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Every particle in this universe attracts every other particle with a force that is proportional to the product of these masses and inversely proportional to the square of the distance between them.

According to Newton,

Force of gravitation, F µ product of masses

directly proportional to 1/square of the separation between them

Thus, F= G(m1m2/r² )

where m₁ and m₂ are the masses of the particles, r is the distance of separation between them and G in Universal Gravitational Constant.

Magnitude (and unit) of G : 6.67 * 10^–11 N m² / kg²

Dimension of G         : M^–1 L³ T².

Characteristics of the Gravitational Force:

(a)   Gravitational force is always attractive and directed along the line joining the particles.

(b)   It is independent of the nature of the medium surrounding the particles.

(c)   It holds good for long distances like inter-planetary distances and also for short distances like inter-atomic distances.

(d)   Interaction means that, both the particles experience forces of equal magnitude in opposite directions. If  F1, F2 are the forces exerted on particle 1 particle 2 and on particle 2 by particle 1 respectively, then F1= -F2. Since the forces F1  and F2  are exerted on different bodies, they are known as action-reaction pair.

(e)   It is a conservation force. Therefore, the work done by the gravitational force on a particle is independent of the path described by the particle. It depends upon the initial and final position of the particle. Therefore no work is done by the gravity if a particle moves in a closed path.

(f)    If a particle 1 is acted by n particles, say, the net force F1  exerted on it must be equal to the vector sum of the forces due to surrounding particles.

Þ F = ΣF

where  F1= force acted on the particle 1, by the i^th particle

Hence, gravitational force between any two particles does not depend upon the presence or absence of other particles (bodies).

Illustrations

1. Three identical particles each of mass mare placed at the vertices of an equilateral triangle of side a. Find the forcer exerted by this system on a particle P of mass m placed at the

(a)            the mid point of a side

(b)            centre of the triangle.

Sol. Using the superposition principle, the net gravitational force on P is

F = FA + FB+FC

(a)   As shown in the figure, when P is at the mid point of a side, FA and FB will be equal in magnitude but opposite in direction. Hence they will cancel each other. So the net force on the particle P will be the force due to the particle placed at C only.

(b)       At the centre of the triangle O, the forces FA, FB, Fc will be in magnitude and will subtend 120º with each other. Hence the resultant force on P at O is

F= FA + FB + FC = 0

You must have seen the night-sky composed of stars, planets, moon and you could have also been lucky enough to see a shooting star or meteor as it is termed. If you make a careful study, you could also observe, that the pattern of night-sky keeps on changing with time as well as with seasons in a seemingly complex manner. Our ancient Indian scientists had made careful observation about this changing pattern and had deduced that planets revolve around the sun, a fact which was later rediscovered by Copernicus.

It took the genius of Newton to reduce this complicated motion into a very simple universal law – the law of Gravitation, a law, which not only applies to celestial bodies but also applies to that famous apple which, Newton saw falling.

OBJECTIVE

After learning this chapter we will be able to understand the law of gravitational force, gravitational potential energy. We will also be able to learn the concept of planetary motion of different planets and other heavenly bodies. In this chapter we will revolve around the Universal Law of Gravitation and its applications and try to gain an insight into one of the basic forces in the Universe.

PRE-REQUISITE

Þ            Concept of acceleration

Þ            Earth’s gravity

Þ            Concept of circular motion

Þ      Concept of centripetal and centrifugal forces

Þ      Newton’s third law of motion (Action and Reaction)

CORE CONCEPTS

When you throw a ball up, the ball goes up, its velocity is retarded and it finally comes to rest. Henceforth, it gains acceleration and returns back to the ground. This retardation and subsequent acceleration is a consequence of the universal gravitational force. There is a specific law, which guides the above phenomena, known as Newton’s Law of Gravitation. It states:

Every particle in this Universe attracts every other particle with a force, which is directly proportional to the product of their masses and inversely proportional to the square of the distance between them.

Let m1 and m2 be the masses of the two particles and r be the separation between them (see figure given below).

“The space around a body within which its force of gravitational attraction is perceptible (by any other body in this space) is called its gravitational field.”

The intensity E, of the gravitational field of a mass ‘m‘ at a point at distance ‘r‘ from it is the force experienced by a unit mass placed at this point in the field. (Assuming that the presence of unit mass does not affect the gravitational field of the mass m)

ü        Energy is defined to be the capacity to do work. Energy is a scalar quantity and measured in Joules (J).

Mechanical energy is of two types:

Þ            Kinetic Energy

Þ            Potential Energy

Kinetic energy is the energy which is possessed by a body by virtue of its motion. Kinetic energy of a body is the energy due to its motion. If a body of mass ‘m’ moves with a velocity ‘v’, its kinetic energy is given by

For an unaccelerated body, the resultant force acting on the body is zero. But for an accelerated body, the resultant force is not zero and will do some work. We can relate the work done by this force with the consequent increase in kinetic energy of the body. Consider a body being acted on by various forces

ü        The above equation states that the work done by the resultant external force on a body is equal to the change in the kinetic energy of the body. This relation is called Work-Energy Theorem.

Illustrations

1.      A 50 g bullet strikes a wooden plank and comes to rest after penetrating 10 cm into wood. If wood offers a resistance of 10000 N, what is the velocity with which the bullet strikes the plank?

Sol. Work done by the resistant force, acting on the bullet opposite in the direction of displacement of bullet, is given by

W = 10000 N * (10 * 10²) m

= 1000 J

According to Work-Energy Theorem this should be equal to the change in kinetic energy.

W = K₂K₁

i.e.,                          1000 = 0  K1

(K₂ = 0, because the final velocity of bullet is zero)

K1 = 1/2(mv²) = 1000

where v is the initial velocity of bullet

v² = (2 * 1000 J)/ (5 * 10^-3 kg)

= 4000 (m/s)²

v = 200 m/s

ü        We all know that it is hard work lifting a heavy box from one platform to another in a railway station. Similarly we know that children need lots of energy as they grow up.

ü        We feel tired if we run a hundred meters in twenty seconds, whereas we could walk that distance easily in a couple of minutes.

ü        These are some common sense notions of work and energy which can however be precisely defined and measured in physics.

ü        These definitions are measurements can be used consistently to describe and predict the behaviour of bodies and thus can form very powerful tool for analysis of physical systems.

OBJECTIVE

ü        In the previous chapter we studied the Newton’s laws of motion to understand how objects move under the influence of force acting on them or the relation between the force and the acceleration produced by this force.

ü        In this chapter however, we will study the situation where one is not interested in such an exhaustive study of object’s motion rather desires to relate the final velocity of an object to the forces acting on it without going into the deeper details how the object acquired that velocity. The work-energy theorem solves this purpose.

PRE-REQUISITE

Velocity

ü        Rate of change of position of an object is known as its velocity.

V = dx /dt

Acceleration

ü        Rate of change of velocity of an object is known as its acceleration.

a= dv / dt = d²x / dt²

Force

ü        Force is a push or pull which tends to change the position of the object on which it is applied.

Newton’s first law of motion

ü        It states that every object persists in its natural state of motion i.e. continues to be at rest or moves in a straight line with uniform (constant) velocity. (This is what is meant by natural state of motion); In the absence of a net external force acting (impressed) on it.

Newton’s Second law of motion

Newton’s Third Law of Motion

ü        When ever a body exerts a force on a second body, the second body always exerts a force of equal magnitude but opposite in direction on the first one.

Conservation of Energy

ü        Whenever one form of energy is transformed into other forms, the total amount of energy before transformation is always equal to the amount of energy after transformation, i.e., the total amount of energy remains conserved.

Centre of mass of system of the discrete particles

The centre of mass of an object is a point that represents the entire body and move in the same way as a point mass having mass equal to that of the object, when subjected to the same external forces that act on the object. That is, if the resultant force acting on an object (or system of objects) of mass m is F, the acceleration of the centre of mass of the object (or system) is given by

Where the sums extend over all masses composing the object. In a uniform gravitational field, the centre of mass and the centre of gravity coincide.

Centre of mass of continuous distribution of particles

Centre of mass of a body having continuous distribution of particles (mass) is given by

Velocity & Acceleration of centre of mass

Position vector of the centre of mass of a system of particle is given by

Conservation of Linear Momentum

System of particles

Conservation of motion of centre of mass

In absence of a net external resultant force along a certain direction on a particulars system will retain its state of motion along that direction. If it had been at rest, it will remain at rest. If it had been moving with uniform velocity then it will continue moving with uniform velocity

Characteristic of linear momentum

If depends on the frame of reference, e.g., the linear momentum of a body at rest in a moving train, is zero relative to a person sitting in the train while it is not zero for a person standing on the ground.

Two bodies of same mass and moving with same speed will have different momenta unless their directions of motion are same.

Relation between kinetic energy and momentum

Each object around us has certain shape and size. When we study motion of such objects, we will have to study motion of all the different particles of the object. Practically speaking, this is not possible and we will have to consider alternative ways to study motion of the object. Scientists have defined a concept called `Centre of Mass’ to take care of this situation. To consider motion of an object having finite size, they consider motion of the centre of mass of the object. The basic objective of this chapter is to understand this concept.

OBJECTIVE

We will attempt to define the concept of a point where the mass of the whole body can be assumed to be concentrated. This helps us in analysis of problems that involve application of forces that result in subsequent motion. We will study systems which have a non uniform distribution of mass. And finally we will understand the motion of centre of mass as the body moves.

PRE-REQUISITE

The following pre-requisites apply

Þ         Vector Representation.

Þ         Integration in one variable.

Þ         Force & Acceleration.

CORE CONCEPTS

Classically each and every object has some finite size, but we have assumed them as particles i.e. having mass but no size so far.

In translatory motion each point on an object undergoes the same displacement as any other point as time goes on, so that the motion of one particle represents the motion of whole object. Thus, the translatory motion of an object of finite size can be studied by analysing the motion of any constituent particle.

But, in cases where the motion is not translatory, rather than considering any point on the object, a point known as the centre of mass is defined, and the motion of this point is studied. The motion of the centre of mass leads to the analysis of the object as a whole

Conceptually, the point where the whole mass of body or system can be assumed to be concentrated for simplified study of its motion is called the centre of mass.

For a discrete system of particles the positions (see figure given below) of the centre of mass is

To find the unit of a given physical quantity in a given system of units:

ü      By expressing a physical quantity in terms of basic quantities we find its dimensions. In the dimensional formula replacing M, L, T by the fundamental units of the required system, we get the unit of physical quantity. However, sometimes we assign a specific name to this unit.

Example

Force is numerically equal to the product of mass and acceleration

i.e. Force = mass × acceleration

Its unit in SI system will be Kgms^-2 which is given a specific name “Newton (N)”.

Similarly, its unit in CGS system will be gm cms^-2 which is called “dyne”.

To find dimensions of physical constants or coefficients

The dimensions of a physical quantity is unique because it is the nature of the physical quantity and the nature does not change. If we write any formula or equation incorporating the given physical constant, we can find the dimensions of the required constant or co-efficient.

Example

From Newton’s law of Gravitation, the force exerted by one mass upon another is

To convert a physical quantity from one system of units to another

This is based on the fact that for a given physical quantity,

magnitude × unit = constant

So, when unit changes, magnitude will also change.

Example

1.         Convert one Newton into dyne

Sol. Dimensional formula for Newton  = [ML/T²]

or 1 N = 1 Kg m/s²    But 1 kg = 10³g

and 1 m = 10²cm

1N = (10³ g) ( 10²cm)/s² = 10^5 g cm/s² = 10^5 dyne

To check the dimensional correctness of a given physical relation

This is based on the principle that the dimensions of the terms on both sides of an equation must be same. This is known as the ‘principle of homogeneity’. If the dimensions of the terms on both sides are same, the equation is dimensionally correct, otherwise not. It is not necessary that a dimensionally correct equation is also physically correct but a physically correct equation has to be dimensionally correct.

Þ        Consider the formula, T = 2Π √(1/g)

where T is the time period of oscillation of a simple pendulum in a simple harmonic motion, l and g are the length of the pendulum and gravitational constants respectively. Check this formula, whether it is correct or not, using the concept of dimension.

Thus the above equation is dimensionally correct (homogeneous) and later you will come to know that it is physically also correct.

Þ        Consider the formula

Check this formula, whether it is correct or not, using the concept of dimension.

Dimensionally

In this case also the formula is dimensionally correct but, you know that it is physically incorrect as the correct formula is given by

Concept of friction:

Component of normal reaction

We have seen that a ball rolling on a floor stops after some time. When we switch off the engine of a car, it stops after traveling some distance. Similarly when we apply brakes, our bicycle comes to rest after traveling some distance. The above examples show that some invisible force is opposing the motion of one body over the other. This opposing force is called friction. Friction is an opposing force that comes into play when one body actually moves (slides or rolls) or even tries to move over the surface of another body.

Cause of friction

Roughness of surfaces is the cause of friction.

When two bodies are in contact with each other, the irregularities on the surfaces get interlocked and oppose any relative motion.

Normal reaction

When a body of mass m is lying on a horizontal surface, it presses the surface due its weight. Contact force F_c = R in case no external force is acting on the body. Also in this case friction is zero and F_c is perpendicular to the surface. In the diagram below, R is the normal reaction.

Static and kinetic friction

The opposing force that comes into play when one body tends to move over the surface of another (but the actual motion has not yet started) is called static friction. Limiting friction is the maximum opposing force that comes into play when one body is at the verge of moving over the surface of another body. It is denoted by f_s, and is called maximum force of static friction or limiting friction.

Laws of friction

The following are the laws of limiting friction:

(a) The magnitude of the force of limiting friction f_s, is directly proportional to the normal reaction R.

(b) The direction of the force of limiting friction is always opposite to the applied force.

(c) It is independent of the apparent area of contact.

(d) It depends on the nature and material of the surfaces in contact.

For example, when two polished metal surfaces are in contact,  = 0.2

When these surfaces are lubricated,  gets reduced. Hence, it depends on the nature of the surfaces.

Rolling friction

Rolling friction is always less than dynamic and static friction. When a body rolls on a level track, the area of contact is very small. This causes a depression in the surface below. This causes rolling friction.

The velocity of the point of contact of the wheel with respect to the floor remains zero all the time. Thus, rolling wheel constantly climbs a hill and has to simultaneously get itself detached from the road. Rolling friction is also directly proportional to the normal reaction and is inversely proportional to the radius of rolling body. Combining the two, we get,

F is directly proportional to R/r

F = μx (R/r)

Where μx is called coefficient of rolling friction.

Advantages and Disadvantages of friction

Friction is a necessary evil. It is necessary because we cannot do any work without it. At the same time, it is also an evil because it involves unnecessary wastage of energy.

There are advantages of friction. It can be understood with the help of the following examples:

Walking will not be possible without friction.

Brakes of vehicles will not work without friction.

Writing on a blackboard or a piece of paper is possible only due to friction between the blackboard and the chalk or the paper and the pen. Cleaning with sand paper will not be possible without friction.

There are disadvantages of friction. Friction is an evil and this can be understood with the following   examples:
Extra energy is required to overcome the friction between moving parts.

ü       Friction causes wear and tear of different parts of machinery.

ü       Frictional forces result in the production of heat, which causes damage to the machinery.

Methods of reducing friction

By polishing, by lubrication, By streamlining, By using ball-bearings,

Angle of friction

It is the angle between normal Reaction and the resultant of the normal reaction and limiting frictional force. i.e. we have a body of mass m which is placed on a table and we say that the body and the surface of the table have a coefficient of friction m between them. (fig. 3.7)

If we apply a small force F, the body will not move. Let us gradually increase the force until the body starts moving. At one stage the applied force will be equal to the frictional force. The coefficient of static friction ms  = F/N where F is the applied force and N = mg is the normal force.

Newton’s third law of motion was discovered and formulated, during the investigation of the fact that in all experiments it appeared that “when ever a body exerts a force on a second body, the second body always exerts a force on the first one”

Let us visualize and understand this phenomena with an experiment.

Suppose, we throw a stone on a surface of good strength; and the surface is made of glass, one finds it broken (the surface). From here one concludes that a force was exerted by stone on the surface and consequently it was broken.

Now, the question is, did that surface also exert a force on the stone. Just to know about it let us change our throwing object from stone to an egg of almost equal mass. Now, one throws this egg on the same surface of good strength with the same throwing force which he used for the stone. What happens? Obviously with your daily experience you know that the egg will be broken (And the damage to the surface will not be visible due to egg’s spoiling the observation).

This is only possible if there was a force acting on the egg at the time it hit the surface. In fact we can now conclude that there is a mutual force acting on the contact point of the surface and the object thrown. The breaking of either one (or may be both) depends on their ability to absorb forces without getting damaged (that is their strength) so in precise words:

Þ  To every action there is always opposite and equal reaction, it is equivalent to say that mutual actions of two bodies upon each other are always equal and directed to contrary parts.

Þ  The most important fact to notice here is that these oppositely directed equal action and reaction can never balance or cancel each other because they always act, on two different point (broadly on two different objects) For balancing any two forces the first requirement is that they should act on one and the same object. (or point, if object can be treated as a point mass, which is a common practice).

Illustrations

1.      Suppose in figure 3.2 we put one more block of 5 kg mass adjacent to 10 kg and a force of 150 N acts as shown in the figure 3.3, then find the forces acting on the interface.

Sol.   The combined acceleration of the two bodies when treated as one is

a= F/ (10+5) = 150 / 15 = 10m/sec²

So each one moves with a = 10m/sec² keeping their contact established.

Here you can feel that due to 150N force the body of 5kg feels as if it is being pushed by the 10 kg mass. There is a force acting on 5kg called R₁, to oppose it by third law this body exerts a force R₂ on 10 kg. The interface is as shown in figure 3.4.

Also, third law tells us that R₁ = R₂ in magnitude and is opposite in direction. (figure 3.5)

R₁ = R₂ = R

Here since 150 N force acts on the 10 kg mass and only R acts on the 5kg mass. For motion in 5 kg only R is responsible. We can write the initial equation as:

F = 150 = (10+5) a           i.e., 150 = 10a + 5a

Here 10a is force experienced by 10 kg mass and 5a is experienced by 5 kg mass.

R = 5a      a = 10m/sec² =>           R = 50 N

Net force experienced by 10kg block is (150 – R) = 10a

150 – R = 10 × 10 = 100 N            =>     R = 50 N

Therefore we get R = 50N for both blocks. Hence we find “action and reaction are equal and opposite”. Now net force on the body of 10 kg mass is 100 N & Net force on the body of 5kg mass is 50N and on the interface action and reaction are both equal and also are equal to force experienced by second body.

It states that rate of change of momentum of a body is equal to the force applied on it, in terms of the magnitude as well as in the sense of direction. Here the momentum is defined as the product of mass and velocity i.e. .

Therefore we can write mathematically.

F = d(mv)/ dt, if ‘m’ remains constant then

Thus acceleration is rate of change of velocity.

Since direction of a is same as F, we can write

F = ma , which is mathematically Newton’s second law of motion.

Here, if F = 0 then we find a = 0. This reminds us of first law of motion. That is, if net external force  is absent, then there will be no change in state of motion, that means its acceleration is zero.

Further we can extend second law of motion, (in fact its decomposition) to three mutually perpendicular directions as per our coordinate system.

If components in x,y, and z directions are Fx, Fy, & Fz respectively, the three accelerations produced when Fx, Fy, & Fz act simultaneously) in the body are ,  Now,

If we add three forces then resultant is called net external force.

Similarly

is called net acceleration produced in the body.

Illustrations

1.             In Figure 3.1. Let us have M = 10 kg and a new net external force in the direction as shown in figure 3.2 is 150 N. Find its acceleration.

Sol.