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Kinematics-1

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Session-1 of Kinematics for Arc-IITJEE study program

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Kinematics-1

1. 1. Physics
2. 2. Session Kinematics - 1
3. 3. 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 ?
4. 4. Session Objectives 1. Scalars and vectors 2. Definitions 3. Vector addition and vector subtraction 4. Components of vectors 5. Multiplication of vectors
5. 5. 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.
6. 6. Vectors OP : Position vector uuur Length OP meter Directed from O to P. Both r and θ are needed to specify P Vectors also obey the law a b b a+ = + r r r r O x y P θ r r
7. 7. Vectors Component of vector ( ) 1 2 2 2 1 OP OX OY OY tan OX − = + θ = OX OP cos OY OP sin = θ = θ O x y P θ r r X Y
8. 8. Unit Vectors O x y P θ r r X Y Unit vector : vector with magnitude of unity = = = θ + θ r r r r ˆ ˆr , r one unit r ˆ ˆr i r cos j r sin O x y θ ˆrˆj ˆi
9. 9. Class Exercise
10. 10. Class Exercise - 6 ∧ ∧ ∧ ∧ = + = + ur ur A 3 i 4 j ; B 7 i 24 j . Find such that C = B and is in the direction of . ur C ur Cur A The unit vector in the direction of , where 5 is the magnitude of . Magnitude of the vector = 25 units. Hence the value of ∧ ∧ + = ur 3 i 4 j A 5ur Aur B ∧ ∧ ∧ ∧   +  = = +     ur 3 i 4 j C 25 15 i 20 j 5 Solution :
11. 11. Vector Definitions Null vector : vector with zero magnitude Equal Vectors : ˆˆa b aa bb= ⇒ = r r If a = b Direction of both are the same ˆˆa b=
12. 12. Addition of vectors OP PQ OQ+ = uuur uuur uuur (displacement O to P, then P to Q : same as displacement O to Q) Triangle law of addition a b c+ = r r r O P Q b r a r c r α θ
13. 13. Addition of vectors Parallelogram law From geometry : ( ) 1/2 2 2 c a b 2abcos= + + θ bsin tan a bcos θ α = + θ a b c+ = r r r b r a r c r αθ a b θ bsinθ bcosθ
14. 14. Subtraction of Vectors Subtracting a vector from vector Is equivalent to adding vector b r a r ( b) to a− r r ( )d a b a b= − = + − r r r r r b r a r b− r d r
15. 15. Class Exercise
16. 16. 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 :
17. 17. Class Exercise - 2 + = ur ur ur A B R , + = ur ur ur A 2B P , ur P is perpendicular to . Then ur 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 B R o Solution :
18. 18. 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? ur A + ur ur A B ur B ∧ ∧ ∧ ∧ ∧ ∧ = + + + + = + ur Let B x i y j .Then (3 x) i (4 y) j 2 i 6 j So x = –1 and y = 2. ∧ ∧ = + ur Hence B – i 2 j Solution :
19. 19. 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.
20. 20. 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. 100 4
21. 21. Multiplication of vector by a real number a multiplied by λ r b a a a= λ ⇒ λ = λ r r r r If is negativeλ c a= −λ r r Direction : opposite a r
22. 22. Scalar multiplication of vectors Scalar product (dot product) x x y yc a.b abcos a b a b= = θ = + r r a.b is scalar. a.b b.a= r r r r r r b r a r bcosθ θ
23. 23. Class Exercise
24. 24. Class Exercise - 8 ∧ ∧ ∧ ∧ ∧ ∧ = + = + ur ur 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 × = ur ur A B 0 × = + + = =− ur ur Therefore A B 10 14 3c 0. So,c 8 Solution :
25. 25. Class Exercise - 9 What is the angle between the two vectors ∧ ∧ ∧ ∧ ∧ ∧ = + + = + ur ur A –2 i 3 j k ; B i 2 j– 4k ? A·B = AB cosθ = –2 + 6 – 4 = 0. Hence the two vectors are perpendicular to each other. Solution :
26. 26. Class Exercise - 10 Find the component of the vector in the direction of the vector . ∧ ∧ ∧ = + + ur A 3 i 4 j 5k ∧ ∧ = + ur B 3 i 4 j A·B = AB cosθ. 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 :
27. 27. Vector multiplication of vectors Vector product (cross product) ˆc a b is vector absin c= × = θ r r r ( )x y y x ˆc k a b a b= − r a b b a× = − × r r r r b r a r θ
28. 28. 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.
29. 29. Vector multiplication of vectors A ur θA ur B ur C A B= × ur ur ur
30. 30. 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 ˆˆj.k 0 ˆ ˆk . i 0 × = θ = ur ur A B AB sin 0(i) (for parallel vectors) = θ = ur ur A.B AB cos 0 (for perpendicular vectors)
31. 31. Class Exercise
32. 32. Class Exercise - 7 What is the torque of a force acting at the point and about the origin? ∧ ∧ ∧  = + ÷ ÷   r F 2 i – 3 j 4k N ∧ ∧ ∧ = + + r r 3 i 2 j 3k  τ = ×  r r r Note r F Solution : ∧ ∧ ∧ ∧ ∧ ∧ = + + = + r r r 3 i 2 j 3k,F 2 i – 3 j 4k ∧ ∧ ∧ ∧ ∧ ∧ ∧ = × = + + + = r F r f –9k– 12 j – 4k 8 i 6 j 9 i 17 i Hence the moment of the force
33. 33. 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 F R=F θ 120∴ θ = o 2 2 2 2 2 R F F 2F.F cos (but R F given) F 2F (1 cos ) = + + θ = = + θ 1 1 cos 2 1 cos 2 ⇒ = + θ ⇒ θ = −
34. 34. Thank you