Angular quantities, in its bsic form, are defined in very geenral context. When we defined, for example, torque in the context of rotation of a rigid body or a single particle attached to a "mass-less" rod, we actually presented a definition in special context of a fixed axis. As a matter of fact, the definition of angular quantities does not require an axis to be defined or interpreted.

It must be understood that we can define and interpret angular quantities very generally with respect to a point in the reference system. For some, it may sound a bit inconsistent to know that we can actually associate all angular quantities even with a straight line motion! For example, we can calculate torque on a particle, which is moving along a straight line. We shall work out with appropriate examples to illustrate the point.

Indeed, angular quantities are found to be suited to rotational motion or where curvature of path is involved. For this reason, we tend to think that angular quantities are applicable only to rotation. There is nothing wrong to think so, but as a student of physics it is important to know the complete picture.

There are many such situations in real time, which can not be classified as a particular motion type. Consider projectile motion along a parabolic path. We employed the concept of independence of motions in mutually perpendicualr directions, which is an experimental fact. This paradigm of analysis suited the problem in best fit manner.

However, we can equivalently think of this motion even in terms of angular quantities like a torque about certain axis, which keeps shifting. If we observe closely a projectile path, then we can see that the parabolic path of a projectile is actually made up of many circular arcs along which the projectile undergoes rotation for brief periods.

We must understand here that the broadening the concept of angular quantities is not without purpose. We shall find that the de-linking of angular concepts like torque and angular momentum from a point (instead of an axis) lets us derive very powerful law known as conservation of angular momentum, which is universally valid unlike Newton's law (for translational or rotational motion). Importantly, we can employ angular quantities to analyze situations, which are not strictly rotational (like motion along a straight line).

Example 1

Problem : A particle is projected with velocity "v" at an angle of "θ" with the horizontal. Find the average angular speed of the particle between point of projection and point of impact.

Solution : The average angular speed is given by :

Figure 1: Average angular speed during the flight of a projectile.

Angular velocity

ω avg = Δ θ Δ t ω avg = Δ θ Δ t

Here,

Δ t = 2 v sin θ g Δ t = 2 v sin θ g

From the figure, magnitude of the total angular displacement is :

Δ θ = 2 θ Δ θ = 2 θ

Putting these values, we have :

ω avg = Δ θ Δ t = 2 θ g 2 v sin θ ω avg = θ g v sin θ rad / s ω avg = Δ θ Δ t = 2 θ g 2 v sin θ ω avg = θ g v sin θ rad / s

From this example, we see that we can indeed associate angular quantity like angular speed with motion like that of projectile, which is not strictly rotational.

Torque in rotation corresponds to force in translation. It is the "cause" that accelerates or decelerates a particle in rotation. Quantitatively, it is defined as a vector given by :

t = r x F t = r x F

The important aspect of this relation for rotation is that vector "r" is constrained to be perpendicular to the axis of rotation as every particle of the body moves around a circular path in the plane perpendicular to the axis of rotation. In the general case of a particle, however, this restriction does not exist at all.

A general interpretation of the vector equation as above tells us that torque may be interpreted even without an axis or a line. An external force on a particle shall constitute a torque with respect to a point. Only condition is that the point should not lie on the line of force, in which case torque is zero. The important aspect of this interpretation is that we are referring torque on the particle with respect to a point - not with respect to a line like axis of rotation. This is the difference. This is the generalized interpretation of torque.

As explained earlier module, we can find the magnitude of torque, using any of the following relations :

Figure 2

Torque on a particle
(a) Torque in terms of angle enclosed. (b) Torque in terms of force perpendicular to position vector.

1: τ = r F sin θ 1: τ = r F sin θ

2: τ = r F ⊥ 2: τ = r F ⊥

3: τ = r ⊥ F 3: τ = r ⊥ F

Figure 3: Torque in terms of moment arm.

Torque on a particle

In order to interpret direction of the vector product, we employ right hand vector product rule. However, the "in place" orientations of vector operands make it difficult to apply this rule. Incidentally, the position vector has its tail at the point about which torque is to be calculated. This fact simplifies to implement the right hand rule.

Figure 4

Vector product
(a) Right hand rule. (b) Application of Right hand rule.

We shift the second vector (F) so that tails of two operand vectors meet at the point about which torque is calculated. The two operand vectors define a plane. The torque (t) is, then, acting along a line perpendicular to this plane and passing through the meeting point. The direction of the orientation is then found by applying right hand rule.

To illustrate the technique, we consider the situation as shown in the figure. Let "r" and "F" vectors lie in the xy-plane. We shift the force vector as shown. Now, we apply right hand rule of vector product to determine the direction of torque. For this, we sweep the closed fingers of the right hand from position vector to force vector. The outstretched thumb, then, indicates that torque is acting in the positive z-direction.

While interpreting the vector product, we must be careful about the sequence of operand. The vector product " F x r F x r " is negative or opposite in sign to that of " r x F r x F ".

Example 2

Problem : A projectile of mass "m" is projected with a speed "v" at an angle "θ" with the horizontal. Calculate the torque on the particle at the maximum height in relation to the point of projection.

Solution : The magnitude of the torque is given by :

Figure 5: Torque on a projectile.

Projectile motion

τ = r ⊥ F = product of moment arm and magnitude of force τ = r ⊥ F = product of moment arm and magnitude of force

where r ⊥ r ⊥ is moment arm. This moment arm is equal to half of the horizontal range of the flight,

r ⊥ = R 2 = v 2 sin 2θ 2 g r ⊥ = R 2 = v 2 sin 2θ 2 g

Now, the force on the projectile is "m" is the force due to gravity :

F = m g F = m g

Putting in the expression of torque, we have :

⇒ τ = m v 2 sin 2θ 2 ⇒ τ = m v 2 sin 2θ 2

Application of right hand rule indicates that torque is clockwise and is directed in to the page.

Example 3

Problem : A particle in xy plane is acted by two forces 5 N and 10 N in xy plane as shown in the figure. Find the net torque on the particle.

Figure 6: Two forces are acting simultaneously on the particle.

Torque on a particle

Solution : In this situation, calculation of the magnitude of torque suits to the form of expression that uses moment arm. Here, moment arms are :

Figure 7: Two forces are acting simultaneously on the particle.

Torque on a particle

For 5 N force, the moment arm "AB" is :

AB = OA cos 30 0 = 5 x 0.866 = 4.33 m AB = OA cos 30 0 = 5 x 0.866 = 4.33 m

For 10 N force, the moment arm "AC" is :

AC = OA sin 30 0 = 10 x 0.5 = 5 m AC = OA sin 30 0 = 10 x 0.5 = 5 m

The torques due to 5 N is :

⇒ τ 1 = = 4.33 x 5 = 21.65 N - m (positive being anti-clockwise) ⇒ τ 1 = = 4.33 x 5 = 21.65 N - m (positive being anti-clockwise)

The torques due to 10 N is :

⇒ τ 2 = - 5 x 10 = - 50 N - m (negative being clockwise) ⇒ τ 2 = - 5 x 10 = - 50 N - m (negative being clockwise)

Net torque is :

⇒ τ net = τ 1 + τ 2 = 21.65 - 50 = - 28.35 N - m (negative being clockwise) ⇒ τ net = τ 1 + τ 2 = 21.65 - 50 = - 28.35 N - m (negative being clockwise)

Note:

We can also first calculate the resultant force and then apply the definition of torque to obtain net torque. Further, we can also use vector notation with unit vectors to solve this problem.

Example 4

Problem : A particle in xy plane is acted by a force 5 N in z-direction as shown in the figure. Find the net torque on the particle.

Figure 8: Two force is acting parallel to z-axis.

Torque on a particle

Solution : In this case, moment arm is equal to the magnitude of position vector. Hence, torque is :

τ = 10 x 5 = 50 N - m τ = 10 x 5 = 50 N - m

However, finding its direction is not as straight forward here. The situation here differs to earlier example in one important respect. The plane containing position vector and force vector is a plane defined by zOA and is perpendicular to xy - plane. Now, torque is passing through the origin and is perpendicular to the plane zoA. Thus, torque is shifted by 30 degree from the y-axis and lies in xy-plane as shown in the figure here.

Figure 9: Two force is acting parallel to z-axis.

Torque on a particle

Like linear momentum, angular momentum is the measure of the "quantity of motion" From Newton's second law, we know that first time derivative of linear momentum gives net external force on a particle. By analogy, we expect that this quantity (angular momentum) should have an expression such that its first time derivative yields torque on the particle.

Angular momentum is defined as a vector, denoted as "ℓ".

ℓ = r x p ℓ = r x p

where "r" is the position vector and "p" is the linear momentum vector. We can interpret the direction of this vector product as in the case of torque. Its magnitude can be obtained using any of the following relations :

Figure 10: Angular momentum in terms of enclosed angle.

Angular momentum of a particle

1: ℓ = r p sin θ 2: ℓ = r p ⊥ 3: ℓ = r ⊥ p 1: ℓ = r p sin θ 2: ℓ = r p ⊥ 3: ℓ = r ⊥ p

If the particle is moving with a velocity "v", then the expression of angular momentum becomes :

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

Again, we can interpret this vector product as in the case of torque. Its magnitude can be obtained using any of the following relations :

1: ℓ = m r v sin θ 2: ℓ = m r v ⊥ 3: ℓ = m r ⊥ v 1: ℓ = m r v sin θ 2: ℓ = m r v ⊥ 3: ℓ = m r ⊥ v

The SI unit of angular momentum is kg-m2/s, which is equivalent to J-s.

Example 5

Problem : A particle of mass, "m", moves with a constant velocity "v" along a straight line parallel to x-axis as shown in the figure. Find the angular momentum of the particle about the origin of the coordinate system. Also discuss the nature of angular momentum in this case.

Figure 11: The particle is moving with a constant velocity.

Angular momentum of a particle

Solution : The magnitude of the angular momentum is given by :

ℓ = m r v sin θ ℓ = m r v sin θ

This expression can rearranged as :

ℓ = m v ( r sin θ ) ℓ = m v ( r sin θ )

From the ΔOAC, it is clear that :

Figure 12: The particle is moving with a constant velocity.

Angular momentum of a particle

r sin θ = AC r sin θ = AC

At another instant, we have :

r ' sin θ ' = BD r ' sin θ ' = BD

But the perpendicular distance between two parallel lines are same (AC = BD). Thus,

r sinθ = a constant ⇒ r sin θ = a constant ⇒ r sin θ = a constant

Also, the quantities "m" and "v" are constants. Therefore, angular momentum of the moving particle about origin "O" is a constant.

ℓ = m v ( r sin θ ) = a constant ℓ = m v ( r sin θ ) = a constant

Since angular momentum is constant, its rate of change with time is zero. But, time rate of change of angular momentum is equal to torque (we shall develop this relation in next module). It means that torque on the particle is zero as time derivate of a constant is zero. Indeed it should be so as the particle is not accelerated. This means that the concept of angular momentum is consistent even for the description of linear motion as set out in the begining of this module.