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# Rotational quantities for general motion

Module by: Sunil Kumar Singh. E-mail the author

Summary: Rotational quantities are not limited to rotation about fixed axis only.

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Rotational quantities like angular displacement, velocity, acceleration and torque etc. have been discussed in earlier modules. We were, however, restricted in interpreting and applying these quantities to either circular motion or pure rotational motion. Essentially, these physical quantities have been visualized in reference to the axis of rotation and a circular path.

The question that we seek to answer now is "can these rotational descriptions be expanded to motions in which paths are not circular or in which axis of rotation is continuously changing ?" We can recall that it needed a great deal of visualization to deal with rolling motion where disk was rotating about a moving axis. Thankfully, it was realized that pure rolling is actually a combination of pure translation and pure rotation. As such, we were able to analyze pure rolling in terms of Newton's laws. What about a motion which is not pure rolling ? We need to deal such situations with more generalized laws like conservation of angular momentum.

## General interpretation of rotational quantities

In very general term, we need to know whether it is possible to de-link concepts of rotational quantities from rotation about a fixed axis and whether it can be applied to situations in which a particle or a body is not exactly rotating ? The answer is yes.

There are many such situations in real time, which can not be subjected to simplified motion types. Consider projectile motion along a parabolic path. We can think of this motion in terms of a torque on the particle about certain axis that keeps shifting. If we observe closely 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 rotational quantities is not without purpose. We shall find that the delinking of angular concepts like torque and angular momentum from a fixed 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 pure motions (rotation about fixed axis or 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 :

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 on a particle

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 may not exist at all. First of all, there may not be a fixed axis of rotation. In this sense, we can realize that we earlier had a restricted interpretation of torque as far as rotation about a fixed axis is concerned.

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.

Mentally, we should develop techniques to quickly interpret the vector product on the right hand side of the equation. We employ right hand vector product rule to determine the direction of resulting product vector. 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 find the direction of the torque.

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 ".

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

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

### 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 :

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 2 g r = R 2 = v 2 sin 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 2 τ = m v 2 sin 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.

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

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.

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.

## Angular momentum of a particle

Angular momentum is the measure of the "quantity of motion" in rotation like linear momentum is the measure of the "quantity of motion" in translation. 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.

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

= r x p = 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 :

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

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

= r x p = m ( r x v ) = 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: = m r v sin θ 2: = m r v 3: = m r v 1: = m r v sin θ 2: = m r v 3: = 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.

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

= m r v sin θ = m r v sin θ

This expression can rearranged as :

= m v ( r sin θ ) = m v ( r sin θ )

From the ΔOAC, it is clear that :

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.

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

Since angular momentum is constant, its rate of change with time is zero. But, time rage 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. 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.

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