Chapter 4: Laws of Motion
In the 1920s, an American physicist named Robert Goddard was quietly building the world’s first liquid-fuelled rockets. He claimed that a rocket could one day travel beyond the atmosphere and even reach the Moon. A leading American newspaper published an editorial mocking him. Its argument sounded like common sense: out in space there is no air, and a rocket needs something to push against in order to move forward. Since a vacuum offers nothing to push against, the writer concluded, Goddard clearly did not understand basic physics.
The newspaper was wrong, and the physics it trusted was wrong. A rocket does not push against the air. It pushes against its own exhaust. The engine throws burnt gases backwards at enormous speed, and those gases, in return, push the rocket forwards with an equal and opposite force. Air is not needed; in fact, air only slows the rocket down.
Nearly fifty years later, in 1969, as Apollo 11 was on its way to the Moon, the same newspaper printed a short and graceful correction, admitting that a rocket does work in a vacuum after all.
That single misunderstanding — about what really makes things move, speed up, or stop — sits at the heart of this chapter.
Diagram
Fig. 4.1 – A rocket in airless space ejecting exhaust gases downward while accelerating upward, with the two opposite force arrows marked on rocket and gases.
By the end of this chapter, you will be able to answer each of these questions using the laws of motion, the idea of momentum, torque and the centre of mass.
4.1 Introduction
An object is said to be in motion if it continuously changes its position. This simple statement is the starting point of a whole branch of Physics called mechanics, which studies motion and the causes behind it.
The Branches of Mechanics
Mechanics is broadly divided into two parts. The first is statics, which deals with objects that are at rest, or in equilibrium under the action of balanced forces. The second is kinetics, which deals with objects that are actually moving.
Kinetics itself is further divided into two branches, and the difference between them is worth understanding clearly, because the whole of this chapter belongs to the second one.
(i) Kinematics: Here we only describe motion. We do not ask what caused it. The quantities we work with are distance, displacement, speed, velocity and acceleration. In Chapter 3 you used exactly these quantities.
(ii) Dynamics: Here we describe the motion along with its cause — and the cause of motion is force and/or torque. In dynamics we therefore need extra quantities: momentum, force, energy, power and so on, in addition to all the kinematic quantities.
Motion is Always Relative
It must be clearly understood that motion is a strictly relative concept. It has no meaning unless we state with respect to what the object is moving. That “what” is called a frame of reference.
For example, if you are sitting in a running bus, then neither you nor your fellow passengers are moving in your reference, that is, with respect to the moving bus. Your co-passenger’s position with respect to you does not change at all. But in the reference of the ground, the bus, you and every passenger inside it are all in motion.
Neither of these two descriptions is “wrong”. They are simply two different, equally valid answers to a question that was incomplete until the frame of reference was named. We will return to frames of reference in detail in Section 4.4, where we shall see that some frames are more convenient than others for applying Newton’s laws.
Types of Motion and What Decides Them
Real-life motions, if they are not completely random, can usually be understood as linear, circular or rotational, oscillatory, or as some combination of these. To describe any of them we must know the forces responsible for them.
The trajectory — the actual path traced by the moving object — is decided by only two things: the acceleration \(\vec{a}\) and the initial velocity \(\vec{u}\). This is a powerful idea. The same acceleration can produce completely different-looking paths, depending only on how the object was launched. Let us see the possibilities one by one.
(a) Linear motion: Here the initial velocity may be zero or non-zero.
If the initial velocity is zero, that is, the object starts from rest, then an acceleration in any direction will result in linear motion. The object simply begins moving along the direction of the acceleration.
If the initial velocity is not zero, then the acceleration must be in line with the initial velocity — either along the same direction or exactly opposite to it — for the resultant motion to be linear. Along the same direction the object speeds up; along the opposite direction it slows down, but in both cases it stays on the same straight line.
(b) Circular motion: If the initial velocity is not zero and the acceleration is throughout perpendicular to the velocity, the resultant motion will be circular. Since the acceleration is always at right angles to the velocity, it can never change the speed of the object; it can only keep bending the direction of the velocity, and the object curls into a circle.
Diagram
Fig. 4.2 – Three panels showing initial velocity vector u and acceleration vector a arranged (i) along the same line, (ii) perpendicular to each other, (iii) at an arbitrary angle, with the resulting straight-line, circular and parabolic paths drawn below each.
(c) Parabolic motion: If the acceleration is constant and the initial velocity is not in line with the acceleration, the motion is parabolic. The trajectory of a projectile, which you have already studied, is the standard example: the initial velocity points at some angle, while the acceleration due to gravity points steadily downward.
(d) Other combinations of \(\vec{u}\) and \(\vec{a}\) will result in different and more complicated motions.
In numerical problems, examiners often test exactly this idea without saying so: you are given the direction of the initial velocity and the direction of the net force, and asked to identify the shape of the path. Look at the angle between them first — that single angle decides the answer.
4.2 Aristotle’s Fallacy
Aristotle (384 BC – 322 BC), one of the greatest thinkers of the ancient world, stated that “an external force is required to keep a body in uniform motion.” In other words, according to him, motion at constant velocity is not something a body can maintain by itself; it must be continuously fed by a force.
This idea was almost certainly built on ordinary, everyday experience. Roll a ball on a floor and it travels some distance, gradually slows down and comes to a stop on its own. So, if you want the ball to keep moving with constant velocity, you clearly have to keep pushing it. Stop pushing, and the motion dies out.
Similar examples are easy to find. A paper plane thrown across a room glides for a while and then drops. A paper boat given some initial velocity in water slowly loses its speed and stops. In every case the moving object seems to “run out” of motion the moment we stop supplying force.
Diagram
Fig. 4.3 – A ball rolling on a floor slowing down and stopping, with the backward-pointing frictional force marked at the contact point; alongside, a paper plane in air and a paper boat in water with backward-pointing resistive force arrows.
Where Aristotle Went Wrong
The correct explanation of this everyday observation was first given by Galileo (1564 – 1642), and it was later used by Newton (1643 – 1727) while formulating the laws of motion.
Galileo showed that all these objects stop moving not because motion needs a force to sustain it, but because some resistive or opposing force is secretly acting on them all the time. Once you notice that hidden force, the puzzle disappears.
For the rolling ball, that hidden force is the force of friction — the opposing force that acts between two surfaces in contact whenever one tends to slide over the other. For the paper plane it is the viscous drag, or viscous force, of the air — the resistance a fluid offers to an object moving through it. For the paper boat it is the viscous force of the water.
The Corrected Statement
So, in reality, for the uninterrupted motion of a body, an additional external force is required — but only for overcoming these opposing forces, not for sustaining the motion itself.
Notice the shift in language, because it is exactly what board examiners look for. Aristotle needed a force for motion. We need a force only against the opposing force. The moment the applied force exactly balances friction, the net force becomes zero — and the body keeps moving at constant velocity without any further help. This is the doorway to Newton’s first law, which we take up next.
4.3 Newton’s Laws of Motion
Galileo cleared away the confusion; Newton turned the cleared ground into three precise laws. These three statements are the foundation of the whole of mechanics, so read each one slowly and notice every condition attached to it.
The Three Laws
First law: Every inanimate object continues to be in its state of rest or of uniform unaccelerated motion unless and until it is acted upon by an external, unbalanced force.
Three words in this statement are doing all the work. Unaccelerated means the velocity is constant in both magnitude and direction. External means the force must come from outside the body — a body cannot push itself. Unbalanced means it is the net force that matters; if several forces act but cancel out, the body behaves exactly as if no force acted at all.
Second law: The rate of change of linear momentum of a rigid body is directly proportional to the applied force and takes place in the direction of the applied force.
The linear momentum of a body, written \(\vec{p}\), is defined as the product of its mass and its velocity:
\[\vec{p} = m\vec{v}\]
Here \(m\) is the mass in kilogram (kg) and \(\vec{v}\) is the velocity in metre per second (m s⁻¹), so the SI unit of momentum is kg m s⁻¹. Momentum is a vector, pointing along the velocity. On selecting suitable units, the second law takes the form
\[\vec{F} = \frac{d\vec{p}}{dt}\]
where \(\vec{F}\) is the force in newton (N) and \(\vec{p} = m\vec{v}\) is the linear momentum.
Third law: To every action (force), there is an equal and opposite reaction (force).
Discussion: How the Second Law Contains the First
The three laws are not three unrelated announcements. Watch what happens when we open up the second law.
From Newton’s second law of motion,
\[\vec{F} = \frac{d\vec{p}}{dt} = \frac{d}{dt}\left(m\vec{v}\right)\]
For a given body, the mass \(m\) is constant, so it can be pulled out of the differentiation:
\[\therefore \vec{F} = m\frac{d\vec{v}}{dt} = m\vec{a} \quad \text{... (for constant mass)}\]
where \(\vec{a} = \dfrac{d\vec{v}}{dt}\) is the acceleration in m s⁻².
Now suppose no force acts. Thus, if \(\vec{F} = 0\), then \(\vec{a} = 0\), which means \(\vec{v}\) is constant. Hence if there is no force, the velocity will not change — neither in size nor in direction. This is nothing but Newton’s first law of motion.
The answer to that question is exactly what the next three subsections are about. Each law carries meaning that the mathematics alone does not reveal, and that is why all three are stated separately.
4.3.1 Importance of Newton’s First Law of Motion
(i) It shows an equivalence between the ‘state of rest’ and the ‘state of uniform motion along a straight line’. Both of these need a net unbalanced force in order to be changed. Because they behave identically in this respect, both are referred to by a single name — the ‘state of motion’. The distinction between rest and uniform motion is not a distinction of physics at all; it lies only in the choice of the frame of reference. A passenger asleep in a moving train is at rest for the train and in uniform motion for the platform, and no experiment inside the train can tell the two apart.
(ii) It defines force. The first law tells us what a force is: an entity, or a physical quantity, that brings about a change in the ‘state of motion’ of a body. In other words, force is something that initiates a motion or controls a motion. The second law then gives its quantitative understanding, that is, its mathematical expression. The first law says what force does; the second says how much.
(iii) It defines inertia. The first law introduces inertia as a fundamental property of every physical object, by which the object resists any change in its state of motion. Inertia is measured as the mass of the object. More specifically it is called inertial mass, which is the ratio of the net force \(\left(\left|\vec{F}\right|\right)\) to the corresponding acceleration \(\left(\left|\vec{a}\right|\right)\).
4.3.2 Importance of Newton’s Second Law of Motion
(i) It gives the mathematical formulation for a quantitative measure of force as the rate of change of linear momentum. Force stops being a vague push and becomes a number we can compute and measure.
(ii) It defines momentum \(\left(\vec{p} = m\vec{v}\right)\), instead of velocity, as the fundamental quantity related to motion. This is a subtle but exam-critical point. What is changed by a force is the momentum, and not necessarily the velocity. A force can act on a body whose mass is also changing, and then the velocity may behave in ways that \(\vec{F} = m\vec{a}\) alone would never predict.
(iii) Aristotle’s fallacy is overcome by considering the resultant unbalanced force. Once we agree that only the net force matters, the rolling ball no longer needs a mysterious “force to keep it going”.
4.3.3 Importance of Newton’s Third Law of Motion
(i) It defines action and reaction as a pair of equal and opposite forces acting along the same line.
(ii) Action and reaction forces are always on different objects. This single line prevents most of the mistakes students make with the third law.
Consequences
The action force exerted by a body \(x\) on a body \(y\), conventionally written as \(\vec{F}_{yx}\), is the force experienced by \(y\).
As a result, body \(y\) exerts a reaction force \(\vec{F}_{xy}\) on body \(x\).
In this case, body \(x\) experiences the force \(\vec{F}_{xy}\) only, while the body \(y\) experiences the force \(\vec{F}_{yx}\) only. Notice carefully: neither body feels both forces.
The forces \(\vec{F}_{xy}\) and \(\vec{F}_{yx}\) are equal in magnitude and opposite in their directions, but there is no question of cancellation of these forces, as those are experienced by different objects. Two forces can cancel only when they act on the same body. Since these two never do, they can never cancel, and that is why a horse can pull a cart at all.
Diagram
Fig. 4.4 – Two bodies x and y in contact, with the action force \(\vec{F}_{yx}\) drawn on body y and the reaction force \(\vec{F}_{xy}\) drawn on body x, clearly on separate bodies and along the same line.
The forces \(\vec{F}_{xy}\) and \(\vec{F}_{yx}\) need not be contact forces. The repulsive forces between two magnets form a pair of action–reaction forces, and in this case the two magnets are not in contact at all. The gravitational force between the Earth and the Moon, or between the Earth and the Sun, are also similar pairs of non-contact action–reaction forces.
4.4 Inertial and Non-Inertial Frames of Reference
Consider yourself standing on a railway platform or at a bus stand, watching a train or a bus go past. According to you, that train or bus is moving. But as per the experience of the passengers inside the train or bus, they are at rest and you are the one moving — backwards. Neither party is lying. As we saw in Section 4.1, motion itself is a relative concept, and to describe a motion you must first define the reference with respect to which you are describing it. Such a reference is called a frame of reference.
In the example discussed above, if you consider the platform as the reference, then the passengers and the train are moving. However, if the train is considered as the reference, then you and the platform are the things that are moving.
Describing a Frame Mathematically
Usually a set of coordinates with a suitable origin is enough to describe a frame of reference. Fix an origin, fix three axes, and you have a frame. If the position coordinates of an object are continuously changing with time in a frame of reference, then that object is in motion in that frame of reference.
So far, every frame is as good as every other. But for doing physics they are not equally good — and Newton’s first law is what separates them.
The Inertial Frame
Any frame of reference in which Newton’s first law of motion is applicable gives us the simplest understanding of an inertial frame of reference. It means: if there is no net force, there is no acceleration. Thus, in an inertial frame, a body will move with constant velocity (which may be zero also) if there is no net force acting upon it.
Now turn the statement around. In the absence of a net force, if an object is nevertheless seen to suffer an acceleration, then that frame of reference is not an inertial frame, and it is called a non-inertial frame of reference.
Notice how the test works. You do not have to look outside and check whether the frame is “really” moving. You simply check, from inside, whether a force-free body stays unaccelerated. If it does not, the frame itself is accelerating.
Measurements made in one inertial frame can be converted to measurements in another inertial frame by a simple transformation, i.e., by simply using some velocity vectors — the relative velocity between the two frames of reference. This is why two observers moving uniformly with respect to each other never disagree about physics; they disagree only about numbers, and those numbers translate easily.
Illustration: The Closed Car
Imagine yourself inside a car with all the windows made opaque, so that you cannot see anything outside. Also consider that there is a pendulum tied inside the car and it is not set into oscillations. You have no external clue at all about the car’s motion — everything you learn must come from inside.
If the car just starts its motion (with reference to the outside or the ground), you will experience a jerk, i.e., an acceleration inside the car, even though there is no force acting upon you. During this time, the string of the pendulum may be steady, but it will not be vertical; it will hang tilted, leaning backwards. During this time of acceleration, the car can be considered to be a non-inertial frame of reference.
Diagram
Fig. 4.5 – A closed car with opaque windows containing a pendulum: (a) car accelerating, string steady but tilted backwards; (b) car moving with constant velocity, string hanging vertical.
Later on, if the car is moving with constant velocity (with reference to the ground), you will not experience any jerky motion within the car, and the car can be considered as an inertial frame of reference. In this case, the pendulum string will be vertical, when not oscillating.
So the tilted string is the give-away. A sealed box that tells you nothing about the outside world still tells you, through that tilt, whether your frame is accelerating.
Does a Perfect Inertial Frame Exist?
Strictly speaking, a truly inertial frame is an idealisation. A rocket in intergalactic space — the gravity-free space between galaxies — with all its engines shut is the closest we can come to an ideal inertial frame. Nothing pulls it, nothing pushes it, and nothing accelerates it.
The Earth is certainly not such a place: it spins on its axis and revolves around the Sun, so it is accelerating. However, the Earth’s acceleration in the reference frame of the Sun is so small that any frame attached to the Earth can be used as an inertial frame for any day-to-day situation or in our laboratories. The error we make by calling the laboratory “inertial” is far too small to be noticed in a school experiment.
4.5 Types of Forces
We have used the word “force” freely so far. It is now time to ask what forces actually exist in nature, how they are classified, and which of them we are really dealing with when we speak of friction, tension or a normal reaction.
4.5.1 Fundamental Forces in Nature
All the forces in nature are classified into the following four interactions, which are termed as fundamental forces.
(i) Gravitational Force
It is the attractive force between two (point) masses separated by a distance. The magnitude of the gravitational force between point masses \(m_1\) and \(m_2\) separated by a distance \(r\) is given by
\[F = \frac{Gm_1m_2}{r^2}\]
where \(G = 6.67 \times 10^{-11}\) SI units is the universal gravitational constant, \(m_1\) and \(m_2\) are in kilogram (kg) and \(r\) is in metre (m), so that \(F\) comes out in newton (N).
Between two point masses (particles) separated by a given distance, this is the weakest force, having infinite range. This force is always attractive — there is no such thing as gravitational repulsion. The structure of the universe, from planets to galaxies, is governed by this force.
The common experience of this force for us is the gravitational force exerted by the Earth on us, which we call our weight \(W\).
\[\therefore W = \frac{GMm}{R^2} = m\left(\frac{GM}{R^2}\right) = mg\]
where \(M\) and \(R\) represent respectively the mass and radius of the Earth. The distance between ourselves and the Earth is taken as the radius of the Earth when we are on the surface of the Earth, because our size is negligible as compared to the radius of the Earth (\(6.4 \times 10^6\) m). In other words, we are so small compared to the planet that our distance from its centre is essentially just \(R\).
Substituting the values,
\[\left(\frac{GM}{R^2}\right) = \frac{6.67\times10^{-11} \times 6\times10^{24}}{\left(6.4\times10^6\right)^2}\]
\[\cong 9.8 \text{ m/s}^2 = g = \text{gravitational acceleration or gravitational field intensity}\]
So the familiar \(g\) is not a separate fact of nature that has to be memorised — it is simply what \(GM/R^2\) works out to for our planet.
We feel this force only due to the normal reaction from the surface of our contact with the Earth. Gravity itself pulls every particle of our body equally and produces no sensation; what we sense as “weight” is the floor pushing up on our feet.
All individual bodies also exert gravitational force on each other, but it is too small compared to that exerted by the Earth. For example, the mutual gravitational force between two SUMO wrestlers, each of mass 300 kg, assuming the distance between them is 0.5 m, will be
\[F = \frac{6.67\times10^{-11}\times300\times300}{0.5^2}\]
\[\cong 24\times10^{-5}\,\text{N} = 24\ \mu\text{N}\]
This force is negligibly small in comparison to the weight of each SUMO wrestler \(\cong 3000\) N. Two very heavy people standing half a metre apart attract each other with a force smaller than the weight of a mosquito — which is why we never notice gravity between ordinary objects at all.
(ii) Electromagnetic (EM) Force
It is an attractive or repulsive force between electrically charged particles. Earlier, electric and magnetic forces were thought to be independent of each other. After the demonstrations by Michael Faraday (1791 – 1867) and James Clerk Maxwell (1831 – 1879), electric and magnetic forces were unified through the theory of electromagnetism.
These forces are much stronger than the gravitational force. Our life is practically governed by these forces. The majority of forces experienced in our daily life — such as the force of friction, normal reaction, tension in strings, collision forces, elastic forces, viscosity (fluid friction), and so on — are EM in nature.
How can a push from a table be “electromagnetic”? Under the action of these forces, there is deformation of objects, which changes the intermolecular distances, thereby resulting in reaction forces. Squeeze the molecules of the table top a little closer than they like to be, and their own electromagnetic repulsion pushes back. That push is what you feel as the normal force.
(iii) Strong (Nuclear) Force
This is the strongest force, which binds the nucleons together inside a nucleus. Though strongest, it is a short range (\(< 10^{-14}\) m) force. It is therefore a very strong attractive force, and it is charge independent — it acts equally between proton–proton, proton–neutron and neutron–neutron pairs. This is what holds a nucleus together in spite of the fierce electrical repulsion between its protons.
(iv) Weak (Nuclear) Force
This is the interaction between subatomic particles that is responsible for the radioactive decay of atoms, in particular beta emission. The weak nuclear force is not as weak as the gravitational force, but it is much weaker than the strong nuclear and EM forces. The range of the weak nuclear force is exceedingly small, of the order of \(10^{-16}\) m.
4.5.2 Contact and Non-Contact Forces
For some forces, like the gravitational force, electrostatic force, magnetostatic force, etc., physical contact is not an essential condition. These forces exist even if the objects are distant or physically separated. Such forces are non-contact forces.
Forces resulting only due to contact are called contact forces. All these are EM in nature, arising due to some deformation. Normal reaction, forces occurring during collision, force of friction, etc., are contact forces.
There are two common categories of contact forces.
Normal force. Two objects in contact, while exerting mutual force, try to push each other away along their common normal. Quite often we call it the ‘normal reaction’ force or simply the ‘normal’ force. While standing on a table, we push the table away from us (downward) and the table pushes us away from it (upward), both being equal in magnitude and acting along the same ‘normal’ line.
Friction. The force of friction is also a contact force, that arises whenever there is a relative motion or a tendency of relative motion between surfaces in contact. This is the parallel (or tangential) component of the reaction force. In this case, the molecules of the surfaces in contact, which have developed a certain equilibrium, are required to be separated — and they resist being separated.
So the single reaction force that a surface offers can be split into two parts: the component perpendicular to the surface, which we call the normal force, and the component along the surface, which we call friction. Notice the word tendency in the definition of friction: friction acts even before anything actually slides.
Diagram
Fig. 4.7 – A block resting on a rough surface with an applied horizontal force; the total reaction from the surface shown resolved into a normal component perpendicular to the surface and a frictional component along the surface.
4.5.3 Real and Pseudo Forces
Consider ourselves inside a lift (or elevator). When the lift just starts moving up (accelerates upward), we feel a bit heavier, as if someone is pushing us down. This is not imaginary, and not just a feeling. If we are standing on a weighing scale inside this lift, during this time the weighing scale records an increase in weight.
During travelling with uniform upward velocity, no such change is recorded — the scale reads exactly what it reads at rest.
While stopping at some upper floor, the lift undergoes downward acceleration in order to decrease the upward velocity. In this case the weighing scale records a loss in weight and we also feel lighter.
These extra upward or downward forces are: (i) Measurable, which means they are not imaginary; (ii) not accountable as per Newton’s second law in the inertial frame; and (iii) not among any of the four fundamental forces.
When we are inside a bus, such forces are experienced when the bus starts to move (forward acceleration), when the bus is about to stop (backward acceleration), or when it takes a turn (centripetal acceleration). In all these cases we are inside an accelerated system, which is our frame of reference.
If a force measuring device is suitably used — like the weighing scale recording the change in weight — these forces can be recorded, and they will be found to be always opposite to the acceleration of your frame of reference. They are also exactly equal to \(-m\vec{a}\), where \(m\) is our mass and \(\vec{a}\) is the acceleration of the system (frame of reference).
We have already defined or described real forces to be those which obey Newton’s laws of motion and are one of the four fundamental forces. The forces in the above illustrations do not satisfy this description and cannot be called real forces. Hence these are called pseudo forces.
Weight Inside an Accelerating Lift
As per the illustration of the lift with downward acceleration \(\vec{a}\), the weight experienced will be
\[\vec{W} = m\vec{g} + \left(-m\vec{a}\right)\]
As \(\vec{g}\) and \(\vec{a}\) are along the same direction in this case (both point downward), the pseudo force \(-m\vec{a}\) points upward, and in magnitude
\[W = mg - ma\]
This explains the feeling of a loss in weight.
During upward acceleration, say \(\vec{a}_1\), we have
\[\vec{W}_1 = m\vec{g} + \left(+m\vec{a}_1\right)\]
In this case, \(\vec{g}\) and \(\vec{a}_1\) are oppositely directed, so the pseudo force \(-m\vec{a}_1\) now points downward, adding to the weight.
\[\therefore W_1 = mg + \left(+ma_1\right) = mg + ma_1\]
That explains the gain in weight, or the existence of an extra downward force.
Diagram
Fig. 4.8 – A person standing on a weighing scale inside a lift, shown in three panels: lift accelerating upward (scale reads more), moving with uniform velocity (scale reads mg), accelerating downward (scale reads less), with mg and the pseudo force marked in each.
4.5.4 Conservative and Non-Conservative Forces and Concept of Potential Energy
Consider an object lying on the ground, which is lifted and kept on a table. Neglecting air resistance, the amount of work done is the work done against the gravitational force, and it is independent of the actual path chosen. (Remember, as there is no air resistance, the gravitational force is the only force acting.) Lift it straight up, or carry it in a wide curve — the work is the same.
Similarly, while keeping the same object back on the ground from the table, the work is done by the gravitational force. In either case the amount of work done is the same and is independent of the actual path chosen.
The work done by a force \(\vec{F}\) in moving the object through a distance \(dx\) can be mathematically represented as
\[dW = \vec{F}.d\vec{x} = -dU \quad\text{or}\quad dU = -\vec{F}.d\vec{x}\]
where \(U\) is the potential energy, measured in joule (J).
If the work done by or against a force is independent of the actual path, the force is said to be a conservative force.
During the work done by a conservative force, the mechanical energy (the sum of kinetic and potential energy) is conserved. In fact, we define the concept of potential at a point, or potential energy (in the topic of gravitation), with the help of conservative forces only.
The work done by or against conservative forces reflects an equal amount of change in the potential energy. The corresponding work done is used in changing the position, or in achieving the new position, in the gravitational field. Hence, potential energy is often referred to as the energy possessed on account of position.
In the illustration given above, the work done is reflected as an increase in the gravitational potential energy when the displacement is against the (gravitational) force. The same amount of potential energy is decreased when the displacement is in the direction of the force. In either case it is independent of the actual path, but depends only upon the initial and final positions. This change in the potential energy takes place in such a way that the mechanical energy is conserved.
As discussed above, the increase in the potential energy is
\[dU = -\vec{F}.d\vec{x} \quad\text{or}\quad U = -\int \vec{F}.d\vec{x}\]
where \(\vec{F}\) is a conservative force. This concept will be described in detail in Chapter 5 on Gravitation, in the context of gravitational potential energy and gravitational potential.
When the Work Cannot Be Recovered
During this process, if friction or air resistance is present, additional work is necessary against the frictional force (for the same displacement). This work is strictly path dependent and not recoverable. Take a longer path and you must do more work against friction; there is no way to get it back.
Such forces (like friction, air drag, etc.) are called non-conservative forces. Work done against these forces appears as heat, sound, light, etc. The work done against non-conservative forces is not recoverable even if the path is exactly reversed.
4.5.5 Work Done by a Variable Force
The popular formula for calculating work done is
\[W = \vec{F}\cdot\vec{s} = F\,s\cos\theta\]
where \(\theta\) is the angle between the applied force \(\vec{F}\) (in newton) and the displacement \(\vec{s}\) (in metre), and \(W\) is in joule (J).
This formula is applicable only if both the force \(\vec{F}\) and the displacement \(\vec{s}\) are constant and finite. That is a much stronger condition than it looks.
In several real-life situations, the force is not constant. For example, while lifting an object through several thousand kilometres, the gravitational force is not constant — it falls off as \(1/r^2\). The viscous forces like fluid resistance depend upon the speed, hence, quite often, they are not constant with time. In order to calculate the work done by such variable forces, we use integration.
Illustration: Splitting the Path into Tiny Pieces
Figure 4.9(a) shows the variation of a force \(\vec{F}\) plotted against the corresponding displacements in its direction, \(\vec{s}\). As the displacement is in the direction of the applied force, the vector nature is not used. We need to calculate the work done by this force during a displacement from \(s_1\) to \(s_2\).
Diagram
Fig. 4.9 (a) – Work done by nonlinearly varying force: a curved F–s graph with a narrow shaded vertical strip of width ds between s₁ and s₂, force axis starting from zero.
As the force is variable, using \(W = F\left(s_2 - s_1\right)\) directly is not possible — there is no single value of \(F\) to use. In order to use integration, let us divide the displacement into a large number of infinitesimal (infinitely small) displacements. One of such displacements is \(ds\).
It is so small that the force \(F\) is practically constant for this displacement. “Practically constant” means the change in the force is so small that the change cannot be recorded. The shaded strip in the figure shows one of such displacements.
As the force is constant over this strip, the area of this strip, \(F.ds\), is the work done \(dW\) for this displacement. The total work done \(W\) for the displacement \(\left(s_2 - s_1\right)\) can then be obtained by using integration — that is, by adding up all such strips.
\[\therefore W = \int_{s_1}^{s_2} F.ds\]
The method of integration is applicable if the exact way of variation in \(\vec{F}\) with \(\vec{s}\) is known, and if that function is integrable.
The area under the curve between \(s_1\) and \(s_2\) also gives the work done \(W\) (if the force axis necessarily starts with zero), as it consists of all the strips of \(ds\) between \(s_1\) and \(s_2\).
In Fig. 4.9(b), the variation in the force is linear. In this case, the area of the trapezium AS₁S₂B gives the total work done \(W\) — and for a straight-line graph you do not need calculus at all; the area of a trapezium is enough.
Diagram
Fig. 4.9 (b) – Work done by linearly varying force: a straight line from A above s₁ sloping down to B above s₂, with the trapezium AS₁S₂B shaded, force axis starting from zero.
The condition in brackets is exam-critical. If the force axis does not start at zero, the “area under the curve” you see on the page is not the true area under the curve, and reading work off the graph will give a wrong answer.
4.6 Work Energy Theorem
Suppose there is a decrease in the potential energy of a body — for instance, a body falling down — due to a conservative force. Then that lost potential energy is entirely converted into kinetic energy. The work done by the force then appears as kinetic energy.
The reverse is equally true. If an object is moving against a conservative force, its kinetic energy decreases by an amount exactly equal to the work done against the force. Nothing is lost; the kinetic energy simply reappears as potential energy.
This principle is called the work–energy theorem for conservative forces.
Let us now prove it, and then see what happens when a non-conservative force such as air resistance enters the picture.
Case I: A Constant Opposing Force
Consider an object of mass \(m\) moving with velocity \(\vec{u}\), experiencing a constant opposing force \(\vec{F}\) which slows it down to \(\vec{\mathrm{v}}\) during a displacement \(\vec{s}\).
As \(\vec{u}\) and \(\vec{F}\) are oppositely directed, the entire motion will be along the same line. In this case we need not use the vector form; just \(\pm\) signs should be good enough. This is a useful simplification and you should look for it in every one-dimensional problem.
Diagram
Fig. 4.10 – A block of mass m moving to the right with initial velocity u, a constant opposing force F acting to the left, and the final velocity v shown after a displacement s.
If \(a = \dfrac{F}{m}\) is the acceleration, we can write the relevant equation of motion as
\[\mathrm{v}^2 - u^2 = 2(-a)s\]
The negative sign appears because the acceleration is negative for an opposing force — the force acts opposite to the displacement.
Multiplying throughout by \(m/2\), we get
\[\frac{1}{2}mu^2 - \frac{1}{2}m\mathrm{v}^2 = \left(ma\right).s = F.s\]
Look carefully at what each side of this equation is saying.
The left-hand side is the decrease in the kinetic energy of the body, since \(\tfrac12 mu^2\) was its initial kinetic energy and \(\tfrac12 m\mathrm{v}^2\) is its final kinetic energy. Kinetic energy is measured in joule (J).
The right-hand side, \(F.s\), is the work done by the force.
Thus, the change in kinetic energy is equal to the work done by the conservative force, which is in accordance with the work–energy theorem.
Case II: An Accelerating Conservative Force along with a Retarding Non-Conservative Force
Now take the more realistic situation. An object dropped from some point at height \(h\) falls down through air.
While coming down, its potential energy decreases. An equal amount of work is done in this case also — gravity does not care that air is present.
However, this time the work is not entirely converted into kinetic energy. Some part of it is used in overcoming the air resistance. This part of the energy appears in some other forms, such as heat, sound, etc. That is why the ball reaches the ground slower than the ideal calculation \(\mathrm{v} = \sqrt{2gh}\) would predict.
In this case, the work–energy theorem can mathematically be written as
\[\Delta PE = \Delta KE + W_{air\ resistance}\]
In words: Decrease in the gravitational P.E. = Increase in the kinetic energy + work done against non-conservative forces.
The energy is not destroyed. It is only pushed into forms — heat in the air, sound in your ears — from which we cannot get it back as motion.
Diagram
Fig. 4.11 – A ball falling from height h through air, with the weight mg acting downward and the air-resistance force acting upward, and a bar chart alongside showing the initial PE splitting into final KE plus energy lost against air resistance.
The magnitude of the air resistance force is not constant, but depends upon the speed. Hence it cannot simply be multiplied by the distance fallen; it has to be written as
\[\int \vec{F}.d\vec{s}\]
as seen during the work done by (or against) a variable force in Section 4.5.5. This is a good illustration of why the integral form of work was worth learning: the moment a real, speed-dependent force enters the problem, the simple product \(F \times s\) stops working.
4.7 Principle of Conservation of Linear Momentum
According to Newton’s second law, the resultant force is equal to the rate of change of linear momentum, or
\[\vec{F} = \frac{d\vec{p}}{dt}\]
Read this equation backwards for a moment. It says that a change in momentum requires a force. In other words, if there is no resultant force, the linear momentum will not change, or will remain constant, or will be conserved.
Mathematically, if
\[\vec{F} = 0,\quad \frac{d\vec{p}}{dt} = 0 \quad\text{or}\quad \vec{p} \text{ is constant (or conserved).}\]
This leads us to the principle of conservation of linear momentum.
Every word of that statement is doing work. “Isolated” means no external force acts. “Total” means the momentum of the whole system, not of the individual pieces — the individual momenta certainly do change. “Any interaction” means it does not matter whether the interaction is a gentle nudge, a violent collision, or an explosion.
Systems and Free Body Diagrams
We now look at the mathematical approach for the application of Newton’s second law to a real arrangement with several connected bodies.
Consider the arrangement shown in Fig. 4.12(a). Pulleys P₁, P₃ and P₄ are fixed, while P₂ is movable. A force \(F = 100\) N, applied at an angle of 60° with the horizontal, is responsible for the motion, if any. The contact surface of the 5 kg mass offers a constant opposing force \(F = 10\) N. Except this, there are no resistive forces anywhere.
Diagram
Fig. 4.12 (a) – System for illustration of free body diagram: fixed pulleys P₁, P₃, P₄ and movable pulley P₂, with 1 kg, 2 kg, 3 kg and 4 kg masses connected by strings and a 5 kg block on a rough horizontal surface pulled by a 100 N force at 60° to the horizontal.
Discussion: Until the 1 kg mass reaches the pulley P₁, the motion of the 1 kg and 2 kg masses is identical. Thus, these two can be considered to be a single system of mass 3 kg, except for knowing the tension \(T_3\). The forces due to the tension in the string joining them are internal forces for this system — and, as we have just seen, internal forces never affect the motion of the system as a whole. They matter only when you want to know \(T_3\) itself, and for that you must break the system open again.
All masses except the 3 kg mass are travelling the same distance in the same time. Thus, their accelerations, if any, have the same magnitudes. If the string S connecting the 1 kg and 4 kg masses moves by \(x\), the lower string S₁ holding the 3 kg mass moves through \(x/2\). This halving comes from the movable pulley P₂: a movable pulley shares the string between its two sides, so its own displacement is half that of the string end.
Free body diagrams (FBD): A free body diagram refers to the forces acting on only one body at a time, and its acceleration.
Free Body Diagram of the 2 kg Mass
Let \(a\) be its upward acceleration. According to Newton’s second law, it must be due to the resultant vertical force on this mass. To know this force, we need to know all the individual forces acting on this mass.
The agencies exerting forces on this mass are the Earth (downward force \(2g\)) and the force due to the tension \(T_3\). Nothing else touches the 2 kg mass, so nothing else appears in its diagram.
In this case, the lower half of the string is connected to the 2 kg mass. The direction of \(T_3\) for the lower part of the string is upwards, as shown in Fig. 4.12(b). The upper part of the string is connected to the 1 kg mass. Thus, the direction of \(T_3\) for the 1 kg mass will be downwards.
However, that downward \(T_3\) will appear only in the free body diagram of the 1 kg mass, and will not appear in the free body diagram of the 2 kg mass. This is the single most common mistake in FBD problems — dragging a force into a diagram where it does not belong. Hence, the free body diagram of the 2 kg mass will be as follows. Its force equation, according to Newton’s second law, will then be
\[T_3 - 2g = 2a\]
Diagram
Fig. 4.12 (b) – Free body diagram for 2 kg mass: tension T₃ acting vertically upward, weight 2g acting vertically downward, with the equation T₃ − 2g = 2a marked.
Free Body Diagram of the 5 kg Mass
Its horizontal acceleration is also \(a\), but towards the right. The force-exerting agencies are: the Earth (force \(5g\) downwards), the contact surface (normal force \(N\), vertically upwards, and the opposing force \(F = 10\) N, towards the left), and the two strings on either side (forces due to their tensions \(T\) and \(T_1\)).
All these are shown in its free body diagram in Fig. 4.12(c). On resolving the force \(F\) along the vertical and horizontal directions, the free body diagram of the 5 kg mass can be drawn as explained below.
Diagram
Fig. 4.12 (c) – Free body diagram for 5 kg mass: weight 5g downward, normal reaction N upward, applied force F = 100 N resolved into F cos60° horizontal and F sin60° vertical components, opposing force f = 10 N towards left, and tensions T and T₁ on either side.
As this mass has only horizontal motion, the vertical forces must cancel. Therefore, along the vertical direction,
\[N + F\sin 60° = 5g\]
Notice that the applied force, being tilted, has a vertical component that presses down on the surface and therefore reduces the normal reaction \(N\) needed to balance the weight.
Along the horizontal direction,
\[T_1 + 10\ \left(\text{opposing force}\right) = F\cos 60^{0} = T\cos 60^{0}\]
Similar equations can be written for all the masses and also for the movable pulley. On solving these equations simultaneously, we can obtain all the necessary quantities. That is the entire strategy: one free body diagram per body, one equation per direction, then solve.
4.8 Collisions
During collisions, a number of objects come together, interact (exert forces on each other) and scatter in different directions. That is the whole story in one sentence — and yet almost everything interesting in mechanics, from carrom strikers to particle accelerators, is contained in it.
Let us fix the picture with the simplest possible case: two masses \(m_1\) and \(m_2\) moving along the same straight line, approaching each other, touching, and moving apart again.
Diagram
Fig. 4.13 (a) – Head on collision, before collision: two masses m₁ and m₂ moving along the same straight line with initial velocities u₁ and u₂.
Diagram
Fig. 4.13 (b) – Head on collision, during impact: masses m₁ and m₂ in contact, with the mutual internal forces F₁₂ and F₂₁ shown acting on each other in opposite directions.
Diagram
Fig. 4.13 (c) – Head on collision, after collision: masses m₁ and m₂ separating with final velocities v₁ and v₂ in opposite directions.
Notice in Fig. 4.13(b) that the two forces during impact, \(F_{12}\) and \(F_{21}\), are an action–reaction pair. They are equal, opposite, and — crucially — internal to the system of the two bodies.
4.8.1 Elastic and Inelastic Collisions
During a collision, the linear momentum of the entire system of particles is always conserved, as there is no external force acting on the system of particles. However, the individual momenta of the particles do change, due to the mutual forces, which are internal forces.
\[\therefore \sum \vec{p}_{initial} = \sum \vec{p}_{final}\]
during any collision (or explosion), where \(\vec{p}\)’s are the linear momenta of the particles.
However, the kinetic energy of the entire system may or may not be conserved. This is the key difference, and it is what splits collisions into two families.
Collisions can be of two types: elastic collisions and inelastic collisions.
Elastic Collision
A collision is said to be elastic if the kinetic energy of the entire system is conserved during the collision (along with the linear momentum). Thus, during an elastic collision,
\[\sum K.E._{initial} = \sum K.E._{final}\]
An elastic collision is impossible in daily life. Some kinetic energy always leaks away as heat or sound, however hard and springy the bodies look. However, in many situations, the interatomic or intermolecular collisions are considered to be elastic (like in the kinetic theory of gases, to be discussed in the next standard).
Inelastic Collision
A collision is said to be inelastic if there is a loss in the kinetic energy during the collision, but the linear momentum is conserved. The loss in kinetic energy is either due to internal friction or the vibrational motion of atoms, causing a heating effect. Thus, during an inelastic collision,
\[\sum K.E._{initial} > \sum K.E._{final}\]
During an explosion, energy is supplied internally. Thus,
\[\sum K.E._{final} > \sum K.E._{initial}\]
As stated earlier, \(\sum \vec{p}_{initial} = \sum \vec{p}_{final}\) for inelastic collisions or explosions also. In fact, this is always the first equation for discussing these interactions, or while solving numerical questions. Whatever else you are unsure about, momentum conservation is always available to you — start there.
4.8.2 Perfectly Inelastic Collision
This is a special case of inelastic collisions. If the colliding bodies join together after the collision, it is said to be a perfectly inelastic collision.
In other words, the colliding bodies have a common final velocity after a perfectly inelastic collision. Being an inelastic collision, obviously there is a loss in the kinetic energy of the system during a perfectly inelastic collision. In fact, the loss in kinetic energy is maximum in a perfectly inelastic collision.
Illustrations
(i) A bullet fired at a block. Consider a bullet fired towards a block kept on a smooth surface.
The collision between the bullet and the block will be elastic if the bullet rebounds with exactly the same initial speed and the block remains stationary.
If the bullet gets embedded into the block and the two move jointly, it is a perfectly inelastic collision.
If the bullet rebounds with a smaller speed, or comes out of the block on the other side with some speed, it is an inelastic (or partially inelastic) collision.
Remember, there is nothing called a partially elastic collision. Elastic collisions are always perfectly elastic. An inelastic collision, however, may be partially or perfectly inelastic. This is a favourite trap in objective questions.
(ii) A ball dropped on a hard surface. Visualise a ball dropped from some height on a hard surface, with the entire system being in an evacuated space (so that air resistance plays no part).
If the ball rebounds exactly to the same height from where it was dropped, the collision between the ball and the surface (in turn, with the Earth) is elastic.
As you know, the ball never reaches the same initial height, or a height greater than the initial height. Rebounding to a smaller height refers to an inelastic collision.
Instead of a ball, if mud or clay is dropped, it sticks to the surface. This is a perfectly inelastic collision.
4.8.3 Coefficient of Restitution e
We now need a single number that tells us where a given collision lies between “perfectly elastic” and “perfectly inelastic”. That number is the coefficient of restitution.
For a collision of two objects, the negative of the ratio of the relative velocity of separation to the relative velocity of approach is defined as the coefficient of restitution \(e\).
One dimensional or head-on collision: A collision is said to be head-on if the colliding objects move along the same straight line, before and after the collision. Here, we use \(u_1\), \(u_2\), \(\mathrm{v}_1\), \(\mathrm{v}_2\) as symbols.
Consider such a head-on collision of two bodies of masses \(m_1\) and \(m_2\) with respective initial velocities \(u_1\) and \(u_2\). As the collision is head-on, the colliding masses are along the same line before and after the collision. Hence, vector treatment is not necessary. (However, velocities must be substituted with proper signs in actual calculation.) This warning is not decorative — a wrong sign is the single commonest source of error in collision numericals.
The relative velocity of approach is then
\[u_a = u_2 - u_1\]
Let \(\mathrm{v}_1\) and \(\mathrm{v}_2\) be their respective velocities after the collision. The relative velocity of recede (or separation) is then
\[\mathrm{v}_s = \mathrm{v}_2 - \mathrm{v}_1\]
\[\therefore \text{Coefficient of restitution, } e = -\frac{\mathrm{v}_s}{u_a} = -\frac{\mathrm{v}_2 - \mathrm{v}_1}{u_2 - u_1} = \frac{\mathrm{v}_1 - \mathrm{v}_2}{u_2 - u_1}\]
\[= \frac{\text{Relative speed of separation}}{\text{Relative speed of approach}} \quad \text{--- (4.1)}\]
For a perfectly inelastic collision, the colliding bodies move jointly after the collision, i.e., \(\mathrm{v}_2 = \mathrm{v}_1\), or \(\mathrm{v}_2 - \mathrm{v}_1 = 0\). Hence, for a perfectly inelastic collision, \(e = 0\). In other words, if \(e = 0\), the head-on collision is a perfectly inelastic collision.
Coefficient of Restitution during a Head-on, Elastic Collision
Consider the collision described above to be elastic. Now we have two conservation laws to use instead of one, and that is exactly what will pin down the value of \(e\).
According to the principle of conservation of linear momentum,
Total initial momentum = Total final momentum.
\[\therefore m_1u_1 + m_2u_2 = m_1\mathrm{v}_1 + m_2\mathrm{v}_2 \quad \text{--- (4.2)}\]
Rearranging so that each body’s change sits on its own side,
\[\therefore m_1\left(u_1 - \mathrm{v}_1\right) = m_2\left(\mathrm{v}_2 - u_2\right) \quad \text{--- (4.3)}\]
As the collision is elastic, the total kinetic energy of the system is also conserved.
\[\therefore \frac{1}{2}m_1u_1^2 + \frac{1}{2}m_2u_2^2 = \frac{1}{2}m_1\mathrm{v}_1^2 + \frac{1}{2}m_2\mathrm{v}_2^2 \quad \text{--- (4.4)}\]
\[\therefore m_1\left(u_1^2 - \mathrm{v}_1^2\right) = m_2\left(\mathrm{v}_2^2 - u_2^2\right)\]
Factorising each difference of squares,
\[\therefore m_1\left(u_1 + \mathrm{v}_1\right)\left(u_1 - \mathrm{v}_1\right) = m_2\left(u_2 + \mathrm{v}_2\right)\left(\mathrm{v}_2 - u_2\right) \quad \text{--- (4.5)}\]
Dividing Eq. (4.5) by Eq. (4.3), the mass factors cancel completely and we are left with a beautifully simple result:
\[u_1 + \mathrm{v}_1 = u_2 + \mathrm{v}_2 \quad \text{--- (4.6)}\]
\[\therefore u_2 - u_1 = \mathrm{v}_1 - \mathrm{v}_2\]
That is, for an elastic collision, the relative speed of separation equals the relative speed of approach. Therefore, for an elastic collision,
\[e = \frac{\mathrm{v}_1 - \mathrm{v}_2}{u_2 - u_1} = 1 \quad \text{--- (4.7)}\]
Thus, for an elastic collision, the coefficient of restitution, \(e = 1\). For a perfectly inelastic collision, \(e = 0\) (by definition). Thus, for any collision, the coefficient of restitution lies between 1 and 0.
Above expressions (Eq. (4.1), Eq. (4.7)) are general. While substituting the values of \(u_1\), \(u_2\), \(\mathrm{v}_1\), \(\mathrm{v}_2\), their algebraic values must be used in actual calculation.
For example, referring to Fig. 4.13 (a), (b) and (c), Eq. (4.1) gives
\[e = -\frac{\mathrm{v}_s}{u_a}\]
Here \(u_a = u_1 - u_2\) since \(u_1 > u_2\),
and \(\mathrm{v}_s = \mathrm{v}_1 + \mathrm{v}_2\) since the objects go in opposite directions.
\[\therefore e = -\frac{\mathrm{v}_1 + \mathrm{v}_2}{u_1 - u_2} \quad \text{--- (a)}\]
Using Eq. (4.6),
\[u_1 + \mathrm{v}_1 = u_2 + \mathrm{v}_2\]
\[\therefore \text{According to Fig. 4.13,}\quad u_1 - \mathrm{v}_1 = u_2 + \mathrm{v}_2\]
\[\therefore \mathrm{v}_1 + \mathrm{v}_2 = u_2 - u_1 \quad \text{--- (b)}\]
By substituting in (a),
\[e = -\frac{\left(u_2 - u_1\right)}{\left(u_1 - u_2\right)} = 1,\]
which is the case of a perfectly elastic collision.
4.8.4 Expressions for Final Velocities after a Head-on, Elastic Collision
We now have two independent equations for an elastic head-on collision — Eq. (4.2) from momentum and Eq. (4.6) from energy — and two unknowns, \(\mathrm{v}_1\) and \(\mathrm{v}_2\). So the final velocities can be found completely.
From Eq. (4.6),
\[\mathrm{v}_2 = u_1 + \mathrm{v}_1 - u_2\]
Using this in Eq. (4.2), we get
\[m_1u_1 + m_2u_2 = m_1\mathrm{v}_1 + m_2\left(u_1 + \mathrm{v}_1 - u_2\right)\]
\[\therefore \mathrm{v}_1 = \left(\frac{m_1 - m_2}{m_1 + m_2}\right)u_1 + \left(\frac{2m_2}{m_1 + m_2}\right)u_2 \quad \text{--- (4.8)}\]
The subscripts 1 and 2 were arbitrarily chosen — there is nothing special about which body we called “the first”. Thus, just interchanging 1 with 2 gives us \(\mathrm{v}_2\) as
\[\mathrm{v}_2 = \left(\frac{m_2 - m_1}{m_1 + m_2}\right)u_2 + \left(\frac{2m_1}{m_1 + m_2}\right)u_1 \quad \text{--- (4.9)}\]
Equation (4.9) can also be obtained by substituting \(\mathrm{v}_1\) from Eq. (4.8) in Eq. (4.6).
Particular Cases
These three cases are worth memorising, because they appear again and again in objective questions — and each one has a picture attached to it.
(i) If the bodies are of equal masses (or identical), \(m_1 = m_2\). Then Eqs. (4.8) and (4.9) give
\[\mathrm{v}_1 = u_2 \quad\text{and}\quad \mathrm{v}_2 = u_1.\]
Thus, the bodies just exchange their velocities. This is exactly what you see on a carrom board when the striker hits a coin of the same size squarely and stops dead while the coin moves off.
(ii) If the colliding body is much heavier and the struck body is initially at rest, i.e.,
\[m_1 \gg m_2 \quad\text{and}\quad u_2 = 0,\]
we can use the approximations
\[m_1 \pm m_2 \cong m_1 \quad\text{and}\quad \frac{m_2}{m_1 + m_2} \cong 0\]
\[\therefore \mathrm{v}_1 \cong u_1 \quad\text{and}\quad \mathrm{v}_2 \cong 2u_1,\]
i.e., the massive striking body is practically unaffected, and the tiny body which is struck travels with double the speed of the massive striking body. A heavy truck brushing a light football sends it flying at twice the truck’s speed, while the truck barely notices.
(iii) The body which is struck is much heavier than the colliding body and is initially at rest, i.e.,
\[m_1 \ll m_2 \quad\text{and}\quad u_2 = 0.\]
Using similar approximations, we get
\[\mathrm{v}_1 \cong -u_1 \quad\text{and}\quad \mathrm{v}_2 \cong 0,\]
i.e., the tiny (lighter) object rebounds with the same speed, while the massive object is unaffected. This is as good as dropping an elastic object on the hard surface of the Earth.
4.8.5 Loss in the Kinetic Energy during a Perfectly Inelastic Head-on Collision
Consider a perfectly inelastic, head-on collision of two bodies of masses \(m_1\) and \(m_2\) with respective initial velocities \(u_1\) and \(u_2\). As the collision is perfectly inelastic, they move jointly after the collision, i.e., their final velocity is the same. Let it be \(\mathrm{v}\).
According to conservation of linear momentum,
\[m_1u_1 + m_2u_2 = \left(m_1 + m_2\right)\mathrm{v}\]
\[\therefore \mathrm{v} = \frac{m_1u_1 + m_2u_2}{m_1 + m_2} \quad \text{--- (4.10)}\]
This is the common velocity after a perfectly inelastic collision. Notice that it is simply the mass-weighted average of the two initial velocities.
Loss in K.E. \(= \Delta\left(K.E.\right)\)
= Total initial K.E. − Total final K.E.
\[\therefore \Delta\left(K.E.\right) = \frac{1}{2}m_1u_1^2 + \frac{1}{2}m_2u_2^2 - \frac{1}{2}\left(m_1 + m_2\right)\mathrm{v}^2\]
Substituting \(\mathrm{v}\) from Eq. (4.10),
\[= \frac{1}{2}m_1u_1^2 + \frac{1}{2}m_2u_2^2 - \frac{1}{2}\left(m_1 + m_2\right)\left(\frac{m_1u_1 + m_2u_2}{m_1 + m_2}\right)^2\]
\[\therefore \Delta\left(K.E.\right) = \frac{1}{2}m_1u_1^2 + \frac{1}{2}m_2u_2^2 - \frac{1}{2}\frac{\left(m_1u_1 + m_2u_2\right)^2}{\left(m_1 + m_2\right)}\]
On simplifying, we get
\[\Delta\left(K.E.\right) = \frac{1}{2}\left(\frac{m_1m_2}{m_1 + m_2}\right)\left(u_1 - u_2\right)^2 \quad \text{--- (4.11)}\]
Now examine this result. Masses are always positive, and \(\left(u_1 - u_2\right)^2\) is also positive, being a square. Hence, there is always a loss in the kinetic energy in a perfectly inelastic collision. The only way to make the loss zero is to make \(u_1 = u_2\) — but then the bodies never collide at all.
4.8.6 Collision in Two Dimensions, i.e., a Non Head-on Collision
In this case, the direction of at least one initial velocity is NOT along the line of impact. The bodies strike each other at an angle and go off at angles — think of a carrom striker hitting a coin off-centre.
In order to discuss such collisions mathematically, it is convenient to use two mutually perpendicular directions, as shown in Fig. 4.14.
One of them is the common tangent at the point of impact, along which there is no force (or, along this direction, there is no change in momentum).
The other direction is perpendicular to this common tangent through the point of impact, in the two-dimensional plane of the initial and final velocities. This is called the line of impact. The internal mutual forces exerted during impact, which are responsible for the change in the momenta, are acting along this line.
That single idea reduces a messy two-dimensional problem into a one-dimensional collision along the line of impact, plus an untouched motion along the tangent.
From Fig. 4.14, \(\vec{u}_1\) and \(\vec{u}_2\), the initial velocities, make angles \(\alpha_1\) and \(\alpha_2\) respectively with the line of impact, while \(\vec{\mathrm{v}}_1\) and \(\vec{\mathrm{v}}_2\), the final velocities, make angles \(\beta_1\) and \(\beta_2\) respectively with the line of impact.
Diagram
Fig. 4.14 – Oblique or non head-on collision: two circular bodies touching, with the common tangent and the line of impact drawn through the point of contact; initial velocities u₁ and u₂ at angles α₁ and α₂, and final velocities v₁ and v₂ at angles β₁ and β₂, all measured from the line of impact.
According to conservation of linear momentum along the line of contact,
\[m_1u_1\cos\alpha_1 + m_2u_2\cos\alpha_2 = m_1\mathrm{v}_1\cos\beta_1 + m_2\mathrm{v}_2\cos\beta_2 \quad \text{--- (4.12)}\]
As there is no force along the common tangent (perpendicular to the line of impact), the momentum of each body separately is unchanged in that direction:
\[m_1u_1\sin\alpha_1 = m_1\mathrm{v}_1\sin\beta_1 \quad \text{--- (4.13)}\]
and
\[m_2u_2\sin\alpha_2 = m_2\mathrm{v}_2\sin\beta_2 \quad \text{--- (4.14)}\]
For the coefficient of restitution, along the line of impact,
\[e = -\left(\frac{\mathrm{v}_2\cos\beta_2 - \mathrm{v}_1\cos\beta_1}{u_2\cos\alpha_2 - u_1\cos\alpha_1}\right)\]
\[= \frac{\mathrm{v}_2\cos\beta_2 - \mathrm{v}_1\cos\beta_1}{u_1\cos\alpha_1 - u_2\cos\alpha_2} \quad \text{--- (4.15)}\]
Equations (4.12), (4.13), (4.14) and (4.15) are to be solved for the four unknowns \(\mathrm{v}_1\), \(\mathrm{v}_2\), \(\beta_1\) and \(\beta_2\). Four equations, four unknowns — the problem is completely determined.
4.9 Impulse of a Force
According to Newton’s first law of motion, any unbalanced force changes the linear momentum of the system. That is to say, the basic effect of an unbalanced force is to change the momentum.
According to Newton’s second law of motion,
\[\vec{F} = \frac{d\vec{p}}{dt}\]
\[\therefore d\vec{p} = \vec{F}.dt\]
Read this last line carefully. It says that the change in momentum depends on two things — how big the force is, and for how long it acts. A gentle force acting for a long time and a huge force acting for a fleeting instant can produce exactly the same change in momentum.
The quantity ‘change in momentum’ is separately named as the Impulse of the force, \(\vec{J}\).
If the force is constant, and is acting for a finite and measurable time, we can write the change in momentum in time \(t\) as
\[\vec{J} = d\vec{p} = \vec{p}_2 - \vec{p}_1 = \vec{F}.t \quad \text{---(4.16)}\]
For a given body of mass \(m\), it becomes
\[\vec{J} = \vec{p}_2 - \vec{p}_1 = m\left(\vec{\mathrm{v}}_2 - \vec{\mathrm{v}}_1\right) = \vec{F}.t \quad \text{---(4.17)}\]
If \(\vec{F}\) is not constant, but we know how it varies with time, then we simply add up \(\vec{F}.dt\) over every instant — which is integration:
\[\vec{J} = \Delta\vec{p} = \int d\vec{p} = \int \vec{F}.dt \quad \text{--- (4.18)}\]
4.9.1 Necessity of Defining Impulse
As discussed above, if a force is constant over a given interval of time, or if we know how it varies with time, we can calculate the corresponding change in momentum directly by multiplying the force and the time. So why invent a new quantity at all?
The reason is practical. In many cases, an appreciable force acts for an extremely small interval of time — too small to measure the force and the time independently. You simply cannot put an instrument between a bat and a ball and read off “force = so many newton” and “time = so many microseconds”.
However, the change in the momentum due to this force is noticeable and can be measured. You can measure how fast the ball was coming and how fast it goes away. This change is defined as the impulse of the force.
Real life illustrations: While (i) hitting a ball with a bat, (ii) giving a kick to a foot-ball, (iii) hammering a nail, and (iv) bouncing a ball from a hard surface, an appreciable amount of force is being exerted.
In such cases the time for which these forces act on the respective objects is negligibly small, and mostly not easily recordable. However, the effect of this force is a recordable change in the momentum of that object. Thus, it is convenient to define the change in momentum itself as a physical quantity.
Impulse as an Area
Figure 4.15 shows the variation of a force as a function of time — for example, for a collision between a bat and a ball — with the force axis starting with zero.
Diagram
Fig. 4.15 – Graphical representation of impulse of a force: a force–time curve rising from zero to a peak and falling back to zero, with the area under the curve over the interval Δt shaded and labelled J = F Δt; force axis starting at zero.
The shaded area, or the area under the curve, gives the product of the force against the corresponding time (in this case, \(\Delta t\)), and hence gives the impulse. For a constant force it is obviously a rectangle.
Generally, the force is zero before the impact, rises to a maximum, and decreases to zero after the impact. It is never a neat rectangle in real life — it is a hump.
Now compare a hard ball with a soft one. For a softer tennis ball, the collision time is larger and the maximum force is less. The area under the \((F - t)\) graph is the same. The same impulse has simply been spread over a longer time, so the peak force comes down.
Indeed, the wicket keeper eases off (by increasing the time of collision) while catching a fast ball. As mentioned earlier, it is absolutely necessary that the force axis must start from zero, otherwise the area you read off the graph is not the true impulse.
Recall from Chapter 3 the analogous concepts using the area under a curve: (i) obtaining the displacement in a given time interval as the area under the curve for a \(v\)–\(t\) graph, with zero origin for the velocity axis; and (ii) obtaining the work done by a force as the area under the curve for an \(F\)–\(s\) graph, with zero origin for the force axis. The pattern is worth noticing — in each case, the area under a graph is the accumulated effect of the quantity on the vertical axis.
4.10 Rotational Analogue of a Force - Moment of a Force or Torque
Until now, every force we discussed produced a linear acceleration. But a force can also make a body turn, and how well it turns is not decided by the size of the force alone.
While opening a door fixed to a frame on hinges, we apply the force away from the hinges and perpendicular to the door, to open it with ease. In this case we are interested in achieving some angular displacement for the door, not in moving the door as a whole from one place to another.
If the force is applied near the hinges, or nearly parallel to the door, it is very difficult to open the door — even though the force you apply is exactly the same. Similarly, if the door is heavier (made up of iron instead of wood or plastic), we need to apply a proportionally larger force for the same angular displacement.
What the Turning Effect Really Depends On
The door tells us something important. It shows that the rotational ability of a force does not depend upon the magnitude of the force alone. It depends upon:
- the mass of the body (a greater force is needed for a greater mass);
- the point of application of the force — the point should be as far away as possible from the axis of rotation;
- the angle between the direction of the force and the line joining the axis of rotation with the point of application. The effect is maximum if this angle is 90°.
Taking into account all these factors, the quantity moment of a force or torque is defined as the rotational analogue of a force.
As rotation refers to a direction (a sense of rotation — clockwise or anticlockwise), torque must be a vector quantity. In its mathematical form, torque or moment of a force is given by
\[\vec{\tau} = \vec{r}\times\vec{F} \quad \text{--- (4.17)}\]
where \(\vec{F}\) is the applied force (in newton, N) and \(\vec{r}\) is the position vector of the point of application of the force from the axis of rotation (in metre, m), as shown in the Figs. 4.16 (b) and 4.6 (c). The SI unit of torque is therefore N m.
The Direction of Torque
Figures 4.6(a), 4.6(b) and 4.6(c) illustrate the directions involved.
Figure 4.16(a) is a 3D drawing indicating a laminar (plane or two-dimensional) object rotating about a (fixed) axis of rotation AOB, the axis being perpendicular to the object and passing through it.
Diagram
Fig. 4.16 (a) – Illustration of moment of force with a laminar object and its fixed axis of rotation AOB shown in 3D view, the axis perpendicular to the object and passing through it.
Figure 4.16(b) indicates the top view of the object when the rotation is in the anticlockwise direction, and Fig. 4.16(c) shows the view from the top if the rotation is in the clockwise direction. (In fact, Figs. 4.16(b) and 4.6(c) are drawn in such a way that the applied force \(\vec{F}\) and the position vector \(\vec{r}\) of the point of application of the force are in the plane of these figures.)
Diagram
Fig. 4.16 (b) – Top view for moment of force F in anticlockwise rotation, with F and r in the plane of the paper and the torque r × F marked as outwards (dot symbol) at O.
Diagram
Fig. 4.16 (c) – Top view of moment of force F in clockwise rotation, with F and r in the plane of the paper and the torque r × F marked as inwards (cross symbol) at O.
The direction of the torque is always perpendicular to the plane containing the vectors \(\vec{r}\) and \(\vec{F}\), and can be obtained from the rule of the cross product, or by using the right-hand thumb rule: curl the fingers of your right hand in the sense of rotation, and the extended thumb points along the torque.
In Fig. 4.16(b), it is perpendicular to the plane of the figure (in this case, perpendicular to the body) and outwards, i.e., coming out of the paper. In Fig. 4.16(c), it is inwards, i.e., going into the paper.
The Dot-and-Cross Convention
In order to indicate directions which are not in the plane of the figure, we use a special convention: \(\odot\) for perpendicular to the plane of the figure and outwards, and \(\otimes\) for perpendicular to the plane of the figure and inwards.
Diagram
Fig. 4.17 – Convention of pictorial representation of vectors: (a) a traditional feathered arrow acting in a direction perpendicular to the plane of paper, (b) a circle with a dot for a vector coming out of the paper, (c) a circle with a cross for a vector going into the paper, (d) a circle with both cross and dot for a line perpendicular to the plane of paper.
This convention depends upon a traditional arrow, shown in Fig. 4.17 (a). Consider yourself looking towards the figure from the top. If this arrow approaches you, the tip of the arrow will be prominently seen. Hence, a circle with a dot in it [Fig. 4.17 (b)] refers to perpendicular and outwards (or towards you).
When you are leaving an arrow, i.e., if an arrow is going away from you, the feathers, like a cross, will be seen. Hence, a circle with a cross [Fig. 4.17 (c)] indicates perpendicular and inwards (or away from you). A circle with both a cross and a dot indicates a line perpendicular to the plane of the figure [Fig. 4.17 (d)].
This little picture — arrow tip versus arrow feathers — is worth remembering, because the same convention is used throughout Physics, especially for magnetic fields and currents.
Magnitude of Torque and Its Consequences
Magnitude of torque,
\[\tau = r\,F\sin\theta \quad \text{--- (4.18)}\]
where \(\theta\) is the smaller angle between the directions of \(\vec{r}\) and \(\vec{F}\).
The quantity \(r\sin\theta\) is worth interpreting on its own: it is the perpendicular distance from the axis of rotation to the line of action of the force. This is why it is sometimes called the “moment arm” of the force.
Consequences:
(i) If \(r\) or \(F\) is greater, the torque (hence the rotational effect) is greater. Thus, it is recommended to apply the force away from the hinges.
(ii) If \(\theta = 90^{0}\), then \(\tau = \tau_{max} = rF\). Thus, the force should be applied along the normal direction for easy rotation. Pushing a door straight, at right angles to its surface, gets the maximum turning effect out of your effort.
(iii) If \(\theta = 0°\) or \(180°\), then \(\tau = \tau_{min} = 0\). Thus, if the force is applied parallel or anti-parallel to \(\vec{r}\), there is no rotation at all. Pushing a door edge-on, straight towards or straight away from the hinges, will never open it, however hard you push.
(iv) The moment of a force depends not only on the magnitude and direction of the force, but also on the point where the force acts, with respect to the axis of rotation. The same force can have a different torque as per its point of application.
4.11 Couple and its Torque
In the discussion of the torque given above, we had considered the rotation of the body about a fixed axis, and due to a single force. In real life, quite often we apply two equal and opposite forces acting along different lines of action in order to cause rotation.
Common illustrations are turning a bicycle handle, turning the steering wheel, opening a common water tap, and opening the lid of a bottle (of the rotation type). In each of these you use two hands, or two fingers, pushing in opposite directions — and the object turns without going anywhere.
Such a pair of forces, consisting of two forces of equal magnitude acting in opposite directions along different lines of action, is called a couple. It is used to realise a purely rotational motion. The moment of a couple, or the rotational effect of a couple, is also called a torque.
The words “different lines of action” are essential. Two equal and opposite forces along the same line would simply cancel and do nothing at all. It is the offset between the two lines that creates the turning.
It may be noted that in the discussion of the rotation of a body about a fixed axis due to a single force, there is a reaction force at the fixed axis. Hence, in a sense, for rotation one always needs two forces acting in opposite directions along different lines of action. Even the door you push is really being turned by a couple — your push, and the reaction at the hinges.
Torque or Moment of a Couple
Figure 4.18 shows a couple consisting of two forces \(\vec{F}_1\) and \(\vec{F}_2\) of equal magnitudes and opposite directions, acting along different lines of action separated by a distance \(r\).
The corresponding position vectors should now be defined with reference to the lines of action of the forces, and not with reference to any axis. This is the essential difference from the previous section, and it is what makes the result so remarkable.
The position vector of any point on the line of action of force \(\vec{F}_1\), drawn from the line of action of force \(\vec{F}_2\), is \(\vec{r}_{12}\). Similarly, the position vector of any point on the line of action of force \(\vec{F}_2\), drawn from the line of action of force \(\vec{F}_1\), is \(\vec{r}_{21}\).
Diagram
Fig. 4.18 – Torque of a couple: two equal and opposite forces F₁ and F₂ acting along parallel lines of action separated by perpendicular distance r, with position vectors r₁₂ and r₂₁ drawn between the lines of action at angles α and β, and r₁₂ sinα = r₂₁ sinβ = r marked.
The torque, or moment of the couple, is then given mathematically as
\[\vec{\tau} = \vec{r}_{12}\times\vec{F}_1 = \vec{r}_{21}\times\vec{F}_2 \quad \text{--- (4.19)}\]
From the figure, it is clear that
\[r_{12}\sin\alpha = r_{21}\sin\beta = r\]
Both of these are simply the perpendicular distance between the two lines of action — the arm of the couple.
If \(\left|\vec{F}_1\right| = \left|\vec{F}_2\right| = F\), the magnitude of the torque is given by
\[\tau = r_{12}\,F_1\sin\alpha = r_{21}\,F_2\sin\beta = r\,F \quad \text{--- (4.20)}\]
Look at what has dropped out of this result. There is no axis of rotation in it. There is no point of application in it. It clearly shows that the torque corresponding to a given couple, i.e., the moment of a given couple is constant — that is, it is independent of the points of application of the forces or the position of the axis of rotation, but depends only upon the magnitude of either force and the separation between their lines of action.
The direction of the moment of a couple can be obtained by using the vector formula of the torque, or by using the right-hand thumb rule. For the couple shown in Fig. 4.18, it is perpendicular to the plane of the figure and inwards. For a given pair of forces, the direction of the torque is fixed.
In many situations the word couple is used synonymously with the moment of the couple or its torque. That is, every time we may not say that a torque due to the couple is acting; we simply say that a couple is acting.
Moment of a Force and Moment of a Couple Compared
| Moment of a force | Moment of a couple | |
|---|---|---|
| 1 | \(\vec{\tau} = \vec{r}\times\vec{f}\) | \(\vec{\tau} = \vec{r}_{12}\times\vec{F}_1 = \vec{r}_{21}\times\vec{F}_2\) |
| 2 | \(\vec{\tau}\) depends upon the axis of rotation and the point of application of the force. | \(\vec{\tau}\) depends only upon the two forces, i.e., it is independent of the axis of rotation or the points of application of the forces. |
| 3 | It can produce translational acceleration also, if the axis of rotation is not fixed or if friction is not enough. | Does not produce any translational acceleration, but produces only rotational or angular acceleration. |
| 4 | Its rotational effect can be balanced by a proper single force or by a proper couple. | Its rotational effect can be balanced only by another couple of equal and opposite torque. |
4.11.1 To Prove that the Moment of a Couple is Independent of the Axis of Rotation
We have already claimed this result from Eq. (4.20). Let us now prove it properly, by computing the torque of the same couple about two different axes and showing that the answer does not change.
Figure 4.19 shows a rectangular sheet (any object would do) free to rotate only about a fixed axis of rotation, perpendicular to the plane of the figure, as shown. A couple consisting of the forces \(\vec{F}\) and \(-\vec{F}\) is acting on the sheet at different locations.
Here we are considering the torque of a couple to be two torques due to the individual forces, each causing rotation about the axis of rotation. We then add them, taking their senses of rotation into account.
Case (a): The Axis Lies Between the Two Forces
In Fig. 4.19(a), the axis of rotation is between the lines of action of the two forces constituting the couple. The perpendicular distances of the axis of rotation from the forces \(\vec{F}\) and \(-\vec{F}\) are \(x\) and \(y\) respectively.
The rotation due to the pair of forces in this case is anticlockwise (from the top view), i.e., the directions of the individual torques due to the two forces are the same. Both forces try to turn the sheet the same way, so their torques add.
\[\therefore \tau = \tau_+ + \tau_- = xF + yF = \left(x + y\right)F = rF \quad \text{(4.21)}\]
since \(x + y = r\), the separation between the two lines of action.
Diagram
Fig. 4.19 (a) – A rectangular sheet with a fixed axis of rotation lying between the lines of action of the couple’s two forces F and −F, with perpendicular distances x and y from the axis marked, and r = x + y.
Case (b): Both Forces on the Same Side of the Axis
In the Fig. 4.19 (b), the lines of action of both the forces are on the same side of the axis of rotation.
Thus, in this case, the rotation of \(+\vec{F}\) is anticlockwise, while that of \(-\vec{F}\) is clockwise (from the top view). As a result, their individual torques are oppositely directed, and now they must be subtracted.
The perpendicular distances of the forces \(F\) and \(-F\) from the axis of rotation are \(q\) and \(p\) respectively.
\[\therefore \tau = \tau_+ - \tau_- = qF - pF\]
\[= \left(q - p\right)F = rF \quad \text{--- (4.22)}\]
since \(q - p = r\), once again the separation between the two lines of action.
Diagram
Fig. 4.19 (b) – The same rectangular sheet and the same couple, but with the fixed axis of rotation now placed outside both lines of action, with perpendicular distances q and p marked from the axis, and r = q − p.
From equations (4.21) and (4.22), it is clear that the torque of a couple is independent of the axis of rotation.
Notice how the two cases arrived at \(rF\) by completely different routes — one by adding the arms, the other by subtracting them — and yet the answer is the same. Wherever you place the axis, the couple delivers the same torque.
4.12 Mechanical Equilibrium
As a consequence of Newton’s second law, the momentum of a system is constant in the absence of an external unbalanced force. This state is called mechanical equilibrium.
A particle is said to be in mechanical equilibrium if no net force is acting upon it.
For a system of bodies to be in mechanical equilibrium, the net force acting on any part of the system should be zero. Those two words — any part — are the whole point. It is not enough for the system as a whole to be balanced; every piece of it must be balanced individually, otherwise some part of it would accelerate.
In other words, the velocity or linear momentum of all parts of the system must be constant (or zero) for the system to be in mechanical equilibrium. Also, there is no acceleration in any part of the system.
Mathematically,
\[\sum \vec{F} = 0\]
for any part of the system, for mechanical equilibrium.
4.12.1 Stable, Unstable and Neutral Equilibrium
Figures 4.10 (a), (b) and (c) show a ball at rest in three situations, under the action of balanced forces. In all these cases, it is under equilibrium — in every one of them, \(\sum \vec{F} = 0\) and the ball stays put.
However, potential energy-wise, the three cases differ, and that difference decides what happens if you nudge the ball.
Diagram
Fig. 4.20 – States of mechanical equilibrium: (a) a ball resting in a valley (stable), (b) a ball balanced on top of a hump (unstable), (c) a ball resting on a flat horizontal plane (neutral).
Stable equilibrium: In Fig. 4.20(a), the ball is most stable and is said to be in stable equilibrium. If it is disturbed slightly from its equilibrium position and released, it tends to recover its position — it rolls back into the valley. In this case, the potential energy of the system is at its local minimum.
Unstable equilibrium: In Fig. 4.20(b), the ball is said to be in unstable equilibrium. If it is slightly disturbed from its equilibrium position, it moves farther from that position — it rolls away down the hump and never comes back.
This happens because initially, the potential energy of the system is at its local maximum. If disturbed, it tries to achieve the configuration of minimum potential energy. Every system, left to itself, slides towards lower potential energy, and a hump-top is the worst place to sit.
Neutral equilibrium: In Fig. 4.20(c), the potential energy of the system is constant over a plane, and remains the same at any position. Thus, even if the ball is disturbed, it still remains in equilibrium at practically any position. This is described as neutral equilibrium — the ball simply stays wherever you leave it.
4.13 Centre of Mass
As discussed earlier, Newton’s laws of motion and many other laws are applicable for point masses only. However, in real life, we always come across finite objects — objects of measurable sizes. A cricket ball, a car, a human body: none of these is a point.
The concept of centre of mass (c.m.) rescues us. It helps us in considering these objects to be point objects at a particular location, thereby allowing us to apply Newton’s laws of motion to them. Once we know where the centre of mass is, we may forget the shape of the object entirely and treat all its mass as sitting at that single point.
4.13.1 Mathematical Understanding of Centre of Mass
(i) System of n Particles
Consider a system of \(n\) particles of masses \(m_1\), \(m_2\) … \(m_n\).
\[\therefore \sum_{1}^{n}m_i = M \text{ the total mass.}\]
Let \(\vec{r}_1\), \(\vec{r}_2\), … \(\vec{r}_n\) be their respective position vectors from a given origin O (Fig. 4.23).
Diagram
Fig. 4.23 – Centre of mass for n particles: several point masses m₁, m₂ … mₙ at positions (x₁,y₁,z₁), (x₂,y₂,z₂) … with position vectors r₁, r₂ … drawn from a common origin O, and the centre of mass C.M. at position vector r.
The position vector \(\vec{r}\) of their centre of mass, from the same origin, is then given by
\[\vec{r} = \frac{\sum_{1}^{n}m_i\vec{r}_i}{\sum_{1}^{n}m_i} = \frac{\sum_{1}^{n}m_i\vec{r}_i}{M}\]
In words: multiply each mass by its position, add them all up, and divide by the total mass. Heavier particles pull the answer towards themselves — the centre of mass is a weighted average of positions, with mass as the weight.
If the origin itself is at the centre of mass,
\[\vec{r} = 0 \quad \therefore \sum_{1}^{n}m_i\vec{r}_i = 0,\]
then \(\sum_{1}^{n}m_i\vec{r}_i\) gives the moment of masses (similar to the moment of force) about the centre of mass.
If \(x_1\), \(x_2\), … \(x_n\) are the respective \(x\)-coordinates of \(\vec{r}_1, \vec{r}_2, ... \vec{r}_n\), the \(x\)-coordinate of the centre of mass is given by
\[x = \frac{\sum_{1}^{n}m_ix_i}{\sum_{1}^{n}m_i} = \frac{\sum_{1}^{n}m_ix_i}{M}\]
Similarly, the \(y\)- and \(z\)-coordinates of the centre of mass are respectively given by
\[y = \frac{\sum_{1}^{n}m_iy_i}{\sum_{1}^{n}m_i} = \frac{\sum_{1}^{n}m_iy_i}{M}\]
and
\[z = \frac{\sum_{1}^{n}m_iz_i}{\sum_{1}^{n}m_i} = \frac{\sum_{1}^{n}m_iz_i}{M}\]
(ii) Continuous Mass Distribution
A solid body is not a handful of separate particles; it is matter spread continuously through a volume. For a continuous mass distribution with uniform density, we need to use integration instead of summation — we chop the body into infinitesimal elements of mass \(dm\) and add them all up.
In this case, the position vector of the centre of mass is given by
\[\vec{r} = \frac{\int \vec{r}\,dm}{\int dm} = \frac{\int \vec{r}\,dm}{M},\]
where \(\int dm = M\) is the total mass of the object. Then the Cartesian coordinates of the c.m. are
\[x = \frac{\int x\,dm}{\int dm} = \frac{\int x\,dm}{M}\]
\[y = \frac{\int y\,dm}{\int dm} = \frac{\int y\,dm}{M}\]
\[z = \frac{\int z\,dm}{\int dm} = \frac{\int z\,dm}{M}\]
Using the expressions given above, the centres of mass of uniform symmetric objects can be obtained. Some of these are listed in Table 4.1 given below. These are standard results and are worth memorising, because numericals routinely expect you to quote them without deriving them.
Table 4.1: Coordinate of the centre of mass (c.m.) for some symmetrical objects
| Coordinates of c.m. | Uniform Symmetric Objects |
|---|---|
| System of two point masses: c.m. divides the distance in inverse proportion of the masses | Two point masses joined by a line |
| Centre of mass at the geometrical centre of the object | Any geometrically symmetric object of uniform density |
| \(x_c = 0,\ y_c = \dfrac{H}{3}\) | Isosceles triangular plate of height \(H\) |
| \(x_c = \dfrac{p}{3},\ y_c = \dfrac{q}{3}\) | Right angled triangular plate with base \(p\) and height \(q\) |
| \(x_c = 0,\ y_c = \dfrac{2R}{\pi}\) | Thin semicircular ring of radius \(R\) |
| \(x_c = 0,\ y_c = \dfrac{4R}{3\pi}\) | Thin semicircular disc of radius \(R\) |
| \(x_c = 0,\ y_c = \dfrac{R}{2}\) | Hemispherical shell of radius \(R\) |
| \(x_c = 0,\ y_c = \dfrac{3R}{8}\) | Solid hemisphere of radius \(R\) |
| \(x_c = 0,\ y_c = \dfrac{h}{3}\) | Hollow right circular cone of height \(h\) |
| \(x_c = 0,\ y_c = \dfrac{h}{4}\) | Solid right circular cone of height \(h\) |
Diagram
Fig. 4.24 – Illustrations accompanying Table 4.1: an isosceles triangular plate, a right-angled triangular plate, a thin semicircular ring, a thin semicircular disc, a hemispherical shell, a solid hemisphere, a hollow cone and a solid cone, each with its centre of mass marked.
4.13.2 Velocity of Centre of Mass
Let \(\mathrm{v}_1\), \(\mathrm{v}_2\), … \(\mathrm{v}_n\) be the velocities of a system of point masses \(m_1\), \(m_2\), … \(m_n\). The velocity of the centre of mass of the system is given by
\[\vec{\mathrm{v}}_{cm} = \frac{\sum_{1}^{n}m_i\vec{\mathrm{v}}_i}{\sum_{1}^{n}m_i} = \frac{\sum_{1}^{n}m_i\vec{\mathrm{v}}_i}{M}\]
\[= \frac{\text{resultant linear momentum}}{\text{total mass}}\]
\[= \text{weighted average of momenta}\]
This is a result of real power. The numerator is nothing but the total momentum of the system. So the whole system, however complicated, moves as though its total mass \(M\) were travelling with the velocity \(\vec{\mathrm{v}}_{cm}\).
The \(x\), \(y\) and \(z\) components of \(\vec{\mathrm{v}}\) can be obtained similarly.
For a continuous distribution,
\[\vec{\mathrm{v}}_{cm} = \frac{\int \vec{\mathrm{v}}\,dm}{M}\]
4.13.3 Acceleration of the Centre of Mass
Let \(\vec{a}_1\), \(\vec{a}_2\), … \(\vec{a}_n\) be the accelerations of a system of point masses \(m_1\), \(m_2\), … \(m_n\). The acceleration of the centre of mass of the system is given by
\[\vec{a}_{cm} = \frac{\sum_{1}^{n}m_i\vec{a}_i}{\sum_{1}^{n}m_i} = \frac{\sum_{1}^{n}m_i\vec{a}_i}{M}\]
\[= \frac{\text{resultant force}}{\text{total mass}}\]
\[= \text{weighted average of forces}\]
The \(x\), \(y\) and \(z\) components of \(\vec{a}\) can be obtained similarly.
For a continuous distribution,
\[\vec{a}_{cm} = \frac{\int \vec{a}\,dm}{M}\]
Notice what this says: the acceleration of the centre of mass depends only on the resultant force. Internal forces come in action–reaction pairs and cancel in the sum, so they contribute nothing. This is why an exploding shell’s centre of mass continues along the original parabola, unmoved by the violence inside it.
4.13.4 Characteristics of Centre of Mass
Centre of mass is a hypothetical point at which the entire mass of the body can be assumed to be concentrated.
Centre of mass is a location, and not a physical quantity.
Centre of mass is the particle equivalent of a given object for applying the laws of motion.
Centre of mass is the point at which, if a force is applied, it causes only linear acceleration and not angular acceleration.
Centre of mass is located at the centroid, for a rigid body of uniform density.
Centre of mass is located at the geometrical centre, for a symmetric rigid body of uniform density.
The location of the centre of mass can be changed only by an external unbalanced force.
Internal forces (like during a collision or an explosion) never change the location of the centre of mass.
The position of the centre of mass depends only upon the distribution of mass. However, to describe its location we may use a coordinate system with a suitable origin. In statistical terms, the centre of mass is decided by the weighted average of the individual masses. This is obtained by giving proper mass weightage to the distance. This should be clear from the mathematical expression for the location of the centre of mass.
For a system of particles, the centre of mass need not coincide with any of the particles.
While balancing an object on a pivot, the line of action of the weight must pass through the centre of mass and the pivot. Quite often, this is an unstable equilibrium.
The centre of mass of a system of only two particles divides the distance between the particles in an inverse ratio of their masses, i.e., it is closer to the heavier mass.
Centre of mass is a point about which the summation of the moments of the masses in the system is zero.
If there is an axial symmetry for a given object, the centre of mass lies on the axis of symmetry.
If there are multiple axes of symmetry for a given object, the centre of mass is at their point of intersection.
The centre of mass need not be within the body (see Fig. 4.28). Other examples are a ring and a horse shoe — in both cases the centre of mass sits in empty space.
Diagram
Fig. 4.28 – A high jumper arching backwards over the bar in a Fosbury Flop, with the dashed path of the centre of gravity shown passing below the bar while the body passes above it.
4.14 Centre of Gravity
The centre of gravity (c.g.) of a body is the point around which the resultant torque due to the force of gravity on the body is zero. Analogous to the centre of mass, it is the weighted average of the gravitational forces (weights) on the individual particles of the body.
Read the two definitions side by side and the difference becomes clear. The centre of mass is a weighted average of masses. The centre of gravity is a weighted average of weights. Since weight is mass times \(g\), the two averages will agree exactly as long as \(g\) is the same everywhere in the body.
For a uniform gravitational field (in simple words, if \(g\) is constant), the c.g. always coincides with the c.m. Obviously, it is true for all the objects on the Earth in our daily life — over the size of a cricket bat or even a building, \(g\) hardly changes at all. Thus, in common usage, the terms c.g. and c.m. are used for the same purpose.
Finding the Centre of Gravity of a Laminar Object
This coincidence gives us a simple experimental method. The property can be used to determine the c.g. (or c.m.) of a laminar object — laminar means like a leaf, that is, two dimensional.
In Fig. 4.29, a laminar object is suspended from a rigid support at two orientations. Lines are to be drawn on the object parallel to the plumb line shown.
Diagram
Fig. 4.29 – An irregular laminar object suspended freely from a rigid support at two different points, with a plumb line hanging from the point of suspension in each case; the two lines drawn on the object intersect at the c.m. or c.g.
The plumb line is always vertical, i.e., parallel to the line of action of the gravitational force. When the object hangs freely and is at rest, the net torque about the point of suspension must be zero — and that can happen only if the weight acts along the vertical line through the point of suspension. Hence the c.g. must lie somewhere on the line you have drawn.
Repeat with a different point of suspension, and you get a second such line. The intersection of the lines drawn is then the point through which the line of action of the gravitational force passes for any orientation. Thus, it gives the location of the c.g. or c.m.
When the Two Points Part Company
The centre of mass is a fixed property for a given rigid body, in spite of any orientation. Turn the body around as much as you like — the c.m. stays at the same point of the body.
The centre of gravity, however, may depend upon the non-uniformity of the gravitational field, and in turn, will depend upon the orientation. Tilt the body, and different parts of it move into regions of different \(g\), so the weighted average of the weights shifts.
For objects on the Earth, this will be possible only if the size of an object is comparable to that of the Earth (a size of at least a few thousand km). In such cases, the c.g. will be slightly lower than the c.m., as on the lower side of an object the gravitational field is stronger — the lower parts are nearer the centre of the Earth, so they are weighted a little more heavily in the average.
Of course, we shall not come across such an object in any practical situation. For everything you will ever handle, calculate or be examined on, the centre of gravity and the centre of mass are the same point.
4.15 Summary
Motion, Mechanics and Frames of Reference
Mechanics is divided into statics (bodies at rest or in equilibrium) and kinetics (bodies actually in motion). Kinetics is further divided into kinematics (describing motion without its cause) and dynamics (describing motion along with its cause — force and torque).
Motion is strictly relative; it must always be described with respect to a frame of reference.
The trajectory of a body is decided by its initial velocity \(\vec{u}\) and its acceleration \(\vec{a}\):
- \(\vec{a}\) in line with \(\vec{u}\) (or \(\vec{u} = 0\)) → linear motion
- \(\vec{a}\) always perpendicular to velocity, with \(\vec{u} \neq 0\) → circular motion
- \(\vec{a}\) constant but not in line with \(\vec{u}\) → parabolic motion (projectile)
Aristotle’s fallacy: the belief that an external force is needed to keep a body in uniform motion. Galileo showed that bodies stop only because of hidden opposing forces — friction, viscous drag. In reality, an external force is needed only to overcome these opposing forces, not to sustain the motion.
Newton’s Laws of Motion
First law: Every inanimate object continues to be in its state of rest or of uniform unaccelerated motion unless and until it is acted upon by an external, unbalanced force.
Second law: The rate of change of linear momentum of a rigid body is directly proportional to the applied force and takes place in the direction of the applied force.
\[\vec{F} = \frac{d\vec{p}}{dt}, \quad \text{where } \vec{p} = m\vec{\mathrm{v}}\]
- Expanding the second law fully:
\[\vec{F} = \frac{dm}{dt}\vec{\mathrm{v}} + m\vec{a}\]
For a body of constant mass, \(\dfrac{dm}{dt} = 0\), and only then does \(\vec{F} = m\vec{a}\) hold. For a rocket, both terms are needed.
Third law: To every action (force) there is an equal and opposite reaction (force). Action and reaction act along the same line but on different objects, and therefore never cancel.
Importance of the laws: the first law defines force and inertia (measured by inertial mass, \(\left|\vec{F}\right| / \left|\vec{a}\right|\)) and shows the equivalence of rest and uniform motion; the second law gives the quantitative measure of force and establishes momentum as the fundamental quantity; the third law defines action–reaction pairs.
Inertial and Non-Inertial Frames
An inertial frame is one in which Newton’s first law holds: with no net force, there is no acceleration.
A non-inertial frame is an accelerating frame, in which a body appears to accelerate without any net force.
Limitations of Newton’s laws: valid only in inertial frames (else a pseudo force \(\left(-m\vec{a}\right)\) must be added); applicable to point objects and rigid bodies; fail for speeds comparable to that of light (use special relativity) and for atomic/molecular sizes (use quantum mechanics).
Types of Forces
Four fundamental forces:
- Gravitational: \(F = \dfrac{Gm_1m_2}{r^2}\), \(G = 6.67\times10^{-11}\) SI units. Weakest, infinite range, always attractive. Weight: \(W = \dfrac{GMm}{R^2} = mg\), with \(\dfrac{GM}{R^2} \cong 9.8\) m/s² \(= g\) for the Earth (\(R = 6.4\times10^6\) m).
- Electromagnetic: attractive or repulsive, between charged particles; much stronger than gravity. Friction, normal reaction, tension, collision, elastic and viscous forces are all EM in nature.
- Strong (nuclear): strongest, binds nucleons, short range (\(< 10^{-14}\) m), charge independent.
- Weak (nuclear): responsible for radioactive beta decay; range of the order of \(10^{-16}\) m.
Contact forces (normal reaction, friction, collision forces) act only on contact and are EM in origin; non-contact forces (gravitational, electrostatic, magnetostatic) act at a distance.
Real forces obey Newton’s laws and are among the four fundamental forces. Pseudo forces \(\left(-m\vec{a}\right)\) are non-real but measurable, and must be added when working in a non-inertial frame of acceleration \(\vec{a}\). In a lift accelerating downward, \(W = mg - ma\); accelerating upward, \(W_1 = mg + ma_1\).
A conservative force does work independent of the path (only the end points matter); mechanical energy is conserved and potential energy can be defined: \(dU = -\vec{F}.d\vec{x}\), or \(U = -\int \vec{F}.d\vec{x}\). A non-conservative force (friction, air drag) does path-dependent, non-recoverable work, which appears as heat, sound or light.
Work Done by a Variable Force
\(W = \vec{F}\cdot\vec{s} = F\,s\cos\theta\) applies only when \(\vec{F}\) and \(\vec{s}\) are constant and finite.
For a variable force,
\[W = \int_{s_1}^{s_2} F.ds\]
which is the area under the \(F\)–\(s\) curve, provided the force axis starts from zero.
Work–Energy Theorem
- The change in kinetic energy equals the work done by the conservative force:
\[\frac{1}{2}mu^2 - \frac{1}{2}m\mathrm{v}^2 = \left(ma\right).s = F.s\]
- When a non-conservative force is also present:
\[\Delta PE = \Delta KE + W_{air\ resistance}\]
that is, decrease in gravitational P.E. = increase in kinetic energy + work done against non-conservative forces.
Conservation of Linear Momentum
If \(\vec{F} = 0\), then \(\dfrac{d\vec{p}}{dt} = 0\) and \(\vec{p}\) is constant.
The total momentum of an isolated system is conserved during any interaction. Forces during a collision or explosion are internal to the system.
A free body diagram shows all the forces acting on a single body, together with its acceleration. Writing one equation per body per direction, and solving them simultaneously, gives all unknowns.
Atwood machine (\(m_2 > m_1\), frictionless pulley, inextensible string):
\[a = \left(\frac{m_2 - m_1}{m_2 + m_1}\right)g, \qquad T = \left(\frac{2m_1m_2}{m_1 + m_2}\right)g\]
Collisions
In every collision (and explosion), linear momentum is conserved: \(\sum \vec{p}_{initial} = \sum \vec{p}_{final}\). Kinetic energy may or may not be conserved.
Elastic collision: kinetic energy conserved, \(\sum K.E._{initial} = \sum K.E._{final}\).
Inelastic collision: kinetic energy decreases, \(\sum K.E._{initial} > \sum K.E._{final}\).
Perfectly inelastic collision: bodies stick together and move with a common velocity; the loss of kinetic energy is maximum.
During an explosion, \(\sum K.E._{final} > \sum K.E._{initial}\).
There is nothing called a partially elastic collision; elastic collisions are always perfectly elastic.
Coefficient of restitution:
\[e = \frac{\mathrm{v}_1 - \mathrm{v}_2}{u_2 - u_1} = \frac{\text{Relative speed of separation}}{\text{Relative speed of approach}} \quad \text{--- (4.1)}\]
\(e = 1\) for an elastic collision, \(e = 0\) for a perfectly inelastic collision; for any collision, \(0 \le e \le 1\).
- Head-on elastic collision — final velocities:
\[\mathrm{v}_1 = \left(\frac{m_1 - m_2}{m_1 + m_2}\right)u_1 + \left(\frac{2m_2}{m_1 + m_2}\right)u_2 \quad \text{--- (4.8)}\]
\[\mathrm{v}_2 = \left(\frac{m_2 - m_1}{m_1 + m_2}\right)u_2 + \left(\frac{2m_1}{m_1 + m_2}\right)u_1 \quad \text{--- (4.9)}\]
- Particular cases (elastic, head-on):
- Equal masses: the bodies exchange velocities, \(\mathrm{v}_1 = u_2\) and \(\mathrm{v}_2 = u_1\).
- \(m_1 \gg m_2\), \(u_2 = 0\): \(\mathrm{v}_1 \cong u_1\) and \(\mathrm{v}_2 \cong 2u_1\).
- \(m_1 \ll m_2\), \(u_2 = 0\): \(\mathrm{v}_1 \cong -u_1\) and \(\mathrm{v}_2 \cong 0\).
- Perfectly inelastic head-on collision: common velocity
\[\mathrm{v} = \frac{m_1u_1 + m_2u_2}{m_1 + m_2} \quad \text{--- (4.10)}\]
and the loss in kinetic energy
\[\Delta\left(K.E.\right) = \frac{1}{2}\left(\frac{m_1m_2}{m_1 + m_2}\right)\left(u_1 - u_2\right)^2 \quad \text{--- (4.11)}\]
which is always positive — there is always a loss.
- For an inelastic collision with coefficient of restitution \(e\),
\[\Delta\left(K.E.\right) = \frac{1}{2}\left(\frac{m_1m_2}{m_1 + m_2}\right)\left(u_1 - u_2\right)^2\left(1 - e^2\right)\]
The quantity \(\mu = \dfrac{m_1m_2}{m_1 + m_2}\) is the reduced mass of the system.
- Non head-on (two-dimensional) collision: the line of impact carries all the mutual forces; along the common tangent there is no force and hence no change in momentum for either body. The equations to be solved are:
\[m_1u_1\cos\alpha_1 + m_2u_2\cos\alpha_2 = m_1\mathrm{v}_1\cos\beta_1 + m_2\mathrm{v}_2\cos\beta_2 \quad \text{--- (4.12)}\]
\[m_1u_1\sin\alpha_1 = m_1\mathrm{v}_1\sin\beta_1 \quad \text{--- (4.13)}\]
\[m_2u_2\sin\alpha_2 = m_2\mathrm{v}_2\sin\beta_2 \quad \text{--- (4.14)}\]
\[e = \frac{\mathrm{v}_2\cos\beta_2 - \mathrm{v}_1\cos\beta_1}{u_1\cos\alpha_1 - u_2\cos\alpha_2} \quad \text{--- (4.15)}\]
Impulse
- Impulse is the change in linear momentum produced by a force:
\[\vec{J} = \vec{p}_2 - \vec{p}_1 = m\left(\vec{\mathrm{v}}_2 - \vec{\mathrm{v}}_1\right) = \vec{F}.t \quad \text{---(4.17)}\]
\[\vec{J} = \Delta\vec{p} = \int d\vec{p} = \int \vec{F}.dt \quad \text{--- (4.18)}\]
Impulse is defined because in many real situations (hitting a ball with a bat, hammering a nail, bouncing a ball) a large force acts for an unmeasurably short time, but the change in momentum is measurable.
Graphically, impulse is the area under the \(F\)–\(t\) curve (the force axis must start from zero). Increasing the collision time reduces the peak force for the same impulse — the principle behind a wicket-keeper drawing back his hands.
Torque and Couple
- Torque (moment of a force) is the rotational analogue of force:
\[\vec{\tau} = \vec{r}\times\vec{F} \quad \text{--- (4.17)}\]
\[\tau = r\,F\sin\theta \quad \text{--- (4.18)}\]
with \(\theta\) the smaller angle between \(\vec{r}\) and \(\vec{F}\). SI unit: N m. Its direction is perpendicular to the plane of \(\vec{r}\) and \(\vec{F}\) (right-hand thumb rule).
Consequences: torque grows with \(r\) and \(F\); \(\tau_{max} = rF\) at \(\theta = 90^{0}\); \(\tau_{min} = 0\) at \(\theta = 0°\) or \(180°\); the same force gives different torques at different points of application.
Convention: \(\odot\) = out of the plane of the paper (arrow tip), \(\otimes\) = into the plane of the paper (arrow feathers).
A couple is a pair of equal and opposite forces acting along different lines of action. It produces purely rotational motion, with no translational acceleration.
\[\vec{\tau} = \vec{r}_{12}\times\vec{F}_1 = \vec{r}_{21}\times\vec{F}_2 \quad \text{--- (4.19)}\]
\[\tau = r_{12}\,F_1\sin\alpha = r_{21}\,F_2\sin\beta = r\,F \quad \text{--- (4.20)}\]
- The moment of a couple is independent of the axis of rotation and of the points of application; it depends only on the magnitude of either force and the separation between their lines of action. It can be balanced only by another couple.
Mechanical Equilibrium
A system is in mechanical equilibrium if \(\sum \vec{F} = 0\) for any part of the system, so that no part accelerates.
Stable equilibrium: potential energy is a local minimum; the body returns on being disturbed.
Unstable equilibrium: potential energy is a local maximum; the body moves further away on being disturbed.
Neutral equilibrium: potential energy is constant; the body stays in equilibrium wherever it is placed.
Using the potential energy function: at equilibrium \(\dfrac{dU}{dx} = 0\); the sign of \(\dfrac{d^2U}{dx^2}\) is positive for stable, negative for unstable, and zero (or non-existent) for neutral equilibrium.
Centre of Mass and Centre of Gravity
- Centre of mass allows a finite object to be treated as a point object, so that Newton’s laws can be applied to it.
\[\vec{r} = \frac{\sum_{1}^{n}m_i\vec{r}_i}{M}, \qquad x = \frac{\sum_{1}^{n}m_ix_i}{M}, \qquad y = \frac{\sum_{1}^{n}m_iy_i}{M}, \qquad z = \frac{\sum_{1}^{n}m_iz_i}{M}\]
- For a continuous distribution of uniform density,
\[\vec{r} = \frac{\int \vec{r}\,dm}{M}\]
The centre of mass is the point about which the summation of the moments of the masses is zero: \(\sum m_i\vec{r}_i = 0\) when the origin is at the c.m.
Velocity and acceleration of the centre of mass:
\[\vec{\mathrm{v}}_{cm} = \frac{\sum m_i\vec{\mathrm{v}}_i}{M} = \frac{\text{resultant linear momentum}}{\text{total mass}}\]
\[\vec{a}_{cm} = \frac{\sum m_i\vec{a}_i}{M} = \frac{\text{resultant force}}{\text{total mass}}\]
The location of the centre of mass can be changed only by an external unbalanced force; internal forces (collisions, explosions) never move it. It is at the geometrical centre for a symmetric body of uniform density, lies on any axis of symmetry, need not coincide with any particle, and need not even lie within the body (a ring, a horse shoe, a high jumper doing the Fosbury Flop).
For two particles, the c.m. divides the distance between them in the inverse ratio of their masses (it is closer to the heavier mass).
Standard results (Table 4.1): thin semicircular ring \(y_c = \dfrac{2R}{\pi}\); thin semicircular disc \(y_c = \dfrac{4R}{3\pi}\); hemispherical shell \(y_c = \dfrac{R}{2}\); solid hemisphere \(y_c = \dfrac{3R}{8}\); hollow cone \(y_c = \dfrac{h}{3}\); solid cone \(y_c = \dfrac{h}{4}\); isosceles triangular plate \(y_c = \dfrac{H}{3}\); right-angled triangular plate \(x_c = \dfrac{p}{3}\), \(y_c = \dfrac{q}{3}\).
Centre of gravity is the point about which the resultant torque due to gravity is zero — the weighted average of the weights of the particles. In a uniform gravitational field (\(g\) constant), the c.g. coincides with the c.m., which is true for all everyday objects on the Earth. The c.g. of a laminar object can be found by suspending it freely from two different points and taking the intersection of the two plumb lines.
The c.m. is a fixed property of a rigid body; the c.g. can shift with orientation if the gravitational field is non-uniform — which requires an object of a size comparable to that of the Earth (at least a few thousand km), and would place the c.g. slightly below the c.m.
4.16 Check Your Concepts
A ball is thrown vertically upward. At the highest point of its path, its velocity is momentarily zero. Is the net force on it also zero at that instant? Explain, and state which of Newton’s laws settles the question.
A book lies at rest on a table. Its weight acts downward and the normal reaction acts upward, and the two are equal and opposite. A student says these two forces are an action–reaction pair. Explain why the student is wrong, and identify the true reaction partner of each of these two forces.
Why must the mathematical form of Newton’s second law be remembered as \(\vec{F} = \dfrac{d\vec{p}}{dt}\) rather than \(\vec{F} = m\vec{a}\)? Give one situation in which the two forms give different answers, and explain why.
A person standing inside a closed, windowless lift observes a plumb line hanging perfectly vertical and steady. Can the person conclude that the lift is at rest? What can, and what cannot, be concluded from this single observation?
Gravity is by far the weakest of the four fundamental forces, and yet it decides the structure of the entire universe. Explain this apparent contradiction by comparing the range and the sign of the gravitational and electromagnetic forces.
A pseudo force is described as “non-real” and yet it can be recorded by a weighing scale. Explain why it is called non-real, and state the two properties of a real force that it fails to satisfy.
You carry a heavy suitcase from the ground floor to the first floor by (a) the staircase and (b) a ramp. Neglecting friction, is the work done against gravity the same in both cases? What if friction is present? Which force is conservative and which is not, and why does the distinction matter here?
The formula \(W = Fs\cos\theta\) cannot be used to find the work done in lifting a satellite through several thousand kilometres. State the reason, and describe the method that must be used instead.
Why is it stated that the force axis must necessarily start from zero if the work done is to be read off as the area under an \(F\)–\(s\) graph, or the impulse as the area under an \(F\)–\(t\) graph? What error would creep in otherwise?
A cracker at rest explodes into several fragments that fly off in all directions. Show that the total momentum after the explosion is still zero, and explain why the kinetic energy is nevertheless greater after the explosion than before.
Two identical balls collide head-on elastically. One was moving, the other was at rest. Predict the outcome, and explain, without doing any calculation, why the moving ball must come to a complete stop.
A truly elastic collision is impossible in daily life, and yet the collisions of gas molecules are treated as elastic. What is lost in an everyday collision that is not lost in a molecular one?
There is no such thing as a “partially elastic” collision, although there is such a thing as a partially inelastic collision. Explain why, in terms of the coefficient of restitution.
Two cars of the same mass, one travelling at 40 km/h and one at 80 km/h, are brought to rest by identical brakes. The second car takes four times the stopping distance. Using the work–energy theorem, explain why the distance goes up as the square of the speed and not in direct proportion to it.
An airbag and a rigid steering wheel both bring a passenger’s body to rest in a crash. Explain, using the definition of impulse, why the airbag is safer, even though the change in momentum of the passenger is exactly the same in both cases.
Why is the handle of a door always fixed at the edge farthest from the hinges? Why does pushing the door in a direction along its surface, straight towards the hinges, never open it, no matter how hard you push?
A couple has zero resultant force and yet it produces rotation. Explain how this is possible, and why two equal and opposite forces acting along the same* line would produce no rotation at all.*
The moment of a couple is independent of the axis of rotation, while the moment of a single force is not. Explain the reason for this difference.
A ball rests (a) at the bottom of a valley, (b) at the top of a hump, and (c) on a flat horizontal plane. All three are equilibrium positions. Distinguish between them in terms of potential energy, and state what the sign of \(\dfrac{d^2U}{dx^2}\) would be in each case.
Cargo ships are loaded so that the heaviest goods lie lowest in the hold. Explain, in terms of centre of mass and stable equilibrium, what would go wrong if the heavy cargo were stacked on the top deck instead.
A shell fired from a gun follows a parabolic path and explodes in mid-air into several fragments. What happens to the centre of mass of the fragments after the explosion? Which property of internal forces makes your answer certain?
The centre of mass of a ring, a horse shoe and a high jumper doing a Fosbury Flop can all lie outside the material of the body. Explain how this is possible, given that the centre of mass is defined as the point where the whole mass may be assumed concentrated.
Under what condition does the centre of gravity of a body coincide with its centre of mass? For what kind of object would the two be measurably different, and which of the two would lie lower?
Describe how you would experimentally locate the centre of gravity of an irregular piece of cardboard using only a nail, a piece of thread and a small weight. Why is a single suspension not enough?
In real life, objects never travel with uniform velocity, even on a horizontal surface, unless something is done. Why is it so? What is to be done?
For the study of any kind of motion, we never use Newton’s first law of motion directly. Why should it be studied?
Are there any situations in which we cannot apply Newton’s laws of motion? Is there any alternative for it?
You are inside a closed capsule from where you are not able to see anything about the outside world. Suddenly you feel that you are pushed towards your right. Can you explain the possible cause(s)? Is it a feeling or a reality? Give at least one more situation like this.
Among the four fundamental forces, only one force governs your daily life almost entirely. Justify the statement by stating that force.
Find the odd man out: (i) Force responsible for a string to become taut on stretching (ii) Weight of an object (iii) The force due to which we can hold an object in hand.
You are sitting next to your friend on the ground. Is there any gravitational force of attraction between you two? If so, why are you not coming together naturally? Is any force other than the gravitational force of the earth coming into the picture?
Distinguish between: (A) Real and pseudo forces, (B) Conservative and non-conservative forces, (C) Contact and non-contact forces, (D) Inertial and non-inertial frames of reference.
State the formula for calculating the work done by a force. Are there any conditions or limitations in using it directly? If so, state those clearly. Is there any mathematical way out for it? Explain.
Justify the statement, “Work and energy are the two sides of a coin”.
From the terrace of a building of height H, you dropped a ball of mass m. It reached the ground with speed v. Is the relation \(mgH = \dfrac{1}{2}m\mathrm{v}^2\) applicable exactly? If not, how can you account for the difference? Will the ball bounce to the same height from where it was dropped?
State the law of conservation of linear momentum. It is a consequence of which law? Give an example from our daily life for conservation of momentum. Does it hold good during the burst of a cracker?
Define coefficient of restitution and obtain its value for an elastic collision and a perfectly inelastic collision.
Discuss the following as special cases of elastic collisions and obtain their exact or approximate final velocities in terms of their initial velocities.
- Colliding bodies are identical.
- A very heavy object collides on a lighter object, initially at rest.
- A very light object collides on a comparatively much massive object, initially at rest.
A bullet of mass \(m_1\) travelling with a velocity \(u\) strikes a stationary wooden block of mass \(m_2\) and gets embedded into it. Determine the expression for the loss in the kinetic energy of the system. Is this violating the principle of conservation of energy? If not, how can you account for this loss?
One of the effects of a force is to change the momentum. Define the quantity related to this and explain it for a variable force. Usually, when do we define it instead of using the force?
While rotating an object, or while opening a door or a water tap, we apply a force or forces. Under which conditions is this process easy for us? Why? Define the vector quantity concerned. How does it differ for a single force and for two opposite forces with different lines of action?
Explain balancing or mechanical equilibrium. The linear velocity of a rotating fan as a whole is generally zero. Is it in mechanical equilibrium? Justify your answer.
Why do we need to know the centre of mass of an object? For which objects may its position differ from that of the centre of gravity?
4.17 Practice with Numericals
Use \(g = 10\) m s⁻², unless otherwise stated.
A body of mass 4 kg moving at 15 m s⁻¹ is brought to rest by a constant retarding force in 6 s. Calculate (a) the retardation, (b) the retarding force, and (c) the distance travelled before stopping.
A hose pipe of cross-sectional area 5 cm² ejects water horizontally at 12 m s⁻¹ against a fixed wall, where the water runs down without rebounding. Taking the density of water as 1000 kg m⁻³, calculate the force exerted on the wall.
A person of mass 70 kg stands on a weighing scale inside a lift. Find the reading of the scale, in newton, when the lift (a) is at rest, (b) accelerates upward at 2.5 m s⁻², (c) moves upward with a uniform velocity of 4 m s⁻¹, and (d) accelerates downward at 2.5 m s⁻².
Calculate the gravitational force of attraction between two spheres, each of mass 500 kg, whose centres are 2 m apart. Take \(G = 6.67\times10^{-11}\) SI units. Compare this force with the weight of one sphere.
A car of mass 1200 kg moving at 20 m s⁻¹ on a horizontal road stops in 40 s when the engine is switched off. On another occasion, moving at the same speed, the driver brakes and stops the car in 25 m. Calculate the braking force applied.
A force acting along the direction of motion of a body varies with position as \(F = 4x + 2\) (newton, with \(x\) in metre). Calculate the work done by this force as the body moves from \(x = 0\) to \(x = 3\) m.
A force increases linearly from 6 N to 14 N while an object is displaced from \(x = 2\) m to \(x = 6\) m along the direction of the force. Using the area under the \(F\)–\(x\) graph, find the work done.
A stone of mass 0.5 kg is dropped from a height of 45 m and strikes the ground with a speed of 28 m s⁻¹. Calculate (a) the loss in potential energy, (b) the kinetic energy on reaching the ground, and (c) the work done against air resistance.
In an Atwood machine, masses of 8 kg and 5 kg hang from the two ends of a light inextensible string passing over a frictionless pulley. Calculate the acceleration of the system and the tension in the string.
Three blocks of masses 2 kg, 3 kg and 5 kg lie in contact on a frictionless horizontal surface, in that order. A horizontal force of 20 N is applied on the 2 kg block. Find the acceleration of the system and the contact force between the 3 kg and 5 kg blocks.
A bullet of mass 20 g moving at 300 m s⁻¹ strikes a wooden block of mass 1.98 kg resting on a frictionless surface and gets embedded in it. Calculate (a) the common velocity after the impact, and (b) the loss in kinetic energy of the system.
A ball dropped from a height of 2.5 m on a hard floor rebounds to a height of 1.6 m. Calculate the coefficient of restitution, and find the fractional loss in the kinetic energy of the ball during the impact.
A body of mass 3 kg moving at 8 m s⁻¹ collides head-on and elastically with a stationary body of mass 5 kg. Calculate the velocities of both the bodies after the collision.
A shell of mass 5 kg moving horizontally at 12 m s⁻¹ explodes into two fragments of masses 2 kg and 3 kg. The 2 kg fragment continues along the original direction with a speed of 30 m s⁻¹. Find the velocity of the 3 kg fragment, and the kinetic energy supplied by the explosion.
A cricket ball of mass 160 g moving at 25 m s⁻¹ is struck by a bat and returns along the same line with a speed of 35 m s⁻¹. If the ball and the bat are in contact for 0.02 s, calculate the impulse received by the ball and the average force exerted by the bat.
Bullets of mass 50 g each are fired at the rate of 4 per second, normally, at a firmly fixed hard wall of area 20 cm². Each bullet strikes at 500 m s⁻¹ and rebounds with a coefficient of restitution of 0.6. Calculate the average force and the average pressure experienced by the wall.
A force of 40 N is applied at a point 60 cm from the axis of rotation of a rigid body. Calculate the torque when the angle between the position vector and the force is (a) 90°, (b) 45°, and (c) 180°.
Two equal and opposite forces of 20 N each act on a rod along parallel lines whose perpendicular separation is 25 cm. Calculate the moment of the couple. If the axis of rotation is shifted so that it now lies outside both lines of action, what is the new moment of the couple?
A uniform plank of mass 40 kg and length 3 m is supported horizontally by two vertical cables tied at points 0.5 m from either end. A boy of mass 30 kg stands on the plank at a distance of 1 m from the left cable. Calculate the tension in each cable.
A ladder of negligible mass with a cross bar rests on a frictionless horizontal floor, the angle between its legs being 60°. Each leg is 1.2 m long and the cross bar is fixed at the mid-points of the legs. Calculate the tension in the cross bar when a person of mass 60 kg stands on the ladder.
Point masses of 2 kg, 3 kg and 5 kg are placed at the points \((1, 0)\), \((0, 4)\) and \((3, 3)\) respectively, all coordinates being in metre. Locate the centre of mass of the system.
Two point masses of 5 kg and 15 kg are placed 80 cm apart. Locate their centre of mass, and state its distance from the heavier mass.
A circular hole of radius \(R/3\) is cut from a uniform disc of radius \(R\), the centre of the hole being at a distance \(R/2\) from the centre of the disc. Locate the centre of mass of the remaining part of the disc.
Three thin uniform rods, each of mass \(m\) and length \(L\), are joined to form the three sides of an equilateral triangle. Taking one vertex as the origin and one side along the \(x\)-axis, locate the centre of mass of the system.
Two skaters of masses 50 kg and 70 kg stand at rest on frictionless ice, 6 m apart, holding the two ends of a light rope. They pull themselves towards each other. At what distance from the initial position of the 50 kg skater will they meet?
A truck of mass 5 ton travelling on a horizontal road with 36 km hr⁻¹ stops after travelling 1 km when its engine fails suddenly. What fraction of its weight is the frictional force exerted by the road? If we assume that the story repeats for a car of mass 1 ton, i.e., a car moving with the same speed stops in a similar distance, how much will the fraction be?
A lighter object A and a heavier object B are initially at rest. Both are imparted the same linear momentum. Which will start with greater kinetic energy: A or B, or will both start with the same energy?
As I was standing on a weighing machine inside a lift, it recorded 50 kg wt. Suddenly, for a few seconds, it recorded 45 kg wt. What must have happened during that time? Explain with complete numerical analysis.
The figure below shows a block of mass 35 kg resting on a table. The table is so rough that it offers a self-adjusting resistive force of 10% of the weight of the block for its sliding motion along the table. A 20 kg wt load is attached to the block and is passed over a pulley to hang freely on the left side. On the right side there is a 2 kg wt pan attached to the block and hung freely. Weights of 1 kg wt each can be added to the pan. Minimum how many, and maximum how many, such weights can be added into the pan so that the block does not slide along the table?
Diagram
Fig. 4.30 – A 35 kg block on a rough table, with a 20 kg wt load hanging over a pulley on the left and a 2 kg wt pan hanging over a pulley on the right.
Power is the rate of doing work, or the rate at which energy is supplied to the system. A constant force \(F\) is applied to a body of mass \(m\). The power delivered by the force at time \(t\) from the start is proportional to
- \(t\)
- \(t^2\)
- \(\sqrt{t}\)
- \(t^0\) Derive the expression for power in terms of \(F\), \(m\) and \(t\).
40000 litre of oil of density 0.9 g cc is pumped from an oil tanker ship into a storage tank at a 10 m higher level than the ship, in half an hour. What should be the power of the pump?
Ten identical masses (\(m\) each) are connected one below the other with 10 strings. Holding the topmost string, the system is accelerated upwards with acceleration \(g/2\). What is the tension in the 6th string from the top (the topmost string being the first string)?
Two galaxies of masses 9 billion solar mass and 4 billion solar mass are 5 million light years apart. If the Sun has to cross the line joining them without being attracted by either of them, through what point should it pass?
While decreasing linearly from 5 N to 3 N, a force displaces an object from 3 m to 5 m. Calculate the work done by this force during this displacement.
The variation of a force in a certain region is given by \(F = 6x^2 - 4x - 8\). It displaces an object from \(x = 1\) m to \(x = 2\) m in this region. Calculate the amount of work done.
A ball of mass 100 g dropped on the ground from 5 m bounces repeatedly. During every bounce, 64% of the potential energy is converted into kinetic energy. Calculate the following:
- Coefficient of restitution.
- Speed with which the ball comes up from the ground after the third bounce.
- Impulse given by the ball to the ground during this bounce.
- Average force exerted by the ground if this impact lasts for 250 ms.
- Average pressure exerted by the ball on the ground during this impact if the contact area of the ball is 0.5 cm².
A spring ball of mass 0.5 kg is dropped from some height. On falling freely for 10 s, it explodes into two fragments of mass ratio 1:2. The lighter fragment continues to travel downwards with a speed of 60 m/s. Calculate the kinetic energy supplied during the explosion.
A marble of mass 2m travelling at 6 cm/s is directly followed by another marble of mass m with double the speed. After collision, the heavier one travels with the average initial speed of the two. Calculate the coefficient of restitution.
A 2 m long wooden plank of mass 20 kg is pivoted (supported from below) at 0.5 m from either end. A person of mass 40 kg starts walking from one of these pivots to the farther end. How far can the person walk before the plank topples?
A 2 m long ladder of mass 10 kg is kept against a wall such that its base is 1.2 m away from the wall. The wall is smooth but the ground is rough. The roughness of the ground is such that it offers a maximum horizontal resistive force (for sliding motion) half that of the normal reaction at the point of contact. A monkey of mass 20 kg starts climbing the ladder. How far can it climb along the ladder? How much is the horizontal reaction at the wall?
Four uniform solid cubes of edges 10 cm, 20 cm, 30 cm and 40 cm are kept on the ground, touching each other in order. Locate the centre of mass of their system.
A uniform solid sphere of radius \(R\) has a hole of radius \(R/2\) drilled inside it. One end of the hole is at the centre of the sphere while the other is at the boundary. Locate the centre of mass of the remaining sphere.
In the following table, every item on the left side can match with any number of items on the right hand side. Select all those.
| Types of collision | Illustrations |
|---|---|
| (a) Elastic collision | (i) A ball hit by a bat. |
| (b) Inelastic collision | (ii) Molecular collisions responsible for pressure exerted by a gas. |
| (c) Perfectly inelastic collision | (iii) A stationary marble A is hit by marble B and the marble B comes to rest. |
| (d) Head on collision | (iv) A blob of clay dropped on the ground sticks to the ground. |
| (v) Out of anger, giving a kick to a wall. | |
| (vi) A striker hits the boundary of a carrom board in a direction perpendicular to the boundary and rebounds. |
4.18 Objective Questions
- Consider the following pair of forces of equal magnitude and opposite directions: (P) Gravitational forces exerted on each other by two point masses separated by a distance. (Q) Couple of forces used to rotate a water tap. (R) Gravitational force and normal force experienced by an object kept on a table. For which of these pair/pairs do the two forces NOT cancel each other’s translational effect?
- Only P
- Only P and Q
- Only R
- Only Q and R
- Consider the following forces: (w) Force due to tension along a string, (x) Normal force given by a surface, (y) Force due to air resistance and (z) Buoyant force or upthrust given by a fluid. Which of these are electromagnetic forces?
- Only w, y and z
- Only w, x and y
- Only x, y and z
- All four.
- At a given instant three point masses m, 2m and 3m are equidistant from each other. Consider only the gravitational forces between them. Select the correct statement/s for this instance only:
- Mass m experiences maximum force.
- Mass 2m experiences maximum force.
- Mass 3m experiences maximum force.
- All masses experience force of the same magnitude.
- The rough surface of a horizontal table offers a definite maximum opposing force to initiate the motion of a block along the table, which is proportional to the resultant normal force given by the table. Forces \(F_1\) and \(F_2\) act at the same angle \(\theta\) with the horizontal and both are just initiating the sliding motion of the block along the table. Force \(F_1\) is a pulling force while the force \(F_2\) is a pushing force. \(F_2 > F_1\), because
- Component of \(F_2\) adds up to weight to increase the normal reaction.
- Component of \(F_1\) adds up to weight to increase the normal reaction.
- Component of \(F_2\) adds up to the opposing force.
- Component of \(F_1\) adds up to the opposing force.
- A mass 2m moving with some speed is directly approaching another mass m moving with double speed. After some time, they collide with coefficient of restitution 0.5. The ratio of their respective speeds after collision is
- 2/3
- 3/2
- 2
- ½
- A uniform rod of mass 2m is held horizontal by two sturdy, practically inextensible vertical strings tied at its ends. A boy of mass 3m hangs himself at one third length of the rod. The ratio of the tension in the string close to the boy to that in the other string is
- 2
- 1.5
- 4/3
- 5/3
- Select the WRONG statement about centre of mass:
- Centre of mass of a ‘C’ shaped uniform rod can never be a point on that rod.
- If the line of action of a force passes through the centre of mass, the moment of that force is zero.
- Centre of mass of our Earth is not at its geometrical centre.
- While balancing an object on a pivot, the line of action of the gravitational force of the earth passes through the centre of mass of the object.
- For which of the following objects will the centre of mass NOT be at their geometrical centre? (I) An egg (II) A cylindrical box full of rice (III) A cubical box containing assorted sweets
- Only (I)
- Only (I) and (II)
- Only (III)
- All, (I), (II) and (III).
- In the following table, every entry on the left column can match with any number of entries on the right side. Pick up all those and write them respectively against A, B, C and D.
| Name of the force | Type of the force |
|---|---|
| A. Force due to tension in a string | P. EM force |
| B. Normal force | Q. Reaction force |
| C. Frictional force | R. Conservative force |
| D. Resistive force offered by air or water for objects moving through it | S. Non-conservative force |