Physics

Session

Kinematics - 1

Session Opener

You fly from Delhi to Mumbai

New Delhi

Mumbai

Hyderabad

•

•

•Then you fly from Mumbai to Hyderabad

Does it mean that you have flown from Delhi to Hyderabad ?

Session Objectives

1. Scalars and vectors

2. Definitions

3. Vector addition and vector subtraction

4. Components of vectors

5. Multiplication of vectors

Scalars And Vectors

Scalars : physical quantities that can be completely specified by just numbers.

Vectors : physical quantities that cannot be completely specified by their magnitude alone. They also need a ‘direction’ specification.

For instance, displacement is a vector quantity.

Vectors

OP : Position vector--------------

Length OP meter

Directed from O to P.

Both r and are needed to specify P

Vectors also obey the law a b b a

O x

yP

r

Vectors

Component of vector

1

2 2 2

1

OP OX OY

OYtan

OX

OX OP cos

OY OP sin

O x

yP

r

X

Y

Unit Vectors

O x

yP

r

X

Y

Unit vector : vector with magnitude of unity

rˆ ˆr , r one unit

r

ˆ ˆr i r cos j r sinO

x

y

rj

i

Class Exercise

Class Exercise - 6

----------------------------A 3 i 4 j ; B 7 i 24 j .

Find such that C = B and is in the direction of .

--------------C

--------------C--------------

A

The unit vector in the direction of ,

where 5 is the magnitude of . Magnitude of the vector = 25 units. Hence the value of

-------------- 3 i 4 jA

5--------------A--------------

B

-------------- 3 i 4 jC 25 15 i 20 j

5

Solution :

Vector Definitions

Null vector : vector with zero magnitude

Equal Vectors : ˆˆa b aa bb

If a = b

Direction of both are the same ˆa b

Addition of vectors

OP PQ OQ ------------------------------------------

(displacement O to P, then P to Q : same as displacement O to Q)

Triangle law of addition

a b c

O

P

Q

b

a

c

Addition of vectors

Parallelogram law

From geometry : 1/ 22 2c a b 2abcos

bsintan

a bcos

a b c

b

a

c

a

b

bsin

bcos

Subtraction of Vectors

Subtracting a vector from vector Is equivalent to adding vector

b

a

( b) to a

d a b a b

b

a

b

d

Class Exercise

Class Exercise - 1

The maximum resultant of 2 vectors is 18 units. The resultant magnitude is 12. If the resultant is perpendicular to the smaller vector , find the magnitude of the two vectors.

R

A

B

A + B = 18 [The maximum amplitude of the resultant is when they are collinear.]

Also 122 + A2 = (18 – A)2.

Solving, we get

A = 5 and B = 13

Solution :

Class Exercise - 2

------------------------------------------A B R ,

------------------------------------------A 2B P ,

--------------P

is perpendicular to . Then--------------A

(a) A = B (b) A + B = R

(c) A = R (d) B = R

Since the given triangle is a right-angled triangle B = R, where point O is the mid point of the hypotenuse.

P

A

B

BR

o

Solution :

Class Exercise - 3

The components along the X and Y-axis of are 3 m and 4 m respectively. The components along X and Y-axis of the vectors is 2 m and 6 m respectively. Find the components of along X and Y-axis?

--------------A

----------------------------A B --------------

B

--------------Let B x i y j .Then (3 x) i (4 y) j 2 i 6 j

So x = –1 and y = 2.

--------------

Hence B – i 2 j

Solution :

Class Exercise - 4

A man swims across a river that flows at 3 m/s. The man moves in a direction directly perpendicular to the flow of the river at 4 m/s. If the width of the river is 100 m, then find the time taken by the man to reach the opposite bank.

Solution - 4

r

R

m

m r R . m, r , R

denote the components of velocities of man, river and resultant during the motion.

The resultant direction in which the man moves is not along the shortest line joining the two banks. But nevertheless the component of the velocities in the Y direction is 4 m/s. Hence the man takes , i.e 25 s to

cross over to the opposite bank.

1004

Multiplication of vector by a real number

a multiplied by

b a a a

Ifi s negative

c a

Direction : opposite a

Scalar multiplication of vectors

Scalar product (dot product)

x x y yc a.b abcos a b a b

a.b is scalar.

a.b b.a

b

a

bcos

Class Exercise

Class Exercise - 8

----------------------------A 5 i 7 j– 3k;B 2 i 2 j– ck .

If the two vectors are perpendicular to each other, then find c.

If the vectors A and B are perpendicular to each other, then

----------------------------A B 0

----------------------------

Therefore A B 10 14 3c 0. So,c 8

Solution :

Class Exercise - 9

What is the angle between the two vectors

----------------------------A –2 i 3 j k ; B i 2 j– 4k ?

A·B = AB cos = –2 + 6 – 4 = 0. Hence the two vectors are perpendicular to each other.

Solution :

Class Exercise - 10

Find the component of the vector in the direction of the vector .

--------------A 3 i 4 j 5k

--------------B 3 i 4 j

A·B = AB cos. This also means that A.B is the product of the magnitude of B and the magnitude of the component of A in the direction of B.

Magnitude of B = 5 units.

A·B = 25. Hence the magnitude of the component of A in the direction of B is 5, which can happen only if the component vector is .

3 i 4 j

Alternately, if we compare the components of the two vectors along the X, Y directions, we find that they are the same. Hence the component of A in the direction of B must be same as the vector B.

Solution :

Vector multiplication of vectors

Vector product (cross product)

ˆc a b is vector absin c

x y y xˆc k a b a b

a b b a b

a

Vector multiplication of vectors

Vector product has an orientation given by the right-hand thumb rule.

Curl palm of your right hand from the first vector to the second vector keeping the thumb upright. The direction of the thumb gives the direction of resultant.

Vector multiplication of vectors

A--------------

A--------------

B--------------

C A B ------------------------------------------

Product of two vectors

ˆˆ ˆi j k

ˆˆ ˆj k i

ˆ ˆ ˆk i j

Some special cases

(ii) For unit vectors

ˆ ˆˆ ˆ ˆ ˆi i 0 j j k k ˆ ˆˆ ˆ ˆ ˆi . i j . j k . k 1

ˆ ˆi. j 0

ˆj .k 0

ˆ ˆk. i 0

----------------------------A B ABsin 0(i) (for parallel vectors)

----------------------------A .B ABcos 0 (for perpendicular vectors)

Class Exercise

Class Exercise - 7

What is the torque of a force

acting

at the point

and about the origin?

F 2 i – 3 j 4k N

r 3 i 2 j 3k

Note r F

Solution :

r 3 i 2 j 3k ,F 2 i – 3 j 4k

F r f –9k– 12 j – 4k 8 i 6 j 9 i 17 i

Hence the moment of the force

Class Exercise - 5

The angle between two forces of equal magnitude acting at a point, such that the resultant force also has the same magnitude is___.

Solution :

F

FR=F

120

2 2 2

2 2

R F F 2F.F cos (but R F given)

F 2F (1 cos )

11 cos

21

cos2

Thank you