PHYSICS I. Lecture 1. Charudatt Kadolkar. Jul-Nov IIT Guwahati

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1 PHYSICS I Lecture 1 Charudatt Kadolkar IIT Guwahati Jul-Nov 2014

2 Section 1 Introduction to the Course

3 Syllabus Topics Classical Mechanics: Kinetic Energy rest mass energy

4 Syllabus Topics Classical Mechanics: Kinetic Energy rest mass energy Relativistic Mechanics: Kinetic Energy rest mass energy and de Broglie wavelength size

5 Syllabus Topics Classical Mechanics: Kinetic Energy rest mass energy Relativistic Mechanics: Kinetic Energy rest mass energy and de Broglie wavelength size Quantum Mechanics: de Broglie wavelength size and Kinetic Energy rest mass energy

6 Syllabus Topics Classical Mechanics: Kinetic Energy rest mass energy Relativistic Mechanics: Kinetic Energy rest mass energy and de Broglie wavelength size Quantum Mechanics: de Broglie wavelength size and Kinetic Energy rest mass energy Texts An Intro to Mechanics Kleppner/Kolenkow Modern Physics Krane Web page:

7 Classes Two classes: Tue (10AM and 3PM) and Wed (11AM and 2PM) Primarily presented with slides. Slides will be available online after the class. Do not hesitate to interrupt if you have doubts during the class.

8 Tutorials The main purpose of the tutorial is to provide you with an opportunity to interact with a teacher. The teacher will assist you in clearing your doubts and answer your queries regarding the course topics. A problem sheet will be given to you for your practice. These problems also indicate the difficulty level of the exams. You are expected to attempt these problems before you come to tutorial class. Ask your doubts regarding these problems to your teacher during tutorial class. The teacher may or may not solve all the problems in the tutorial class. In case you find a problem very difficult, do ask your teacher to help you.

9 Evaluations Two Quizzes each of 10% weightage Mid-semester Exam of 30% weightage End-Semester Exam of 50% weightage Pass mark > 30

10 Class Manners Maintain Silence Attendance Rule Mobile Phone Policy

11 Section 2 Review of Kinematics in 1 and 2 Dimensions

12 Kinematics in One Dimensions Notion of a particle

13 Kinematics in One Dimensions Notion of a particle Trajectory of the particle: Position of the particle as a function of time, say, x (t)

14 Kinematics in One Dimensions Notion of a particle Trajectory of the particle: Position of the particle as a function of time, say, x (t) Instantaneous velocity is defined as v (t) = dx (t) (1) dt

15 Kinematics in One Dimensions Notion of a particle Trajectory of the particle: Position of the particle as a function of time, say, x (t) Instantaneous velocity is defined as Instantaneous acceleration, as v (t) = dx (t) (1) dt a (t) = dv dt (t) = d 2 x (t) (2) dt2

16 Simple Example If x (t) = sin(t), then v (t) = cos(t) and a (t) = sin(t). 1D Motion Position Velocity Acceleration

17 Kinematics in One Dimension Usually the x (t) is not known in advance!

18 Kinematics in One Dimension Usually the x (t) is not known in advance! However, the acceleration a (t) is known at all times t > t 0 and at t 0, position x (t 0 ) and velocity v (t 0 ) are known.

19 Kinematics in One Dimension Usually the x (t) is not known in advance! However, the acceleration a (t) is known at all times t > t 0 and at t 0, position x (t 0 ) and velocity v (t 0 ) are known. The formal solution to this problem is t v (t) = v (t 0 ) + a (t ) dt t 0 x (t) = x (t 0 ) + t t 0 v (t ) dt

20 Constant Acceleration Let the acceleration of a particle be a 0, a constant at all times. If, at t = 0 velocity of the particle is v 0, then

21 Constant Acceleration Let the acceleration of a particle be a 0, a constant at all times. If, at t = 0 velocity of the particle is v 0, then t v (t) = v 0 + a 0 dt 0 = v 0 + a 0 t (3)

22 Constant Acceleration Let the acceleration of a particle be a 0, a constant at all times. If, at t = 0 velocity of the particle is v 0, then t v (t) = v 0 + a 0 dt 0 = v 0 + a 0 t (3) And if the position at t = 0 is x 0, t x (t) = x 0 + (v 0 + a 0 t ) dt 0 = x 0 + v 0 t a 0t 2 (4) These are familiar formulae.

23 Kinematics in One Dimension More complex situations may arise, where an acceleration is specified as a function of position, velocity and time. a (x, ẋ, t).

24 Kinematics in One Dimension More complex situations may arise, where an acceleration is specified as a function of position, velocity and time. a (x, ẋ, t). In this case, we need to solve a differential equation which may or may not be simple. d 2 x = a (x, ẋ, t) dt2

25 Some Tricks Simple integration

26 Some Tricks Simple integration Acceleration is function of t alone dv dt = a(t)

27 Some Tricks Simple integration Acceleration is function of t alone dv dt = a(t) dv Acceleration is function of x alone dt = dx dt dv dx = v dv dx = a(x)

28 Some Tricks Simple integration Acceleration is function of t alone dv dt = a(t) dv Acceleration is function of x alone dt = dx dt Acceleration is function of v alone dv dt dv dx = v dv = a(v) dt = dv dx = a(x) a(v)

29 Some Tricks Simple integration Acceleration is function of t alone dv dt = a(t) dv Acceleration is function of x alone dt = dx dt Acceleration is function of v alone dv dt First ordered differential equation dv dx = v dv = a(v) dt = dv dx = a(x) a(v)

30 Some Tricks Simple integration Acceleration is function of t alone dv dt = a(t) dv Acceleration is function of x alone dt = dx dt Acceleration is function of v alone dv dt First ordered differential equation Acceleration is function of v and t dv dt dv dx = v dv = a(v) dt = dv = a(v, t) dx = a(x) a(v)

31 Some Tricks Simple integration Acceleration is function of t alone dv dt = a(t) dv Acceleration is function of x alone dt = dx dt Acceleration is function of v alone dv dt First ordered differential equation dv dx = v dv = a(v) dt = dv dv Acceleration is function of v and t = a(v, t) dt dv Acceleration is function of v and x v = a(v, x) dx dx = a(x) a(v)

32 Some Tricks Simple integration Acceleration is function of t alone dv dt = a(t) dv Acceleration is function of x alone dt = dx dt Acceleration is function of v alone dv dt First ordered differential equation dv dx = v dv = a(v) dt = dv dv Acceleration is function of v and t = a(v, t) dt dv Acceleration is function of v and x v = a(v, x) dx Second ordered differential equation: other cases dx = a(x) a(v)

33 Kinematics in Two Dimensions The position now is specified by a vector in a plane. r (t) = x (t) i + y (t) j. 6 4 y dy y 2 r t dt r t Position at t+dt Position at t x dx x

34 Kinematics in Two Dimensions The instantaneous velocity vector is defined as v (t) = d dt r = dx dt i + dy dt j

35 Kinematics in Two Dimensions The instantaneous velocity vector is defined as Simillarly, we can show that v (t) = d dt r = dx dt i + dy dt j a (t) = d 2 x dt 2 i + d 2 y dt 2 j

36 Kinematics in Two Dimensions Velocity and Acceleration are vector quantities. Formal Solutions can be written in exactly the same way as in case of one dimensional motion.

37 Kinematics in Two Dimensions Velocity and Acceleration are vector quantities. Formal Solutions can be written in exactly the same way as in case of one dimensional motion. Typical problem, however, may specify acceleration as a function of coordinates, velocities etc. In this case, we have to solve two coupled differential equations d 2 x dt 2 = a x (x, y, ẋ, ẏ, t) d 2 y dt 2 = a y (x, y, ẋ, ẏ, t)

38 Projectile Motion A ball is projected at an angle θ with a speed u. The net acceleration is in downward direction. Then a x = 0 and a y = g. The equations are We know the solutions d 2 x dt 2 = 0 d 2 y dt 2 = g

39 Charged Particle in Magnetic Field A particle has a velocity v in XY plane. Magnetic field is in z direction The acceleration is given by q m v B d 2 x dt 2 = qb m v y d 2 y dt 2 = qb m v x Solution is rather simple, that is circular motion in xy plane.

40 Section 3 Polar Coordinate System

41 Cartesian Coordinate System Coordinates: A unique pair of numbers for each point Y x P y X

42 Cartesian Coordinate System Coordinates: A unique pair of numbers for each point Choose origin and coordinate axes Y x P y X

43 Cartesian Coordinate System Coordinates: A unique pair of numbers for each point Choose origin and coordinate axes x coordinate of P = distance of P from Y axis (- if left side) Y x P y X

44 Cartesian Coordinate System Coordinates: A unique pair of numbers for each point Choose origin and coordinate axes x coordinate of P = distance of P from Y axis (- if left side) y coordinate of P = distance of P from X axis (- if below) Y x P y X

45 Cartesian Coordinate System Coordinates: A unique pair of numbers for each point Choose origin and coordinate axes x coordinate of P = distance of P from Y axis (- if left side) y coordinate of P = distance of P from X axis (- if below) Cartesian coordinates are unique Y x P y X

46 Cartesian Coordinate System Coordinates: A unique pair of numbers for each point Choose origin and coordinate axes x coordinate of P = distance of P from Y axis (- if left side) y coordinate of P = distance of P from X axis (- if below) Cartesian coordinates are unique Ranges: x, y (, ) Y x P y X

47 Cartesian Coordinate System Coordinates: A unique pair of numbers for each point Choose origin and coordinate axes x coordinate of P = distance of P from Y axis (- if left side) y coordinate of P = distance of P from X axis (- if below) Cartesian coordinates are unique Ranges: x, y (, ) Operational definition Y x P y X

48 Constant Coordinate Lines Y Constant y lines X

49 Constant Coordinate Lines Y Constant y lines Constant x lines X

50 Constant Coordinate Lines Y Constant y lines Constant x lines Clearly no two points can have same coordinates X

51 Unit Vectors Position vector to P : r = xî + yĵ P x

52 Unit Vectors P Q Position vector to P : r = xî + yĵ Change x keeping y constant, then P moves to Q. Let PQ = d r. x dx x

53 Unit Vectors P Q Position vector to P : r = xî + yĵ Change x keeping y constant, then P moves to Q. Let PQ = d r. 1 r î = r/ x x x dx x

54 Unit Vectors x P Q dx x Position vector to P : r = xî + yĵ Change x keeping y constant, then P moves to Q. Let PQ = d r. 1 r î = r/ x x 1 r ĵ = r/ y y

55 Unit Vectors P i Position vector to P : r = xî + yĵ Change x keeping y constant, then P moves to Q. Let PQ = d r. 1 r î = r/ x x 1 r ĵ = r/ y y The unit vector for a coordinate is the direction in which a point moves when only that coordinate chages.

56 Unit Vectors P i Position vector to P : r = xî + yĵ Change x keeping y constant, then P moves to Q. Let PQ = d r. 1 r î = r/ x x 1 r ĵ = r/ y y The unit vector for a coordinate is the direction in which a point moves when only that coordinate chages. Unit vectors are perpendicular to the constant coordinate lines

57 Polar Coordinates Each point P = (x, y) of a plane can also be specified by its distance from the origin, O and the angle that the line OP makes with the x axis. P r sin Θ r Θ r cos Θ

58 Polar Coordinates Each point P = (x, y) of a plane can also be specified by its distance from the origin, O and the angle that the line OP makes with the x axis. Transformations: r sin Θ r Θ P r = x 2 + ( y 2 θ = tan 1 y ) x The pair of numbers (r, θ) are called Polar Coordinates r cos Θ

59 Polar Coordinates Each point P = (x, y) of a plane can also be specified by its distance from the origin, O and the angle that the line OP makes with the x axis. Transformations: r sin Θ r Θ r cos Θ P r = x 2 + ( y 2 θ = tan 1 y ) x The pair of numbers (r, θ) are called Polar Coordinates Inverse Transformations: x = r cos θ y = r sin θ

60 Polar Coordinates Each point P = (x, y) of a plane can also be specified by its distance from the origin, O and the angle that the line OP makes with the x axis. Transformations: r sin Θ r Θ r cos Θ P r = x 2 + ( y 2 θ = tan 1 y ) x The pair of numbers (r, θ) are called Polar Coordinates Inverse Transformations: x = r cos θ y = r sin θ Ranges: r [0, ) and θ [0, 2π)

61 Polar Coordinates Each point P = (x, y) of a plane can also be specified by its distance from the origin, O and the angle that the line OP makes with the x axis. Transformations: r sin Θ r Θ r cos Θ P r = x 2 + ( y 2 θ = tan 1 y ) x The pair of numbers (r, θ) are called Polar Coordinates Inverse Transformations: x = r cos θ y = r sin θ Ranges: r [0, ) and θ [0, 2π) Discontinuity in θ at origin

62 Constant Coordinate Curves constant r curves are circles

63 Π 8 Constant Coordinate Curves constant r curves are circles 1.6 Π 2 3 Π 8 Π 4 constant θ curves are straight rays originating at origin Π Π 4 3 Π 8

64 Π 8 Constant Coordinate Curves constant r curves are circles Π 2 3 Π 8 Π 4 Π 8 constant θ curves are straight rays originating at origin. Unit vectors are perpendicular to constant coordinate curves 0 Π 4 3 Π 8

65 Unit Vectors

66 Unit Vectors

67 Unit Vectors Θ

68 Unit Vectors Θ

69 Unit Vectors Θ r

70 Unit Vectors Θ j r Θ Θ i Θ

71 Unit Vectors Position Vector to P = (r, θ) Θ j r = xî ( + yĵ ) = r cos θî + sin θĵ r Θ Θ i Θ

72 Unit Vectors Position Vector to P = (r, θ) Θ j r = xî ( + yĵ ) = r cos θî + sin θĵ r Unit vectors at P Θ Θ i ^r = 1 r r r r = i cos θ + j sin θ Θ

73 Unit Vectors Position Vector to P = (r, θ) Θ j r = xî ( + yĵ ) = r cos θî + sin θĵ r Unit vectors at P Θ Θ i ^r = 1 r r r r = i cos θ + j sin θ Θ And ^θ = 1 r r θ θ = i sin θ + j cos θ

74 Unit Vectors Unit vectors at a point P depend on θ coordnate of P Θ Θ r r 45 30

75 Unit Vectors Θ Unit vectors at a point P depend on θ coordnate of P Example: If P (1, π/6), then Θ r r ˆr = 1 ( ) 3 î + 2 ĵ ˆθ = 1 ( î 2 + ) 3ĵ 45 30

76 Unit Vectors Θ Unit vectors at a point P depend on θ coordnate of P Example: If P (1, π/6), then Θ r r ˆr = 1 ( ) 3 î + 2 ĵ ˆθ = 1 ( î 2 + ) 3ĵ Example: Let P (1, π/4), then ˆr = 1 ) (î + ĵ 2 ˆθ = 1 ) ( î + ĵ 2

77 Unit Vectors Θ Unit vectors at a point P depend on θ coordnate of P Example: If P (1, π/6), then Θ r r ˆr = 1 ( ) 3 î + 2 ĵ ˆθ = 1 ( î 2 + ) 3ĵ Example: Let P (1, π/4), then ˆr = 1 ) (î + ĵ 2 ˆθ = 1 ) ( î + ĵ 2 Note that everywhere, ˆr ˆθ.

78 r Unit Vectors These unit vectors are functions of the polar coordinates, only of θ in fact. ^r = i cos θ + j sin θ ^θ = i sin θ + j cos θ Θ r r Θ

79 r Unit Vectors These unit vectors are functions of the polar coordinates, only of θ in fact. ^r = i cos θ + j sin θ ^θ = i sin θ + j cos θ Θ r r ^r θ ^θ θ = sin θî + cos θĵ = ^θ = cos θî sin θĵ = ^r Θ

80 Motion in Plane: Polar Coordinates Suppose a particle is travelling along a trajectory given by r (t). Now the position vector 6 y dy 4 Position at t+dt y 2 r t dt r t Position at t x dx x

81 Motion in Plane: Polar Coordinates Suppose a particle is travelling along a trajectory given by r (t). Now the position vector 6 r (t) = ix (t) + jy (t) y dy 4 Position at t+dt y 2 r t dt r t Position at t x dx x

82 Motion in Plane: Polar Coordinates Suppose a particle is travelling along a trajectory given by r (t). Now the position vector 6 r (t) = ix (t) + jy (t) = r (t) (i cos (θ (t)) + j sin (θ (t))) y dy 4 Position at t+dt y 2 r t dt r t Position at t x dx x

83 Motion in Plane: Polar Coordinates Suppose a particle is travelling along a trajectory given by r (t). Now the position vector y dy 6 4 Position at t+dt r (t) = ix (t) + jy (t) = r (t) (i cos (θ (t)) + j sin (θ (t))) = r (t)^r (θ (t)) y 2 r t dt r t Position at t x dx x

84 Motion in Plane: Polar Coordinates Suppose a particle is travelling along a trajectory given by r (t). Now the position vector 6 4 y dy y 2 r t dt r t Position at t+dt Position at t r (t) = ix (t) + jy (t) = r (t) (i cos (θ (t)) + j sin (θ (t))) = r (t)^r (θ (t)) = rˆr x dx x

85 Motion in Plane: Polar Coordinates Suppose a particle is travelling along a trajectory given by r (t). Now the position vector 6 4 y dy y 2 r t dt r t Position at t+dt Position at t r (t) = ix (t) + jy (t) = r (t) (i cos (θ (t)) + j sin (θ (t))) = r (t)^r (θ (t)) = rˆr x dx x clearly the position vector depends on θ through ^r vector.

86 Motion in Plane: Polar Coordinates Then the velocity vector v = dr dt

87 Motion in Plane: Polar Coordinates Then the velocity vector v = dr dt = d (r (t)^r (θ (t))) dt

88 Motion in Plane: Polar Coordinates Then the velocity vector v = dr dt = d (r (t)^r (θ (t))) dt = dr d^r dθ + r dt^r dθ dt

89 Motion in Plane: Polar Coordinates Then the velocity vector v = dr dt = d (r (t)^r (θ (t))) dt = dr d^r dθ + r dt^r dθ dt = ṙ^r+r θ^θ

90 Motion in Plane: Polar Coordinates Then the velocity vector Radial velocity = ṙˆr v = dr dt = d (r (t)^r (θ (t))) dt = dr d^r dθ + r dt^r dθ dt = ṙ^r+r θ^θ

91 Motion in Plane: Polar Coordinates Then the velocity vector Radial velocity = ṙˆr v = dr dt = d (r (t)^r (θ (t))) dt = dr d^r dθ + r dt^r dθ dt = ṙ^r+r θ^θ tangential velocity = r θ^θ.

92 Motion in Plane: Polar Coordinates Then the velocity vector Radial velocity = ṙˆr v = dr dt = d (r (t)^r (θ (t))) dt = dr d^r dθ + r dt^r dθ dt = ṙ^r+r θ^θ tangential velocity = r θ^θ. θ is called the angular speed.

93 Motion in Polar Coordinates The Acceleration Vector a = dv dt

94 Motion in Polar Coordinates The Acceleration Vector a = dv dt = d dt ( ṙ^r+r θ^θ )

95 Motion in Polar Coordinates The Acceleration Vector a = dv dt = d dt = ( ṙ^r+r θ^θ ) ( d d ṙˆr + r θ dt dtˆr ) ( + ṙ dt^r+ d d ( θ) ) r ^θ dt

96 Motion in Polar Coordinates The Acceleration Vector a = dv dt = d ( ṙ^r+r dt θ^θ ) ( ) d d = ṙˆr + r θ dt dtˆr = ( r r θ 2)^r + ( + ṙ dt^r+ d d ( θ) ) r ^θ dt ( r θ + 2ṙ θ) ^θ

97 Motion in Polar Coordinates The Acceleration Vector a = dv dt = d ( ṙ^r+r dt θ^θ ) ( ) d d = ṙˆr + r θ dt dtˆr = ( r r θ 2)^r + ( + ṙ dt^r+ d d ( θ) ) r ^θ dt ( r θ + 2ṙ θ) ^θ The term r θ 2ˆr is usual centripetal acceleration.

98 Motion in Polar Coordinates The Acceleration Vector a = dv dt = d ( ṙ^r+r dt θ^θ ) ( ) d d = ṙˆr + r θ dt dtˆr = ( r r θ 2)^r + ( + ṙ dt^r+ d d ( θ) ) r ^θ dt ( r θ + 2ṙ θ) ^θ The term r θ 2ˆr is usual centripetal acceleration. The term 2ṙ θ is called as coriolis acceleration.

99 Uniform Linear Motion Position Vector: r(t) = 2tî + ĵ. Polar coordinates r(t) = x(t) 2 + y(t) 2 = 4t Radial Tangential and tan θ(t) = 1/2t 1

100 Uniform Linear Motion Velocity Vector: v(t) = 2î. Radial ṙ = θ = v(t) = 2t 4t t t 4t2 + 1ˆr 2 ˆθ 4t2 + 1 Tangential

101 Uniform Linear Motion Velocity Vector: v(t) = 2î. Radial ṙ = θ = v(t) = 2t 4t t t 4t2 + 1ˆr 2 ˆθ 4t2 + 1 Tangential

102 Uniform Linear Motion Velocity Vector: v(t) = 2î. Radial ṙ = θ = v(t) = 2t 4t t t 4t2 + 1ˆr 2 ˆθ 4t2 + 1 Tangential

103 Uniform Linear Motion Velocity Vector: v(t) = 2î. Radial ṙ = θ = v(t) = 2t 4t t t 4t2 + 1ˆr 2 ˆθ 4t2 + 1 Tangential

104 Uniform Linear Motion Velocity Vector: v(t) = 2î. Radial ṙ = θ = v(t) = 2t 4t t t 4t2 + 1ˆr 2 ˆθ 4t2 + 1 Tangential

105 Uniform Linear Motion Velocity Vector: v(t) = 2î. Radial ṙ = θ = v(t) = 2t 4t t t 4t2 + 1ˆr 2 ˆθ 4t2 + 1 Tangential

106 Circular Motion In a circular motion, r = R = Constant. Then, ṙ = r = 0. Thus v = R θˆθ a = R θ 2ˆr + R θˆθ Consider a special case of uniform circular motion: θ = 0. Speed is constant, and velocity is tangential Acceleration is radial (central)

107 Circular Motion Consider a now a case of a non-uniform circular motion, in which θ = α = constant Here, v = R θˆθ = Rαt ˆθ a = R θ 2ˆr + R θˆθ = Rα 2 t 2ˆr + Rαˆθ movie

108 Circular Motion Consider a now a case of a non-uniform circular motion, in which θ = α = constant Here, v = R θˆθ = Rαt ˆθ a = R θ 2ˆr + R θˆθ = Rα 2 t 2ˆr + Rαˆθ

109 Circular Motion: Hammer Throw

110 Circular Motion: Hammer Throw

111 Circular Motion: Hammer Throw

112 Circular Motion: Hammer Throw

113 Circular Motion: Hammer Throw

114 Circular Motion: Hammer Throw

115 Spiral Motion Consider a particle moving on a spiral gi ven by r = aθ with a uniform angular speed ω. Then ṙ = a θ = aω. v = aωˆr + aω 2 t ˆθ and a = aω 3 tˆr + 2aω 2 ˆθ

116 Central Accelerations Central Accelerations When the acceleration of a particle points to the origin at all times and is only function of distance of the particle from the origin, the acceleration is called central acceleration. a = f (r)^r Examples are Gravitational Force, Electrostatic Force, Van-der-Waals Force, Spring tied to a point etc.. Angular momentum of the particle remains constant. r θ + 2ṙ θ = d dt (r 2 θ) = 0 r 2 θ = constant during motion.

117 An Example Acceleration of a particle is given by a (x) = ω 2 x. And at time t = 0, position is A and velocity is zero. dv dx After integrating this, we get dv dt dx dt v dv dx = ω 2 x = ω 2 x = ω 2 x v = ω A 2 x 2 (5)

118 An Example Finally the solution for x is We have considered this motion earlier. dx dt = ω A 2 x 2 (6) x(t) = A sin(ωt) (7)

119 Air Resistance Suppose a ball is falling under gravity in air, resistance of which is proportional to the velocity of the ball. a (ẏ) = g kẏ If the ball was just dropped, velocity of the ball after time then v(t) = g k ( 1 e kt )

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