Chapter 4 MotioN In A Plane

CHAPTER NO.4 MOTION IN A PLANE

4.1 INTRODUCTION

In the last chapter we developed the concepts of position,displacement, velocity and acceleration that are needed to describe the motion of an object along a straight line. We

found that the directional aspect of these quantities can be taken care of by + and - signs, as in one dimension only two directions are possible. But in order to describe motion of an

object in two dimensions (a plane) or three dimensions (space), we need to use vectors to deacribe the above-

mentioned physical quantities. Therefore, it is first necessary

to learn the language of vectors. What is a vector? How to add, subtract and multiply vectors ? What is the result of

multiplying a vector by a real number ? We shall learn this te enable us to use vectors for defining velocity and

acceleration in a plane. We then discuss motion of an object ina plane. As a simple case of motion in a plane, we shall discuss motion with constant acceleration and treat in detail

the projectile motion. Circular motion is a familiar class of motion that has a special significance in daily-life situations.We shall discuss uniform circular motion in some detail.The equations developed in this chapter for motion in a plane can be easily extended to the case of three dimensions.

4.2 SCALARS AND VECTORA

In physics, we can classify quantities as scalars or vectors. Basically, the difference is that a direction is

assoctated with a vector but not with a scalar. A scalar quantity is a quantity with magnitude only. It is specified

completely by a single number, along with the proper unit. Examples are : the distance between two points,mass of an object, the temperature of a body and the time at which a certain event happened. The rules for combining scalars are the rules of ordinary algebra.Scalara can be added,aubtracted, multiplied and divided just as the ordinary numbers*. For example,if the length and breadth of a rectangle are

1.0 m and 0.5 m respectively, then its

perimeter is the sum of the lengths of the four sides, 1.0m+05m+10m+05m=

3.0 m. The length of each side is a scalar and the perimeter is also a scalar. Take another example: the maximum and minimum temperatures on a particular day are 35.6 °C and 24.2 °C respectively. Then,the difference between the two temperatures

is 11.4 °C. Similarly, if a uniform solid cube of aluminium of side 10 cm has a mass of 2.7 kg, then its volume is 10° m® (a scalar)and its density is 2.7x10° kg m= (a scalar).

Avector quantity is a quantity that has both a magnitude and a direction and obeys the triangle law of addition or equivalently the parallelogram law of addition. So, a vector is specified by giving its magnitude by a number

and its direction. Some physical quantities that are represented by vectors are displacement,velocity, acceleration and force.

To represent a vector, we use a bold face type in this book. Thus, a velocity vector can be represented by a symbol v. Since bold face is difficult to produce, when written by hand, a vector is often represented by an arrow placed

over a letter, say U. Thus, both w and v

represent the velocity vector. The magnitude of a vector is often called ita absolute value,indicated by Iw! = v. Thus, a vector is represented by a bold face, e.g. by A, a, p,q, & «..x, y, with respective magnitudes denoted by light

face A, @, p, GT, ...% Y.

4.2.1 Position and Displacement Vectors

To describe the position of an object moving in a plane, we need to choose a convenient point,say O as origin. Let P and P’ be the positions of the object at time tand f’, respectively [Fig. 4.1(a)].We join O and P by a straight line. Then, OP is the position vector of the object at time t. An arrow is marked at the head of this line. It is represented by a symbol r, i.e.OP=r. Point P’ is

Tepresented by another position vector, OP’ denoted by r’. The length of the vector r Tepresents the magnitude of the vector and its direction is the direction in which P les as seen

from ©. If the object moves from P to P’, the vector PF’ (with tail at P and tip at P’) is called the displacement vector corresponding to motion from paint P {at time 9 to point P’ fat time 7).

It is important to note that displacement vector is the straight line joining the initial and final positions and does not depend on the actual

path undertaken by the object between the two positions. For exampke, in Fig. 4.1b, given the initial and final positions as P and Q, the displacement vector is the same Pg for different

paths of journey, say PABCQ, PDQ, and PBEFQ.Therefore, the magnitude of displacement is either less or equal to the path length of an abject between two points. This fact was emphasised in the previous chapter also while discussing motion along a straight line.

4.2.2 Equality of Vectors

Two vectors A and B are said to be equal if, and only if, they have the same magnitude and the same direction.**

Figure 4.2(a} shows two equal vectors A and B. We can easily check their equality. Shift B parallel to itself until its tail Q coincides with that

of A, i.e. Q coincides with O. Then, since their tips S and P also coincide, the two vectors are said to be equal. In general, equality ia indicated

as A=B., Note that in Mig. 4.2(b). vectora A’ and B’ have the same magnitude but they are not

equal because they have different directions.Even if we shift B’ paralle] to ttself so that its tail 0 coincides with the tail O’ of A’, the tip S’ of B’

does not coincide with the tip P’ of A’.

4.3 MULTIPLICATION OF VECTORS BY REAL

NUMBERS Multiplying a vector A with a positive number 1.gives a vector whose magnitude is changed by the factor 4 but the direction is the same as that

ofA:laal =a lal tta>o.

For example, ifAis multiplied by 2, the resultant vector 2A is in the same direction as Aand has a magnitude twice of |Al as shown in Fig, 4.3(a).

Multiplying a vector A by a negative number 1 gives a vector 1A whose direction is opposite to the direction of A and whose magnitude ia -\ times IAI.

Multiplying a given vector A by negative

numbers, say -1 and -1.5, gives vectors aa shown in Fig 4.3(b).

The factor 4 by which a vector A is multiplied could be a scalar having {ts own physical dimension. Then, the dimension of 4 A ia the product of the dimensions of 4 and A. For example, ifwe multiply a constant velocity vector by duration (of time), we get a displacement vector.

4.4 ADDITION AND SUBTRACTION OF

VECTORS — GRAPHICAL METHOD As mentioned in section 4.2, vectors, by definition, obey the triangle law or equivalently,

the parallelogram law of addition. We shall now describe this law of addition using the graphical method. Let 113 consider two vectors Aand B that

lie in. a plane as shown in Fig. 4.4(a). The lengths of the line segments representing these vectors are proportional to the magnitude of the vectors.To find the sum A+ B, we place vector B so that its tail is at the head of the vector A, as in Fig. 4.4(b). Then, we join the tafl of A to the head

of B. This line OC represents a vector R, that is,the sum of the vectors A and B. Since, in this procedure of vector addition, vectors are

arranged head to tail, this graphical method ts called the head-to-tail method. The two vectors and their resultant form three sides ofa triangle.so this method fs also known as triangle method

of vector addition. If we find the resultant of B + Aas in Fig. 4.4(c), the same vector R is obtained. Thus, vector addition is communtative:

A+B=BrtaA (4.1)

The addition of vectors also obeys the associative law as illustrated in Fig. 4.4(d). The result of adding vectors A and B first and then adding vector C is the same as the result of adding B

and C first and then adding vector A:

(A+B) +C=A+ (B+C) (4.2)

What is the result of adding two equal and opposite vectors ? Consider two vectors Aand -A shown in Fig. 4.3(b). Their aum is A + (-A).Since the magnitudes of the two vectors are the

same, but the directions are opposite, the resultant vector haa zero magnitude and is represented by 0 called a null vector or a zero vector :

A-A=0 lol=0 (4.3)

Stoce the magnitude of a null vector is zero, its direction cannot be specified.

The null vector also results when we multiply a vector A by the number zero. The main properties of O are :

AtO=A

140=0

OA=0 (4.4)

What ts the physical meaning of a zero vector?

Consider the position and displacement vectors ina plane as shown in Fig. 4.1 fa). Now suppose that an object which fs at P at time t, moves to P and then comes back to P. Then, what is its

displacement? Since the initial and final positions coincide, the displacement {s a “null vector",

Subtraction of vectors can be defined in terms of addition of vectors. We define the difference of two vectors A and B as the sum of two vectors Aand -8:

A-B=A+(3B) (4.5)

It is shown in Fig 4.5. The vector -B is added to vector A to get R, = (A- 8). The vector R= A+B is also shown in the same figure for comparison.We can also use the parallelogram method to

find the sum of two vectors. Suppose we have two vectors A and B. To add these vectors, we bring their tails to a common origin O as shown in Fig. 4.6{a). Then we draw a line from the head of A parallel to B and another line from

the head of B parallel to A to complete a parallelogram OQSP. Now we join the point of the intersection of these two lines to the origin O, The resultant vector R is directed from the

common origin 6 along the diagonal (OS) of the parallelogram [Fig. 4.6(b)]. In Fig.4.6(c), the triangle law is used to obtain the resultant of A and 8 and we see that the two methods yield the

same result. Thus, the two methods are

equivalent.

Example 4.1 Rain is falling vertically with a speed of 35 m s°. Winds starts blowing after sometime with a speed of 12 ms" in east to west direction. In which direction should a boy waiting at a bus stop hold his umbrella ?

Answer The velocity of the rain and the wind are represented by the vectorsv,andv_,in Fig.

4.7 and are in the direction specified by the problem. Using the rule of vector addition, we see that the resultant ofv, and v, is R as shown.in the figure. The magnitude of R is Rev + ts ¥357 +127 ms! =37ms"! The direction @ that R makes with the vertical is given by

tang= 2 = 12 0343

v, 35

Orn ¢=tan{0.343)=19°

Therefore, the boy should hold his umbrella in the vertical plane at an angle of about 19° with the vertical towards the east. 5

4.5 RESOLUTION OF VECTORS

Let a and b be any two non-zero vectors in a plane with different directions and let A be another vector in the same planefFig. 4.8). Acan De expressed as a sum of two vectors -— one obtained by multiplying a by a real number and

the other obtained by multiplying b by another Teal number. To see this, let O and P be the tail and head of the vector A. Then, through O, draw a straight line paralle] to a, and through P, a straight line parallel to b. Let them intersect at Q. Then, we have

A=OP =-09+9P (4.6)

But since OG is parallel to a, and QP is parallel to b, we can write :

O09 = Aa, and @P=pb (4.7)

where A and pare real numbers.Therefore, A=Aa+pb (4.8)

We say that A haa been resolved into two

component vectors 4a and ub along a and b Tespectively. Using this method one can resolve a given vector into two component vectors along a set of two vectors — all the three lie in the same

plane. It is convenient to resolve a general vector along the axes of a rectangular coordinate system using vectors of unit magnitude. These

are called unit vectors that we discuss now.

Unit vectors: A unit vector is a vector of unit Magnitude and points in a particular direction.It has no dimension and unit. Itis used to specily

a direction only. Unit vectors along the x-, y-and z-axes of a rectangular coordinate system are denoted by i, j and k, respectively, as shown in Fig. 4.9(a).

Since these are unit vectors, we have

lil =lil=lel= 4.9)

These unit vectors are perpendicular to each other. In this text, they are printed in bold face with a cap (A) to distinguish them from other vectors. Since we are dealing with motion in two dimensions in this chapter, we require use of only two unit vectors. If we multiply a unit vector,say m bya scalar, the result ie a vector 4=101. Ingeneral, a vector Acan be written as A=IAla (4.10) where ® is a unit vector along A.

We can now resolve a vector A in terma

of component vectors that lie along unit vectors i and j. Consider a vector A that Hes in xy plane as shown in Fig. 4.9(b). We draw lines from the head of A perpendicular to the coordinate

axes as in Fig. 4.9(b), and get vectors A, and A,such that A, + A,=A Since A, is parallel to i and A, is parallel to j, we have :

A=A,i, A=A,j (4.11)where A, and A, are real numbers.Thus, A=A,i+A, j (4.12)

This is represented in Fig. 4.9(c). The quantities A, and A, are called x-, and y- components of the vector A. Note that A, is tiself not a vector, but

A, i 18 a vector, and so is A, j. Using simple trigonometry, we can express A, and 4 in tems of the magnitude of A and the angle 6 it makes with the x-axis :

A, =Acos @

A,=Asin 6 (4.13)

As is clear from Eg. (4.13), a component of a vector can be positive, negative or zero depending on the value of 0.

Now, we have two ways to specify a vector A in a plane. It can be specified by :

(} ite magnitude A and the direction 6 it makes with the x-axis; or

(i) its components A, and A,IfAand 6 are given, A,and A, can be obtained using Eq. (4.13). If A, and A, are given, Aand 8 can be obtained as follows :

AP sa? = A*cos’0+ A’sin”@

=A! :

Or, As (a? + Ar (4.14)

And tane= tt, 6a tan {e (4.15)

A, A, °

So far we have considered a vector lying in an xy plane. The same procedure can be used to resolve a general vector A into three components along x-, y-, and z-axes in three dimensions. If a, f, and ¥ are the angles*between A.and the x-, y-, and z-axes, respectively Fig. 4.9(d), we have

“AY =Acos@, A, =Acosf. A, =Acosy (4.16a)In general, we have

A=AAi+Aj+ Ak (4.16b)The magnitude of vector A is A= {# + Ai +A; (4.16c)

A position vector r can be expressed as

r=xityj+zk (4.17) where x, y, and zare the components ofr along x, Yy-, zZ-axes, respectively.

4.6 VECTOR ADDITION - ANALYTICAL

METHOD Although the graphical method of adding vectors helps us in visualising the vectors and the resultant vector, it is sometimes tedious and has limited accuracy. It is much easier to add vectors by combining their respective components.Consider two vectors A and B in x-y plane with components A, A, and B,, B,:A=Ai+A,) (4.18)

B= B,i+B,j

Let R be their sum. We have

R=-A+B

=(a,i+ A,j)+(B it Bj) (4.198)

Since vectors obey the commutative and

associative laws, we can arrange and regroup the vectors in Eq. (4.19) as convenient to us :R=(A, +B,)}i+ (4, + B,)j (4.19b)SinceR = R,i+R,j (4.20)

we have, R, = A, + B,. R, =A, +B, (4.21)

Thus, each component of the resultant

vector R is the sum of the corresponding

components of Aand B.In three dimensions, we have A=A,i+A,j+ Ak

B= B,i+B,j+ Bk R= A+B= Ri+R,jt Rk

with R, =A, +B,(Ry = Ay + By

‘R,=A,+B, (4.22)

This method can be extended to addition and subtraction of any number of vectors. For example, ff vectors a, b and ¢ are given as as a,ita,j+ ak

b=b,it+b, J+ bk ce Cyd + Cy + ck (4.23a)

then, a vector T = a + b - c has componenta :Ty Ba,+b,-c,Ty Sy t by Cy (4.23b)T, =@, 4b, -C,

Example 4.2 Find the magnitude and

direction of the resultant of two vectors A and B in terms of their magnttudes and angle @ between them.

Answer Let OP and O@ represent the two vectors Aand B making an angle 6 (Fig. 4.10). Then,using the parallelogram method of vector addition, 08 represents the resultant vector R :R=-A+B

SN is normal to OPand Pi is normal to OS.From the geometry of the figure,

OS = ON + SNP but ON=OP+ PN=A+ Bcos @

SN=Bain @OS = (A+ Bcos @? + (Bain g?

or, R= A’ + B’ + 2AB cos @ R= A?+B + 2ABcos0 (4.24a)INnAOSN, SN= OS sine =Rsing, and in APSN, SN= PSsin @=B sin 6

Therefore, Rsin a = Bain @R B

ing sina (4.24b)Similarly,PM=A sine =B ainf A |B Or, Sind sina (4.24c)

Combining Eqs. (4.24b) and (4.24c), we get RA LB sing sin ~ sina (424d)

Using Eq. (4.24d), we get:

sin a= —sin 8 (4.24e)

where R ts given by Eq. (4.24).

_ SN ____siBsin@é on tan@=05 > PN ~ A+ Bcos@ (4.244 Equation (4.24a} gives the magnitude of the resultant and Eqs. (4.24¢) and (4.249 fts direction.

Equation (4.24) is known as the Law of cosines and Eq. (4.24d) as the Law of sines. <

Example 4.3 A motorboat is racing

towards north at 25 km/h and the water

current in that region 1s 10 km/h in the

direction of 60° east of south. Find the

resultant velocity of the boat.

Answer The vector v, representing the velocity of the motorboat and the vector v, representing the water current are shown in Fig. 4.11 in directions specified by the problem. Using the

parallelogram method of addition, the resultant R is obtained in the direction shown in the figure.

We can obtain the magnitude of R using the Law of cosine :pe fds 242 R= Yfupyt vet Ayu, 008l120° = 25? +107 +2x25x10(-1/2) = 22 km/h

To obtain the direction, we apply the Law of sines sin 8 = sing or, sin ¢= zo 6 ~ 1oxsini20’ —10g3 = = ——_ £ 0.397

21.8 2x21.8 $= 23.4 <

4.7 MOTION IN A PLANE

In this section we shall see how to describe motion in two dimensions using vectors.

4.7.1 Position Vector and Displacement

The poattion vector r of a particle P located ina plane with reference to the origin of an x-y reference frame (Fig. 4.12) is given by rexityj where xand y are components of r along x-, and

y- axes or simply they are the coordinates of the object.

Suppose a particle moves along the curve shown by the thick line and is at P at time ¢t and P’ at time ¢ [Fig. 4.12(b)]. Then, the displacement is :ar =P -F (4.25) and is directed from P to P’.

We can write Eq. (4.25) in a component form:ar =(vity'j)-(xi+y))

= iAx+ jay

where Ax=x'-x, ay=yf-y (4.28)

Velocity The average velocity (v} of an object is the ratio of the displacement and the corresponding time interval :

vee I aan

At At At At

Or v-t.i+ b,j

—_ 4

Since = 7. the direction of the average velocity is the same as that of ar (Fig. 4.12). The velocity (instantaneous Velocity} is given by the limiting

value of the average velocity as the time interval approaches zero :

Ar_ dr

v~ Phat dt 4.28)

The meaning of the limiting procesa can be easily understood with the help of Fig 4.13(a) to (d). In these figures, the thick line represents the path

of an object, which is at Pat timet P,, P, and P, represent the positions of the object after times Af At,, and At,. Ar,, Ar,, and Ar, are the displacements of the object in times Af,, Ai,, and

At,, respectively. The direction of the average velocity V 1s shown in figurea {a). (b) and (c) for three decreasing valuca of st, Le. At,.at,, and At,

{at, > at, > at). As at > 0, ar + O

and is along the tangent to the path Fig. 4.13()].Therefore, the direction of velocity at any point on the path of an object is tangential to the path at that point and is in the direction of

motion.We can express v in a component form :

/ dr

v=—

, Ax, Ay:

= lim|—i+—j (4.29)

Atm Al Al

=i lim aj lim Ay

Ate At At-0 At

dx dy | fan

Or v=i di +j dt vAt vj

, dx dy

where Us Yu = ae (4.30a)

So, if the expressions for the coordinates x and y are known as functions of time, we can use these equations to find v, and v,,The magnitude of v is then ve yext uy (4.30b)and the direction of v is given by the angle 6:, ly -1 by

tan@=——, @=tan |— (4.30c)

uy by,

v, ¥, and angle @are shown in Fig. 4.14 for a velocity vector v.

Acceleration

The average acceleration a of an object fora time interval At moving in x-y plane is the change in velocity divided by the time interval :

A vite, j > . Avy,

poor Medtend) ae Ate (4.31a)

Ai At At At

* in terms of x and y, a, and a, can be expressed as

| f(y Peg fan) ey

fer alal a2 ala) ae

Or, aza,ita,j.

{4.31b)

The acceleration (instantaneous acceleration)is the Itmiting value of the average acceleration as the time interval approaches zero :

, li Av

a= lim — 4,32:

Aimo At ¢ a)

Since Av = Av,i + Av ybwe have

oo, Av.» Av,

a=i lim See yj lim —*

dio At aio At

Or, a=a,ita,j (4.32)

, dv, dv,

where, a,=—, a, =— (4.32c)*

dt ~ dt

As in the case of velocity, we can understand graphically the limiting process used in defining acceleration on a graph showing the path of the

object's motion. This is shown in Figs. 4.15(a) to (@). P represents the posttion of the object at time ¢ and P,, P., P, positions after time At, Ai,

Af,, respectively (Af > At>At). The velocity vectors at points P, P,. P.. P, are also shown in Figs. 4.15 (a), (b) and (c). fn each case of At, Avia

obtained using the triangle law of vector addition.By definition, the direction of average acceleration is the same as that of Av. We see that as At decreases, the direction of Av changes

and consequently, the direction of the

acceleration changes. Finally, in the limit Af0

Fig. 4.15(d), the average acceleration becomes the instantaneous acceleration and has the direction as shown.Note that in one dimension, the velocity and

the acceleration of an object are always along the same straight line (either in the same di-rection or in the oppoaite direction). However,for motion in two or three dimensions, veloc-ity and acceleration vectors may have any

angle between 0° and 180° between them.

Example 4.4 The posttion of a particle is given by r= 3.0ti+2.0t77+5.0E

where ¢ is tn seconds and the coefficients have the proper units for r to be tn metres.(a) Find v(§ and a( of the particle. (b) Find the magnitude and direction of v(§ at t=1.0s.Answer : 1 1 . yr .w(0=—=—(3.0 142.007 545.0 i)

dt dt = 3.0i+ 4.04 a (Qa 242.0)

dt a=4.0m s° along y- direction

At t=1.08, v=3.0i+4.0j It's magnitude is v=¥3° +4 =5.0ms"!and direction ts

4.8 MOTION IN A PLANE WITH CONSTANT

ACCELERATION

Suppose that an object is moving in x-y plane and its acceleration a is constant. Over an interval of time, the average acceleration will equal this constant value. Now, let the velocity

of the object be vy at time t= 0 and vw at time £ Then, by definttion

azvivo_¥-vo

aet-0

Or, v=v,tat (4.33a)

In terma of components :

‘Ly = Vox + Axl

ty HU yy Fl (4.33)

Let us now find how the position r changes with time. We follow the method used im the one-dimensional case. Let r, and r be the position vectors of the particle at time 0 and t and let the

velocities at these instants be v, and v. Then,over this time interval t, the average velocity is (v, + v)/2. The displacement is the average velocity multiplied by the time interval :

rn -( je-[ +at)ivg le

2) 4 2 }

= Vol + 5 al"

, 1,

Or, FetytVol+7 al” (4.34a)

It can be easily verified that the derivative of Eq. (4.34a), ie. = gives Eq.(4.33a) and it also satisfies the condition that at tO, r = r,.

Equation (4.34a) can be written in component

form as

X= Xqtvytt+ iat"

Y= Yo + v,,0+ 5a,t° (4.34b)

One immediate interpretation of Eq.(4.34b) 1s that the motions in ~ and y-directions can he treated independently of each other. That is, motion in

a plane (two-dimensions) can be treated as two separate simultaneous one-dimensional motions with constant acceleration along two perpendicular directions. This is an important

result and is useful in analysing motion of objects in two dimensions. A similar result holds for three dimensions. The choice of perpendicular directions is convenient in many physical sthuations, as we shall see in section 4.10 for

projectile motion.Example 4.5 A particle starts from origin at t= O with a velocity 5.0fm/s and moves in xy plane under action of a force which

produces a constant acceleration of

+ | m/s?. (a) What is the y-coordinate of the particle at the instant

its x-coordinate is 84 m ? (bh) What is the speed of the particle at this time ?

Answer The position of the particle is given by r (0) = vot + sat’=5.0it + (1/2)(3.0i + 2.05) @ =(5.0t+1.57)i+1.005

Therefore, x ({)= 5.0¢+ 1.52 ‘y(t)=+1.08

Given x(§ = 84m, t=?

§.0t+150C?=&>t=6s

Att=68, y= 1.0 (6)? =36.0m

Now, the velocity v= Z-(5.0 + 3.00)4+2.08 j

At t=68, v= 23.014+12.0j

speed =|v/=¥23° +12" =26 ms" , 4

4.9 RELATIVE VELOCITY IN TWO

DIMENSIONS

The concept of relative velocity, introduced in section 3.7 for motion along a straight line, can be eastly extended to include motion in a plane

or in three dimensions. Suppose that two objects A and B are moving with velocities v, and v,(each with respect to some common frame of reference, say ground.). Then, velocity of object

Arelative to that of B is :V.n=¥,— ps (4.35a)and similarly, the velocity of object B relative to that of Ais:

Van = VV Therefore, v,, =— Vi» (4.35b)

and, |v 4,|={Vpa| {4.38c)p> Example 4.6 Rain is falling vertically with

a speed of 35 ms". Awoman rides a bicycle with a speed of 12 ms" in east to west direction. What is the direction in which she should hold her umbrella ?

Answer In Fig. 4.16 v, represents the velocity of rain and v, , the velocity of the bicycle, the woman is riding. Both these velocities are with

respect to the ground. Since the woman is riding a bicycle, the velocity of rain as experienced by

her is the velocity of rain relative to the velocity of the bicycle she is riding. That is v, =v_- ¥,

This relative velocity vector as shown in Wig. 4.16 makes an angle @ with the vertical. It is given by

, vu, 12

tan @=—— =— = 0.343

v, 35

Or, 6219

Therefore, the woman should hold her

umbrella at an angle of about 19° with the vertical towards the west.

Note carefolly the difference between this Example and the Example 4, 1. In Example 4.1, the boy experiences the resultant (vector sum) of two velocities while in this example,the woman experiences the velocity of rain

relative to the bicycle (the vector difference of the two velocities). <

4.10 PROJECTILE MOTION

As an application of the ideas developed in the previous sections, we consider the motion of a projectile. An object that is in flight after being

thrown or projected is called a projectile. Such @ projectile might be a football, a cricket ball, a baseball or any other object. The motion of a

projectile may be thought of as the result of two separate, simultaneously occurring components of motions. One component is along a horizontal

direction without any acceleration and the other along the vertical direction with constant acceleration due to the force of gravity. It was Galfleo who first stated this tndependency of the

horizontal and the vertical components of projectile motion in his Dialogue on the great world systems (1632).

In our discussion, we shall asaume that the air resistance has negligible effect on the motion of the projectile. Suppose that the projectile is launched with velocity v, that makes an angle

8, with the x-axis as shown in Fig. 4.17.

After the object has been projected, the

acceleration acting on it fs that due to gravity which is directed vertically downward:as-gj

Or a,=0,q=-¢ (4.36)

The components of initial velocity v, are :u,, = U, C08 4,

U, = V, sin 6, (4.87)

If we take the initial position to be the origin of the reference frame as shown in Fig. 4.17, we

have :

x= 0, Yu, = 0

Then, Eq.{4.47b) becomes :

x=v,t=(u, cos 8) t

and y=(v,sin @,)t-(%)gl (4.38)

The components of velocity at time ¢ can be obtained using Eq.(4.33b) :

UL =U, =U, C08 O,

v, =v, sin @,-gt (4.39)

Equation (4.38) gives the x-, and y-coordinates of the position of a projectile at time ¢ in terms of

two parameters — initial speed v, and projection angle @,. Notice that the choice of mutually perpendicular x-, and y-directions for the analysis of the projectile motion has resulted tn

a simplification. One of the components of velocity, i.e. x-component remains constant throughout the motion and only the y- component changes, like an object in free fall in vertical direction. This is shown graphically at few instants in Fig. 4.18. Note that at the point

of maximum height, v= 0 and therefore,

q=tan''—“=0

vy,

Equation of path of a projectile

What is the shape of the path followed by the projectile? This can be seen by eliminating the time between the expreasions for x and y as given in Eq. (4.38). We obtain:

, g :

y = (tan @, )x« ———@———- x

( o) 2 (v,cos@, ) (4.40)

Now, since g, 9, and v, are constants, Eq. (4.40)ts of the form y = a.x +b, in which aand b are constants. This is the equation of a parabola,f.e. the path of the projectile is a parabola (Fig. 4.18).

Time of maximum height

How much time does the projectile take to reach the maximum height ? Let this time be denoted by ¢,. Since at this point, v.= 0, we have from Eq. (4.39):ov, =v, siné,— gt, =0

Or, t,=u,siné/g (4.41a)The total time T, during which the projectile is

in flight can be obtained by putting y = O in Eq. (4.38). We get:T22 {u, sin 8, )/g (4.41b)T, is known as the time of flight of the projectile.We note that T, = 2 £,, which is expected because of the symmetry of the parabolic path.Maxtmum hetght of a projectile The maximum height h, reached by the projectile can be calculated by substituting t=t, in Eq. (4.38) :yan bonnet) fsa)g 2 g

rysind, ?Or, hy, = {erosingo) (4.42)

29 Horizontal range of a projectile

The hortontal distance travelled by a projectile from fis initial position (x = y =0) to the position where it passes y = O during {ts fall is called the

horizontal range, R. It is the distance travelled during the time of flight T, . Therefore, the range Ris R =(v, cos 6) (7)=(u, cos 8) (2 uv, sin 6)/g

. ve sin 20, Or, R= 97 (4.432)

Equation (4.43a) shows that for a given

projection velocity v,, Ris maximum when ain 26, ia maximum, 1.c., when 6, = 45°.

The maximum horizontal range is, therefore,. «2 Ry = g (4.43b)

Exampie 4.7 Galileo, in his book Two new

sciences, stated that “for elevations which exceed or fall short of 45° by equal amounts, the ranges are equal". Prove this statement.Answer For a projectile launched with velocity

v, at an angle @, , the range fs given by R= v2 sin2@ g Now, for anglea, (45° + a) and { 45°- a, 20, ia (90° + 2a) and (90° - 2a) . respectively. The

values of sin (90° + 2a) and sin (G0° - 2a) are the same, equal to that of cos 20 Therefore,ranges are equal for elevations which exceed or fall short of 45° by equal amounts a. <>» Example 4.8 A hiker stands on the edge of a cHff 490 m above the ground and throws a stone horizontally with an initial speed of 15m s!. Neglecting air resistance,

find the time taken by the stone to reach the ground, and the apeed with which it hits the ground. (Take g= 9.8m 8%).

Answer We choose the origin of the x-,and y-axis at the edge of the cliff and t= 0 a at the instant the stone is thrown. Choose the positive direction of x-axis to be along the initial velocity

and the positive direction of y-axis to be the vertically upward direction. The x-, and y-components of the motion can be treated independently. The equations of motion are :x} =x, +u,,t uf) =y,t+u, t+1/2 a?Here, x, = y, = 0, v,,=0, a,=-g =-9.8m 8%,o,=15ms".The stone hits the ground when y(t) = — 490 m.-490 m =-{1/2)(9.8) #.This gives t=10s8.

The velocity components are v, =v, and

vj=o,-gt so that when the atone hits the ground :v,.= 15ms7? vu, =0-9.8x 10=-98ms"Therefore, the speed of the stone is fel +02 =V15° +98" =99 ms? <

Example 4.9 A cricket ball is thrown at a speed of 28 m s* ina direction 30° above the horizontal. Calculate (a) the maximum height, fb) the time taken by the ball to return to the same level, and (c) the distance from the thrower to the point where the ball returns to the same vel.Answer (a} The maximum height is given by (ve, sine, (28sin 30°)

Mm "3g 8)14x14 “3x9.5" 10.0 m

(b) The time taken to return to the same level is T= (@ v, sin @,)/g = 2x 28 x sin 30°)/9.8 =28/9.85=2.98(c) The distance from the thrower to the point

where the ball returns to the same level is R= (vesin 28, ) _ 28x 28x sin 60" -69m g 9.8 <

Neglecting air resiatance - what does

the assimmption really mean?

Whike treating the topic of projectile motion,we have stated that we assume that the air resistance has no effect on the motion of the projectile. You must understand what the statement really means. Friction, force due to viscosity, air resistance are all dissipative forces. In the presence of any of

such forces opposing motion, any object will lose some part of its initial energy and consequently, momentum too. Thus, a projectile that traverses a parabolic path would certainly show deviation from its idealised trajectory in the presence of air resistance. It will not hit the ground with

the same speed with which it was projected from it. In the absence of air resistance, the x-component of the velocity remains constant and it is only the y-component that undergoes a continuous change. However,in the presence of air resistance, both of

these would get affected. That would mean that the range would be less than the one given by Eq. (4.43). Maximum height attained would also be less than that predicted by Eq. (4.42). Can you then,anticipate the change in the time of flight?In order to avoid air resistance, we will have to perform the experiment tn vacuum or under low pressure, which 1s not easy.When we use a phrase like ‘neglect air resistance’, we imply that the change in parameters such as range, height etc. is much smaller than their values without after resistance. The calculation without air

resistance {8 much simpler than that with air resistance.

4.11 UNIFORM CIRCULAR MOTION

When an object follows a circular path at a constant speed, the motion of the object is called uniform circular motion. The word “uniform” Tefera to the speed, which is uniform (constant}

throughout the motion. Suppose an object is moving with uniform speed v ina circle of radius Ras shown in Fig. 4.19. Since the velocity of the object is changing continuously in direction, the

object undergoes acceleration. Let us find the magnitude and the direction of this acceleration.

Let r and r’ be the postition vectors and wand¥ the velocities of the object when tt is at point P and as shown in Fig. 4.19{a). By definition,velocity at a point is along the tangent at that

point in the direction of motion. The velocity vectors v and w’ are as shown in Fig. 4.19fa1).Avis obtained tn Fig. 4.19 (22) using the trianglelaw of vector addition. Since the path is circular,v is perpendicular to r and so is w’ to r’.Therefore, Av is perpendicular to Ar. Since average acceleration is along A¥ (-*"} the

average acceleration a is perpendicular to Ar. If we place Av on the line that bisects the angle between r and r’, we see that tt is directed towards

the centre of the circle. Figure 4.19{(b) shows the same quanitities for smaller time tnterval. Aw and

hence 4 1s again directed towarda the centre.In Fig. 4.19(c), At-~O and the average acceleration becomes the instantaneous acceleration, It is directed towards the centre*.Thus, we find that the acceleration of an object

in uniform circular motion is always directed.towards the centre of the circle. Let us now find the magnitude of the acceleration.

The magnitude ofa is, by definition, given by

lal = |Av|

m= As 0 at

Let the angle between position vectors r and

¥ be A@. Since the velocity vectors v and w are always perpendicular to the position vectors, the angle between them is also A@ . Therefore, the triangle CPP’ formed by the position vectors and

the triangle GHI formed by the velocity vectors v. v and Av are similar (Fig. 4.19a). Therefore,the ratio of the base-length to side-length for

one of the triangles is equal to that of the other triangle. That ta :

Jav| | Ar|

Vv - R

| |ar|

Or, |Av| = e—

Therefore,lal- avi eae] oe |x|

= ft 9 Al = Ro RAt Ral o Al

If At is small, A6 will also be small and then arc PP’ can be approximately taken to be | Ar|:

[Arle vat

larly

tr lal

On AlsoAL

Therefore, the centripetal acceleration a, is :(v a, = [gJe=wve (4.44)

Thus, the acceleration of an object moving with speed v in a circle of radius R has a magnitude v /Rand is always directed towards the centre.

This is why this acceleration is called centripetal acceleration (a term proposed by Newton). A thorough analysis of centripetal acceleration was

first published in 1673 by the Dutch scientist Christiaan Huygens (1629-1695) but it was probably known to Newton alao some years earlier.“Centripetal” comes from a Greek term which means ‘centre-seeking’. Since v and R are constant, the magnitude of the centripetal acceleration is also constant. However, the direction changes —pointing always towards the centre. Therefore, a

centripetal acceleration is not a constant vector.We have another way of describing the velocity and the acceleration of an object in uniform circular motion. As the object moves

from P to FP’ in time At (= ¢ - 4, the line CP (Fig. 4.19) turns through an angle A@ as shown in the figure. Aé is called angular distance. We define the angular speed @ (Greek letter omega)

as the time rate of change of angular

displacement ;@= 28 [4.45 At (4.45)

Now, if the distance travelled by the object during the time At is As, i.e. PP’is As, then :

pe AS

but As = RAO. Therefore :

= R—=

v At Ro

v= Rae (4.46)

We can express centripetal acceleration @, in terms of angular speed :

Fr wR? 2

a, =—=——=@'R

R R a, =o'R (4.47)

The time taken by an object to make one revolution.is known as fts time period Tand the number of revolution made in one second is called ita frequency v (=1/T). However, during this time the

distance moved by the object is s = 2nR.

Therefore, v = 2nR/T=2nRv (4.48)

In terms of frequency v, we have

@=2nv

v=2nkv

@,= 4° VR (4.49)

> Example 4.10 An insect trapped in a

circular groove of radius 12 cm moves along the groove steadily and completes 7 revolutions in 100 s. (a) What is the

angular speed, and the Hnear speed of the motion? (b) Is the acceleration vector a constant vector ? What is its magnitude ?

Answer This is an example of uniform circular motion. Here R= 12 cm. The angular speed o is given by

= 2n/T= 2nx 7/100 = 0.44 rad/s

The linear speed v is :

v=o R=0.448'x 12cm= 5.3cms'!

The direction of velocity vis along the tangent to the circle at every point. The acceleration is directed towards the centre of the circle. Since this direction changes continuously,

acceleration here ia not a constant vector.However, the magnitude of acceleration is constant:a= o R= (0.44 9")? (12 cm)= 2.3 cm 8? <

SUMMARY

1. Scalar quantities are quantities with magnitudes only. Examples are distance, speed,masa and temperature.

2. Vector quantities are quantities with magnitude and direction both. Examples are displacement, velocity and acceleration. They obey special rules of vector algebra.

3. Avector Amultiplied by a real number A 1s also a vector, whose magnitude is A times the magnitude of the vector Aand whose direction is the same or opposite depending upon whether 1 is positive or negative.

4, Two vectors Aand B may be added graphically using head-to-tail method or parallelogram method,

5. = Vector addition is commutative :

A+B=B+A It also obeys the associative taw :(A+B +CuA+ B+

6. Anal or zero vector is a vector with zero magnitude. Since the magnitude Is zero, we don't have to specify its direction. It has the properties :

AtOnA

A0=0

0A=0

7. The subtraction of vector B from A is defined as the sum of A and -B :

A-B = A+ (-B)

8. Avector Acan be resolved into component along two given vectors a and b lying in the same plane :Az=Aa+yub where A.and pt are real numbers.

9. A-unit vector associated with a vector A has magnitude one and is along the vector A:

. A

i=—

|4|

The unit vectora i, j, _k are vectors of unit magnitude and point in the direction of the x, y-, and zaxes, respectively in a right-handed coordinate system.

10. A-vector A can be expressed as

A=A,i+ Ajj where A,, A, are its componente along x-. and y -axes. If vector A makes an angle 6 , — A,

With the x-axis, then A,= A cos 6, A=Asin 6 and A=|Al= fal + 45. tan@ ==.

1l. Vectora can be conveniently added using analytical method. If aum of two vectors A and B, that lie in xy plane, is R, then :R=R,i+R,j. where, ReA+B,and R«A,+B,12. The position. vector of an object in xy plane is given by 5 = xit+ uj and the displacement from position r to position r’ ts given by 4rer-r

=(- x) it(y’—y) j = Axi+Ayj

13. If an object undergoes a displacement Ar in time At its average velocity is given by ‘Ar Va,” The veloctty of an object at time tie the limiting valuc of the average velocity

as Attends to zero :

Ar dr

v= Um —=—., It can be written in unit vector notation as :

At—>Oar dt . dx d de

. : : < L vec dte,j+e,k where Uy Grey = Geet, =e When position of an object is plotted on a coordinate system, v ia always tangent to the curve representing the path of the object.14. If the velocity of an object changes from v to w/in time At, then its average acceleration _ vev' Av we aven ty: BA ae

The acceleration a at any time tis the limiting value of a as At—0:

lim Av dv

a= —_—=—

At>O0ar dt

In component form, we have : a= c,i+a,j+a,k , dv, dey dv,where, 4,.= at tu ar at

15. If an object ‘s moving tn a plane with constant acceleration a=|al= fae + a3 and ite position vector at time {= 0 fa r,, then at any other time ¢, it will be at a point given by:

r=4r,+V,f+ lat

oO Oo 2

and its velocity is given by :vev tat where v, is the velocity at time t= 0 In component form :1 2

X=X%, +0, 04 2a

o TPal to Ge

. ; 1 2

yay, +d,, foe

Uy Bly, $ayt

by = Coy tat

Motion in a plane can be treated as superpasition of two separate simultaneous one-dimenstonal mations along two perpendicular directions

16. An object that ts in flight after being projected 1s called a projectile. If an object is projected with initial velocity ¥, making an angle 0, with x-axis and if we assume its initial position to coincide with the origin of the coordinate system, then the position

and velocity of the projectile at time ¢ are given by :

x= (v,coa a)t

y = (vu, sin @) £- (1/2) g &

v,= v= 0, cos 6,

v, =v, ain 6,-gt

The path of a projectile is parabolic and is gtven by :y =(tan@,)x-—__“&___.

2(v, cos 6, )

The maximum height that a projectile attaina is :h, = (uv, sine, Y

The time taken to reach thie height ia :

tote sing,,

t, =

The horizontal distance travelled by a projectile from its initial poaition to the poaition it passes y = O during its fall is called the range, R of the projectile. It is:R= 2sin2e,g

17. When an object follows a circular path at constant speed, the motian of the object ia called uniform circular motion. The magnitude of its acceleration ia a, = v? /R. The

direction of a, is alwaya towards the centre of the circle.The angular speed «, is the rate of change of angular distance. It is related to velocity

uby v=@R. The acceleration ia a, = @°R.

If T ia the time period of revolution of the object in circular motion and y 1s its frequency, we have = 2nv, v= 2nvR, a,= 40° °R

POINTS TO PONDER

1. The path length traversed by an object between two points is, in gencral, not the same as the magnitude of displacement. The displacement depends only on the end points; the

path length (as the name implies) depends on the actual path. The two quantities are equal only if the object does not change its direction during the course of motion. In all other cases, the path length is greater than the magnitude of displacement.

2. In view of point 1 above, the avernge speed of an object is greater than or equal to the magnitude of the average velocity over a given time interval. The two are equal only if the path length is equal to the magnitude of displacement.

3. The vector equations (4.33a) and (4.34) do not involve any choice of axes. Of courage,you can always resolve them along any two independent axea.

4. The kinematic equatiana for uniform acceleration do not apply to the case of uniform circular motion aince in this case the magnitude of acceleration ia constant but ita direction is changing.

5. An object subjected to two velocitics v, and v, has a resultant velocity vw = v, + v,. Take care to distinguish it from velocity of object 1 relative to velocity of object 2 : v,,= v, — ¥,.

Here v, and v, are velocities with reference to some common reference frame.

6. The resultant acceleration of an object in circular motion is towards the centre only if the speed is constant.

7. The shape of the trajectory of the motion of an object is not determined by the acceleration alone but also depends on the intftial conditions of motion ( initial position and mitial velocity). For example, the trajectory of an object moving under the aame acceleration

due to gravity can be a straight line or a parabola depending on the imitial conditiona.

EXERCISES

4.1 State, for each of the following physical quantities, if it ie a scalar or a vector :volume, mass, speed, acceleration, density, number of moles, velocity, angular frequency, displacement, angular velocity.

4.2 Pick out the two scalar quantities in the following list :force, angular momentum, work, current, linear momentum, electric field, average

velocity, magnetic moment, relative velocity.

4.3 Pick out the only vector quantity in the following list :

Temperature, pressure, impulse, time, power, total path length, energy, gravitational potential, cocfiicient of friction, charge.

4.4 State with reasons, whether the following algebraic operations with scalar and vector phyaical quantities are meaningful :

(a) adding any two scalars, (b) adding a scalar to a vector of the same dimensions ,(9 multiplying any vector by any scalar, (d) multiplying any two scalars, (ec) adding any two vectors, (9 adding a component of a vector to the same vector.

4.5 Read each statement below carefully and state with reasons, if it is true or false :

(a) The magnitude of a vector is always a scalar, (b) each component of a vector is always ea scalar, (c) the total path length is always equal to the magnitude of the displacement vector of a particle. (d) the average speed of a particle (defined as total peth length divided by the time taken to cover the path) is either greater or equal to

the magnitude of average velocity of the particle over the same interval of time, (e)Three vectors not lying in a plane can never add up to give a null vector.

4.8 Establish the following vector inequalities geometrically or otherwise :(a) «slatbl < Ial + Ibi

(b) lasbl > Ilal-IbI!

(d) la-bl > Ilal—IbII

When does the equality sign above apply?

4.7 Givena+b+e+d=0, which of the folowing statements are correct :

(a) a, b, c, and d must each be a null vector,

(b) The magnitude of (a+c) equals the magnitude of (b+),

(d) b + & must lie in the plane of a and d if a and d are not collinear, and in the line of a and @, if they are

collinear ?

4.8 Three girls skating on a circular ice ground of radius 200 m start from a point P on the edge of the ground

and reach a point Q diametrically opposite to Pfollowing different paths as shown in Fig. 4.20. What is the

magnitude of the displacement vector for each ? For which girl ia this equal to the actual length of path skate?

4.9 Acyciiat starts from the centre O of a circular park of radiue 1 kon, reaches the edge P of the park, then cycles along the circumference, and returns to the centre along QO as shown in Fig. 4.21. If the round trip takes 10 min, what is the (a) net displacement,

(b} average velocity, and (c) average speed of the cyclist 7

4.10 On an open ground, a motorist follows a track that turns to his left by an angle of 60° after every 500 m. Starting from a given turn, specify the displacement of the motorist at the third, sixth and eighth turn. Compare the magnitude of the displacement with

the total path length covered by the motoriat in each case.

4.11 A passenger arriving in a new town wishes to go from the station to a hotel located 10 km away on a straight road from the station. A dishonest cabman takes him along a circuitous path 23 km long and reaches the hotel in 28 min. What is (a) the average speed of the taxi, (b) the magnitude of average velocity ? Are the two equal ?

4.12 Rain ts falling vertically with a speed of 30 m 5". A woman rides a bicycle with a speed of 10 ms" in the north to south direction. What is the direction in which she should hold her umbrella ?

4.13 A man can swim with a speed of 4.0 km/h in still water. How long does he take to crosa a river 1.0 km wide if the river flows steadily at 3.0 km/h and he makes hia

strokes normal to the river current? How far down the river does he go when he

reaches the other bank ?

4.14 In a harbour, wind is blowing at the speed of 72 km/h and the flag on the mast of a boat ++++++++++++++

.....+

.+ anchored m the harbour fiutters along the N-E direction. If the boat starts moving at a speed of 51 km/h to the north, what is the direction of the flag on the mast of the boat ?

4.16 The ceiling of a long hall is 25 m high. What is the maximum horizontal distance that a ball thrown with a speed of 40 ms" can go without hitting the ceiling of the hall ?

4.16 A cricketer can throw a ball to a maximum horizontal distance of 100 m. How much high above the ground can the cricketer throw the same ball ?

4.17 A stone tied to the end of a string 80 cm long is whirled in a horizontal circle with a constant speed. If the stone makes 14 revohitions m 25 «, what is the magnitude and direction of acceleration of the stone ?

4.18 An aircraft executes a horizontal loop of radius 1.00 km with a steady speed of 900 km/h. Compare ita centripetal acceleration with the acceleration due to gravity.

4.19 Read each statement below carefully and state, with reasons, if it is true or false :

fa) The net acceleration of a particle in circular motion is always along the radius of the circle towards the centre

(b) The velocity vector ofa particle at a point is ahways along the tangent to the path of the particle at that point

4.20 The position of a particle is given by

r= 3.0(1-2.07 j+4.0kim

where ¢is in seconds and the coefficienta have the proper unite for rto be in metres.

{a) Find the v and a of the particle? (b) What is the magnitude and direction of velocity of the particle at t= 2.0 5 ?

4.21 A particle starts from the origin at t= 0 a with a velocity of 10.0 j m/s and moves in the x-y plane with a constant acceleration of (8.01 +2.0)) m s*, (a) At what time is the x coordinate of the particle 16 m? What is the y-coordinate of the particle at that time? (b) What is the speed of the particle at the time ?

4.22 | and j are unit vectors along x- and y axia respectively. What is the magnitude and direction of the vectors i+}: and j-j ? What are the components of a vector Az 2 i+3j along the directions of i+j and i-j? [You may use graphical method]

4.23 For any arbitrary motion in apace, which of the following relations are true :

fa) Vacage = (1/2) ( (4) + ¥ (G)

{D) Yee = Inte) - (4) 1 /(Q-t)

fc) ¥(G@ =" (O)+at

{2 a ae = YG) - v0 - 6)

{The ‘average’ standa for average of the quantity over the time mterval ¢, to 4)

4.24 Read cach statement below carefully and state, with reasons and cxamples, if it is true or false :

A scalar quantity is one that

{a} is conserved in a process

{b) can never take negative values

{c) must be dimensionless

{d) does not vary from one point to another in space

{Q has the same value for observers with different orientations of axes.

4.26 An aircraft is flying at a height of 3400 m above the ground. If the angle subtended at a ground observation point by the aircraft poaitions 10.0 5 apart is 30°, what is the speed of the aircraft ?

Additional Exercises

4.26 A vector has magnitude and direction. Does ft have a location in space ? Can it vary with time ? Will two equal vectors a and b at different locations in space necessarily have identical] physical effects ? Give examples in support of your answer.

4.27 A vector has both magnitude and direction. Does it mean that anything that has Magnitude and direction is necessarily a vector ? The rotation of a bedy can be specified by the direction of the axis of rotation, and the angle of rotation about the axis. Doea

that make any rotation a vector ?

4.28 Can you associate vectors with (a) the length of a wire bent into a loop, (b) a plane area, (c} a sphere ? Explain.

4.20 A bullet fired at an angle of 30° with the horizontal hits the ground 3.0 km away. By adjusting its angle of projection, can one hope to hit a target 5.0 lan away ? Assume the muzzle speed to be fixed, and neglect air resistance.

4.30 A fighter plane flying horizontally at an altitude of 1.5 km with speed 720 km/h passes directly overhead an anti-aircraft gun. At what angle from the vertical ahould the gun be fired for the shell with muzzle speed 600 m s" to hit the plane ? At what minimum altitude should the pilot fly the plane to avoid being hit ? (Take g = 10 m 8).

4.31 A cyclist is riding with a speed of 27 km/h. As he approaches a circular turn on the road of radtue 80m, he applies brakes and reduces his speed at the constant rate of 0.50 m/a every second. What is the magnitude and direction of the net acceleration of

the cyclist on the circular turn ?

4,32 (a) Show that for a projectile the angle between the velocity and the x-axis as a function of time is given by

a= wane")

Cox

(b) Shows that the projection angle 06, for a projectile launched from the origin is given by

af 4h

= tan!

anew (Te)

where the symbole have their usual meaning.

8.2 ARISTOTLE’S FALLACY

The question posed above appears to be simple.However, it took ages to answer it. Indeed, the correct answer to this question piven by Galfleo in the seventeenth century was the foundation

of Newtonian mechanics, which signalled the birth of modern science.

The Greek thinker, Aristotle (384 B.C- 322 B.C,), heki the view that if a body is moving,something extemal is required to keep it moving.According to this view, for example, an arrow shot from a bow keeps flying since the air behind

the arrow keeps pushing it. The view was part of an elaborate framework of ideas developed by Aristotle on the motion of bodies in the universe.Most of the Aristotelian ideas on motion are now

known to be wrong and need not concern us.For our purpose here, the Aristotelian law of motion may be phrased thus: An external force

is required to keep a body in motion.

Aristotelian law of motion is awed, as we shall see. However, It is a natural view that anyone would hold from common experience. Even a small child playing with a simple (non-electric)toy-car on a floor knows intuttively that it needs

to constantly drag the string attached to the toy-car with some force to keep it gaing. If it releases the string, it comes to rest. This experience is

common to most terrestrial motion. External forces seem to be needed to keep bodies in motion. Left to themselves, all bodies eventually

come to rest.

What is the flaw in Aristotle's argument? The answer is: a moving toy car comes to rest because the external force of friction on the car by the floor opposes its motion. To counter this force, the child has to apply an external force on the car in the

direction of motion. When the car is in uniform motion, there is no net external force acting on tft:the force by the child cancels the force { friction)

by the floor. The corollary is: tithere were no friction,the child would not be required to apply any force to keep the toy car in uniform motion.

The opposing forces such as friction {solids)and viscous forces {for flukis) are always present in the natural workd. This explains why forces by external agencies are necessary to overcame

the frictional forces to keep bodies in untform motion. Now we understand where Aristotle went wrong. He coded this practical experience in the form of a basic argument. To get at the

Plus One

Wednesday 3 February 2021

Chapter 4 MotioN In A Plane

Chapter 4 MotioN In A Plane

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