UNIT 4 MOTION IN A STRAIGHT LINE

Structure

4.1 Introduction Objectives

4.2 Basic Concepts

4.3 Free Fall of a Body

4.4 Upward Motion Under Gravity

4.5 Simple Harmonic Motion

4.6 Projectile Motion

4.7 Summary

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Page No.

4.1 INTRODUCTION

Motion of a particle in a straight line forms the backbone of all types of particle motion. Modelling this situation, you will be in a position to predict the outcome of a great variety of experiments -.time to reach your office from house, or the speed with which to move to cover a given distance in a given time, or in a slightly more complex situation, the dista:lce covered by a ball when thrown with a given velocity at a given angle to the horizontal. We shall, in this unit, model the motion of a particle in a straight line.

We begin this unit by equipping ourselves in Sec. 4.2 with the basic concepts which are required for modelling any problem related t o mechanics in general and dynamics in particular, i.e., velocity, acceleration, resistive forces etc. In Sec. 4.3, we shall model free fall of a body under gravity with and without resistance. Upward motion under gravity shall be modelled in Sec. 4.4 Again, we shall discuss here the two formulations i.e. motion with and without resistance. This is followed by a simple model for simple harmonic motion (SHM) in Sec. 4.5, Models for damped and forced SHM are also discussed here. 'Au shall end the unit by extending our discussion in Sec. 4.6 to the motion in a plane, that of a projectile. Plenty of exercises are given at the end of each section to illustrate the theory.

For some of you, the problem discussed in this unit may not be new. You might have studied them in your school in the course on mechanics. Rut the objective here is to look at these problems in mechanics as problems in modelling. Thus, always keep the steps given in the flow chart of block summary of Block-1, in your mind while going through the unit.

Objectives

After reading this unit, you should be able to @ write down the equation modelling the motion of a particle;

a write down the boundary conditions associated with the motion;

e solve the resulting equation;

explain the solution obtained and give its physical interpretation. Motion in a Straight Line

4.2 BASIC CONCEPTS

It is necessary to have a .complete understanding of the physics of the situation-before modelling it. It is only then, we will be able to express the situation/problem in proper mathematical form. As we have already mentioned in the introduction, we start our discussion with a few concepts needed for discussions in this unit.

ds You may recall that in Unit 1, Block-1, we modelled velocity as v = - and

d t dv

acceleration as a = -. In general v = (vl , vz, va), a = (al, aa, as) and d t

s = (sl, s2, s3) where (vl, vz, va), (al , az, a3) and (sl, s i r ss) are the components of velocity, acceleration and distance respectively, in x, y and a directions. So if the motion is in straight line, say, along the x-axis then velocity can be written as v = vli and acceleration as a = al i , whereas, for a motion in a plane, we have v = vli + vzj and a = ali -I- a2j

N o t e that in the above discussion, all the letters in bold represent vector quantities. In this unit, where the modelling is i~iostly for the motion in a straight line, whenever we talk about velocity, acceleration or distance for the sale of simplicity, we may not carry on with the vector notation. But, it should be understood that we are talki~lg about one dimensio~lal vector cpantities.

With the above notion in mind we shall now define the physical entities in relation to the motion in one dimension.

Motion In O n e Dimension

Suppose P and Q are the positions of a particle at time t and t + dt as shown x ( t + 6t) in Fig. 1. The average velocity of a particle is then given by

PQ - ~ ( t -t- bt) - ~ ( t ) Vave = - - d t St

and the velocity a t time t will be x(t + dt) - x(t) dx

v = lim - -- 6t+o 6 t dt

In a similar way, the acceleration is given by v(t +St) - v(t)

a = lim 6t+O b t dv = - dt

dt2 Also,

W e shall now derive some elementary relations between these quantities when the particle is movilig in a straight line with uniform velocity/acceleration.

Displacement of a part icle i n t e rms of velocity in a given time:

Let a particle move a distance x in time t with a uniform velocity v. Then by definition of velocity,

X v = - o r x = v t

t (5)

Fig. 1

Physical E13vironment Velocity of a particle moving with a uniform acceleration a and wi th an initial velocity t ~ :

Let a particle P with initial velocity u move under uniform acceleration dv a = - dt

Integrating both sides of Eyn.(6), w.r,t. t we obtain

where A is the constant of integration and can be obtained using the initial condition that v =L u at t = 0. Therefore,

v = u + a t (7) Distance covered by a particle movilig with uniform acceleration a and initial velocity u:

Let the uniform acceleration be a.

Then by definition d2x - dt2 = a

Integrating this differential equation twice, we get

where cl and cz are integration constants. Using the initial conciitioals, that is, x = 0 at t = 0 and velocity = u at t = 0, we obtain

Velocity under uniform acceleration:

Let tlle particle P move with uniform acceleration a starting with initial velocity u. Suppose at time t , itis velocity is v. Then

dv V- = a

dx (11)

Integrating this equation, and using the initid condition that when x = 0, v = u, we obtain the relation

We now take up a few examples to help you have a better understanding of the relation betwean these physical quamtitiea.

Example 1: The velocity v m/sec. of a particle moving in a straight line at time t sec. is given by

v = 2 t + 3 (13)

Find i) acceleration at time t ii) the displacement a t time t = 5 minutes, if the displacement a t t = 1 minute is 4 metres.

Solution: i) Given v = 2t + 3, the acceleration a is dv 2

a = - = 2 rn/sec . , dt

Hence the acceleration is constant.

ii) To find the displacement at time t = 5 sec., let us assume that the displacement at time t is x

Integrating this equation, we get x = t2 + 3t + A, where A is the constailt of integration.

To determine the const ant A, we use the boundary condition. We are given that at t = 1, x = 4. Using this condition, we obtain A == 0

Hence, the displacement a t any time t is given by

x = t 2 + 3t

To obtain the displaceinent at t = 5, we substitute this value o f t iii the above equation, aricl obtain

x(t = 5) = 5% 3.5 = 25 + 15 = 40 metres. (16) Let us take up another example.

Example 2: A train stops at; two stations A and B, 2 liilornetres apart, and takes 4 seconds to travel this distance. Initially it inwes with a uniform accel.eration a1 kilometres/sec2 and later with b, unifurm deceleration (negative accelernt,ion) a3 lriloinetres/scc2. Sliow t l l d

1 1 2 - + - = 4 sec / h n . a1 a2

(17)

S01ution:Let tlie train illove with uiiiforn~ acceleration a1 for time t l and with deceleration a2 for tirne ta. Also, snppose in tiinc tl , the train covers a dis1;ance of xl km. aiid acquires a velocity v. Then, siiice it attains this velocity in tiine t l starting with zcro vclocity and moving with uniform acceleration a l , we have tlie following relations:

The remaining distant:^ x2 kni. is covered in tiine t2. In this time, the velocity of thc: train changas from v to ~c.r.0. Hcnce, we ]lave the following relations:

0 = v - a2t2 01- v = a2t2, (initrial velocity is v ancl final velocity is zero, and the deceleration is az)

1 , l Also xa = 2 - x l == -aata = -vta

2 2 Combining Eqns. (18) arid (19), we obtain

1 -v.4, (since the total time of travel is 4 seconds.) 2

This gives, v = 1 kilonlel;re/sec.

Also, from Eclns (18) and (19), vie obtain v v - + - = 4 secs. a1 a2

1 1 2 . - + - = 4 sec /km. a1 a 2

And now some exercises for you.

lMotion in a Straight Line

El) A body starts from rest and its displa,cement x at tirne t is given by t2 + 12t. . Find the velocity a t time t = 2 see. Find also the time at which velocity will be zcro, if at all.

E2) Tlle acceleration of a particle at a distance x m. is given by

Physical Environment a, = 6m/sec2 If the-velocity at the initial position is 4m/sec, find the velocity when the body is at a distance 10 m.

E3) A bus moves with an acceleration of 10 m/sec2. A passenger can run with a maximum velocity of 40 m/sec. What is the maximum distance between the passenger and the bus, when the bus starts, so that the passenger can board the bus by running at his maximum velocity?

E4) A bullet fired into a target loses half its velocity after penetrating 3 cm. How much further will it penetrate before coming to rest?

E5) The distance covered by a particle from rest varies as the square of the time. Show that the acceleration of the particle is constant.

So far we have considered the motion of a particle without bothering about what caused the motion. We did not worry about how the motion took place. This part of mechanics is called Kinematics. Obviously, for any particle to move, there should be some force to give it a 'push'. When you are riding a bicycle, unless you pedal, you are not going to move. In the case of scooter, or any motorised vehicle, the engine gives the required push. Now, we shall consider this aspect of mechanics, which is called Kinetics.

As we have discussed in Block-1, while modelling any physical situation, the initial step should be to look for the empirical laws describing the system. You may be aware that most of the problems in mechanics have the Newton's laws as their starting point. These laws are useful as they give the influence of any force on the motion of a particle. They relate how and 'why a particle moves.

Before the enunciation of these laws, people did not have any clue as to why small stones moved faster and to a greater distance compared to larger stones for the same amount of push. But once you understand these laws, you are in a position to predict the motion of particles under the influence of

Galileo.(1546-1643) different forces. In fact, sitting in your drawing room, you can predict the motion of different planets and that of the moon.

Galileo took a giant step in understanding motion. He enunciated the principle of inertia. This principle says that if a body or a particle is in motion and left alone to itself, it will continue to be in motion. If, however, the particle is at rest and left to itself, it will continue to be at rest, Isn't this applicable to us also, for, when we are sitting lazily in our drawing rooms, we need some interference to wake us up from the lazy state and make us active? I

However, you might say that this is not true, for not all shots hit by the I

batsmen reach thc boundary ropes! Well, the reason for this is that the ball is not left to itself - no, we are not talking of the fielders coming in between- i the ball is moving on the ground, which introduces some resistance. Thus, to 1 enunciate this law, it needed great imagination and vision which Galileo displayed. Once the ball is in motion, the next question is how and why does

l it change its speed. The answer to this was given by Sir Isaac Newton.

I

These laws cannot be proved a priori, but were enunciated by Sir Isaac I

Newton(1642-1727) Newton in 1686, after careful observations. They hold good in most situations, except in the innermost core of the atom, or when the velocities 11 of the particle approach that of light. In these situations, Newton's laws of I

I motion fail, and should be replaced by quantum mechanical model and relativistic model respectively. However, for our purposes, Newton's laws of I 1 motion are more than adequate. I

8 i 1

I

Using these simple looking laws, we can even predict the niotion of various Motion in a Straight planets or the moon. 'We shall be doing this in the next unit. For the Line present, let us try to model some everyday phenomena considering the motion of particles under the influence of the commonest force in nature, the force of gravity.

4.3 FREE FALL OF A BODY

Here we shall consider the one dimensional motion of a body moving vertically under the influence of gravity. But before formulating the model, let us recall the steps involved in the modelling process as discussed in Unit-2. For modelling the free fall of a body, we must

i) identify the essentials of a problem and draw a neat diagram indicating the direction of the various forces involved.

ii) establish relationship between these essentials and obtain the basic equations using en~pirical laws.

iii) solve the basic equations

iv) interpret and eyaluate the solution obtained The essential features of this physical situation are, the accelcration due to gravity, the height frorn whidi the body is dropped and the resistance to the motion due to air. We shall first formulate a model for the nlotion neglecting air resistance and then mala this model more sophisticated by incorpoiating the effects of air resistance.

Formulation-1: Consider a body to be dropped from a height h. To rnodcl the motion of the body, let us choosc the point 0, from where tlie body is dropped, as the origiri of reference. Due to tlie force of gravity, the body will move along a st,raight line, vertically dowriwards. We choose this direction as x-axis (sec Fig. 2).

Fig. 2

Suppose the body is dropped at time t = O and at somc other time t, it has reached the position P, at a distance x from 0. Suppose its velocity at that time is v. The force acting on the body is its weight, whicli is equal to mg, g being the acceleration due to gravity, and m thc: mass of tlie body . The .

acceleration due t o gravity will be assumed constant. Newton's second law of \ ,

Phy~ic:al Enviroi~ment m.otion then enables us to write down the following equation

nlv gives the momenLunr of the body at that time and hence, $(rn\r) gives the rate of change of m.ome~rl;um. We shill assilrne throughout our discussions that the mass ul the body is constant. This will not, be alwa,ys true, pa;rticuZarly in s.it~.nations lili~: rocket prcrpulslon, where because of the fuel depletion, the mass keeps decreasixlg.

Silzce nlir.ss is constant,, Eqn. (20) s.irnplifiBs to

Integrating once, we o1)l;ain the velocity in L;ermzs of time as,

v r- gt -t- C

where, CI is the constant of integration.

Since the body is dropped from a height h its initial velocity is zero, i.e., v t= O a(; 1; = O a,nd this gives

I=-@ (23)

Putting this value of C in Eqn. (22), we o'l)f;ain the velocity af .the. part icle at any t ime t a,s,

v = gt, (24)

'ro find the distanrz traversed in tirne t, VJ,.~ shall write tho ve2ocit;y in Eqn.(24) as rd;s of change of distance; i.e.,

1nl;egrating this equation, we ohlain

To obtain the value of irlteffr~tioil r:onstsnt B, we use the condition that x = O i1.t 1;ime t = 0: i.e., ini.Lial.l!y the part;icle has not travelled any distance, and find

O = B

Tllus, the distance traversed in time t becomes

which, as axpected is same as given I y Eqn.(lO) for the nlotioil now ic ur~del: uniform acceleration g with iujtial velocity u =. 0.

The time taken by the body to hit t11.c ground can also he obtained from %qn.(27). Since the ppaticlc is dropped from a height h, it milst; cover a distance h to 91it the ground. Tllus, substitul;i~ig h for x in fiqn.(27), we obtain the equation giving the time as

I

1 t.1 - 2

I 2h or, t2 = --

1 g

Lll or. t2 = --

1 g

This gives t = &/' g

(29)

Since g is a real quantit.~ and h is positive, being the height from which the particle: is dropped, 2h/g is always positive. Ilence, one value of t is positive a n d one negative. Since, negative time does not make any sense, we

10

disregard that value and say that the t ime taken t o hit the ground is

This situation reinforces the fact that the mathematical solutions obtained in any model formulation should be related to the real situations before final conclusions are arrived at.

The velocity of the particle as it reaches the earth can also be determined by rewriting Eqn. (21) in the form

Integrating and using the initial condition, that 'v = 0 when x = 0, ,gives the velocity on reaching t h e ear th (when x = h) as

This velocity v,,th is called the terminal velocity.

Note that here we assumed g to be constant, lu reality acceleration due to gravity varies inversely a.5 the square of the distance from the centre of the earth. However, this variation does not make much impact on our study of falling bodies, for our distance measures me very small compared to the radius of the earth itself.

Also notice, that we assumed m to be a con~tant. We found that the motion is independent of mass. This fact was indeed observed by Galileo, when he dropped stones of varying sizes and masses from the Leaning Tower of Pisa in Italy, and found that all of them hit the ground at the same time. Now, if you try to drop a stone and a feather, or a piece of paper at the same time from the same height, what do you observe? T h ~ y hit the ground at different times. ,Does this go against the model we had set up? As fnentioned in Block 1, if the model is not validated as in this case, then we should look at the assumptions, and see whether they are really correct or not. Many a time, some assumptions are made to make our model simple, but that will lead to a compromise in the accuracy of the inodel predictions. In this model, we have completely omitted the resistance due to the air. However, for feathers or paper pieces, it is significant. This is the reason for the discrepancy between the model prediction and the observations. However, if the experiment is done in pure vacuum, then the mndel prediction will be true. In fact, this experiment was done on moon, whcre there is no atmosphere, by dropping.feathers and rocks. The times of fall were found to be exactly same, as predicted by the model. But beware, the value of g or1 the moon is different from that on the earth. The difference in the times taken by stone and feather to reach the ground when dropped from the same height can be explained, if we make our model more sophisticated, Themost important aspect,which we have neglected in the above model is the resistance due to air. This must remind you of the problem of the rain drop we formulated in Block-1. A rain drop cannot be modelled as a falling body since resistance plays a vital role in the dynamics of the raindrop fall.

You know that the air gives rise to resistance. In fact, if you try to run with an open umbrella, you will feel the resistance due to air. However, if you keep the umbrella closed, then you do not feel the resistance. This is because the resistance due to air is proportional to the area exposed. We shall now reformulate the motion of a particle with air resistance.

Formulation 2: The very term resistance means some opposing force - i.e., force in the direction opposite to the direction of motion, Thus, for

Motion in a Straight Line

Physical Environment downwards motion, the model Eqn.(20) modifies to

where R represents the resistance to motion.

The expression for resistance should be found out from experiments. Usually, it is taken proportional to the square of the velocity or to the velocity itself. We shall first consider the case of resistance being proportional t o t h e square of t h e velocity. Thus, if v is the velocity of the particle at time t , then the resistance can be taken as

R = K V ~ (34)

where K, the constant of proportionality, is to be determined from experimeilts.

With this expression for resistance, the model equation for downward motion under gravity with resistance is

Assuming m to be constant, this reduces to

dv or, V- dx = g - - k V 2

where k = K/m.

From Eqn.(36), we observe that when v = vt = d m , there is no acceleration. The motion would then be without resistance (acceleration is zero), and the velocity will remain vt. This velocity is called the terminal velocity.

Eqns. (36) and (37) can be rewritten in terms of this terminal velocity as dv - dt = g (1 -v2/v)) (38)

To determine velocity as a function of time t , we integrate Eqn.(38) and proceed as follows:

Substituting, v/vt = tan hO in Eqn.(38), it reduces to

On integration, this gives,

or, v = vt tau h -t + A (:t )

where A is the constant of integration.

To determine the constant A, we use the initial condition. Since the particle is falling freely under gravity, the initial velocity is zero, and hence,

v = O a t t =o. (42)

Using this condition in Eqn.(41), we obtain

Similarly, to integrate Eqn.(39), substitute

Eqn.(39) is then transformed into dy 2g -- - - T d ~ Y V t

and integrating Eqn. (46), we obtain

where B is the constant of integration to be determined.

We can rewrite Eqn.(47) in the form

To determine B, we use the initial condition, that when the particle is dropped, v = 0, x = 0, and obtilin B = 0.

With this value of B, Ecln.(48) reduces to v2

I - - = exp (3) vt2

which simplifies to

Eliminating v between Eqns.(44) and (SO), we nbtilin the relation between t h e distance a n d t i m e as,

Just as we have obtained relationship bctween velocity and distance in the case when resistance is proportional to the square of the velocity, you can obtain relationship between them when the resisljance is proportional to the velocity. You may now try the following exercises:

I

E6) Derive relations similar to the one given by the Eqns.(44), (50) and (51), when the resistance is proportiorla1 to the velocity.

E7) What Imppens when vt in relation (51) is very large. Give physical

I interpretation for your result. I

E8) A particle falls from 1000 metres under gravity. The air resistance is mDv2, where D = 0.1. How much time does i t take for the particle to reach the ground?

Motion in a Straight Line

We shall now consider in the next section one-dimensional upward motion of ' a body under gravity.

4.4 UPWARD MOTION UNDER GRAVITY P.

Just as in the case of a free falling body we shall give here two formulations, we shall first consider the motion under gravity neglecting air resistance and then refine the model obtained, by incorporating the effects of air resistance.

Physical Environment Formulation-1 : Suppose a particle is thrown vertically upward with a velocity uo metres/sec. Let 0 be the point from where the particle is thrown upwards. We choose this point as origin and the vertica,l line directed upwards through 0, along which the particle moves, as the x axis. Suppose at some time t , the particle is at P, at a distance x from O(see Fig. 3).

Fig. 3 From Newton's second law of motion,

d (mv) -- d t

- -rug (52)

Since m is a constant. As before the solution to this equation gives the velocity at, any time t as

V = UO - gt (53) To find the velocity at any position x from 0, we write Eqn.(52) in the form

The initial conditions for solving this equation are that at time t = 0, when the particle is thrcfwn, it has a velocity uo, and the distance traversed is zero, that is

x = O when t = 0

Integrating Eqn. (54)' we obtain

v2 -. = -gx+B, 2

where B is the constant of integration.

Using the initial condition, u2 o-

I 2 - B (56)

and hence, v2 u; - = -gx+ - 2 2 ' (57) ,

which on simplification gives the velocity at any position x, as

v = 4- (58)

Maximum height of flight:

The natural queslion which may bother you now is, how high the particle

will. go? The! particle will rise kill its 'vcl.ocity is ~:cd.uc.ed to zc?ro. 11 flhoulcl be Mot,ion in a Si;raitr;ht noted from Eyns (53) and (SO), that; the velocity of tlle particlc keep:.; :Liz10 decreasing as tirne inc~eases, or a.s the pa,rticle goes farth.er ancl fa;trl;hc?r From 0. This is hecal~ee the gravitational force is acting i11 a, direckion opposite to t,hat, of niotion, khnc; resisting motion. To get th.c imnxix~i~na-mn height of flight, substitute v -= 0, ill Eqn. (58). This gives 11s

Thus, greater the initial velocity, greater i s the maximum height attained.

Now whal; is the time taken by t11c particle .I;o ~.t?a,c:!l this rllaxin~unr height;? To deter~niile the t i rns needed to reach the maximum heighit, we substitute v = 0 in Eqn. (53), and obtain.

uo - l---'---'--- t,, =: = [2xnIax/g) (6U) 6

Once th.e velocity br?comes zern, Ithe p~irticIe sta,rt,s ft~,lling freely undcs gravity. We have already shown in Fcr.mulsl;ion 1 (11' Sclc. 4.3 that, i.'r),r :.L free falling body tlic time of dowixward flight as given by Ecjn.(30) is

--- tadowrl = V'%& l&~lnax /g ) (61)

and the terminal velocity or1 reaclliilg tlic groluld is

If i11 Eqn. (00), we substitute the value of x,,, from Eqn. (Ijilj, v~i! Blid thal,

Wliat does this equa,tion show'? This ~;'ilows tha.k tho partic:lc couies I.)sc:ls t o the earth wi.t,h the sarne vc?l.ocity wil;h .which r t \v:i,s I,lirc~~vn up. .Also, irolll Eqns.(60) ant1 (GI), we observe that t,l~e I;inleci nf R;fl;l~t,s u1.j nud down ;WF:

exactly t,he same. Irt this sense, we find that the rmotiorn i.s symn~.etria: about t h e highest point reached. Wc ciLn sunn up tlne observn1;ions lnstle from tho. above study as follows:

0 bservat ions

Iii the case of an i.lpward rnot,i.on of a particle undt:r gritvi.t,y

1) A particle returns to its point of prqject;ion with 1;he sn,rne velocity a2s the one with w11icl.l it was projcct;ed.

2) The time of upward flight and dowriwnrcl flight of a pr~sticlc arc: equa,l.

3) Motion under gravity is same as urider uni.ljrm acc:eles;ztion, it6 .value being the acceleration du~e to gravity.

Note that the above observatiorls are valid as lung as g reaxlaifis constnl.lt. But is it the situatio11 always'? No! you may comc: across problems rela,ttltl to astronomy where g cloes not remain corrstant. There may be a situatioll when a particle projecterl xnay uot return back to cart,lr. In such a, situation the concept of asca cape vclocit,~' comes illto picturc. We sl~dcll ta11c nbovat it; in Unit 5 when we discuss planetary motion. For the tliscusuion in this unit g continues to rcniain constant. Let us now il1ustr;tt;e the t;heory discussed at)ove tl.~rough t;bc following examples.

Example 3: A. ball is clropped from a 20 mctrc high tower. If its spc?ecl is lialved on impact;, how Eiigb does it bounce'?

Solution: Since the ball is dropped from a tower its initial vc1ocil;y is zero.

Physical ~nvironment Using Eqn.(32) the velocity of the ball at the time of impact is .

After impact, since the velocity is halved, this becomes 1

uo = - &Op metres/sec. 2

Suppose the ball bounces upto height s, that means at height s, the velocity becomes zero.

Hence, from Eqn. (58) we obtain 2 0 = uo - 2gs

or, s = 5 metres.

Example 4: A particle is dropped fiom a point at a height H metres above the ground, and at the same time another particle is thrown upwards with velocity uo metres/sec. f~om a point vertically below the first point. When and where will the two particles collide?

Solution: Suppose they collide at a distance x &om the top after time t. Then in time t the particle dropped would have travelled a distance x metres and that thrown upward would have travelled H - x metres.

Then 1

x = -gt2 (Using Eqn. 27) 2

and 1 dx

H - x = uot - -gr "using Eqn.(53) after integration w.r.t; t, with v = -). 2 dt

H Combining the two equations, we obtain t = - secs. The position of

uo collision is then obtained as

Xcollision = -

You may now try this exercise.

E9) A particle is thrown vertically upwards with a velocity 100 metres/sec. How high does it go, and how much time does it take to reach tliis height? (g = 9.8m/sec2)

We now take into consideration the effects of air resistance to the upward motion under gravity and give another improved formulation of this motion.

Formulation 2: For upward motion under gravity with a velocity of projection uo and resistance mkv2, Eqn. (52) of motion modifies to

Since m is a constant this equation reduces to

v Integrating Eqn.(65), using the substitution - = tan9 and using the

Vt condition v = uo alt t = 0, we obtain the velocity t ime relation as

uo t = v' (tan-' - - tan-l g vt

('36) 1 6

dv dv Again in Eqn.(G5), if we write the acceleration -- as v- and integrate the Motion in a Straight

dt dx resulting equation using the condition v = uo, x = 0 at t = 0, we obtain the Line

velocity distance relation as

You know that velocity v of a particle is zero when it attains its maximum flight. So relation (67) gives us the maximum height possible of a particle as

What happens if k tends to zero? Physically, this means that resistance to the motion of a particle tends to zero. Now if you find tlie limit of the right hand side of Eqn.(68) as k tends to zero, you would see that maximum height tends to u;/2g, the value obtained for lion resisting medium as given by Eqn. (59).

Note that you can get back Eqn.(52) i.e., equation of motion in a non resisting medium directly from Eqn.(G4) by putting k = 0. But you can not obtain Eqn. (59) froin Eqn. (68) by just putting k = 0 (why?). You need to take limit k + O in this case.

E10) Verify that tlie liniit k T, 0 in Eqn.(68) gives you x,, = ~ $ 1 2 ~ .

You all niust have experienced vibratio~~s/oscillations of various magnitudes - may be the to and fro jerlts in a bus, the ding dong movements of a pendulum of a clock or the vibrations in any n-usical instrument like sitar, veena etc. All these vibratory motions have one thing in common. They all have a to and fro niotion without any body movement. These vibrations play an important role in many applications - from c'onstructio~i of bridges to electrical and electronic circuits. These motions can be modeled using Newton's laws of motion. One of the fundamental vibratory/oscillatory motion is the simple harmonic motion (SHM). which we shall discuss now.

4r5 SIMPLE HARMONIC MOTION

Of all the oscillatory motions, the most important is simple harmonic motion because, besides being the simplest motion to describe mathematically, it constitutes a rather accurate description of many oscillations found in nature. We shall now concentrate on this kind of motion.

Definition: If the force on a particle is proportional to its distance from a fixed point and is directed towards it, then the particle will execute a simple harmonic motion.

There are two types of SHM. i) Linear, when the body moves in a linear path under the action of a

constant force. For example, up and down oscillations of the piston of a cylinder containing gas, when suddenly pressed and released, and oscillations of an elastic spring suspended vertically and loaded at its lower end etc.

ii) Angular, when the body rotates about an axis under the aciion of a torque or couple; examples are pendulum oscillations and torsional oscillations etc.

Physical Environment I11 order to visualise linear SHM consider a particle P moving uniformly in a circle (see Fig. 4). From P draw a line PA perpendicular to the Y axis. As P moves towards M along the circle, A moves towards M along OY. When P starts moving along the arc MN, A Inoves from M to 0 as P traces arc NQ, A traces 0 Q and finally as P reaches R from Q, A reaches O from &. Thus Ihe point A keeps moving about 0, on either side along the Y-axis. Phis particle thus executes a linear SHM along YOY';

Fig. 4 We shall now model this SHM, Remember that SHM r.:prcsents one-dimensional motion or motion in a straight line. IOU lnay recall that in Unit 2 we formulated a model for the oscillatory motion of a simple peizdulum which is a simple application of the SHM. In this section we shall again take up the motion of simple pendulum and give two more formulations of it, one when the motion is resisted by a force and another when the oscillation is,induced by external forces. But first, consider the '

modelling of SHM.

Formulation 1: Suppose a particle of mass m at time t is at A, at a distance x from 0 . Also, let A moves under the influence of a force F = px towards O(see Fig, 5),

Fig, 5 Then Newton's second-law of motion yield the following equation.

where p is positive. The negative sign in Eqn,(69) is because the force F is directed towards 0. Eqn.(69) can also be written as

. ,, = - - px - -w 2 d"x dt2 m (70)

where w2 = ,u/m. . Tl~us, when A is on the right side of 0, the force is towards the left. This is the prototype equation for simple harmonic motion. Since both p and m are 1

18 positive w is real. But you may wonder what does this w represent

physically? To get an answer to this question let us write Eq11.(70) in lhe Motion in a Straight form Line

You know that the solution to this differential equation is written in thc form

x(t) = A cos wt + B sill wt (7'1) where A and B are arbitrary corlsta~its of integration and their values can be obtained by using initiallboundary conditions. Eqa.(71) is the model equation for SHM.

Observe that, Eqn.(71) can be written in the form

(A cos wt + B sill wt) x(t) =

If we define A

= sin$, ~~

Then, Eqn.(72) reduces to

x(t) = Rsin(k~t + 0) where $ = tan-' (A/B) Siilcc sin H = sin(@ + 2x.1~) for all integral values of n, we obtain from Eqi1.(73)

x(t) = x (t + $) This shows tllat the positio~l of the pi~rticle is same ixft,er intervals of ill + 2x

27r 2 7 ~ ;;-. This value, , is cixllecl the t i m e period of the SHM, and w is calletl the frequency of the oscillation (see Fig. 6). Fig. 6: Siriaple

Harmo~lir. Blom Eqn.(73), we find that tihe rll~xiinum displacement of the particle Motion occurs when the sin term is a nlaxirrium which is 1, and lli~nce the maximum displacement is R. This illsximum displaccrnent is called the amplitude of the SHM. The quantity wt -t- $ is crtlled tlle phase of tlic: oscillation and thus 6, is the initial pllase i.e., its value for t -- 0.

You may notice here that, by fluitably deiiiiing siric/q ant1 cos cj,l Eqn. (71) can also be reduced Lo thc form x = R cos(wt + dl). We are leaving it, as an exercise for you to try it.

E l l ) Show that model Eqn.(71) for SI-IM can be written in the form x = Rcos (wt + (PI).

L

Let us now consider the following examples.

Example 5: A particle is executing $HM about the origin 0. Find its clx

displacement x(t) at any time t if at t = 0, x = 3 and - = 0. d t

Solution: In Eqn. (73), using the above conditions, wc obtiirl

3 = Rsin$ (75)

0 = Rwcosq!l (76) n- Eqns.(75) and (76) imply that R = 3 and # = - 2 '

Pllysical Environment Hellcc t;lle displacement at any time t is given b,Y

x( t) = 3 sin (wt + :) = 3 cos ut

Now if lye want to obtain the velocity in terms of the displacement, we ccnsider tlic model equation in the form

dv niv - = -fix

dx dv 2 I L x = - w x SO that, v- = -- dx m

Iutegratir~g this equation and assuming that x = xo when v = 0, i.e., the niaxi~~ium displacemeilt or amplitude is xo, we obtain the relation

v2 = w2 (x; - x2) (77) Sirice v2 2 0, Eqn. (77) gives

1x1 5 xo. This means that the displacement is always less than xu. But, we assumed xo to be the position at which the velocity is zero. Hence, we conclude that the velocity of the particle becomes zero, when i t s displacement is maximum. But when is the velocity of the particle maximum? You may try to find the answer to this question.

E12) When and where does a particle executing SHM in a straight line lias maximum velocity?

Example 6: The maximum velocity of a particle moving in SHM is 10 metres/sec and its period is 5 sec. Find its amplitude.

Solution: Let the SHM be x - Asin(wt + 4 ) dx

Then the velocity is - = Aw cos(wt t $) dt

Since maximum velocity is 10 metreslsec, at that time, acceleration has to be zero and hence

dx This gives wt + q5 = 0, and at this time - = 10.

d t Hence, Aw = 10

27T Also, we are given that period = - . . = 5.

W 10 50

Herice the amplitude of the SHM = A = - = - metres. w 2Ir

Example 7: The amplitude of a particle executing SHM is b and its period is T. Show that the maximum velocity of the particle is p. Solution: x = b +in(& + 4) = b sin -t + 4 (: )

dx 2nb Therefore, - = -

21rb dt

T c ~ s ( $ t + ~ ) . H e n c e , $ 1 max =-

You may now try this exercise. .

E13) A particle executing SHM passes two points a and b with velocities v and v'. Show that the amplitude is i

As we have mentioned earlier simple harmonic motion is the most important 20 I

of all the oscillatory motions. There are many applicatious of this motion in nature. You have already come across one of the simplest applications of this type of motion in Unit-2 the simple pendulum. We shall once again consider it here and try to improve the model obtained there, Applications of a Simple Harmonic Motion - Simple Penduluni A simple pendulum is a point mass tied to an inextensible string and suspended from a fixed point. The point mass is allowed to oscillate in its own plane with small amplitude. In Unit-2, using the Newton's second law of motion we set up the model equation for the motion of this point mass and obtained its solution in Unit 3. We shall not repeat it here. We only recapitulate that the motion of a pendulum is a SISM with time period

T = 2 . 8 .

However in real situation if you perform this experiment with a ball tied to a string and observe the oscillations, you will scc? that the oscillations die down after some time. This is because of the air rcsistai~ce and the frictional forces at the point of suspension, which have bee11 ignored in thc previous model. Of course, the model can be iniproved to incorporate tliese cffects, but that will take 11s beyond the scope of this course. However, wc shall give the model equation for d a m p e d SHM, i.e., when the motion is resisted by a force.

Formulation 2: Lei; a body be attached to a spring at the poirit 0 as shown in Fig. 7. Then the forces acting on thc mass are i) the tension T along the string from A towards 0 ii) the weight mg of the hocly acting vertically

downwards. Let n resistance force R = -f , proportional to velocity, acts

on the body apart from the tension T = where both f and p are positive constants.

Using Newton's law of motion we obtain

d2x dx n1- = -px - f -

dt2 d t f

Let a = - and u2 = k. Then certainly a 2 0 and for this a and w tlre rn

above equation reduces to

dx If you compare Eqn.(78) with Eqn.(70), you would noticc that 2~vu- is the

dt additional term in Eqn.(78). Wliat do you expect when a == O'? Let us solve this equation and find out an answer to this.,

To solve the second order linem differential Eqn.(78), we ~uhstitute a triad solution

x = CePt

in it where C and p are the constants to he determined.

With this choice of x, Eqn.(78) reduces to

p2CePt + 2awpCePt + w2cePt = 0

Now for x # 0, we obtain the quadratic equatiou in p, i.e.,

p 2 + 2 a ~ P + ~ 2 = o

Motion in a Straight Line

'E'hc siinple h~trmonic r~iotion of' s particle, whose motion is resisted by a, force is called a damped SHM.

Pig. 7

physical Environment

Fig. 8: Underdamped motion

Fig. 9: Overdamped motion

This yields two roots for p,

p = [ -aw wd-] and hence the solution of Eqn.(78) will be

+B exp {I-crwt - w t ~ ' ] ) (82)

where, A and B are constants of integration. Now in Eqn.(82) if you put a = 0 then you obtain x = Aexp(itw) + Bexp(-itw) = A1 cos wt + B1 sin wt, A1 and B1 being constants arid if you compare this equation with Eqn. (71), then it is the model equation for SHM as expected.

For other values of,cr, the three cases of the solution of Eqn.(80) are of special interestrwhich we shall discuss now.

i) I f a: < 1 ,' then roots of Eqn. (80) are imaginary and the system is called underdamped (see Fig. 8). In this case we have

+B exp {[-awt - iut d m ) ] ) (83)

where i is the imaginary number m). Eqn.(83) can be further simplified to the form

= exp(- -awt) [ ~ c o u w + iA sin ( wt T O (1 - cr

+i(A - B) sin wt d m ) ] (85 )

= exp(-awt ) [A' cos (wt d m + dl)]

where A' = 2m and = tan- (A + B) i(A - B)

ii)If a > 1, roots of Eqn.(80) are real and system is overdamged, (see Fig. 9) then

iii)If a = 1, roots of Eqn.(80) are real and equal and system is critically d a m p e d (see Fig. lo), then

In Eqn.(86), if we let t tend to infinity we will see that x which is oscillatory tends to 0, and thus the motion dies down eventually. However, in Eqn. (87), even though the motion is cot oscillatory, still as t tends to infinity, the motion dies down eventually. If a, = 1, there is no vibration. Physically, this means that if the variable a has value which is different f rom 1, t h e n t h e motion dies down eventually, t ha t is, t h e motion is completely damped .

However, there are cases, when the oscillation is induced by external forces. Consider the simple swing in a park, bhere you sit and induce a pendulum type of motion. If at any time, somebody gives a push, then, depending

upon the relationship between the push given ;tnd the ilatural oscillatioll already induced, your further oscillations will ei';her die down, or will be lnore violent. This is an example of forced oscillation, which we shall considcr now and give another fornlulation of the SIIM.

Formulation 3: Suppose tliat thc oscillations axe under tile influence of an external force F. Tlien Eqn. (69) of motion talces the form

Motion in a Straight Line

Note that here F = 0 corresponds to SHM as given by Eqr1.(69) F positive will help enhance tlie rnotion and F negative will be resistive and rlampeii the motioli as you have seen in Forrnula1;ion 2 where I" was talcen as

Depending upon the i1al;ure of F, tJio s~lutioii oT Eqii.(89) car1 bc writtcn Fig. 10: Critically down. Wc sha,ll liere considcr an illlportarit c1a.s~ of forcing fuiictioris, the damped

periodic functions. So, Ict F = L cos pt,. mot,ion

Now what does this forcing: function correspolid t,o'! This correspoilds to 1,hc situatioii when another oscillatory force with frecyucncy p is suporposecl on t,be initially oscil1al;ory inotiori of the pendulum. With this choice of forcing function F; Eqn.(89) reduccs to

d2x - - P L G - -x + - cospt

m rn (acceleration) = (at,tra,ctiou force) -l- (forcing term)

Eqri.(90) can be rewritten 2s

d2x ~1 L - + -:c = - cos pt dt-n in

Do you recognise this equation? Yes! This is a 1ine:lr hccoiid order non-liomogeneous differential ecluatioil wit,h cons taut coefIicients. You know that complete solutior~ Lo this eclualion coiriprises of two parts, the

' eIlCOI1S compleinentary function i.c. solution of the cwrrcspoiiding homog equation and the particular integral of tlie lion-l~omogerieous term (refer Uliit 8 of MTE-08).

Solution wf the hornoganeoas equat,ioti is - -

which car1 also be written as

Also, the particular integral is obtained as

(i wliere D = - is a linear differential operatlor. Tlie coxnplele solutio~i to

dx ' Eqn.(91) may then be wibitt,en 38

1 L - COB ~h

fl For deterrninirig a particular integral, two cases arise, iiainely, - # -P2 and 111

P - = -p" In 23

P Physical Environment If - # -p2 then solution (93) reduces to rn

L cospt x = ~ c o s [{(i)t+$] +-

In (i;l; - p2) Now if a t t = 0, the particle is at rest at x = a, then the complete solution will be given by

Therefore the motion of the particle is the resultant of two SHM, of periods 21rdm (natural) and (forced)

If 2 is nearly equal to p2, then the forcing term is very great compared to thc natural oscillations.

If 5 = p2, then the solution of ~ ~ n . ( 9 1 ) will be given by

x =acos ( \ m t ) + ,"fitsin (/at) (96)

Thus, in this case, as t increases, the displacement keeps increasing without limit. That is, there is no oscillations. This is of great importance in '

engineering applications, for if the forcing function is periodic with the period which is equal to the natural period, then the effects ca be devastating. This

I Fig. 11: The is called resonance effect (see Fig. 11). And now a few exercises for you.

phenomenon of resonance

E14) What happens to the motion of simple pendulum if the forcing function is F = L sin pt?

E15) In formulation 3 above, if along with the forcing function F = Lcos yt,

you also consider the effect of damping force = -f - then what will 1:l be the form and type of equation of motion? Are Eqns. (78) and (89) particular cases of this equation? Solve this equation of motion in the case of underdamped oscillations.

So far, we considered the motion of a particle under gravity, with and without resistance. However, in Secs. 4.3 and 4.4 we a1 ways considered the particle to move in a straight line. This is very much simplified compared to the real life situations.

I Before we conclude this unit, we shall discuss in the next section a slight and I ' immediate extension of motion in a straight line to the projectile motion, - .

Y which is 'the motion of a particle in a plane, moving under gravity. Since this

y(t + 69 involves velocity/acceleration in two dimensions, let us first define them as we did in Sec. 4.2 for the motion in a straight line.

Mot ion In Two Dimensions y(tq-/'J 1 consider a particle moving in a plane. Suppose r( t ) = (x(t), y(t)) and , r (t) ;

X r ( t + dt) = (x(t + bt),y(t + bt)) are the positions of the particle at times t

x (t) and t + 6t as shown in Fig. 12. .b x (L + st)+ Then the displacement tr of the particle in time bt is given by

Fig. 12 Sr = r ( t + dt) - r( t ) This vector equation is equivalent to the two scalar equations

The velocity v of the particle as i t moves along the path is defined to be Motion in a Straight 6r Line

v = ;jyo

which is equivalent to the two scalar equations J ' 1 . 6x dx

v1 = lim - = - at-to 6t . dt

Eqns.(98) and (99) give the components of velocity in x and y directions.

In a similar way the acceleration a is given by

In terms of the components a l , ag of the acceleration, Eqn.(100) gives

4.6 PROJECTILE MOTION

The free motion of a body which is projected in a non-vertical direction under gravity is called projectile motion. Numerl?us examples of projectile motion can be had from our everyday experiences. When you throw a ball from the boundary in a cricket ground, the path of the ball is a projectile. Same is true of the flight of a football when kicked, or even the motion of a jet of water from a hose pipe. Even in artillery applications, the path of a cannon when fired is a projectile. Now suppose a ball is thrown from a point 0 with a velocity uo at an angle a with.the hor~zonal. Then, you may like to know the answers to the following questions:

i) How far does the ball reach before i t touches the ground?

ii) How high does it rise? I I iii) What is the maximum distance it can cover for a given velocity, and

what is the angle of projection to achieve this distance?

iv) What is the total time of flight? To kno& the answers to these questions we will have to model the projectile motion. But before we do that we shall describe a small experiment to show the independence of horizontal and vertical motions.

Experiment 1: Place three marbles A, B and C at the edge of a table (see Fig. 13) At the same time, hit B and C horizontally with different forces, so that their horizontal velocities are different, while allow A to fall vertically downwards. You will observe that

i) horizontal distances traversed by A, B and C are different. They depend on the horizontal velocities

ii) they all hit the ground at the same instance.

iii) vertical distances covered by all the particles are same, equal to the height of the table.

Physical Environment

Fig, 13 What does this show? From this we conclude that

a) the effect pf gravity is not altered by horizontal motion

b) when a particle with an initial horizontal velociky moves under gravity, the motion can be studied by considering the Lori-contal and vertical mot ions independently.

Having got this much i~iformation horn the experiment about the way the two components of motion are effected by graxity, we are now in a position to set up a model equation for the motion of a projectile.

I Formulation 1: We shall consider the projectile without any resistance. Let a particle be projected from 0 with ail initial velocity uo making an angle 8 with the horizontal. Since, we do not coiisider any force other than gravity i the particle will be confined to the pl?rie in which the particle is thrown. We call this plane XOY, with OX and OY the two axes.(see Fig. 14). I I

Fig. 14 Suppose at time t, the particle is at P(x, y) . We have already shown though Experiment-1, that the motion can be studied in the vertical horizontal directions independently. Thus, writing down the Newtorls motion, we have the following equations of motion.

d2x m-

dt2 = 0, since there is no force in x-direction

and laws of

(104) Motion in a Straight

Line d2x d2y

where - and - are the co~nponents of the acceleration in x and y dt2 df

directions respectively. dx

The initial conditions are that a t t = 0,x = y = 0 and - = uo cos 0, while dt

- uo sin 8. With these conditions, the solmtion to Eqns. (103) and (104) is dt -

Eqns. (105) and (106) describe the motioil of a projectile. Elinlinating t between these equations, we obtain the equation to t h e t rajectory as

Do you recognise this equation? Yes! This is clearly the equation to a parabola. To determine the distance the particle reaches, that is the range R., we substitute y = 0 in tlie above eclualiotl. This gives us

This equation gives two values for x, one x = 0, tha? is the point from where the particle was thrown, and the other point is where it hits the ground. Thus, the range R of the particle is

R = ( ) sin o cns 0 = ($) sin 20

Using this value of R,, Eqn.(l07) to the trajectory can be written in a neat form as

Thus using Eqn.(lOS), the to t a l t ime of flight is given by

But this is the total time of flight of a particle moving under gravity with an initial velocity 110 sir18 . This again corroborates the observations of Experiment -1 where we found that the vertical and horizontal motions are independent. Of course, this is true only so long as there is no resistance.

Now in order to obtain the maximum height attained by the particle we proceed as follows:

For y to be maximum, we know that $ = 0, and hence from Eqn.(llO)

R giving us the value of x for which y is maximum as x = -. To mc&e sure 2

that this corresponds to a maximum, take the second derivative of y and

d2Y Environment ensure that, - is negative for this value of X. dx2

Thus the maximum y is obtained from the Eqn. (110) as R

y,, = - tan9 4 (113)

Also from Eqn.(lOS), we see that for a given velocity of projection, the maximum value of R i.e., the maximum range is given by

- And the angles of projection for this range is

and, 6 = --

Thus, for a given velocity of projection, to reach a certain range, within the maximurn possible with that velocity, there are two po:,sible afigles of projection.

To be more clear about whatever we have discussed above let us consider the , following example:

A " Example 8: Find the velocity and direction of project,ion of a ball which O +I 00 rn+\

moves in thc horizontal direction just over the top of a wall 100 metres high Fig. 15 at a di'stance of 100 metres.

Solution: Suppose velocity of projection is uo and the angle of projection is 9. Let PA be the wall and 0 the point of projection (see Fig. 15). Then = OA = PA = 100 metres.

Suppose at time t, the ball is at P. Then,

x = OA = 100 =: uo cos 9 t and

y = PA = 100 1

= uo sin6 t - -gt2 2

Also, since the ball crosses the wall horizontally, the vertical velocity of the dy ball at P i.e. - is zero. Hence, d t

uo sin 6 - gt = 0, thus giving t = sing. Substituting this vadue o f t in the above equations, wi can obtain the values of 9 and uo

And now an exercise for you,

-

3 -

E16) A projectile, when thrown at an angle tan-'(;;) falls 40 metres short of 'i

the'carget. When it is fired at an angle of 45O, it falls 50 metres beyond the target. Find the distance of the target from the point of projection.

1 I What do you obscrrve from the above discussion? Let us now try to sum up

these observations.

Observations

If you look at Eqn.(lO7) of the trajectory, i.e.,

you will observe that this can be transformed to

This being a quadratik in tan 8 will give for a given x, y and uo, Lwo roots for 8, provided

If this condition is not satisfied, then you cannot reach the point (x,y), with the velocity of projecti~n uo. When this condition is satisfied, you can reach the given point through two trajectories - one M hen the particle is moving upwards, and the other when it is moving downwards.

In the beginning of this section we mentioned about the application of the projectile in games, artillery etc. But you inay recall that in Block 1, we warned you to validate the model before you go for predictions using the model. For instance, if you are a goalkeeper, do not move to stop a shot calculating on the basis of what was presented here - you might he in for a rude shock. The reason is tibat, we have completely ignored the air resistance. Besides, if the ball is kiclced with an initial spin, it will swing towards a particular direction depending upon the initial spin. This fact can be easily illustrated by throwing a table tennis ball. This being very small and light, the air resistance effect will be dominant, and you will observe that more often than not, the motion will not be along predicted lines. In artillery also, for accurate prediction, the effect of air resistance has to be considered. '

Thus, our discussion would be incomplete if we do not consider thc effect of air resistance on {;Be motion of u projectil~ and improve Formulation-1.

Formulation 2: Let us consider the projectile ~ ~ ~ n t i o n under gravity with air resistance R proportional to velocity. Eqns.(lOS) arid (104) of motion then take the form

where k > 0 is a constant of proportionality, Using the initial conditions dx dy

x = 0 = y at, f = 0 and - = uo cos 0, - = uo sin 9, the solution to Eqns. dt d t

(117) and (118) can be written as

mg "7 - e-?it) y + -t = 2 (uo sine + - k k k

Eqns. (119) and (120) describe the mot ion of a projectile in a resisting medium. Eliminating t between these equations, we obtain the equation t o t h e t ra jec tory as

m2

kx ) +-A (uosine+ - mu0 cos 9 uo cos 0 mg) k

dy For y to be maximum, ,we know - =E 0, this gives us dx

Motion in a Straight Line

mug sin 9 cos 6' x =

kuo sin 0 + mg (122)

Physical Environment and for this value of x we get from Eqn.(l21), maximum y as mu0 sin 9 kuo sin 9

Ymax = k - & l n ( l + k2 mg ) (123) mu0 cos 0

It is clear from Eqns.(llS) and (120) that when t -+ w, x = k and

y -t -CQ. Hence the path has a vertical asymptote at a horizontal distance mu0 cos 9

k from the point of projection (see Fig. 16).

b

I

Fig. 16

You may observe here that as usual you can obtain tke Eqns.(l03) and (104) of motion directly from Eqns.(ll7) and (118) by putting k '= 0 whereas, to obtain Eqn. (107) of the trajectory you have to take the limit k + 0 in Eqn.(l21). We are leaving it to you to check it.

E17) Expand the right hand side of Eqn.(l21) in powers of k and then put k equal to zero to obtain the equation to the trajectory for uilresisted motion.

I We now end this unit by giving the summary of what we have covered in it. I

I 4.7 SUMMARY I I

i In this unit, we have covered the following: 1. Basic steps involved in modelling any physical situation are I

i) Draw a clear diagram of the problem at hand in a suitable 1 coordinate system indicating the various forces involved.

ii) Choose the essentials of a problem. 1 iii) Set up model equation using empirical laws of motion, which are

mostly Newtbn's laws in the case of dynamics. I

iv) Solve the model equations and interpret the results. 1 I

I 2. The downward and upward motion of a particle in a straight line moving vertically under the influence of gravity with no resistance is 1 modelled respectively by the following equations. *

dv dv d2x - =v-=- - 1

dt dx dtz - g i 30 1 I

- I

dv dv d2x - = v - = - - dt dx dt2 - -'

However, if there is resistance R to the motion these equations reduce to

dv dv d2x and - = v- = - = -g - R. respecti 1

dt dx dt2 Ve Y,

3. If the force acting on a particle is proportional to its distance from a fixed point and is directed 1;owards it, then the particle executes a siinple harmonic m,otion (SHM). The model equation for SHM can be written ia the form x = R cos(wt -t 4) or x = R sin(wt + 4) where w is the frequency, R the amplitudc and q5 the phase of the oscillation. The simplest application of SHM is the simple pcndulum.

4. Ecluation modelling the SHM of a simple pendulum under the influence , of an ezternal force F is given by

where, the positive F will enhance the notion and negative F will dampen the motion

5. The free motioll of a body projected in a lion-vertical direction under gravity is called projectile motion. The prr\jectile motion of a particle with an initial velocity uo, making an angle 0 with the horizontal can be studied in the vertical and llorizontal directions independently. The eql~ations of motion without and with resistance (proportional to the velocity) are given by

and

where lc > 0 is the constttnt of proportionality. dx

The initial conditions are that at t = 0, x = y = 0, - = uo cos 0 and dt

dy - = uo sin 0. dt

Motion in a Straight Line

Differentiating once, we obtain the velocity as

v = 2t $- 12, and hence, when t = 2, velocity is 16.

For velocity to be zero, we should have t as -6, which is meaningless. This means that the velocity can never be zero.

E2) metres/sec

E3) Let the distance between the passenger and the bus when the bus starts be x metres. Suppose the passenger reaches the bus after time t,

Physical Environment and suppose th'e bus has moved a distance y metres in that time. Then, 1

y = - x 10 x t2 metres (since motion starts from rest) 2

E7) When vt is very large, you have to take the limits using LJHospital's 1

rule and the result is -gt2. Physically this can be explained as follows. 2

Also, in this time, the passenger covers a distance of x + y. Hence,

x + y = 40t.

Thus, combining these two equations, 40t = x + 5t2

i.e. 5t2 - 40t + x = 0.

This is a qua,dratic in t. If the passenger has to reach the bus, then the roots of this quadratic equation have to be real. The condition for this, I

is

40' - 4x x 5 > 0 1 which gives, I

x 5 80 metres. I I

E4) Let the deceleration in the block be a. If the initial velocity is u, then !

U after penetrating 3 cm, the velocity becomes -. Hence, I

2 I

which gives uL

a = S' Suppose the bullet penetrates a distance y further before coming to

U rest. Then, the initial velocity is - and the final velocity is 0. Then

2

-

On substitut'ng the value of a in terms of u and solving we obtain y = 1 cm.

E5) Let the distance travelled in time be s = kt2, where k is the constant of. proportionality. ~

1 Also, s is given by s = -at2, where a is the acceleration. Comparing the

2 two expressions for s, we obtain a = 2k, i.e., the acceleration is a , constant . 1

E6) Let the resistance be kv. I i

Then the eq~~ations for motion will be I

dv - = g-kv dt dv

v~ = g - kv.

Solve these differential equations and you will get

When ;;-is large expression (36) means k and hence the resistance is . zero. This corresponds to free fall under gravity and hence the distance

1 1

covered in time t should be -gt2. 2

E8) Using Eqn.(51) with vt = with D = 0.1 we obtain t = cos h-' (eloooD).

E9) Let the height covered be h. Then

Hence,

h = loo3

metres. 2 x 980

Also, if t is the time taken to reach this hl,ight, then

loo2 and t = -

980 Set-

E10) Expand the In term in Eqn.(68) with vt = and take the limit k+O.

E l l ) Suitably define sin dl and cos 4' and reduce Eqn.(71) to the required form.

E12) From Eq11.(77), we obtain the point where v is maximum as x = 0.

E13) Write down the SHM equations and use the given conditions to determine the constants, and proceed further.

d% p L E14) Solve the equation - + -x = - cos pt and discuss its solution when

dt2 m m P 2 P - = -p and when - # -p2 m m

E15) Equation of motion will be of the form d2x dx

m-+f -+px=Lcosp t dt2 d t

(resistance) (external force)

f where a = - P and u2 = -

2 m m General solution to this equation is of the form x = xc 4- xp, where

x, = eMUut (A cos w t Jl+ B sinsinwtJ1-,P)

where R = 4- and 4 = tan" (k) L (w2 - p2) cost + psinpt

X p , = - m (w2 - p2)2 + 4a2p2u2

where = tan-1 (W?::~)

Motion in a Straight Line

Physical Environment E16) We have two equations here .-,

In the first equation, writing sin 20 in terms of tan 0 and solving the two we obtain

24 (R - 40) = (R + 50) x - 2 5 this gives R = 2200 m.

E17) Expanding the r.h.s. of Eqn.(l21) in powers of k, it becomes

+ ( u o s i n ~ + - U" COS I3 .Y)T

1 .gkx3 gx2 - - 1 gk2x4 i.e., y = x tan I3 - - - . . .

2u; cos2 0 3 mu: cos3 I3 4 m2u; cos4 I3

on putting k = 0, we obtain

y = x tan 19 - sec2 0 , which is the equation to the trajectory for 2 4

unresisted motion.

UNIT 5 PLANETARY MOTION

Structure Page No.

5.1 Introduction Objectives

5.2 Newton's Law of Gravitation 3 6

5.3 Particle Projected From The Earth -Finding tlze Escape Velocity

5.4 Central Forces-Basic Concepts 40

5.5 Modelling Planetary Motion 47

5.6 Kepler's Laws Lead To Newton's Law Of Gravitation 5 3

5.7 Limitations

5.8 Summary

5.9 Solutions/Answers

5.10 Appendix

5.1 INTRODUCTION

In Unit 4, you were introduced to the idea of motion in a straight line. This was illust,rated by motion under gravity and sihple harmoriic motion with special reference to simple pendulum. At the end of the unit, we also introduced you to motion in a plane through a discussion of projectile motion. In the case of projectiles, the path was a parabola, a two dimensional curve. But all these motions had one thing in common - the only forces causing the motion were in a fixed direction. Gravity acts towards the centre of the Earth, but for motion under gravity, as in Unit 4, the displacements are so small compared to the Earth's radius, that thc direction of the force can be considered to be constant. In this Unit, we introduce motion under a force directed towards a fixed point - this is called motion unde r cent ra l forces. An example of this type of motion is planetary motion. In this case also the motiori is in the same plane. However, the direction of the force is no longer a constant.( See fig. 1.) Fig. 1

Historically, celestial bodies have been of interest to mankind from time immemorial. There are records of Chinese, Indian and Greek astronomers' observations of celestial motion. Since, to us, it appears as if the Sun moves round the Earth, early man thought that Earth was the centre of the universe, and all planets moved round the Earth. However, in 1543, Copernicus, a Polish monk, proposed the 'helioccnlric theory' in his book De revolutionibus o r b i u m Coelestium. According lo this theory, Sun was the centre of the universe, and that all planets moved round the Sun in circular orbits.

Tycho Brahe, a Danish astronomer, made systematic observations of the heavenly bodies for about 25 years. These observations were so accurate that they were used by navigators for a long time. Johannes Kepler, Tycho