NotesShala Class 11 · Physics

Chapter03

Motion in a Plane

Scalars and Vectors · Motion in a Plane · Projectile and Circular Motion

Reading key Definition Note Remember Derivation Strategy Formulas sit on graph paper · coloured words are the key terms

Part 1

Scalars and Vectors

Chapter 2 dealt with motion along a straight line. This chapter goes further, to motion in a plane (2-D) and motion in space (3-D). Such motion needs vectors, so we learn them first. After that comes motion with constant acceleration in a plane, mainly projectile motion. The last topic is uniform circular motion.

1.1Scalar and Vector Quantities

Many quantities help us describe the physical world. Some of them are mass, distance, displacement, speed, velocity, acceleration, momentum, force, energy, work and power. Motion in two or three dimensions cannot be described without vectors. So we start with them.

Every physical quantity has a magnitude (size). Some quantities also have a direction. This gives two groups:

iScalars

Scalars have only a size (magnitude). They have no direction.

Magnitude✓ Yes

Direction✗ No

iiVectors

Vectors have a size (magnitude) and also a direction.

Magnitude✓ Yes

Direction✓ Yes

Scalar Quantities

Definition

A physical quantity that has only magnitude and no direction is called a scalar quantity. A scalar is fully described by a single number with its proper unit.

Scalars follow the ordinary rules of algebra:

iAlgebra of Scalars

Scalars are added, subtracted, multiplied and divided just like ordinary numbers.

iiCondition of Units

Only quantities with the same unit can be added or subtracted. Quantities with different units can be multiplied or divided.

Vector Quantities

Definition

A physical quantity that has both magnitude and direction is called a vector quantity. Vectors are added and subtracted by the triangle law and the parallelogram law.

e.g. Force, velocity, displacement, acceleration, momentum, etc.

A vector is printed in bold type, like $\mathbf{F},\ \mathbf{a},\ \mathbf{b}$ . It is also written with an arrow over the letter, like $\vec{F},\ \vec{a},\ \vec{b}$ . We draw a vector as a straight line with an arrowhead:

How long the arrow isShows the magnitude

Which way the arrowhead pointsShows the direction

The magnitude is also called the absolute value

|\vec{v}| = v

Example

Representing a Vector

A body moves with a velocity of 60 m/s due East. Take a scale of 1 cm = 10 m/s. Draw a line $OP$ , 6 cm long, towards East. Put an arrowhead at $P$ . This line fully represents the velocity.

Point $P$ is the head (terminal point). Point $O$ is the tail (initial point) of $\overrightarrow{OP}$ .

Vector

OP

is 6 cm long towards East: tail

O

, head

P

Vectors are of two types:

1Polar Vectors

A vector that acts from a starting point, or at a point of application, is called a polar vector.

e.g. Displacement, force.

2Axial Vectors

A vector that describes a turning (rotational) effect is called an axial vector. It lies along the axis of rotation. Which way it points on the axis depends on the turning: clockwise or anticlockwise.

e.g. Torque, angular velocity.

An axial vector lies along the axis of rotation (right-hand rule)

Good to know

Tensors

Some quantities do not have one fixed direction. Yet their value changes from one direction to another. They are known as tensors.

e.g. Stress, pressure, surface tension, moment of inertia, etc.

1.2Important Definitions Related to Vectors

These terms are used again and again with vectors.

iModulus of a Vector

The modulus of a vector is simply its magnitude.

How we write it for

\vec{A}

|\vec{A}|

A

iiUnit Vector

A unit vector is a vector of magnitude one (unity) that points in a definite direction. The unit vector along $\vec{A}$ is $\hat{A}$ , read as "A cap".

So every vector = its magnitude × the unit vector in its own direction.

Unit Vector

\hat{A} = \dfrac{\vec{A}}{|\vec{A}|} = \dfrac{\vec{A}}{A} \qquad\Rightarrow\qquad \vec{A} = |\vec{A}|\,\hat{A}

Unit vector $= \dfrac{\text{vector}}{\text{magnitude of the vector}}$

Cartesian Unit Vectors

$\hat{i},\,\hat{j}$ and $\hat{k}$ point along the x, y and z axes. Each has magnitude one.

Magnitude of a Unit Vector

It is equal to one. A unit vector has no unit and no dimensions.

iiiNull Vector (Zero Vector)

A null vector (or zero vector) is a vector of zero magnitude. Its direction can be anything (arbitrary). It is written as $\vec{0}$ .

Velocity of a body at restNull vector

Acceleration of a body moving at constant velocityNull vector

ivEqual Vectors

Equal vectors have the same magnitude and point in the same direction.

\overrightarrow{OP}=\overrightarrow{SQ}

\vec{A}=\vec{B}

Equal vectors

\vec{A}

and

\vec{B}

vNegative Vector

A vector with the same magnitude as $\vec{A}$ but the opposite direction is called the negative of $\vec{A}$ .

Negative of

\vec{A}

-\vec{A}

\vec{A}

and its negative

-\vec{A}

Note

Free Vectors

A vector is not tied to one place. Moving it parallel to itself leaves it unchanged. We call such vectors free vectors.

e.g. The velocity of a particle in straight-line motion.

viCollinear Vectors

Collinear vectors lie on one line or on parallel lines. They can point the same way or opposite ways.

e.g. The pulls in a tug of war.

Collinear vectors

Parallel Vectors

Collinear vectors that point the same way are called parallel vectors.

Angle between the two

\theta=0^\circ

Parallel vectors

Anti-parallel Vectors

Collinear vectors that point opposite ways are called anti-parallel vectors.

Angle between the two

\theta=180^\circ

Anti-parallel vectors

viiCoplanar Vectors

Vectors lying in one plane are called coplanar vectors. In the figure, $\vec{A}$ , $\vec{B}$ and $\vec{C}$ are all in the plane of the paper.

Coplanar vectors

viiiCo-initial Vectors

Vectors that start from one common point are called co-initial vectors. In the figure, $\vec{A}$ and $\vec{B}$ both begin at $O$ .

Co-initial vectors

ixOrthogonal Unit Vectors

Two or three unit vectors at right angles to one another are called orthogonal unit vectors. A common example is $\hat{i}$ , $\hat{j}$ and $\hat{k}$ , which lie along the x, y and z axes.

\hat{i}=\dfrac{\vec{x}}{x},\quad \hat{j}=\dfrac{\vec{y}}{y},\quad \hat{k}=\dfrac{\vec{z}}{z}

Orthogonal unit vectors

\hat{i}

\hat{j}

\hat{k}

xLocalised Vectors

A vector that must start from a fixed point is called a localised vector.

e.g. Position vector of a particle. It always starts at the origin.

xiNon-localised Vectors

A vector whose starting point is free to change is called a non-localised vector.

e.g. Velocity of a particle in straight-line motion.

1.3Position and Displacement Vectors

Position Vector

Definition

The position vector of an object is the vector drawn from the origin of the coordinate system to the object. Its symbol is $\vec{r}$ .

Let an object move in the xy-plane, with the origin at $O$ . At an instant $t$ , the object is at point $A$ . Its position vector at $A$ is:

\overrightarrow{OA}=\vec{r}

A position vector gives two pieces of information:

the minimum distance of the object from the origin $O$ , and
the direction of the object from the origin.

Position vector

\vec{r}

of point

A

Displacement Vector

Definition

The vector that shows how much and in which direction the position of an object changes in a time interval is called the displacement vector.

It is the straight line from the initial position to the final position. The path actually followed in between does not matter.

Let an object move in the xy-plane. At an instant $t$ , it is at point $A$ . At a later instant $t'$ , it is at point $B$ .

Its displacement from $t$ to $t'$ is $\overrightarrow{AB}$ . Let the coordinates of $A$ and $B$ be $(x_1, y_1)$ and $(x_2, y_2)$ .

Displacement vector

\Delta\vec{r}

from

A

B

Derivation

The position vectors of the object at $A$ and $B$ are:
$\vec{r}_1=x_1\hat{i}+y_1\hat{j} \qquad\text{and}\qquad \vec{r}_2=x_2\hat{i}+y_2\hat{j}$
The displacement $\overrightarrow{AB}$ is the change in the position vector:
$\Delta\vec{r}=\vec{r}_2-\vec{r}_1$
Put in the two position vectors:
$\Delta\vec{r}=(x_2\hat{i}+y_2\hat{j})-(x_1\hat{i}+y_1\hat{j})$

Displacement Vector in a Plane

\Delta\vec{r}=(x_2-x_1)\,\hat{i}+(y_2-y_1)\,\hat{j}

Magnitude of Displacement

|\Delta\vec{r}|=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}

|\Delta\vec{r}|=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2}

First line: in a plane. Second line: in space (3-D), where the $z$ term is added.

Magnitude of displacement compared with the path length

\leq

path length

1.4Multiplication of a Vector by a Real Number

Definition

Multiplying a vector $\vec{A}$ by a real number (scalar) $\lambda$ gives a new vector along the direction of $\vec{A}$ . Its magnitude is $\lambda$ times the magnitude of $\vec{A}$ .

The answer depends on whether the number is positive or negative:

Positive Number

\lambda

Multiplying $\vec{A}$ by a positive number $\lambda$ gives a vector of magnitude $\lambda|\vec{A}|$ . It points in the same direction as $\vec{A}$ .

Direction of

\lambda\vec{A}

Same as

\vec{A}

Negative Number

-\lambda

Multiplying $\vec{A}$ by a negative number $-\lambda$ gives a vector of magnitude $\lambda|\vec{A}|$ . It points in the opposite direction to $\vec{A}$ .

Direction of

-\lambda\vec{A}

Opposite to

\vec{A}

Magnitude after Multiplication

|\lambda\vec{A}| = \lambda\,|\vec{A}| \qquad\text{and}\qquad |-\lambda\vec{A}| = \lambda\,|\vec{A}|

In both cases, the magnitude becomes $\lambda$ times. For a negative number, only the direction reverses.

Examples

Example

Multiply $\vec{A}$ by $\lambda = 3$ . We get $3\vec{A}$ . It has the same direction as $\vec{A}$ , but it is three times as long.

Multiply $\vec{A}$ by $\lambda = -3$ . We get $-3\vec{A}$ . It is also three times as long, but it points in the opposite direction.

\vec{A}

3\vec{A}

and

-3\vec{A}

Example

A constant velocity vector multiplied by time gives a displacement vector. This displacement is in the direction of the velocity.

Velocity

\times

timeDisplacement

1.5Resultant Vector

Definition

A single vector that has the same effect as two or more vectors acting together is called their resultant vector.

Two simple cases are given below.

Case I · Vectors in the Same Direction

Let $\vec{A}$ and $\vec{B}$ act in the same direction. The resultant also points in that direction. Its magnitude is the sum of their magnitudes, $(A+B)$ .

Resultant vector

\vec{R}=\vec{A}+\vec{B}

Same direction:

R = A + B

Case II · Vectors in Opposite Directions

Let $\vec{A}$ and $\vec{B}$ act in opposite directions. The resultant points along the vector with the bigger magnitude. Its magnitude is the difference of their magnitudes, $(A-B)$ .

Resultant vector

\vec{R}=\vec{A}+\vec{B}

Its magnitude (when

A>B

)

R=A-B

Opposite directions:

R = A - B

, along

\vec{A}

B>A

\vec{R}

is along

\vec{B}

A>B

\vec{R}

is along

\vec{A}

Conditions for Zero Resultant Vector

Triangle Condition

Three vectors act on a point object. Suppose they can be drawn head to tail to form the three sides of a triangle, going round in one order. Then they add up to zero, and the object stays in equilibrium.

\vec{A}+\vec{B}+\vec{C}=\vec{0}

\vec{A}

\vec{B}

\vec{C}

along

OP

PQ

QO

Polygon Condition

Any number of vectors drawn head to tail as the sides of a closed polygon, in one order, also add up to zero. The object is then in equilibrium.

\overrightarrow{OA}+\overrightarrow{AB}+\overrightarrow{BC}+\cdots+\overrightarrow{GO}=\vec{0}

Eight vectors around a closed figure add up to zero

Conditions for Equilibrium

iNo Linear Motion

The net force acting on the body is zero.

iiNo Rotational Motion

The net torque of all the forces on it is zero.

iiiStable Equilibrium

Its potential energy is the least (minimum).

1.6Addition of Vectors — Graphical Method

We can add two vectors only when they are of the same kind (nature). A displacement adds to another displacement. It never adds to a force.

Vectors of the same nature (displacement + displacement)Can be added

Vectors of different nature (displacement + force)Cannot be added

In the graphical method, we draw the vectors and see their resultant. Three laws are used.

1 · Triangle Law of Vector Addition

Law

Place the tail of the second vector at the head of the first. The two vectors now form two sides of a triangle. The third side, drawn from the tail of the first to the head of the second, is their resultant in magnitude and direction.

Draw $\overrightarrow{OM}=\vec{A}$ . From the head $M$ , draw $\overrightarrow{MN}=\vec{B}$ . The resultant $\overrightarrow{ON}$ joins the tail of $\vec{A}$ to the head of $\vec{B}$ .

\overrightarrow{ON}=\overrightarrow{OM}+\overrightarrow{MN}

\vec{R}=\vec{A}+\vec{B}

\vec{R}

goes from the tail of

\vec{A}

to the head of

\vec{B}

2 · Parallelogram Law of Vector Addition

Law

Draw two vectors from one common point as the two adjacent sides of a parallelogram. Complete the parallelogram. Its diagonal that starts at the common point is the resultant, in magnitude and direction.

Draw $\overrightarrow{OP}=\vec{A}$ and $\overrightarrow{OR}=\vec{B}$ from the common point $O$ . Complete the parallelogram $OPQR$ . The diagonal $\overrightarrow{OQ}$ is the resultant.

\overrightarrow{OQ}=\overrightarrow{OP}+\overrightarrow{OR}

\vec{R}=\vec{A}+\vec{B}

The diagonal

OQ

is the resultant

\vec{R}

3 · Polygon Law of Vector Addition

Law

Draw many vectors head to tail. They form the sides of an open polygon. The side that closes the polygon, drawn from the first tail to the last head, is their resultant in magnitude and direction.

Here $\vec{A}$ , $\vec{B}$ , $\vec{C}$ and $\vec{D}$ run head to tail from $O$ to $T$ . So $\overrightarrow{OT}$ is the resultant.

Closing side

\overrightarrow{OT}

\vec{R}=\vec{A}+\vec{B}+\vec{C}+\vec{D}

The closing side

OT

gives

\vec{R}

Properties of Vector Addition

iCommutative Law

\vec{A}+\vec{B}=\vec{B}+\vec{A}

iiAssociative Law

(\vec{A}+\vec{B})+\vec{C}=\vec{A}+(\vec{B}+\vec{C})

iiiDistributive Law

\lambda(\vec{A}+\vec{B})=\lambda\vec{A}+\lambda\vec{B}

ivAdditive Identity

\vec{A}+\vec{0}=\vec{A}

1.7Subtraction of Vectors — Graphical Method

Definition

Subtracting $\vec{B}$ from $\vec{A}$ means adding the negative of $\vec{B}$ , that is $(-\vec{B})$ , to $\vec{A}$ . So first invert $\vec{B}$ . Then add it to $\vec{A}$ by the laws of vector addition.

Subtraction of Vectors

\vec{R}=\vec{A}+(-\vec{B})=\vec{A}-\vec{B}

Graphical Method

Reverse the direction of $\vec{B}$ to get $-\vec{B}$ .
Shift $-\vec{B}$ parallel to itself. Place its tail at the head of $\vec{A}$ .
Join the tail of $\vec{A}$ to the head of $-\vec{B}$ . This line is $\vec{R}=\vec{A}-\vec{B}$ .

We can also add $-\vec{B}$ to $\vec{A}$ by the parallelogram law. The result is the same.

Subtraction:

\vec{R} = \vec{A} + (-\vec{B})

Properties of Vector Subtraction

iNot Commutative

\vec{A}-\vec{B}\neq\vec{B}-\vec{A}

iiNot Associative

{\vec{A}{-}(\vec{B}{-}\vec{C})\neq(\vec{A}{-}\vec{B}){-}\vec{C}}

iiiDistributive

\lambda(\vec{A}-\vec{B})=\lambda\vec{A}-\lambda\vec{B}

1.8Resolution of a Vector in a Plane

Definition

Resolution of a vector means breaking one vector into two or more vectors along different directions. These parts, acting together, have the same effect as the original vector.

Components of a Vector

The parts obtained by resolving a vector are called its components.

Oblique Components

When a vector is resolved along the directions of two given vectors, the two parts are called oblique components.

To resolve a vector, we write it with unit vectors. Any vector $\vec{r}$ in a plane can be written as a linear combination of the unit vectors $\hat{i}$ and $\hat{j}$ . These two unit vectors are perpendicular to each other.

Vector in Terms of Components

\vec{r}=x\,\hat{i}+y\,\hat{j}

$x\hat{i}$ and $y\hat{j}$ are the perpendicular components of $\vec{r}$ . $x$ and $y$ are the resolved parts of $\vec{r}$ along the x-axis and the y-axis.

Magnitude

|\vec{r}|=\sqrt{x^{2}+y^{2}}

Direction (Angle with x-axis)

\theta=\tan^{-1}\left(\dfrac{y}{x}\right)

\vec{r}

resolved into

x\hat{i}

and

y\hat{j}

Resolving a Vector Along Two Given Directions

Let $\overrightarrow{OQ}$ stand for $\vec{R}$ . The triangle law gives:

\overrightarrow{OQ}=\overrightarrow{OP}+\overrightarrow{PQ}

Here $\overrightarrow{OP}$ and $\overrightarrow{PQ}$ are the components of $\vec{R}$ along $\vec{A}$ and $\vec{B}$ . Take $\overrightarrow{OP}=\lambda\vec{A}$ and $\overrightarrow{PQ}=\mu\vec{B}$ , where $\lambda$ and $\mu$ are real numbers. Then $\vec{R}$ becomes a linear combination of $\vec{A}$ and $\vec{B}$ .

Components of

\vec{R}

along

\vec{A}

and

\vec{B}

Resultant along Two Given Directions

\vec{R}=\lambda\vec{A}+\mu\vec{B}

Rectangular Components

Definition

The two components of a vector that are at right angles to each other are called its rectangular components.

Magnitude

A=\sqrt{A_x^{2}+A_y^{2}}

Direction (Angle with x-axis)

\theta=\tan^{-1}\left(\dfrac{A_y}{A_x}\right)

Rectangular components of

\vec{A}

Note

Here $A_x$ and $A_y$ are the components of $\vec{A}$ along the x-axis ( $\hat{i}$ ) and the y-axis ( $\hat{j}$ ).

1.9Resolution of a Vector in Space

In the same way, a vector $\vec{A}$ in three dimensions (space) can be resolved into three components. These are along the x, y and z axes.

Let $\alpha$ , $\beta$ and $\gamma$ be the angles that $\vec{A}$ makes with the x, y and z axes. Let $\hat{i}$ , $\hat{j}$ and $\hat{k}$ be the unit vectors along these axes. Resolving $\vec{A}$ gives its three components.

Components of

\vec{A}

in space

x-component

A_x=A\cos\alpha

y-component

A_y=A\cos\beta

z-component

A_z=A\cos\gamma

Vector in Three Dimensions

\vec{A}=A_x\,\hat{i}+A_y\,\hat{j}+A_z\,\hat{k}

Magnitude

A=\sqrt{A_x^{2}+A_y^{2}+A_z^{2}}

Position Vector in Space

\vec{r}=x\,\hat{i}+y\,\hat{j}+z\,\hat{k}

Direction Cosines

Definition

The cosines of the angles $\alpha$ , $\beta$ and $\gamma$ that a vector makes with the x, y and z axes are called its direction cosines. They are written as $l$ , $m$ and $n$ .

With x-axis

\cos\alpha=\dfrac{A_x}{A}=l

With y-axis

\cos\beta=\dfrac{A_y}{A}=m

With z-axis

\cos\gamma=\dfrac{A_z}{A}=n

Derivation

Put $A_x=A\cos\alpha$ , $A_y=A\cos\beta$ and $A_z=A\cos\gamma$ in $A^2=A_x^2+A_y^2+A_z^2$ :

$A^{2}=A^{2}\cos^{2}\alpha+A^{2}\cos^{2}\beta+A^{2}\cos^{2}\gamma$
Take $A^2$ common:
$A^{2}=A^{2}\left(\cos^{2}\alpha+\cos^{2}\beta+\cos^{2}\gamma\right)$
Divide both sides by $A^2$ :
$\cos^{2}\alpha+\cos^{2}\beta+\cos^{2}\gamma=1$

Property of Direction Cosines

l^{2}+m^{2}+n^{2}=1

Add the squares of the three direction cosines of any vector. The total is always one.

Note

$\alpha$ , $\beta$ and $\gamma$ are space angles. The pairs of lines that form them do not lie in one plane.

1.10Addition of Vectors — Analytical Method

Setup — Given Information

Two vectors $\vec{A}$ and $\vec{B}$ act on a particle. The angle between them is $\theta$ .
They are drawn head to tail as the sides $\overrightarrow{OP}$ and $\overrightarrow{PQ}$ of $\triangle OPQ$ .
The closing side $\overrightarrow{OQ}$ is then their resultant $\vec{R}$ (triangle law).
Draw $QN \perp OP$ produced.

Adding

\vec{A}

and

\vec{B}

analytically

Derivation

In the right triangle $QPN$ , the side $PQ=B$ . So:
$PN=B\cos\theta \quad \text{…(i)} \qquad\text{and}\qquad QN=B\sin\theta \quad \text{…(ii)}$
In the right triangle $ONQ$ , $OQ^2=ON^2+NQ^2=(OP+PN)^2+(QN)^2$ . Hence:
$R^{2}=(A+B\cos\theta)^{2}+(B\sin\theta)^{2}$
Expand the brackets and use $\cos^{2}\theta+\sin^{2}\theta=1$ :
$R^{2}=A^{2}+B^{2}\cos^{2}\theta+2AB\cos\theta+B^{2}\sin^{2}\theta=A^{2}+B^{2}+2AB\cos\theta$

Magnitude of the Resultant

R=|\vec{R}|=\sqrt{A^{2}+B^{2}+2AB\cos\theta}

Let $\vec{R}$ make an angle $\beta$ with the direction of $\vec{A}$ . In the right triangle $QNO$ :

\tan\beta=\dfrac{QN}{ON}=\dfrac{QN}{OP+PN}=\dfrac{B\sin\theta}{A+B\cos\theta}

Direction of the Resultant

\beta=\tan^{-1}\left(\dfrac{B\sin\theta}{A+B\cos\theta}\right)

Special Cases for the Magnitude of R

\theta=0^\circ

R=A+B

This is the maximum value.

\theta=90^\circ

R=\sqrt{A^{2}+B^{2}}

The two vectors are perpendicular.

\theta=180^\circ

R=A-B

This is the minimum value.

More Than Two Vectors

The same method works for any number of vectors. Take three of them:

\vec{a}=a_x\hat{i}+a_y\hat{j}+a_z\hat{k},\quad \vec{b}=b_x\hat{i}+b_y\hat{j}+b_z\hat{k},\quad \vec{c}=c_x\hat{i}+c_y\hat{j}+c_z\hat{k}

Their resultant $\vec{r}=\vec{a}+\vec{b}+\vec{c}$ has these components:

r_x=a_x+b_x+c_x,\quad r_y=a_y+b_y+c_y,\quad r_z=a_z+b_z+c_z

1.11Subtraction of Vectors — Analytical Method

Let $\vec{A}$ and $\vec{B}$ make an angle $\theta$ with each other. To subtract $\vec{B}$ from $\vec{A}$ , we first invert $\vec{B}$ . Then we add it to $\vec{A}$ .

Resultant vector

\vec{R}=\vec{A}+(-\vec{B})=\vec{A}-\vec{B}

\vec{R} = \vec{A} - \vec{B}

by the parallelogram method

Derivation

The angle between $\vec{A}$ and the inverted vector $(-\vec{B})$ is $(180^\circ-\theta)$ . Use the formula for addition:
$R=\sqrt{A^{2}+B^{2}+2AB\cos(180^\circ-\theta)}$
Since $\cos(180^\circ-\theta)=-\cos\theta$ , the magnitude of the difference becomes:
$R=\sqrt{A^{2}+B^{2}-2AB\cos\theta}$

Magnitude of the Difference

R=\sqrt{A^{2}+B^{2}-2AB\cos\theta}

where, $\theta$ = angle between $\vec{A}$ and $\vec{B}$

Special Cases for the Magnitude of R

\theta=0^\circ

R=A-B

This is the minimum value.

\theta=90^\circ

R=\sqrt{A^{2}+B^{2}}

The two vectors are perpendicular.

\theta=180^\circ

R=A+B

This is the maximum value.

Sum and Difference Side by Side

Magnitude of the Sum

R=\sqrt{A^{2}+B^{2}+2AB\cos\theta}

Magnitude of the Difference

R=\sqrt{A^{2}+B^{2}-2AB\cos\theta}

1.12Scalar Product (Dot Product)

Definition

The scalar product (dot product) of $\vec{A}$ and $\vec{B}$ is found by multiplying their magnitudes by the cosine of the angle $\theta$ between them. Its value is a scalar.

Scalar Product

\vec{A}\cdot\vec{B}=AB\cos\theta

Geometrical Meaning · Projection

In other words, $\vec{A}\cdot\vec{B}$ = (size of one vector) × (projection of the second vector on the first).

(a)

B\cos\theta

Projection of

\vec{B}

\vec{A}

(b)

A\cos\theta

Projection of

\vec{A}

\vec{B}

The scalar product as a projection

Special Cases

IParallel

Here $\theta=0^\circ$ .

\vec{A}\cdot\vec{B}=AB

IIPerpendicular

Here $\theta=90^\circ$ .

\vec{A}\cdot\vec{B}=0

IIIAnti-parallel

Here $\theta=180^\circ$ .

\vec{A}\cdot\vec{B}=-AB

Dot Product in Terms of Components

Write both vectors in component form. Then use the distributive law:

\vec{A}\cdot\vec{B}=(A_x\hat{i}+A_y\hat{j}+A_z\hat{k})\cdot(B_x\hat{i}+B_y\hat{j}+B_z\hat{k})

Same Unit Vectors

\hat{i}\cdot\hat{i}=\hat{j}\cdot\hat{j}=\hat{k}\cdot\hat{k}=1

Perpendicular Unit Vectors

\hat{i}\cdot\hat{j}=\hat{j}\cdot\hat{k}=\hat{k}\cdot\hat{i}=0

Scalar Product in Component Form

\vec{A}\cdot\vec{B}=A_xB_x+A_yB_y+A_zB_z

Properties of the Dot Product

iSelf Product

\vec{a}\cdot\vec{a}=|\vec{a}|^{2}

iiCommutative Law

\vec{a}\cdot\vec{b}=\vec{b}\cdot\vec{a}

iiiDistributive Law

\vec{a}\cdot(\vec{b}+\vec{c})=\vec{a}\cdot\vec{b}+\vec{a}\cdot\vec{c}

ivScalar Multiple

(\lambda\vec{a})\cdot\vec{b}=\lambda(\vec{a}\cdot\vec{b})=\vec{a}\cdot(\lambda\vec{b})

1.13Vector Product (Cross Product)

Definition

The vector product (cross product) of $\vec{A}$ and $\vec{B}$ is a vector. Its magnitude is $A$ times $B$ times the sine of the angle $\theta$ between them.

Vector Product

\vec{A}\times\vec{B}=AB\sin\theta\;\hat{n}

where, $\hat{n}$ = unit vector perpendicular to the plane of $\vec{A}$ and $\vec{B}$ , in the direction of $\vec{A}\times\vec{B}$

Principle

Right-Hand Rule

Hold the right hand so that its fingers turn from $\vec{A}$ towards $\vec{B}$ . The thumb then gives the direction of both $\hat{n}$ and $\vec{A}\times\vec{B}$ .

\vec{A}\times\vec{B}

is perpendicular to the plane of

\vec{A}

and

\vec{B}

. A screw turned from

\vec{A}

\vec{B}

, or the right thumb, points along it

Cross Product in Terms of Components

Let $\vec{a}=a_x\hat{i}+a_y\hat{j}+a_z\hat{k}$ and $\vec{b}=b_x\hat{i}+b_y\hat{j}+b_z\hat{k}$ . Then:

Determinant Form

\vec{a}\times\vec{b}=\begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\ a_x&a_y&a_z\\ b_x&b_y&b_z\end{vmatrix}=\hat{i}(a_yb_z-a_zb_y)-\hat{j}(a_xb_z-a_zb_x)+\hat{k}(a_xb_y-a_yb_x)

Same Unit Vectors

\hat{i}\times\hat{i}=\hat{j}\times\hat{j}=\hat{k}\times\hat{k}=\vec{0}

Cyclic Unit Vectors

\hat{i}\times\hat{j}=\hat{k},\;\;\hat{j}\times\hat{k}=\hat{i},\;\;\hat{k}\times\hat{i}=\hat{j}

Special Cases

Parallel or Anti-parallel

\vec{A}\times\vec{B}=\vec{0}

Perpendicular

Here $\theta=90^\circ$ .

|\vec{A}\times\vec{B}|=AB

Area of a Triangle

\text{Area}=\tfrac{1}{2}\,|\vec{A}\times\vec{B}|

$\vec{A}$ and $\vec{B}$ are two sides of the triangle.

Properties of the Cross Product

iNot Commutative (Anti-commutative)

\vec{a}\times\vec{b}=-(\vec{b}\times\vec{a})

iiSelf Product

\vec{a}\times\vec{a}=\vec{0}

iiiDistributive Law

\vec{a}\times(\vec{b}+\vec{c})=\vec{a}\times\vec{b}+\vec{a}\times\vec{c}

ivScalar Multiple

(\lambda\vec{a})\times\vec{b}=\lambda(\vec{a}\times\vec{b})=\vec{a}\times(\lambda\vec{b})

Unit Vectors on a Circle

Place $\hat{i}$ , $\hat{j}$ and $\hat{k}$ on a circle. The cross product of two neighbouring unit vectors taken anticlockwise gives the third unit vector. Taken clockwise, it gives the negative of the third one.

Anticlockwise

\hat{i}\times\hat{j}=\hat{k},\;\hat{j}\times\hat{k}=\hat{i},\;\hat{k}\times\hat{i}=\hat{j}

Clockwise (reverse order)

\hat{j}\times\hat{i}=-\hat{k},\;\hat{k}\times\hat{j}=-\hat{i},\;\hat{i}\times\hat{k}=-\hat{j}

Part 2

Motion in a Plane

Vectors now help us study two-dimensional motion. With them, we describe the position, displacement, velocity and acceleration of an object.

2.1Position, Displacement and Velocity Vectors

Position Vector

Definition

The vector drawn from a reference point to the position of a particle is called its position vector.

For a particle at $P$ in the xy-plane, the parallelogram law gives $\overrightarrow{OP}=\overrightarrow{OA}+\overrightarrow{OB}$ .

Position Vector in a Plane

\vec{r}=x\,\hat{i}+y\,\hat{j}

Position Vector in Space

\vec{r}=x\,\hat{i}+y\,\hat{j}+z\,\hat{k}

Position vector

\vec{r}

of a particle at

P(x, y)

Displacement

A particle moves from $A$ to $B$ . At time $t_1$ it is at $A$ , with position vector $\vec{r}_1$ . At time $t_2$ it is at $B$ , with position vector $\vec{r}_2$ .

By the triangle law, $\overrightarrow{OA}+\overrightarrow{AB}=\overrightarrow{OB}$ . So the displacement is:

\overrightarrow{AB}=\overrightarrow{OB}-\overrightarrow{OA}=\vec{r}_2-\vec{r}_1

Displacement

\Delta\vec{r}

, with components

\Delta x

and

\Delta y

Displacement in a Plane and in Space

\overrightarrow{AB}=(x_2-x_1)\hat{i}+(y_2-y_1)\hat{j}

\Delta\vec{r}=(x_2-x_1)\hat{i}+(y_2-y_1)\hat{j}+(z_2-z_1)\hat{k}

First line: in a plane. Second line: in space.

Velocity

Definition

Velocity is the rate at which the displacement of an object changes with time, in a given direction. It has two forms: average velocity and instantaneous velocity.

Average Velocity

Definition

The average velocity of an object is its displacement divided by the time interval taken for it.

Average Velocity

\vec{v}_{av}=\dfrac{\Delta\vec{r}}{\Delta t}=\dfrac{\vec{r}_2-\vec{r}_1}{t_2-t_1}

In component form, the average velocity, its magnitude and its direction are:

Component Form

\vec{v}_{av}=v_x\hat{i}+v_y\hat{j}

Magnitude

v_{av}=\sqrt{v_x^{2}+v_y^{2}}

Direction

\theta=\tan^{-1}\left(\dfrac{v_y}{v_x}\right)

2.2Instantaneous Velocity

Definition

The velocity of an object at a particular instant of time $(t)$ is called its instantaneous velocity. It is the limit of the average velocity as $\Delta t$ approaches zero.

Instantaneous Velocity

\vec{v}=\lim_{\Delta t\to 0}\dfrac{\Delta\vec{r}}{\Delta t}=\dfrac{d\vec{r}}{dt}

Understanding the Limiting Process

The curve shows the path of the object. At time $t$ , the object is at point $P$ . $P_1$ , $P_2$ and $P_3$ are its positions after the time intervals $\Delta t_1$ , $\Delta t_2$ and $\Delta t_3$ . Here $\Delta t_1>\Delta t_2>\Delta t_3$ .

\Delta t

becomes smaller, the chord

\Delta\vec{r}

turns towards the tangent at

P

(a) $\Delta t_1$ average

\vec{v}

(b) $\Delta t_2$ average

\vec{v}

(c) $\Delta t_3$ average

\vec{v}

(d) $\Delta t\to 0$ instantaneous

\vec{v}

When $\Delta t$ becomes very small (tends to zero), the average velocity becomes the velocity $\vec{v}$ at $P$ . Then $\vec{v}$ lies along the tangent to the path.

Note

The instantaneous velocity at any point is along the tangent to the path there. It shows the direction of motion of the object at that point.

2.3Acceleration in a Plane

Definition

Acceleration is the change in velocity divided by the time taken for that change. It has two forms: average and instantaneous acceleration.

Acceleration

\vec{a}=\dfrac{\Delta\vec{v}}{\Delta t}=\dfrac{\vec{v}_2-\vec{v}_1}{t_2-t_1}

Average Acceleration

The average acceleration is the change in velocity $\Delta\vec{v}$ divided by the time interval $\Delta t$ :

\vec{a}_{av}=\dfrac{\Delta\vec{v}}{\Delta t}=\dfrac{\Delta v_x\hat{i}+\Delta v_y\hat{j}}{\Delta t}=\dfrac{\Delta v_x}{\Delta t}\hat{i}+\dfrac{\Delta v_y}{\Delta t}\hat{j}

The figure shows a velocity $\vec{v}$ split into its components $v_x\hat{i}$ and $v_y\hat{j}$ .

Components of velocity along the x and y axes

Average Acceleration in Component Form

\vec{a}_{av}=a_x\hat{i}+a_y\hat{j}

2.4Instantaneous Acceleration

Definition

Instantaneous acceleration is the limiting value of the average acceleration when the time interval tends to zero.

Instantaneous Acceleration

\vec{a}=\lim_{\Delta t\to 0}\dfrac{\Delta\vec{v}}{\Delta t}=\dfrac{d\vec{v}}{dt}

In component form, the instantaneous acceleration and its magnitude are:

Component Form

\vec{a}=a_x\hat{i}+a_y\hat{j}

Magnitude

a=\sqrt{a_x^{2}+a_y^{2}}

Each component is the rate of change of the matching velocity component. In terms of $x$ and $y$ :

x-component

a_x=\dfrac{dv_x}{dt}=\dfrac{d}{dt}\left(\dfrac{dx}{dt}\right)=\dfrac{d^{2}x}{dt^{2}}

y-component

a_y=\dfrac{dv_y}{dt}=\dfrac{d}{dt}\left(\dfrac{dy}{dt}\right)=\dfrac{d^{2}y}{dt^{2}}

Understanding the Limiting Process

At time $t$ , the object is at point $P$ . $P_1$ , $P_2$ and $P_3$ are its positions after the time intervals $\Delta t_1$ , $\Delta t_2$ and $\Delta t_3$ , where $\Delta t_1>\Delta t_2>\Delta t_3$ . The figure also shows the velocity vectors at these points. We find $\Delta\vec{v}$ by the triangle law. The direction of $\vec{a}$ is parallel to $\Delta\vec{v}$ .

The direction of

\Delta\vec{v}

, and so of

\vec{a}

, changes as

\Delta t

becomes smaller

(a) $\Delta t_1$ average

\vec{a}

(b) $\Delta t_2$ average

\vec{a}

(c) $\Delta t_3$ average

\vec{a}

(d) $\Delta t\to 0$ instantaneous

\vec{a}

Going from (a) to (d), $\Delta t$ gets smaller and $\Delta\vec{v}$ turns. So $\vec{a}$ turns too. In (d), $\Delta t\to 0$ , and the average acceleration becomes the instantaneous acceleration. We simply call it acceleration.

Note

In a plane or in space, the angle between $\vec{v}$ and $\vec{a}$ can be anything from 0° to 180°. In one-dimensional motion, they always lie on one straight line. They are either along each other or opposite to each other.

2.5Motion in a Plane with Uniform Velocity

Definition

Uniform velocity means that a body has equal displacements in equal time intervals, however short the intervals.

Let an object move with uniform velocity $\vec{v}$ in the xy-plane. Let its position vector be $\vec{r}(0)$ at $t=0$ and $\vec{r}(t)$ at time $t$ . Then:

Position–Time Relation

\vec{v}=\dfrac{\vec{r}(t)-\vec{r}(0)}{t-0}\;\;\Rightarrow\;\;\vec{r}(t)=\vec{r}(0)+\vec{v}\,t

In terms of rectangular components:

\vec{v}=v_x\hat{i}+v_y\hat{j},\quad v=\sqrt{v_x^{2}+v_y^{2}}

\vec{r}(0)=x(0)\hat{i}+y(0)\hat{j},\;\; \vec{r}(t)=x(t)\hat{i}+y(t)\hat{j}

Along x-axis

x(t)=x(0)+v_x\,t

Along y-axis

y(t)=y(0)+v_y\,t

Concept

So uniform motion in a plane is made of two independent uniform motions, one along x and one along y. These two directions are perpendicular to each other.

2.6Motion in a Plane with Constant Acceleration

Definition

Uniform (constant) acceleration means that the velocity of a body changes by equal amounts in equal time intervals, however short the intervals.

Setup — Given Information

An object moves in the xy-plane with constant acceleration $\vec{a}$ .
Its velocity is $\vec{v}_0$ at $t=0$ and $\vec{v}$ at time $t$ .
Its position vectors are $\vec{r}_0$ at $t=0$ and $\vec{r}$ at time $t$ .

Velocity–Time Relation

Derivation

From the definition of average acceleration:
$\vec{a}=\dfrac{\vec{v}-\vec{v}_0}{t-0}=\dfrac{\vec{v}-\vec{v}_0}{t}$
Rearrange the terms:
$\vec{v}=\vec{v}_0+\vec{a}\,t$

Velocity–Time Relation

\vec{v}=\vec{v}_0+\vec{a}\,t

Written for the x and y directions separately:

x-component of Velocity

v_x=v_{0x}+a_x\,t

y-component of Velocity

v_y=v_{0y}+a_y\,t

Result

Each rectangular component of velocity changes with time exactly like the velocity in one-dimensional uniformly accelerated motion.

Position–Time Relation

Derivation

For constant acceleration, the average velocity is:
$\vec{v}_{av}=\dfrac{\vec{v}_0+\vec{v}}{2}$
Displacement = average velocity × time interval:
$\vec{r}-\vec{r}_0=\left(\dfrac{\vec{v}+\vec{v}_0}{2}\right)t=\left[\dfrac{(\vec{v}_0+\vec{a}t)+\vec{v}_0}{2}\right]t$
Simplify:
$\vec{r}-\vec{r}_0=\vec{v}_0\,t+\tfrac{1}{2}\vec{a}\,t^{2}$
So the position–time relation is:
$\vec{r}=\vec{r}_0+\vec{v}_0\,t+\tfrac{1}{2}\vec{a}\,t^{2}$

Position–Time Relation

\vec{r}=\vec{r}_0+\vec{v}_0\,t+\tfrac{1}{2}\vec{a}\,t^{2}

In terms of rectangular components:

x\hat{i}+y\hat{j}=x_0\hat{i}+y_0\hat{j}+(v_{0x}\hat{i}+v_{0y}\hat{j})\,t+\tfrac{1}{2}(a_x\hat{i}+a_y\hat{j})\,t^{2}

Equate the coefficients of $\hat{i}$ and $\hat{j}$ on both sides:

Along x-axis

x=x_0+v_{0x}\,t+\tfrac{1}{2}a_x\,t^{2}

Along y-axis

y=y_0+v_{0y}\,t+\tfrac{1}{2}a_y\,t^{2}

Note

So two-dimensional motion with constant acceleration is just two one-dimensional motions happening together. One is along x and the other along y, at right angles to each other.

Part 3

Projectile and Circular Motion

Here we study two important motions in a plane. A projectile moves under gravity with a constant acceleration. A body in uniform circular motion moves on a circle with a constant speed.

3.1Projectile Motion

Definition

Projectile motion is the motion of an object that is thrown near the Earth's surface with some initial velocity and then moves only under gravity. Its path is a curve.

Such a moving object is called a projectile. Its curved path is called its trajectory.

Examples of Projectiles

A tennis ball or baseball moving through the air.
A bullet after it leaves a rifle.
An object let go from the window of a moving train.
Water shooting out from a hole in the side of a tank.
A javelin after an athlete throws it.

Assumptions Made in This Study

Air resistance is ignored (there is no friction due to air).
The Earth's rotation and its curvature have a negligible effect.
The acceleration due to gravity $(g)$ has the same size and direction everywhere along the path.

Mathematical Analysis

Throw an object from $O$ with speed $u$ , at an angle $\theta$ above the horizontal. Split $u$ into two parts:

Horizontal (along

OX

u\cos\theta

Stays constant

Vertical (along

OY

u\sin\theta

Decreases

The vertical component becomes zero at the highest point $H$ . At $H$ , the velocity is fully horizontal.

A projectile thrown at angle

\theta

with the horizontal

3.2Equation of Path of a Projectile

Setup — Given Information

The object reaches the point $P(x, y)$ at time $t$ . We treat its motion as two motions along perpendicular directions:

Horizontal (OX): $a_x=0,\;\; u_x=u\cos\theta$ . The horizontal velocity stays constant.
Vertical (OY): $a_y=-g,\;\; u_y=u\sin\theta$ . Gravity slows down the upward motion.

Derivation

Horizontal motion: the distance covered in time $t$ is:
$x=u\cos\theta\cdot t \;\;\Rightarrow\;\; t=\dfrac{x}{u\cos\theta}\qquad \text{…(i)}$
Vertical motion: use $y=u_y t+\tfrac12 a_y t^{2}$ with $u_y=u\sin\theta$ and $a_y=-g$ :
$y=u\sin\theta\cdot t-\tfrac12 g\,t^{2}$
Put the value of $t$ from Eq. (i):
$y=u\sin\theta\left(\dfrac{x}{u\cos\theta}\right)-\tfrac12 g\left(\dfrac{x}{u\cos\theta}\right)^{2}$
Simplify to get the equation of the path:
$y=x\tan\theta-\dfrac{g\,x^{2}}{2u^{2}\cos^{2}\theta}$

Equation of Trajectory

y=x\tan\theta-\dfrac{g\,x^{2}}{2u^{2}\cos^{2}\theta}

Result

This equation has the form $y=ax-bx^{2}$ . It is the equation of a parabola. So the path of a projectile is a parabola.

3.3Time of Flight

Definition

The total time for which a projectile stays in flight is called its time of flight $(T)$ .

It is the time from $O$ to $H$ (time of ascent) plus the time from $H$ to $B$ (time of descent). So $T=2t$ , where $t=T/2$ is the time of ascent.

Derivation

At the highest point $H$ , the vertical velocity is zero. Here $u_y=u\sin\theta$ , $a_y=-g$ and $t=T/2$ . Use $v_y=u_y+a_y t$ :
$0=u\sin\theta-g\cdot\dfrac{T}{2}$
Solve for $T$ :
$\dfrac{T}{2}=\dfrac{u\sin\theta}{g}\;\;\Rightarrow\;\;T=\dfrac{2u\sin\theta}{g}$

Time of Flight

T=\dfrac{2u\sin\theta}{g}

Time to rise (time of ascent) = time to fall (time of descent)

\dfrac{u\sin\theta}{g}

3.4Maximum Height

Definition

The greatest vertical height reached by a projectile above the point of projection is called its maximum height $(H)$ .

Derivation

Take the upward motion from $O$ to $H$ . Here $u_y=u\sin\theta$ , $a_y=-g$ and $v_y=0$ . Use $v_y^{2}=u_y^{2}+2a_y H$ :
$0=(u\sin\theta)^{2}-2gH$
Solve for $H$ :
$H=\dfrac{u^{2}\sin^{2}\theta}{2g}$

Maximum Height

H=\dfrac{u^{2}\sin^{2}\theta}{2g}

Maximum height

H

of a projectile

3.5Horizontal Range

Definition

The horizontal distance covered by a projectile during its time of flight is called its horizontal range $(R)$ .

Derivation

Range = constant horizontal velocity × time of flight:
$R=u\cos\theta\times T=u\cos\theta\times\dfrac{2u\sin\theta}{g}$
Use $2\sin\theta\cos\theta=\sin 2\theta$ :
$R=\dfrac{u^{2}(2\sin\theta\cos\theta)}{g}=\dfrac{u^{2}\sin 2\theta}{g}$

Horizontal Range

R=\dfrac{u^{2}\sin 2\theta}{g}

Condition for Maximum Range

$R$ is maximum when $\sin 2\theta=1$ . That is, $2\theta=90^\circ\Rightarrow\theta=45^\circ$ .

Maximum Range

R_{max}=\dfrac{u^{2}}{g}

3.6Projectile Thrown at an Angle to the Vertical

Let a projectile be fired at an angle $\theta$ with the vertical. Then it makes an angle $(90^\circ-\theta)$ with the horizontal.

Horizontal component

u\sin\theta

Vertical component

u\cos\theta

So we use $\theta\to(90^\circ-\theta)$ in the formulas above.

Projectile fired at angle

\theta

with the vertical

Time of Flight

T=\dfrac{2u\cos\theta}{g}

Maximum Height

H=\dfrac{u^{2}\cos^{2}\theta}{2g}

Horizontal Range

R=\dfrac{u^{2}\sin 2\theta}{g}

3.7Uniform Circular Motion

Definition

Uniform circular motion is the motion of an object along a circle at constant speed. Here "uniform" tells us that the speed does not change.

Remember

Why Is It Accelerated?

The speed is constant, but the particle still accelerates. This is because the direction of its velocity changes at every point of the circular path.

\vec{v}

is along the tangent.

\vec{a}

points towards the centre

O

Examples

The tip of a clock's second hand as it goes round.

Any point on the rim of a wheel that turns at a steady rate.

3.8Terms Related to Circular Motion

1Angular Displacement

The angular displacement $(\Delta\theta)$ is the angle through which the radius vector turns at the centre in a given time.

It is dimensionless and it is a vector quantity. Its unit is the radian. Its direction is given by the right-hand rule.

2Angular Velocity

The time rate of change of angular position is called the angular velocity. It is a vector. Its unit is rad s⁻¹ and its dimensions are $[\mathrm{T^{-1}}]$ .

\omega=\dfrac{\Delta\theta}{\Delta t}=\dfrac{d\theta}{dt}

(a) Anticlockwise rotation:

\vec{\omega}

points up. (b) Clockwise rotation:

\vec{\omega}

points down (right-hand rule)

3Time Period (T)

The time taken to complete one revolution is called the time period. Its unit is the second.

4Frequency (f)

The number of revolutions completed per unit time is called the frequency. Its unit is the hertz (Hz).

5Angular Acceleration

The time rate of change of angular velocity is called the angular acceleration. Unit: rad s⁻². Dimensions: $[\mathrm{T^{-2}}]$ .

\alpha=\dfrac{\Delta\omega}{\Delta t}=\dfrac{d\omega}{dt}

Important Relations

Time Period and Frequency

T=\dfrac{1}{f}

Angular Velocity and Frequency

\omega=\dfrac{\theta}{t}=\dfrac{2\pi}{T}=2\pi f

Note

Why These Relations Hold

In one complete revolution, the radius vector turns through $\theta=2\pi$ radian in time $t=T$ . So $\omega=\dfrac{2\pi}{T}=2\pi f$ . Also, one revolution takes $\dfrac{1}{f}$ second. So the time period is $T=\dfrac{1}{f}$ .

3.9Relation between Linear and Angular Velocity

A particle moves on a circular track of radius $r$ . In time $\Delta t$ , it turns through an angle $\Delta\theta$ . It covers an arc of length $\Delta s$ along the track.

The linear velocity $\vec{v}$ is along the tangent to the path. The angular velocity $\omega$ is along the axis of rotation.

The particle moves from

A

B

in time

\Delta t

Derivation

Angular velocity:
$\omega=\dfrac{\Delta\theta}{\Delta t}\qquad \text{…(i)}$
Linear velocity:
$v=\dfrac{\Delta s}{\Delta t}\qquad \text{…(ii)}$
Angle $=\dfrac{\text{arc}}{\text{radius}}$ . So the arc length is:
$\Delta s=r\,\Delta\theta\qquad \text{…(iii)}$
From (i), (ii) and (iii):
$v=\dfrac{\Delta s}{\Delta t}=r\,\dfrac{\Delta\theta}{\Delta t}=r\,\omega$

Linear and Angular Velocity

v=r\,\dfrac{\Delta\theta}{\Delta t}=r\,\omega

where, $r$ = radius of the circle

3.10Centripetal Acceleration

Definition

Centripetal acceleration is the acceleration of a body in uniform circular motion. It always points towards the centre of the circle.

The velocity of the particle keeps changing its direction. So the particle has an acceleration that points radially inwards.

Setup — Given Information

A particle of mass $m$ moves on a circle of radius $r$ and centre $O$ . Its speed $v$ is constant. Its angular velocity $\omega$ is uniform.
At time $t$ it is at $P$ , with $\overrightarrow{OP}=\vec{r}_1$ . At time $t+\Delta t$ it is at $Q$ , with $\overrightarrow{OQ}=\vec{r}_2$ . Here $\angle POQ=\Delta\theta$ and $|\vec{r}_1|=|\vec{r}_2|=r$ .
Angular speed of the particle: $\omega=\dfrac{\Delta\theta}{\Delta t}\quad \text{…(i)}$
The velocities $\vec{v}_1$ (along the tangent $PA$ ) and $\vec{v}_2$ (along the tangent $QB$ ) have equal magnitudes: $|PA|=|QB|=|\vec{v}|$ .

The velocities at

P

and

Q

are along the tangents

Now draw the two velocity vectors from one point $P'$ . They form a triangle of vectors:

$\overrightarrow{P'A'}=\vec{v}_1$ and $\overrightarrow{P'B'}=\vec{v}_2$ .
The angle between them is also $\Delta\theta$ .
The side $\overrightarrow{A'B'}$ gives the change in velocity.

A triangle of velocity vectors

Derivation

By the triangle law of vectors, $\overrightarrow{P'A'}+\overrightarrow{A'B'}=\overrightarrow{P'B'}$ . So the change in velocity is:
$\overrightarrow{A'B'}=\overrightarrow{P'B'}-\overrightarrow{P'A'}=\vec{v}_2-\vec{v}_1=\Delta\vec{v}$
When $\Delta t\to 0$ , $A'$ is very close to $B'$ . Then $A'B'$ is almost an arc of a circle whose radius is $P'A'=|\vec{v}|$ . So:
$\Delta\theta=\dfrac{A'B'}{P'A'}=\dfrac{|\Delta\vec{v}|}{|\vec{v}|}$
From Eq. (i), $\Delta\theta=\omega\,\Delta t$ . Therefore:
$\omega\,\Delta t=\dfrac{|\Delta\vec{v}|}{|\vec{v}|}\;\;\Rightarrow\;\;\omega\,|\vec{v}|=\dfrac{|\Delta\vec{v}|}{\Delta t}$
Use $v=\omega r$ :
$\dfrac{|\Delta\vec{v}|}{\Delta t}=\omega\,|\vec{v}|=(\omega r)\,\omega=\omega^{2}r$
In the limit $\Delta t\to 0$ , the ratio $\dfrac{|\Delta\vec{v}|}{\Delta t}$ is the size of the centripetal acceleration at $P$ :
$|\vec{a}|=\dfrac{|\Delta\vec{v}|}{\Delta t}=\omega^{2}r=\left(\dfrac{v}{r}\right)^{2}r=\dfrac{v^{2}}{r}$

Centripetal Acceleration

a=\dfrac{v^{2}}{r}=r\omega^{2}=v\omega

It points towards the centre of the circle.

Note

Radius of Curvature

A small piece of any curved path looks like a part of a circle. The radius of the circle that fits the curve best at a point is called the radius of curvature at that point. In $a=\dfrac{v^{2}}{r}$ , the $r$ stands for this radius of curvature.

End of Chapter 3 · notesshala.in