆ ( Iω ) ⅆ t

Since particle moves around a circular path of a constant radius, I = m r ⊥ 2 I = m r ⊥ 2 is a constant and the same can be taken out of differentiation,

t net = ⅆ ( Iω ) ⅆ t = I ⅆ ( Iω ) ⅆ t = I α t net = ⅆ ( Iω ) ⅆ t = I ⅆ ( Iω ) ⅆ t = I α

This derivation underlines that the above relation is specific to rotation about an axis - not for general motion. On the other hand, the definition of torque as the time rate of angular momentum about a point is a general form of Newton's second law.

Since discussion for angular momentum and corresponding Newton's second law are considered for different situations, it would be of great help to summarize the main results and special points for each of the cases so far covered. For clarity, we summarize as follows :