4.0 Preview

In this chapter, you will learn about some of the salient features of Scalars and Vectors. In addition, you will learn how to describe the position and displacement of objects with respect to the origin as position and displacement vectors respectively.

Next about Equality of vectors; two vectors are said to be equal if they have the same magnitude and are oriented exactly in the same direction. You will also learn how to multiply, add and subtract vectors by graphical method; Triangle law of vector addition as one of the graphical methods where you find the resultant by adding the vectors (arranging them in a head-to-tail order). Similarly, you will learn about subtraction of vectors by the same method, then by parallelogram law of vector addition and learn to apply these laws in problems.

You will learn how to resolve vectors into its components, then about properties of a unit vector, addition of vectors by analytical method, average velocity (graphically, the tangent at any point on the circle intersected by the position vector gives the angular velocity at that point), instantaneous velocity by limiting process (as we did in the previous chapter), acceleration (as a change in velocity to time interval) and average acceleration (again by calculus, you will be asked to arrive at it through differentiation).

Further, you will find it very interesting to learn about Projectile Motion (as motion of an object thrown in a direction other than vertical, at some angle to the horizontal) and obtain equations for various aspects associated with projectile motion like, expressions for horizontal motion, equation for trajectory of the projectile, time of flight (the total time taken for the entire journey) of the projectile (the object in air thrown at some angle to the horizontal under projectile motion), time taken for a projectile to reach maximum height, Range (the total horizontal distance traversed by the projectile) and for the maximum height reached by the projectile during its traverse.

You would be deriving equations for an object moving with constant acceleration. Then, you will learn how to establish a relation between linear and angular velocity (Linear Velocity = radius x angular velocity).

4.1 Introduction:

In the last chapter we learnt about motion of an object along a straight line. As you may recollect, directional aspects of these quantities could be taken care of with +ve and –ve signs because in one dimension, only two directions are possible. But in order to describe motion of an object in two dimensions (as you would for objects in a plane) or three dimensions (for objects in space), we need to use vectors.

In this chapter, you will learn how to use vectors since we are dealing with motion of objects in a plane or in two dimensions. You will learn several operations like, adding, subtracting and multiplying vectors. We shall learn these to use vectors for defining velocity and acceleration in a plane.

As a simple case of motion in a plane, we shall discuss motion with constant acceleration and learn in detail, the projectile motion. Circular motion is one that would be familiar to you as you would experience 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 as well.

4.2 Scalars and Vectors

As you are already familiar with scalars and vectors, you must know that they can be used to classify quantities. Let us now list the differences between the two:

• Direction is associated with a vector but not with a scalar.
• A scalar quantity is a quantity with magnitude only while a vector quantity has both magnitude and direction.
• A scalar quantity is always positive or zero, whereas, a vector can be positive, negative or zero.
• The rules for combining scalars are the rules of ordinary algebra. Thus, they can be added, subtracted, multiplied or divided.
• Distance between two points is a scalar quantity because we disregard the directional aspect while displacement between two points would be a vector quantity since direction comes into play, mass and weight of an object would obviously be scalar as they are independent of direction of the body in motion. Similarly, temperature of an object and time at which an event occurred are also scalars. It follows that derivatives of the above parameters like, perimeter of a rectangle, difference in temperatures, volume (which is product of the three dimensions of an object for a rectangular solid), density (which is defined by ratio of mass to volume of an object) are also all scalars.
• A vector quantity is a quantity that has both a magnitude and a direction. Thus, a vector is represented by a number (i.e, magnitude) and its direction. It obeys the Triangle law of addition or the Parallelogram law of addition (about which you will study in detail in later sessions) and cannot simply be added algebraically like scalars. Displacement, acceleration and velocity are all vectors.

Concept box 1:   How do you denote a vector mathematically? We will be representing vectors using bold phases (like, v) in the sessions that will follow. This bold phase is as good as writing the number with an arrow—always pointing towards right, on its head! Thus, v and $ v ⃗$are both same. The magnitude of a vector is called its absolute value and is represented as: |v|= v. Therefore, for vectors, A, a, p, q, r, … x, y, the respective magnitudes could be denoted by lightface A, a, p, q, r, … x, y.

Graphical representation of a vector:

• A line segment with an arrow head can be used to denote a vector graphically.
• The length of the line segment is proportionate to the magnitude of vector.
• The direction of line segment is made to coincide (or represent) the direction of vector.
• The arrow-head of the segment is called tip or head and the other end is called the tail.

Note box 1:   Absolute value of any quantity is mathematically “0” or positive and never negative because it is a measure of distance on the number line from “0” to any specified point.

4.2.1 Position and Displacement Vectors

Concept box 2:   Position vector is a term used to describe a vector which represents the position of an object with respect to the origin. The direction and magnitude of this arrow represents the direction and magnitude of displacement of the object in motion, in a plane.

Now, try to use this concept to understand how to represent position vector for an object moving in a plane.

• Look at figure, 4.1(a); we choose a convenient point, say O as origin. P and $P′ $ represent the positions of the object at time t and $ t′ $, respectively. We join O and P by a straight line. Then, OP becomes the position vector of the object at time t. An arrow is marked at the head of this line to represent the orientation of the object at P with respect to the origin. As you can see, “r” is the measure of this vector. Thus, it represents the magnitude of the vector, OP.
• When the object moves from P to $P′ $ , point P′ is represented by another position vector, $ OP′ $ denoted by $ r’ $. The length of the vector $ r’ $ represents the magnitude of the vector and its direction is the direction in which P’ lies as seen from O.
• The vector $PP′ $ (with tail at P and tip at $P′ $ ) is called the displacement vector corresponding to motion from point P (at time t) to point $P′ $ (at time $ t′ $).

To get a clear picture of displacement vector, refer figure 4.1(b);

• 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.
• As you can see in fig 4.1b, the initial and final positions are P and Q. The displacement vector is the same, that is, PQ for any path traversed, say for PABCQ, PDQ or PBEFQ.
• It would be right to conclude that, magnitude of displacement is either less than or equal to the path length of an object between two points.

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.

• Look at figure, 4.2(a), the two vectors A and B are equal. We can check their equality by shifting B parallel to itself, until its tail Q coincides with the tail of A, i.e. Q coincides with O. When we do this, we find that their tips S and P also coincide. Since, the arrow heads also point towards same direction, we can say that A = B.
• In Fig. 4.2(b), vectors $A′$ and $ B′ $ have the same magnitude but they are not equal because they clearly have different directions. Even if we shift $ B′ $ parallel to itself until Q′ coincides with $ O′ $, the tip $ S′ $of $ B′ $ does not coincide with the tip $P′ $ of A′. Thus, $ A’≠B’ $.

Questions from 4.2 and 4.2.1:

1. List the differences between Scalars and Vectors.
2. Describe Position Vector with graphical representation.
3. When do you say two vectors are equal?

4.3 Multiplication of Vectors by Real Numbers

Figure 4.3 (a) Vector A and the resultant vector after multiplying A by a positive number 2. (b) Vector A and resultant vectors after multiplying it by a negative number -1 and -1.5

• Multiplying a vector A with a positive number, λ gives a vector whose magnitude is changed by the multiplication factor. However, the direction is the same as that of A as indicated below:
  $$|λA|=λ|A|, if λ>0.$$

For example, if A is multiplied by 2, the resultant vector is 2A. The direction remains same as A. The magnitude becomes twice of $ |A|$ as shown in Fig. 4.3(a).

• Multiplying a vector A by a negative number, λ gives a vector λA whose direction is opposite to the direction of A and whose magnitude is –λ. $ |A|$ as indicated below:

For example, if a given vector A is multiplied by negative numbers, say -1 and -1.5, it gives vectors –A and -1.5A as shown in Fig 4.3(b). As you may already know, -ve sign here only signifies an opposite direction as that of the original vector.

• The factor λ by which a vector A is multiplied could be a scalar having its own physical dimension. Then, the dimension of λ A is the product of the dimensions of λ and A. This is because a scalar quantity just has a magnitude, which is just a number.

For example, if we multiply a constant velocity vector by duration (of time), we get a displacement vector.

4.4 Addition and Subtraction of Vectors by Graphical Method

As mentioned in section 4.2, vectors, by definition, obey the triangle law or equivalently, the parallelogram law of addition.

Figure 3.4 (a) Vectors A and B (b) Vectors A and B added graphically (c) Vectors B and A added graphically (d) Illustrating the associate law of vector addition

• We shall now describe this law of addition using the graphical method. Let us consider two vectors A and 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.
• Referring figure, 4.4(a&b), to find the sum $ A + B$, we place vector B so that its tail is at the head of the vector A. Then, we join the tail of A to the head of B. This line OQ 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 is called the head-to-tail method. Since, the two vectors and their resultant form three sides of a triangle, this method is also known as triangle method of vector addition.
• If we find the resultant of $ B + A $ as in Fig. 4.4(c), by changing the arrangement of vectors, the same vector R is obtained. Thus, vector addition is cumulative, that is:
  $$ A + B = B + A \space 4.1 $$
• 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 as expressed below:
  $$ (A + B) + C = A + (B + C) \space 4.2 $$
• The result of adding two vectors, equal in magnitude and opposite in direction would be 0. That is to say, the magnitude of the resultant vector is 0; as you will understand from the expression below:
  $ A + (-A) = 0 $ (or) $ |0|= 0 \space 4.3 $
The 0 got here is called a null vector or a zero vector.

Properties associated with a null vector:

Apart from the case obtained in equation, 4.3, The Null vector also results when we multiply a vector A by the number zero, that is,

$$ 0.A = 0 \space 4.3$$

It thus follows that,

$$ A + 0 = A $$
$$ λ.0 = 0$$

Under what scenario do you think we physically obtain a zero vector?

Consider the position and displacement vectors in a plane as shown in Fig. 4.1(a). Now suppose that an object which is at “P” at time “t”, moves to “$ P’ $ “ and then comes back to P. Then, its displacement would be 0 because the initial and final positions coincide. Thus, the displacement would be a null vector.

Subtraction of vectors

It 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 A and –B as shown below,
$$ A – B = A + (-B) \space 4.5

Figure 4.5 (a) Two vectors A and B, – B is also shown (b) Subtracting vector B from vector A – the result is $R_s$. For comparison, addition of vectors A and B i.e $R_i$ is also shown

• From figure, 4.5(a and b), using triangle law of addition, the vector -B is added to vector A to get $ R_2 = (A-B) $ .The vector $ R_1= (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, QS and PS to the origin O. This would be the diagonal (OS) of the parallelogram as you can see in Fig. 4.6(b). The diagonal serves as the resultant vector R.
• In Fig.4.6(c), the triangle law is used to obtain the resultant of A and B and we see that the two methods yield the same result. Thus, triangle law and parallelogram law of addition are equivalent.

Figure 4.6 (a) Two vectors A and B with their tails brought to a common orgin (b) The sum A+B obtained using the parallelogram method (c) The parallelogram method of vector addition is equivalent to the triangle method

Now that you have learnt how to use the two laws, parallelogram and triangle laws, let us try to define them.

Definition box 1:   1. Triangle law of addition of vectors: Triangle law of vector addition states that if two vectors are represented by two sides of a triangle (in direction and in magnitude), then the resultant vector is represented by the third side of the triangle taken in opposite direction (in direction and in magnitude).   2. Parallelogram law of addition of vectors: Parallelogram law of vector addition states that if two vectors are considered to be the adjacent sides of a Parallelogram, then the resultant of two vectors is given by the vector which is the diagonal passing through the point of contact of two vectors* of the completed parallelogram.   *The “two vectors” here refer to the reflection of the original vectors used to complete the parallelogram. This is just for your understanding and need not become a part of the definition.

Example 4.1: Rain is falling vertically with a speed of 35 m $s^{–1}$. Winds starts blowing after sometime with a speed of 12 m $s^{–1}$ 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 vectors $v_r$ and $v_w$ in Fig. 4.7 and are in the direction specified by the problem. Using the rule of vector addition, we see that the resultant of $v_r$ and $v_w$ is R as shown in the figure. The magnitude of R is

$$ R = \sqrt { v_r^2 + v_w^2 } = \sqrt { 35^2+12^2 } \space m s^{-1} = 37 \space m \space s^{-1} $$

The direction θ that R makes with the vertical is given by $ tan \pi = \frac {v_w}{v_r} = \frac {12}{35} = 0.343 $

Or, $ θ=tan^{-1} (0.343 )= 19^o $

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

Questions from section 4.5:

1. Define Triangle law of addition of vectors.
2. Define Parallelogram law of addition of vectors.

4.5 Resolution of Vectors

When you say you are resolving a vector, it means that you are expressing the vector as a sum of two vectors.

Consider vectors, “a” and “b”— two non-zero vectors in a plane with different directions. Let A be another vector in the same plane (Fig. 4.8).
Since the three vectors, a, b and A all lie in the same plane, A can be expressed as a sum of the two vectors with a multiplication factor as follows:

• One of the vectors is obtained by multiplying “a” by a real number and the other is obtained by multiplying “b” by another real number. As you can see in figure 4.8, if “O” and “P” are the tail and head of the vector, A. Then, through O, draw a straight line parallel to a, and through P, a straight line parallel to b. Let them intersect at Q. Then, we have

$$ A = OP = OQ + QP \space 4.6$$

Figure 4.8 (a) Two non-colinear vectors a and b (b) Resolving a vector A in terms of vectors a and b

• Since, OQ is parallel to a, and QP is parallel to b, we can write :

OQ = λ. a and QP = μ. B 4.7

Where, λ and μ are real numbers.

Therefore, A = λ.a + μ. 4.8

• We say that A has been resolved into two component vectors λ.a and µ.b along a and b respectively.
• 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 as we will discuss now.

Definition box 2: Unit vectors: A unit vector is a vector of unit magnitude and points in a particular direction.

Properties of unit vector:

• It has neither dimension nor unit.
• It is used to specify a direction only.
• Unit vectors along the x, y and z – axes of a rectangular coordinate system are denoted by, $\hat{i}$, $\hat{j}$ and $\hat{k}$ (read as i-cap, j-cap, k-cap) respectively, as shown in Fig. 4.9(a) below.
• Since they are unit vectors, their magnitude is 1 as expressed below:
  $| i ̂ | = | j ̂ | = | (k ) ̂|=1 $ (4.9)
• These unit vectors are perpendicular to each other.
• A cap (^) is used 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 n by a scalar, λ the result is a vector, λ= λ n. (notice how the scalar, λ changes into a vector upon multiplication with a vector, n to give λ) In general, a vector A can be written as:
  A = |A|.n $$ A = |A|.n $$ (4.10)
  Where, n is a unit vector along A.

Coming to the resolution part,

• We resolve a vector A in terms of component vectors that lie along two unit vectors (because we are dealing with motion in two dimensions), $\hat{i}$ and $\hat{j}$
• Consider a vector A that lies in x-y 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_1 $ and $ A_2 $  such that A1 + A2 = A. Since $ A_1 $ is parallel to $\hat{i}$ and $ A_2 $ is parallel to $\hat{i}$, we have:

$$ A_1= A_x i ̂,  A_2 = A_y j ̂ \space 4.11 $$ 

Where, $A_x$  and $A_y$  are real numbers.

Thus, $$ A = A_x i ̂+ A_y j ̂  \space 4.12 $$,as it was got in the last section in the form of equation, 4.8.

• This is represented in Fig. 4.9(c). The quantities $A_x$   and $A_y$ are called x and y components of the vector A. $A_x$   itself is not a vector, but $A_xi ̂ $  is a vector, since $\hat{i}$ is a vector; and so is $ A_yj ̂$   for the same reason.
• Using simple trigonometry, we can express $A_x$  and $A_y$  in terms of the magnitude of A and the angle, θ it makes with the x-axis:
  $$ A_x  = A sin θ \space 4.13$$         

From Eq. (4.13), you can say that, a component of a vector can be positive, negative or zero depending on the value of θ.

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

(i)  Its magnitude A and the direction θ it makes with the x-axis; or

(ii) Its components $A_x$   and and $A_y$  .

If A and θ are given, $A_x$    and and $A_y$  can be obtained using Eq. (4.13). If $A_x$   and  $A_y$  are given, A and θ can be obtained as follows :

From equation, 4.13:

$$A_x^2+A_y^2=A^2  cos^2⁡ \big(θ \big) A^2  sin^2⁡θ =A^2 \space 4.14$$ 

And, $ tanθ= \frac{A_y}{A_x} $$ ,     $$ θ=tan^-1 \frac{A_y}{A_x} \space 4.15$

So far we have considered a vector lying in an x-y 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 α, β, and γ are the angles between A and the x, y, and z-axes, respectively as shown in Fig. 4.9(d), we have,

$$ A_x =A \space cosα, A_y =A \space cosβ, A_z=A \space cosγ $$

In general, we have $ A=A_xi ̂ +A_yj ̂ +A_zk ̂   $

The magnitude of vector A is $ A= \sqrt {A^2_x+A^2_y+A^2_z}  \space 4.16b$

A position vector r can be expressed as $ r=x i ̂ +yj ̂  +z k ̂ $ where x, y, and z are the components of r along x-, y-, z-axes, respectively

Questions from section 4.4 and 4.5:

1. Define unit vector, list its properties.

2. Obtain equation for a vector, A; in the form, $ A = A_x^2 + A_y^2 $

4.6 Vector Addition-Analytical Method

• In analytical method of vector addition, we simply combine the individual components as sum of equations of these components.
• Though the graphical method of adding vectors helps us in visualising the vectors and the resultant vector, it is sometimes tedious and has limited accuracy.
• Thus, it is much easier to add vectors by combining their respective components.

Consider two vectors A and B in x-y plane with components $A_x$, $A_y$ and $B_x$, $B_y$:

$$ A = A_x i ̂+ A_y j ̂  $$       

$$ B= B_x i ̂+ B_y j ̂  $$   

Let R be their sum,we have R = A + B   

= $ \big( A_x i ̂+ A_y j ̂  \big) + \big( B_x i ̂+ B_y j ̂  \big)$  

Since vectors obey the commutative and associative laws, we can arrange and regroup the vectors in Eq. (4.19a) as convenient to us:            

$$ R=(A_x+B_x)i ̂ +(A_y+B_y)j ̂ $$

Since, $$ R=R_xi ̂ +R_yj ̂  $$

We have, $ R_x=A_x+B_x , R_y=A_y+B_y$

Thus, each component of the resultant vector R is the sum of the corresponding components of A and B.

In three dimensions, we have $ A=A_xi ̂ +A_yj ̂ +A_zk ̂$ 

$$ B=B_xi ̂ +B_yj ̂ +B_zk ̂  $$

$$ R=A+B=R_xi ̂ +R_yj ̂ +R_zk ̂  $$

With $ R_x=A_x+B_x $

$$ R_y=A_y+B_y $$

$$ R_z=A_z+B_z  $$

This method can be extended to addition and subtraction of any number of vectors. For example, if vectors a, b and c are given as

$$ a=a_xi ̂ +a_yj ̂ +a_zk ̂  $$

$$ b=b_xi ̂ +b_yj ̂ +b_zk ̂   $$

$$ c=c_xi ̂ +c_yj ̂ +c_zk ̂    $$

then, a vector T = a + b – c has components:

$$T_x =a_x+b_x-c_x$$

$$T_y=a_y+b_y−c_y $$

$$T_z=a_z+b_z−c_z$$

Example 4.2 Find the magnitude and direction of the resultant of two vectors A And B in terms of their magnitudes and angle θ between them.     

                                    

Answer: Let OP and OQ represent the two vectors A and B making an angle θ (Fig. 4.10). Then, using the parallelogram method of vector addition, OS represents the resultant vector R :

R = A + B

SN is normal to OP and PM is normal to OS.

From the geometry of the figure,

$$ OS^2 = ON^2 + SN^2$$

but ON = OP + PN = A + B cos θ

SN = B sin θ

$$ OS^2 = (A + B \space cos \space θ)^2 + (B \space sin \space θ)^2 $$

or, $$ R^2 = A^2 + B^2 + 2AB \space cos \space θ $$

$$ R= \sqrt {A^2+B^2+2AB \space cos \space θ} $$            (4.24a)

In Δ OSN, SN = OS sinα = R sinα, and in Δ PSN, SN = PS sin θ = B sin θ,

Therefore, R sin α = B sin θ

Or, $$ \frac{R}{sin \space θ}= \frac {B}{ sin \space α} $$     (4.24b)

Similarly,

PM = A sin α = B sin β

Or, $$ \frac {A}{sin \space β }= \frac{B}{ sin \space α}       \hspace{10mm}        (4.24c)$$      

Combining Eqs. (4.24b) and (4.24c), we get

$$ \frac{ R}{ sin \space θ}=\frac{A}{ sin \space β}=\frac {B}{ sin \space α}      \hspace{10mm}  (4.24d)$$      

Using Eq. (4.24d), we get:

$$ sin \space α = \frac{B}{R} sin \space θ  \hspace{10mm}            (4.24e)$$                                                                

where R is given by Eq. (4.24a).

or,$$ tan \space α =\frac {SN}{OP+PN}= \frac {B \space sin \space θ}{A+B \space cosθ} $$

Equation (4.24a) gives the magnitude of theresultant and Eqs. (4.24e) and (4.24f) its direction. Equation (4.24a) 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 is 10 km/h in the direction of 60° east of south. Find the resultant velocity of the boat.     

Answer: The vector $ v_b $ representing the velocity of the motorboat and the vector $ v_c $ 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:

$$ R= \sqrt {v^2_b+v^2_c+2v_bv_c cos120^o } $$

$$ R= \sqrt {25^2+10^2+2*25*10  \bigg( \frac {-1}{2} \bigg) } 2)≅ 22 km/h $$

To obtain the direction, we apply the Law of sines

$$ \frac {R}{sin \space θ} = \frac{v_c}{sin \space α} or,sin \space α =\frac {v_c}{R}sin \space  φ $$

$$ \frac{10* \sin 120^o}{21.8} = \frac{10 \sqrt 3}{2*21.8} = 0.397 $$

$$ φ ≅ 23.4^o $$

4.7 Motion in a Plane

In this section you will learn how to describe motion in two dimensions using vectors.

4.7.1 Position Vector and Displacement

The position vector, r of a particle, P located in a plane with reference to the origin of an x-y coordinate system as reference frame as shown in (Fig. 4.12), is given by: $\vec{r} = x\hat{i} + y\hat{j}$

Where, x and 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 $t’$ [Fig. 4.12(b)]. Then, the displacement is directed from P to $P’$ as represented by the blue line:

Then, the displacement is: Δr = r′ – r    4.25  and is directed from P to P’

We can write Eq. (4.25) in a component form: $ Δr =(x’i ̂ +y’j ̂ )−(x i ̂ +y j ̂ ) =i ̂ Δx+j ̂ Δy$

where Δx = x ′ – x, Δy = y′ – y    4.26      

4.7.1.1 Velocity

In the study of vectors, the average velocity $\vec{v}$ of an object is defined as the ratio of the displacement to the corresponding time interval, denoted as below:

$ \bar v =  \frac {Δr}{Δt}= \frac {Δxi ̂ +Δyj ̂ }{Δt}=   i ̂  \frac{Δx}{Δt}+j ̂\frac { Δy}{Δt} $

Or $ \bar v =\bar v_xi ̂ + \bar v_yj ̂  \space 4.27$

As you have studied in the previous chapter, by definition of calculus, the instantaneous velocity is given by the limiting value of the average velocity as the time interval approaches zero and is denoted as below:

$$ v= \lim _{Δt \to 0} \frac{Δr}{Δt} = \frac{dr}{dt} \space 4.28$$

Let us try to express this limiting process graphically as shown in Fig 4.13(a) to (d) as shown below.

In these figures, the blue line represents the path of an object, starting from “P” at time t. $P_1$, $P_2$ and $P_3$ represent the positions of the object after short time intervals, $Δt_1$, $Δt_2$, and $Δt_3$. $Δr_1$, $Δr_2$, and $Δr_3$ are the displacements of the object in time intervals, $Δt_1$, $Δt_2$, and C respectively.

The direction of the average velocity $\vec{v}$ is shown in figures (a), (b) and (c) for three decreasing values of Δt, i.e., in the order, $Δt_1$, > $Δt_2$ > $Δt_3$.

As Δt → 0, Δr → 0 and the average velocity is along the tangent to the path as shown in Fig. 4.13(d). Therefore, the direction of velocity at any point is given by tangent at that point (in the direction of motion of the object).

We will now express v in a component form as we learnt during resolution of vectors;

$$v= \frac{dr}{dt}$$

$$ = \lim_{Δt \to 0} \bigg( \frac{Δx}{Δt}i ̂+ \frac {Δy}{Δt}j ̂\bigg) \space 4.29$$

$$ = i ̂ \lim_{Δt \to 0}  \frac{Δx}{Δt}+ j ̂\lim_{Δt \to 0} \frac {Δy}{Δt}$$

Or ,$$ v = i ̂\frac {dx}{dt} +j ̂ \frac{dy}{dt} =v_xi ̂+ v_yj ̂$$

Where, $$ v_x= \frac {dx}{dt} ,v_y = \frac {dy}{dt} \space 4.30a$$

So, if the expressions for the coordinates x andy are known as functions of time, we can use

these equations to find $v_x$  and $v_y$.

The magnitude of v is then , $ v = \sqrt v^2_x+v^2_v \space 4.30 b $ and the direction of v is given by the angle θ: $ tanθ= \frac{v_y}{v_x} $,$ θ=tan^- 1 \bigg( \frac{v_y}{v_x} \bigg) \space 4.30c $

vx, vy and angle θ are shown in Fig. 4.14 for a velocity vector v.

4.7.1.2 Acceleration

The average acceleration, a of an object for a time interval Δt moving in x-y plane is the ratio of change in velocity to the time interval as denoted:

$$ \bar a = \frac{Δv}{Δt } = Δ \frac {\big(v_xi ̂+v_yj ̂\big)}{Δt}= \frac {Δv_x}{Δt}i ̂+\frac {Δv_y }{Δt} j ̂ \space 4.31a$$

Fig 4.14 The components $v_x$ and $v_y$ of velocity v and the angle $\theta$   it makes with x-axis. Note that $v_x = v \space cos \space $\theta$, v_y = v \space sin \space $\theta$

$$\bar a=a_xi ̂+a_yj ̂ \space 4.31b$$

The acceleration (instantaneous acceleration) is the limiting value of the average acceleration

as the time interval approaches zero:

$$ a =\lim_{Δt \to 0} \frac {Δv}{Δt} \space 4.32a$$

Since $$Δv=Δv_xi ̂+Δv_yj ̂$$ ,we have $ a=i ̂\lim _{Δt \to 0} \frac{Δv_x}{Δt} +j ̂\lim_{ Δt \to 0} \frac {Δv_y}{Δt}$

$$ a=a_xi ̂+a_yj ̂\space 4.32b$$

Where,$ a_x= \frac {dv_x}{dt} , a_y= \frac {dv_y}{dt} \space 4.32c$

By breaking down the equation further, you would get an equation in terms of x and y. Then, $ a_x$ and $ a_y$can be expressed as follows:

$$ a_x = \frac {d}{d} \bigg( \frac {dx}{dt} \bigg) = \frac{d^2x}{dt^2},a_y = \frac {d}{dt} \bigg (\frac {dy}{dt} \bigg) = \frac {d^2y}{dt^2}$$

• Like velocity, we can graphically express the limiting process in defining acceleration (on a graph showing the path of the object in motion) as shown in fig. 4.15(a) to (d). P represents the position of the object at time t and     $P_1$,$P_2$ ,$P_3 $ positions after time $Δt_1, Δt_2$ and $Δt_3$, respectively in decreasing order, $Δt_1> Δt_2>Δt_3$.
• The velocity vectors at points $P_1$,$P_2$,$P_3 $ and P are also shown in Figs. 4.15 (a), (b),(c) and (d) respectively. In each case of Δt (that is, for different instants of time represented by different figures), Δv is obtained using the triangle law of vector addition.
• By definition of average acceleration, the direction of average acceleration is the same as that of Δv. We see that as Δt decreases (as you may see the diminishing interval from figure, 4.15 a to d), the direction of Δv changes and consequently, the direction of the acceleration changes.

Finally, in the limiting process, as $\Delta t \to 0$, the average acceleration becomes the instantaneous acceleration and has the direction as shown in figure 4.15d.

Note box: In one dimension, the velocity and the acceleration of an object are always along the same straight line (either in the same direction or in the opposite direction). Hence, the angle formed between them is either $0^o$ or $180^o$. However, for motion in two or three dimensions, velocity and acceleration vectors may have angles anywhere between $0^o$ or $180^o$ between them.

Example 4.4 The position of a particle is given by $ r=3.0t \hat i +2.0t ^2 \hat j +5.0 \hat k $where t is in seconds and the coefficients have the proper units for r to be in metres. (a) Find v(t) and a(t) of the particle. (b) Find the magnitude and direction of v(t) at = 1.0s.            

Answer:               

$$ v \big( t \big)= \frac{dr}{dt}= \frac{d}{dt} \big( 3.0 t ˆi+2.0t ^2ˆj+5.0kˆ \big) $$

$$ = 3.0  \hat i+4.0 t \hat j $$

$$ a \big( t \big) = \frac{dv}{dt} = \big (+4.0 \big) \hat j $$

a = 4.0 m $s^{–2}$ along y- direction

At t = 1.0 s,$ v=3.0 \hat i+4.0 \hat j$

It’s magnitude is $ v = \sqrt{3^2+4^2}  =5.0 \space m \space s^{-1} $ and direction is

$$ θ = tan^−1  \bigg( \frac{v_y}{v_x} \bigg) =tan^{−1}  \bigg( \frac{4}{3} \bigg) ≅ 53°  \space with  \space x- axis $$.

4.8 Motion in a Plane with Constant Acceleration

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

Now, let the velocity of the object be $v_0$ at time, t = 0 and v at time t. Then, by definition of acceleration, we have:

$$ a= \frac {v-v_o} {t-0}= \frac {v-v_0 }{t} $$

$$v= v_0+at \space 4.33a$$

In the form of components;

$$v_x= v_{ox}+a_{xt}$$

$$v_y= v_{oy}+a_{yt} \space 4.33b$$

$$r-r_o = \bigg( \frac {v+v_o}{2} \bigg) t = \bigg( \frac { (v_o+at)+v_o}{2} \bigg)t$$

$$ = v_ot + \frac {1}{2}at^2$$

$$r =r_o +v_ot +\frac {1}{2}at^2 \space 4.34a$$

$$ x =x_o +v_{ax}t + \frac {1}{2}a_xt^2$$

$$ y=y_o +v_{oy}t + \frac {1}{2}a_yt^2 \space 4.34b$$

• Referring the above figure, let $r_o$ and r be the position vectors of the particle at time 0 and t.
• Let the velocities at these instants be $v_o$ and v. Then, over this time interval t₁, the average velocity is$ \frac {v_1 + v_o}{2}$. Since, the displacement is the average velocity multiplied by the time interval:

$$r-r_o= \bigg( \frac {v+v_o}{2} \bigg) t= \bigg( \frac{v_o+at+v_o}{2} \bigg) t $$

$$ = \bigg( \frac {2v_o+at}{2} \bigg) t= v_ot+ \frac{1}{2} at^2 $$

$$ r=r_o +v_o+ \frac{1}{2} at^2 $$

x-component of it is $ x=x_o + \big( v_{ox} \big)t+ \frac{1}{2} a_xt^2 $ and y-component is $$ y=y_o + \big( v_{oy} \big)t+ \frac{1}{2} a_yt^2 $$

As you can see, you obtain two expressions as in the study of one-dimensional motion.

Thus, motion in a plane can be treated as two, one-dimensional motions with constant acceleration along two perpendicular directions.

Example 4.5:  A particle starts from origin at t = 0 with a velocity 5.0 î m/s and moves in x-y plane under action of a force which produces a constant acceleration of $ \big( 3.0t ˆi +2.0^2ˆj  \big) m/s^2$. (a) What is the y-coordinate of the particle at the instant its x-coordinate is 84 m ? (b) What is the speed of the particle at this time?

Answer: The position of the particle is given by

$$ r \big( t \big) = v_ot+ \frac{1}{2} at^2 $$

$$ =5.0 \hat I t+(1/2)(3.0 \hat i+2.0 \hat j)t^2 $$

$$ =(5.0t+1.5t^2) \hat i+1.0t^2 \hat j  $$

Therefore, $$ x (t)=5.0t+1.5t ^2$$

$$ y (t ) = +1.0t ^2 $$

Given x (t) = 84 m, t = ?

$$ 5.0 t+1.5t^2=84⇒ t= 6 s$$

$$ At \space t = 6 s, y = 1.0 (6)^2 = 36.0 m $$

$$ Now, the \space velocity \space  v= \frac {dr}{dt} =(5.0 +3.0 t) \hat i +2.0 \hat j  $$

$$ At \space t = 6 s, v=23.0 \hat i +12.0 \hat j $$

 $$=|v| =\sqrt {23^2+12^2}≅26 m s^{−1}  $$

4.9 Relative Velocity in Two Dimensions

Suppose that two objects A and B are moving with velocities $v_A  $ and $v_B$ (each with respect to some common frame of reference, say ground.). Then, velocity of object A relative to that of B is:

$ v_{AB} = v_A -v_B    \space 4.35a $ and similarly, the velocity of object B relative to that of A is:

$$ v_{BA} = v_B -v_A   $$

Therefore, $ v_{AB} =     v_{BA} \space 4.35b $

$$ |v_{AB}|=|v_{BA}| \space 4.35c $$   

Example 4.6: Rain is falling vertically with a speed of $35ms^{-1}$. A women rides a bicycle with a speed of $12ms^{-1}$ in east to west direction. What is the direction in which she should hold her umbrella?

Answer: In Fig. 4.16 $v_r$ represents the velocity of rain and $v_b$ , 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_{rb} = $v_r$ – $v_b$ This relative velocity vector as shown in Fig. 4.16 makes an angle θ with the vertical. It is given by

$$tan \space \theta =\frac {v_b}{v_r} = \frac{12}{35} =0.343 $$

$$ θ ≅ 19^o$$

Therefore, the woman should hold her umbrella at an angle of about 19° with the vertical towards the west. Note carefully 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

Concept box: Any plane parallel to the ground is said to be a horizontal plane while any plane perpendicular to the ground is said to be a vertical plane.   An object that is in flight after being thrown or projected is called a projectile.

To understand projectile motion, picture these scenarios:

• In the absence of wind, if a particle is thrown at an angle with the horizontal axis, that is, in a direction other than the vertical plane, its motion is restricted to a vertical plane.
• Further, if resistance of air is negligible, the object moves with a constant acceleration, g. This type of motion of particle is called projectile motion.
• For instance, a ball or a Frisbee being thrown in air are said to follow a projectile motion.

Now, let us try to define projectile motion,

Definition box :   1. A small object thrown in a direction other than the vertical, that is, at some angle with horizontal is called a projectile motion. 2. The initial velocity of projectile is known as velocity of projection. 3. The angle between velocity of projection and x-axis (which is generally horizontal) is called angle of projection. 4. The path described by the projectile is called its trajectory. 5. The point of the origin of projectile motion Point of projection.

Mathematical analysis of projectile motion:

Questions for the section 4.10 and 4.10.1:

1. Define:

a) Projectile

b) Projectile Motion

c) Velocity of Projection

d) Angle of projection

e) Trajectory

f) Point of Projection

2. Obtain expressions for the following:

a) Horizontal Motion in the form: $y = v_{_0y }t- \frac{1}{2}(gt^2)$

b) Equation of Trajectory of a projectile

c) Time of flight, $T_{_}f$

d) Time taken to reach maximum height, $t_m$

e) Maximum height, $h_m$

f) Range of projectile, R and maximum range, $R_{max}$

4.10.1 Mathematical analysis of Projectile motion

Let us consider a projectile projected from the origin, ‘O’ with an initial velocity, at an angle of projection and ‘θ’ with the horizontal surface.

At t = 0 the rectangular components of $v_o$ are $v_{ox}= v_o \space cos \space θ$ along horizontal and $v_{oy} =v_o \space sin \space θ$ along vertical ( from resolution of vector).

At t = 0,$x_o=0$ and $y_o=0$ because the projectile is at origin.

At any time ‘t’ the projectile is at p(x,y) i.e., it has covered a distance of x along x-axis and y along y-axis from origin.

$$ v_{ox}=v_o \space cos \space θ;a_x=0$$
$$ v_{oy}=v_o \space sin \space θ;a_y=-g$$

Horizontal motion (along x-axis)

As $a_x=0$

$v_x = v_{ox} + a_xt$

$ \implies v_x = v_{ox} cos \space θ $

and

$ x = v_{ox}t + \frac {1}{2} a_x t^2$

$ \implies x = v_{ox}t = v_ot \space cos \space θ $

i.e., velocity of x-component remains constant as particle moves ($v_o \space cos \space θ)$

Vertical Motion ( along y-axis)

As $a_y = -g$

$ v_y = v_{oy} – gt $ and $ y = v_{oy} t – \frac {1}{2} gt^2$

also $ v^2_y = v^2_{oy} -2gy$

i.e., vertical motion is identical to the motion of a particle vertically upward with speed $v_o \space sin \space θ)$

Characteristics of projectile motion

1. Equation of trajectory

Path described by the projectile is called trajectory

From equation $ x= v_{o}t \space cos \space θ \implies t = \frac {x}{v_o \space cos \space θ} $

Substitute in equation $ y = v_{oy} t – \frac {1}{2} gt^2$

$$y = v_o \space sin \space θ \bigg( \frac {x}{v_o \space cos \space θ} \bigg) – \frac {1}{2} g\bigg( \frac {x^2}{v^2_o cos^2 θ} \bigg) $$

$ \implies y =ax-bx^2$ is the equation of parabola where a = tan θ and $ b = \frac {g}{2v^2_o cos^2θ}$ are constants for any given projectile.

$ \therefore $ The trajectory of the projectile is parabolic.

2. Time of flight ($T_f$)

It is the time elapsing from the launching to the time the projectile returns to the ground i.e., to same level again.

$$ y = (v_o \space sin \space θ)t – \frac {1}{2} gt^2 $$

$$ 0 = (v_o \space sin \space θ)t – \frac {1}{2} gt^2 $$

$$ \implies t=0 \space or \space t = \frac {2v_o \space sin \space θ}{g}$$

t=0 corresponds to the time instant of projection

$ \therefore $ Time of flight,

$$ T_f = \frac {2v_o \space sin \space θ}{g}$$

Time taken to reach maximum height $’t_m’$v.

At maximum height,y component of velocity $v_y=0$

$ \therefore $ Put  $v_y=0$ and $t=t_m$ in

$$ v_y = v_{oy} + at$$

$$ 0 = v_o sin \space θ – gt_m$$

$$ t_m = \frac {v_o \space sin \space θ}{g}$$

$ \therefore  t_m$ is half the time of flight

$$ t_m = \frac {T_f}{2}$$

This proves the symmetry of parabola

3. Maximum Height ( $h_m$)

It is the maximum vertical distance attained by the projectile.

$$ y = v_{oy}t- \frac {1}{2} gt^2 $$

$$ y = (v_o \space sin \space θ)t – \frac {1}{2} gt^2$$

When $t_m = \frac {T_f}{2} = \frac {v_o \space sin \space θ}{g} $,then $y=h_m$

$$h_m = (v_o \space sin \space θ) \frac {v_o \space sin \space θ}{g} – \frac {1}{2}g \bigg [ \frac {v_o \space sin \space θ}{g} \bigg]^2 $$

$$ h_m = \frac {v^2_o sin^2 θ}{2g}$$

4. Range (R)

It is the horizontal distance covered by the projectile during its flight.

i.e., horizontal distance covered by projectile until it reaches the y=0 level.

$x= v_{ox}t$

i.e., $R= v_{ox}T_f$

$$R = v_o \space cos \space θ \bigg [ \frac {2 v_o \space sin \space θ}{g} \bigg] $$

$$R =  \frac {v_o^2 \space sin \space 2θ}{g}  $$

For fixed,$ R \propto v^2_o$;doubling the projection speed will increase the range four times.

sin 2θ = 1 $ \implies 2θ = sin^{-1}1$

2θ = $90^o$

Θ = $45^o$

When θ = $45^o$ range is maximum

i.e., $R_{max} = \frac {v^2_o}{g}$

Note box The equations $T_f = \frac {2 v_o sin \space θ}{g} ,h_m = \frac {v^2_o sin^2 θ}{2g} $ and $R = \frac {v^2_o \space sin \space 2θ}{g}$ can be used when the porjection is on the horizontal surface.

Example 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 $v_o$ at an angle θ_o , the range is given by

$$ R= \frac {v^2_osin2 \theta_o}{g} $$

Now, for angles, (45° + α) and (45° – α), 2θo is (90° + 2α) and ( 90° – 2α) , respectively.

The values of sin (90° + 2α) and sin (90° – 2α) are the same, equal to that of cos 2α. Therefore, ranges are equal for elevations which exceed or fall short of 45° by equal amounts α.

Example 4.8 A hiker stands on the edge of a cliff 490 m above the ground and throws a stone horizontally with an initial speed of 15 m $s^{-1}$. Neglecting air resistance, find the time taken by the stone to reach the ground, and the speed with which it hits the ground. (Take g = 9.8 m $s^{-2}$ ).                                                                                                                                                   

Answer:  We choose the origin of the x-,and yaxis at the edge of the cliff and t = 0 s 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:

$$ x (t) = x_o + v_{ox} t  $$

$$ y (t) = y_o + v_{oy} t +(1/2) a_y t^2 $$

Here,$ x_o$  = $y_o$ = 0, $v_{oy} = 0, $a_y$ =–g = –9.8 m $s^{-2}$,

$$ v_{ox} = 15 m s-1. $$

 The stone hits the ground when y(t) = – 490 m

$$ – 490 m = –(1/2)(9.8) t^2.$$

This gives t =10 s.

The velocity components are $ v_x $ = $ v_{ox}$ and

$$ v_y = v_{oy} – g t  $$

so that when the stone hits the ground :

$$ v_{ox} = 15 m s^{–1} $$

$$ v_{oy} = 0 – 9.8 × 10 = – 98 m s^{–1}          $$

Therefore, the speed of the stone is

$$ \sqrt {v^2_x+v^2_y} = \sqrt {15^2+98^2} = 99ms^{-1} $$

Example 4.9:  A cricket ball is thrown at a speed of 28 m $s^{–1}$ in a direction 30° above the horizontal. Calculate (a) the maximum height, (b) 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 level.  

Answer: (a) The maximum height is given by

$$ h_m= \frac {\big( v_o sin \theta_o \big)^2}{2g} = \frac {\big( 28 sin30^o \big)^2}{2×9.8} m $$

$$ = \frac{14 ×14}{2×9.8} =10.0 m$$ 

(b) The time taken to return to the same level is

$$ T_f = \frac { \big (2v_o sin \theta_o \big)}{g} =\frac {\big( 2×28×sin30^o \big) }{9.8} $$

(c) The distance from the thrower to the point where the ball returns to the same level is

$$ R = \frac {\big( v_o ^2 sin2 \theta_o \big)}{g} = \frac {\big( 28×28× sin60^o \big)}{9.8} = 69m $$

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 linear 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 ω is given by

ω = 2π/T = 2π × 7/100 = 0.44 rad/s

The linear speed v is:

v =ω R = 0.44 $s^{-1}$ × 12 cm = 5.3 cm $s^{-1}$

The direction of velocity v is 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 is not a constant vector.However, the magnitude of acceleration is constant:

$$ a = ω^2 \space R = (0.44 \space  s^{–1})^2 (12  \space cm) $$

$$ = 2.3 \space  cm \space s^{-2} $$

A read to visualize and understand projectile motion;

• While 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 in vacuum or under low pressure, which is 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 air resistance. The calculation without air resistance is much simpler than that with air resistance.

4.11 Uniform Circular Motion

• An object moving along the circumference of a circle at a constant speed is said to be in uniform circular motion.
• The time taken by the object to go round the circular path once is called Time Period (T) of the object. In other words, the time taken to complete one revolution of the circle is called Time period. It is given by:
  $$ T = \frac {2πr}{v} = \frac{2π}{ω} $$
• The number of times the object goes round the circle in one second is known as Frequency (n) of the particle. In other words, the number of revolutions made by the object in one second is called frequency. It is given by: $$ n = \frac {1}{T}$$

Note box: Position vector is also called as radius vector.

4.11.1 Angular Variables:

Fig showing angular variables for an object in uniform circular motion

Consider a particle moving with a constant speed along the circumference of a circle. Suppose, the object moves from point, P, making an angle, $\theta_1$ to point, Q, making an angle $\theta_2$  with the horizontal, in a time interval, Δt.

• Then, position vector rotates by an angle, $θ_2-θ_1 $. Let this be equal to Δθ. This angle described by the position vector in finite interval of time is called angular displacement of the object.
• Consequently, Angular Velocity can be defined as the rate of change of angular displacement.
• The Average Angular Velocity $ω{av}$ of the object is the ratio. Given by: $$ω{av}= \frac {Δθ}{Δt}$$
• The Instantaneous Angular Velocity (ω) is the limiting value of average angular velocity. Given by:
  $$ ω = \lim_{Δt \to 0}⁡ \frac {∆θ}{∆t}= \frac {dθ}{dt} $$
• The average Angular Acceleration ( ) of the object is the ratio of change in angular speed to time interval. Given by:
  $ a ̅ = \frac {Δω}{Δt}; where, Δω = w_2-w_1 $ is the change in angular speed in the interval, Δt.
• Instantaneous Angular Acceleration is the limiting value of angular acceleration. Its magnitude is given by:
  $$ |α|= \lim_{∆t\to 0}⁡ \frac {∆ω}{∆t} = \frac {dω}{dt} $$

4.11.2 Relation between linear velocity and angular velocity

Consider an object moving along a circle of radius, R with a constant speed as shown in figure 4.11.1. As the object moves along the circumference of the circle, at any point, its velocity is along the tangent drawn at that point. Thus, the direction of velocity changes continuously as the point, P keeps changing during the circular motion.

As discussed earlier, time taken for one complete rotation is known as the period, denoted by T.

The angle described by the radius vector for one complete revolution is 2π radian. Thus,

Angular velocity is$ ω= \frac {2π}{T} $     (“ω” denotes Greek letter omega)

Linear Velocity, $v= \frac {2πr}{T} = r.ω $      (4.11.2)

(That is, Linear Velocity = radius x angular velocity)

In vector form, $\vec{v} = \vec{\omega} \times \vec{r}$ 

Equation, 4.11.2 thus gives the relation between Linear Velocity and Angular velocity.

4.11.3 Acceleration in uniform circular motion

Consider an object in uniform circular motion.

• Let $\vec{r} $ and  $\vec{r}\,’$  be position vectors and $\vec{v} $ and  $\vec{v}\,’$  be the velocities at positions, P and P’ of the object in circular motion (as shown in the above figure). We know that, velocity of an object at any point in its path is directed along the tangent at that point in direction of motion. Vectors $\vec{v} $ and  $\vec{v}\,’$ are shown in figure, 4.19 (a1).
• Δv is obtained in Fig. 4.19 (a2) using the triangle law of vector addition. Since the path is circular, v is perpendicular to r and so is v′ to r′. Therefore, Δv is perpendicular to Δr. Since, average acceleration is along Δv (as acceleration is directly proportional to velocity, from the relation, $\vec{a}$ = Δv/Δt), the average acceleration a, is perpendicular to Δr.
• If we place Δv on the line that bisects the angle between r and r′, we see that Δv is directed towards the centre of the circle, so will be acceleration, “a”. Figure 4.19(b) shows the same quantities for smaller time interval.
• In Fig. 4.19(c), $\Delta t \to 0$ , 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. (In the limit $\Delta t \to 0$, Δr becomes perpendicular to r. In this limit $\Delta t \to 0$ and is consequently also perpendicular to V)
• Therefore, the acceleration is directed towards the centre, at each point of the circular path. Let us now find the magnitude of the acceleration. The magnitude of “a” is, by definition, given by:
  $$ |α|= \lim_{∆t\to 0}⁡\bigg( \frac {∆v}{∆t} \bigg) = \frac {dv}{dt} $$
• Let the angle between position vectors r and r′ be Δθ. Since the velocity vectors v and v′ are always perpendicular to the position vectors, the angle between them is also Δθ. Therefore, the triangle CPP′ formed by the position vectors and the triangle GHI formed by the velocity vectors v, v′ and Δv 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 is:
  $$ \frac {|\Delta v|}{v}= \frac {|\Delta r|}{R} $$
  $$or,|\Delta v| = v \frac {| \Delta r|}{R}$$
  Therefore,
  $$|a| = \lim_{ \Delta t\to 0} \frac {| \Delta v|}{\Delta t} =  \lim_{ \Delta t\to 0} \frac {v |\Delta r|}{R \Delta t} = \frac {v}{R}  \lim_{ \Delta t\to 0} \frac {|\Delta r|}{\Delta t}$$
  If Δt is small, Δθ will also be small and then arc PP′ can be approximately taken to be|Δr|:
  $$ | \Delta r | ≅ v \Delta t $$
  $$ \frac {|\Delta r|}{\Delta t } ≅ v $$
  $$or, \lim_{ \Delta t\to 0} \frac {|\Delta r|}{\Delta t} = v$$
  Therefore, the centripetal acceleration $a_c$ is :
  $$a_c = \bigg( \frac {v}{R} \bigg) v = v^2/R \space….. 4.44 $$
• Thus, the acceleration of an object moving with speed, “v” in a circle of radius, “R” has a magnitude, $v^²/R$ and is always directed towards the centre. Hence, this acceleration is called Centripetal Acceleration (a term proposed by Newton). Since “v” and “R” are constant, the magnitude of the centripetal acceleration is also constant.
• However, the direction changes, pointing 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 P′ in time Δt (which is = t′-t), the line CP (Fig. 4.19) turns through an angle Δθ as shown in the figure. Δθ is called angular displacement. As defined earlier, the angular speed, “ω” is the time rate of change of angular displacement, given by:
  $$ω = \frac {Δθ}{Δt} $$     (4.45)

Now, if the distance travelled by the object during the time Δt is Δs, i.e. PP′ is Δs, then:

$$v = \frac {Δs}{Δt} $$        

but, Δs = R Δθ. Therefore,

$$ v= R \frac  {∆θ}{∆t}= Rω  $$

We can express centripetal acceleration  $a_c$ in terms of angular speed :

$$ a_c= \frac {v^2}{R}=\frac {ω^2R^2}{R}=ω^2R $$

$$ a_c=ω^2R $$

The time taken by an object to make one revolution is known as its time period T and the number of revolution made in one second is called its frequency $ν \big(=\frac {1}{T} \big) $. However, during this time the distance moved by the object is s = 2πR.

Therefore, $ v = \frac {2πr}{T} = 2πRv $

In terms of frequency ν, we have

$$ ω = 2πν $$

 $$ v = 2πRν $$  $$a_c = 4π^2 ν^2R $$      

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 linear speed of the motion? (b) Is the acceleration vector a constant vector? What is its magnitude?

Solution:

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

ω = 2π/T = 2π × 7/100 = 0.44 rad/s

The linear speed v is : v =ω R = 0.44 $s^{-1}$ × 12 cm = 5.3 cm $s^{-1}$

The direction of velocity v is 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 is not a constant vector.

However, the magnitude of acceleration is constant:

$$a = ω^2 R = (0.44 \space s^{–1})^2 (12 \space cm)$$

$$ = 2.3 \space cm  \space s^{-2}$$

Questions for the sections, 4.11, 4.11.1, 4.11.2 and 4.11.3:

1. Define:

a) Uniform Circular Motion

b) Time period

c) Angular Displacement

d) Angular Velocity. Also, give its mathematical relation.

e) Average Angular Velocity. Also, give its mathematical relation.

f) Instantaneous Angular Velocity. Also, give its calculus relation.

g) Angular Acceleration

h) Instantaneous Angular Acceleration. Also, give its calculus relation.

2. Obtain the relation between Angular and Linear Velocity.

3. Obtain the expression for centripetal acceleration (equation 4.44), centripetal acceleration in terms of angular velocity (equation 4.47) and in terms of frequency (equation 4.49).