Motion types and associated terms

Following description is a brief of motion types and associated terms :

1: Particle

It is a hypothetical object construct occupying a single coordinate point. All mass is concentrated at that point.

2: Particle like objects

All particles composing the object under go same motion (parallel trajectories, same velocity and acceleration).

3: Extended or rigid bodies

The bodies, which are not a particle, are extended bodies. The important feature of rigid body is that mass distribution does not change with motion. Different parts of the body are locked or fixed with respect to other parts.

4: Pure translational motion

It is a motion along a straight line. We must note that motion such as projectile motion under gravity is not a pure translational motion. In general, we drop the prefix "pure" and refer it by the term "translation" alone. The idea is that if we have to refer "impure" translation, we shall qualify the same.

5: Translational kinematics

The description of motion in translation forms the subject of translational kinematics. In case of rigid or extended objects, kinematics describes motion of "center of mass" instead of individual particles or particle like objects. This study also covers translation, which are strictly not pure translational i.e. we also study translation along non-linear paths.

6: Translational dynamics

The study of cause and effect for translational motion constitutes the subject here. The Laws of motion as applied directly is valid for a particle or particle like objects only. Translational motion of a system of particles or rigid bodies is studied in conjunction with the concept of "center of mass". The law of motion, in this case, is applied not on individual particles, but on the "center of mass" of the rigid bodies.

7: Pure rotational motion

It is a motion of a particle or rigid body about a fixed axis. Each particle of the rigid body moves along a circular path. A central force called centripetal force is the requirement of rotational motion. Each of the particle constituting the body under go circular motion with same angular velocity and acceleration. Like in the case of translation, we may choose to drop the prefix "pure" and refer this motion simply by the term "rotation".

Figure 1: Each particle of the body follows a circular path about axis in pure rotational motion. .

Pure rotational motion

8: Rotational kinematics

The description of motion in rotation about a fixed axis forms the subject of rotational kinematics. If a rigid body is involved, then the mass distribution of the body should not change during rotation.

9: Rotational dynamics

The study of cause and effect for rotational motion constitutes the subject here. This is the subject matter of the discussion in this module and modules following it. Here, we shall study rotation of a particle or particle like object and rigid body. Our discussion will start with answering questions such as what is the "cause" of rotation ?

Torque in rotation

A particle or a rigid body in rotation about a fixed axis should continue to rotate with the given angular velocity indefinitely unless obstructed externally. Like translational motion, however, angular velocity of rotation can be changed by external cause. In the case of translation motion, the external cause is "force". We have to investigate what is the equivalent cause in the case of rotational motion.

Let us consider a rigid block placed over a smooth horizontal surface as shown in the figure which is subjected to a force across one of its face. What is expected? The center of mass will move with linear acceleration following Newton's second law. However, the force is not passing through the center of mass. As such, the face of the block will also be turned in the direction shown. The resulting motion will be a composite motion constituting both translation and rotation.

Figure 2: The resulting motion is a composite motion constituting both translation and rotation.

Composite motion

Let us now imagine if the body is pierced through in the middle by a vertical bar with some clearance. The vertical bar will inhibit translation and the block will only rotate about the vertical bar. The question is what caused the block to turn around? Indeed it is the force that caused the angular motion. However, it is not only the force that determined the outcome (magnitude and direction of angular velocity). There are other considerations as well.

Figure 3: The vertical bar inhibits translation and the block only rotates about the vertical bar. .

Rotational motion

We all have the experience of opening and closing a hinged door in our house. It takes lesser effort (force) to open the door when force is applied farther from the hinge. We can further reduce the effort by applying force normal to the plane of the door. On the other hand, we would require greater effort (force) if we push or pull the door from a point closer to the hinge. If we apply force in the radial direction (in the plane of door) towards the hinge, then the door does not rotate a bit - whatever be the magnitude of force. In the nutshell, rotation of the door depends on :

These factors, which cause rotation, are captured by a quantity known as torque, which is defined as :

where "r" is the position vector. There is a slight difficulty in interpreting this vector equation as applicable to rotation. The position vector,"r", as we can recall, is measured from a point. Now, the question is what is that point in the case of rotation about a fixed axis? Here, we observe that there is a unique point of application of the force. By the nature of motion, this point rotates in a plane perpendicular to the axis of rotation. We can, therefore, uniquely define the position of the point of application of force by measuring it from the point on the axis in that plane of rotation.

Figure 4: Rotation about z-axis. .

Rotational motion

The force may be directed in any possible direction. However, the body rotates about a fixed axis. It is not allowed to rotate or move about any other axes. What it means that we are limited only to the component of torque along the axis of rotation. It, then, means that we are only interested to calculate component of torque about the axis of rotation.

We can implicitly ensure calculation of the component of torque in the direction of rotation, by ensuring to only consider the component of force which is perpendicular to the axis of rotation. In the figure below, we have tried to capture this aspect. We consider only the component of force in xy-plane while evaluating the vector torque equation.

Figure 5: Rotation about z-axis.

Rotational motion

In the nutshell, we interpret the vector product with two specific conditions :

Clearly, the plane of rotation of the point of application of force is perpendicular to the axis of rotation.

Magnitude of torque

With the understandings as deliberated above, the magnitude of torque for rotation is given by :

Figure 6: Torque causes rotation or change in rotation.

Torque

τ = r F sin θ τ = r F sin θ (2)

where "r","F" and "θ" are quantities measured in the plane perpendicular to the axis of rotation.

Torque can be evaluated in two other alternative ways :

τ = r ( F sin θ ) = r F ⊥ τ = r ( F sin θ ) = r F ⊥ (3)

The magnitude is equal to the product of the magnitude of position vector and component of force perpendicular to position vector. This is an important interpretation as it highlights that it is the tangential component (perpendicular to radial direction) of force, which is capable to produce change in the rotation. The component of force in the radial direction (F cosθ) does not affect rotation.

Also,

τ = ( r sin θ ) F = r ⊥ F τ = ( r sin θ ) F = r ⊥ F (4)

The magnitude is equal to the product of the perpendicular distance and magnitude of the force. Perpendicular distance is obtained by drawing a perpendicular on the extended line of application of force as shown in the figure below. We may note here that this line is perpendicular to both the axis of rotation and force. This perpendicular distance is also known as "moment arm" of the force and is denoted as " r ⊥ r ⊥ ".

Figure 7: The magnitude is equal to the product of moment arm and magnitude of the force.

Torque

Note:

In this section, we have explained how to evaluate vector definition of torque by considering radius vector and force in the plane of rotation of the point of application of force. In case, however, if we choose to evaluate the vector definition in general terms, using vector notation in component form with unit vectors, then we should realize that torque causing angular acceleration is the component of torque in the direction of axis of rotation. Other components of torque are not considered.

Direction of torque

The nature of cross (or vector) product of two vectors, conveys a great deal about the direction of cross product i.e. torque in our case. It tells us that (i) torque vector is perpendicular to the plane formed by operand vectors i.e. "r" and "F" and (ii) torque vector is individually perpendicular to each of the operand vectors. Applying this explanation to the case in hand, we realize here that torque is perpendicular to the plane formed by radius and force vectors. Since this plane is defined by "xy" plane in the figure shown, the torque vector acts along z-axis. However, we do not know which side of the plane i.e +z or -z direction, the torque is directed.

Figure 8: Direction of torque is perpendicular to "xy" - plane.

Direction of torque

We apply right hand rule to determine the remaining piece of information, regarding direction of torque. We have two options here. Either we can shift radius vector such that tails of two vectors meet at the position of particle or we can shift the force vector (parallel shifting) so that tails of two vectors meet at the axis. Second approach has the advantage that direction of torque vector along the axis also gives the sense of rotation about that axis. Thus, following the second approach, we shift the force vector parallely to the origin as shown here.

Figure 9: Force vector is shifted to origin in order to apply right hand rule. .

Direction of torque

Now, the direction of rotation is obtained by applying rule of vector cross product. We place right hand with closed fingers such that the curl of fingers point in the direction as we transverse from the direction of position vector (first vector) to the force vector (second vector). Then, the direction of extended thumb points in the direction of torque. A counter clock-wise torque is positive, whereas clock-wise torque is negative. In this case, torque is counter-clockwise and is positive. Therefore, we conclude that torque is acting in +z-direction.

Pure rotation about a fixed axis gives us an incredible advantage in determining torque. We work with only two directions (positive and negative). In general, however, we can consider other directions of torque as well with different orientation of axis of rotation. It is worthy to note that torque follows the superposition principle. Mathematically, we can add torque vectors to obtain net or resultant torque. In words, it means that if a rigid body is subjected to more than one torque, then we can represent the torques by a single torque, which has the same effect on rotation.

Torque has the unit of Newton - meter.

Torque - acceleration relation in rotation

The relation that connects external torque to the angular acceleration has the same form as that of translational motion. In translational motion, this relationship is given as :

F = m a F = m a

Intuitively, we may conclude that relation between torque (cause) and angular acceleration (effect) should have the following form :

τ = m α τ = m α

However, we shall find that the role of inertia is not represented by mass alone in rotation. As there are corresponding quantities for force and acceleration, there is also a separate corresponding term or quantity that represents inertia to rotation. This quantity is termed as "moment of inertia", represented by symbol, "I". The equivalent Newton's second law for rotation is, thus :

We can reinforce the idea that mass alone does not determine inertia to the external torque in rotation. Often we find that a change in the orientation of the axis or alternatively the change in the distribution of mass with respect to axis of rotation, changes the requirement of external torque to produce a given angular acceleration. To prove the point, let us consider a cylindrical body and its rotation about two different axes as shown in the figure.

Figure 10: Distribution of mass about the axis affects inertia to external torque. .

Rotation

It is found that it takes lesser torque to produce a given angular acceleration in the second case. In this case, mass distribution is closer to the axis than that in the first case and, therefore, presents lesser resistance to external torque.

Torque - acceleration relation for a particle in rotation

We introduced torque and second law of motion in respect of a rigid body. This is actually an inverted sequence of study. We should have started with a particle and then should have extended the concept to the rigid body, which is an aggregation of particles with locked positions. We did this inversion purposely, because physical meaning of torque and rotation is best visualized in terms of rigid body.

For the rotation of a particle, we simulate a situation in which a particle like sphere of mass "m" is rotated by applying external force. We consider a particle of mass "m" attached to one end of a mass-less (it is a mere theoretical consideration) rod of length "r". The other hand of the rod is hinged at "O" such that the particle can be rotated in a horizontal plane about a vertical axis passing through center of circle, "O" .

Figure 11: The path rotates following a circular path about axis of rotation. .

Rotation

Now let us consider that a force "F" is applied to the particle in the plane of the circular path as shown in the figure. The radial component does not produce any acceleration as the rod is hinged at "O". The tangential component of force produces tangential acceleration, a T a T . The particle follows a circular path of radius "r" with its center at "O". From second law of motion,

F T = m a T F T = m a T

On the other hand, torque on the particle is :

τ = r F T = r m a T τ = r F T = r m a T

But we know that tangential acceleration is related to angular acceleration as :

a T = α r a T = α r

Substituting in the expression of torque, we have :

τ = r m α r = ( m r 2 ) α τ = r m α r = ( m r 2 ) α

where I = m r 2 I = m r 2 . Its unit is kg - m 2 . kg - m 2 .

τ = I α τ = I α (6)

Note:

Here, we have provided the basis of Newton's second law of rotation for a particle, using translational concept. We shall, however, revisit this topic after introducing the concept of angular momentum. Then, we shall provide the theoretical basis using angular momentum and drive the relation again.

Example 1

Problem : A particle of mass 0.1 kg attached to a light rod that rotates about an axis at a perpendicular distance of 1 m. If the angular speed of the particle is 60 rpm, find the constant torque that can stop the rotation of the particle in one minute.

Solution : The idea here is to apply Newton's second law of rotation for a particle. For this, we need to find angular deceleration.

In order to find angular deceleration, we first convert angular velocity in rad/s :

ω 1 = 60 rpm = 60 x 2 π 60 = 2 π rad / s ω 1 = 60 rpm = 60 x 2 π 60 = 2 π rad / s

Since torque "τ" is constant as given in the question, this is obviously the case of uniform deceleration. As such, we can apply equation of rotational motion for constant acceleration,

ω 1 = ω 2 + α t ω 1 = ω 2 + α t

Here,

ω 1 = 2 π rad / s ; ω 2 = 0 ; t = 60 s ω 1 = 2 π rad / s ; ω 2 = 0 ; t = 60 s

Putting values in the equation of motion, we have :

⇒ 0 = 2 π + α x 60 ⇒ α = - 2 π 60 rad / s 2 ⇒ 0 = 2 π + α x 60 ⇒ α = - 2 π 60 rad / s 2

Now, applying Newton's second law of motion for rotation,

τ = I α τ = I α

Here,

I = m r 2 = 0.1 x 1 2 = 0.1 kg - m 2 I = m r 2 = 0.1 x 1 2 = 0.1 kg - m 2

Dropping the sign of deceleration and putting values in the expression of torque, we have the magnitude of torque :

τ = I α = 0.1 x 2 π 60 = 0.01 N - m τ = I α = 0.1 x 2 π 60 = 0.01 N - m

Summary

1. The particles in a rigid body are locked and are placed at fixed linear distances from others.

2. Each particle constituting a rigid body executes circular motion about a fixed axis in pure rotation.

3. The cause of change in angular velocity in rotation is torque. It is defined as :

τ = r x F τ = r x F

where position vector,"r", and the force vector," (F)" are in the plane of rotation of the point of application of force, which is perpendicular to the axis.

4. The magnitude of torque

The magnitude of torque is determined in three equivalent ways :

(i) In terms of angle between position vector (r) and force (F)

τ = r F sin θ τ = r F sin θ

(ii) In terms of tangential component of force ( F T F T or F ⊥ F ⊥ )

τ = r ( F sin θ ) = r F ⊥ τ = r ( F sin θ ) = r F ⊥

(iii) In terms of moment arm ( r ⊥ r ⊥ )

τ = ( r sin θ ) F = r ⊥ F τ = ( r sin θ ) F = r ⊥ F

4. Direction of torque

The vector equation of torque reveals that torque is perpendicular to the plane formed by position and force vectors and is also perpendicular to each of them individually. In order to know the sign of torque (positive for anticlockwise and negative for clockwise rotation), we apply right hand rule.

5. Newton’s second law of motion in angular form for a rotating rigid body

It is given as :

τ = I α τ = I α

6. Newton’s second law of motion in angular form for a rotating particle

It is given as :

τ = I α τ = I α

where I is m r 2 m r 2 .

7. Rotation about an axis can be analyzed in terms of a simplified vector notation format. Since only two directions are involved, we can use Newton’s second law of motion for rotation in terms of equivalent scalar angular quantities with appropriate sign : positive for anticlockwise rotation and negative for clockwise rotation.