Physics Remedial Module.pdf

PRE-UNIVERSITY REMEDIAL PROGRAM FOR THE 2014 E.C. ESSLCE EXAMINEES PHYSICS MODULE Pre-University Remedial Program for the Year 2014 E.C., Physics Module Addis Ababa, Ethiopia Contents Chapter one: Vectors (2hrs)..................................................................................................................... 4 1.1 Vector and Scalar quantities....................................................................................................... 4 1.2 Vector addition and subtraction................................................................................................ 4 1.3 Multiplication of vectors............................................................................................................. 4 Chapter Two: Kinematics (9hrs)............................................................................................................. 4 2.1 Motion in one dimension............................................................................................................ 4 2.2 Motion in two dimensions....................................................................................................... 4 Chapter Three: Angular motion (3hrs).................................................................................................. 4 3.1 Angular Position.......................................................................................................................... 4 3.2 Angular velocity........................................................................................................................... 4 3.3 Angular Acceleration................................................................................................................... 4 3.4 Relation between linear and angular motion........................................................................... 4 Chapter Four: Dynamics (10hrs)............................................................................................................. 4 4.1 Types of forces (contact force, Normal force friction force, applied force, gravitational force, restoring force…)..................................................................................................................... 4 4.2 Newtonian’s laws of motion....................................................................................................... 4 4.3 Applications of Newton's Laws................................................................................................. 4 4.4 Linear momentum (elastic and non-elastic collision)............................................................. 4 4.5 Center of mass and moment of inertia.................................................................................. 4 4.6 Torque and angular momentum................................................................................................ 4 4.7 Conditions of Equilibrium (First and second)......................................................................... 4 Chapter Five: Work, energy and power (6hrs)..................................................................................... 4 5.1 Work done by constant and variable forces............................................................................. 4 5.2 Conservation of energy............................................................................................................... 4 5.3 Work energy theorem.............................................................................................................. 4 5.4 Conservative forces...................................................................................................................... 4 5.5 Power............................................................................................................................................. 5 Chapter Six: Oscillation and Waves (9hrs)........................................................................................... 5 6.1 Oscillatory motion........................................................................................................................ 5 6.2 Properties of wave (frequency, wave length, period)............................................................. 5 6.3 Types of Waves............................................................................................................................. 5 6.4 Wave behavior (reflection, refraction, interference, diffraction)........................................... 5 Pre-University Remedial Program for the Year 2014 E.C., Physics Module Addis Ababa, Ethiopia 6.5 Wave equation.............................................................................................................................. 5 Chapter Seven: Heat and thermodynamics (7hrs).............................................................................. 5 7.1 Temperature and Heat................................................................................................................ 5 7.2 The effect of heat on materials (change of Temperature, expansion, change of phase, heat capacity)...................................................................................................................................... 5 7.3Laws of thermodynamics (zeros, first and second Laws)....................................................... 5 Chapter Eight: Electrostatics and Magnetism (9hrs).......................................................................... 5 8.1 Coulomb Law............................................................................................................................... 5 8.2 Electric field due to point charges............................................................................................. 5 8.3 Electric field lines...................................................................................................................... 5 8.4Electric Potential due to point charges....................................................................................... 5 8.5 Capacitors (capacitance and Capacitor networks).................................................................. 5 Chapter Nine: Electric current and Magnetism (10hrs)..................................................................... 6 9.1 Electric current (ohm’s law, resistance & Resistivity, measuring instruments).................. 6 9.2 Electric Circuit (series, parallel)................................................................................................. 6 9.3 Sources of magnetic field (Bar magnet, Earth magnetic field, moving charge, electric current)................................................................................................................................................ 6 9.4 Magnetic forces (on charged particles and current carrying conductor, two current carrying wires).................................................................................................................................... 6 Chapter Ten: Electromagnetic Induction and AC current (3hrs)..................................................... 6 10.1 Magnetic flux and Gauss law................................................................................................... 6 10.2 Faradays Law.............................................................................................................................. 6 10.3 AC current................................................................................................................................... 6 10.4 Transformer................................................................................................................................ 6 Pre-University Remedial Program for the Year 2014 E.C., Physics Module Addis Ababa, Ethiopia Chapter one: Vectors (2 hrs.) 1.1 Vector and Scalar quantities 1.2 Vector addition and subtraction 1.3 Multiplication of vectors Chapter Two: Kinematics (9 hrs.) 2.1 Motion in one dimension 2.1.1 Position and reference frame 2.1.2 Distance and displacement 2.1.3 Speed and velocity 2.1.4 Uniform motion 2.1.5 Uniformly accelerated motion 2.2 Motion in two dimensions 2.2.1 Projectile motion 2.2.2 Circular motion Chapter Three: Angular motion (3 hrs.) 3.1 Angular Position 3.2 Angular velocity 3.3 Angular Acceleration 3.4 Relation between linear and angular motion Chapter Four: Dynamics (10 hrs.) 4.1 Types of forces (contact force, Normal force friction force, applied force, gravitational force, restoring force…) 4.2 Newtonian’s laws of motion 4.3 Applications of Newton's Laws. 4.4 Linear momentum (elastic and non-elastic collision) 4.5 Center of mass and moment of inertia 4.6 Torque and angular momentum 4.7 Conditions of Equilibrium (First and second) Chapter Five: Work, energy and power (6 hrs.) 5.1 Work done by constant and variable forces 5.2 Conservation of energy 5.3 Work energy theorem 5.4 Conservative forces Pre-University Remedial Program for the Year 2014 E.C., Physics Module Addis Ababa, Ethiopia 5.5 Power Chapter Six: Oscillation and Waves (9 hrs.) 6.1 Oscillatory motion 6.1.1 Harmonic Motion 6.1.2 Damped and Forced Oscillation 6.2 Properties of wave (frequency, wave length, period) 6.3 Types of Waves 6.3.1 Transverse and longitudinal 6.3.2 Mechanical and Electromagnetic wave 6.4 Wave behavior (reflection, refraction, interference, diffraction) 6.5 Wave equation Chapter Seven: Heat and thermodynamics (7 hrs.) 7.1 Temperature and Heat 7.2 The effect of heat on materials (change of Temperature, expansion, change of phase, heat capacity) 7.3Laws of thermodynamics (zeros, first and second Laws) Chapter Eight: Electrostatics and Magnetism (9 hrs.) 8.1 Coulomb Law 8.2 Electric field due to point charges 8.3 Electric field lines 8.4Electric Potential due to point charges. 8.5 Capacitors (capacitance and Capacitor networks) Pre-University Remedial Program for the Year 2014 E.C., Physics Module Addis Ababa, Ethiopia Chapter Nine: Electric current and Magnetism (10 hrs.) 9.1 Electric current (ohm’s law, resistance & Resistivity, measuring instruments) 9.2 Electric Circuit (series, parallel) 9.3 Sources of magnetic field (Bar magnet, Earth magnetic field, moving charge, electric current) 9.4 Magnetic forces (on charged particles and current carrying conductor, two current carrying wires) Chapter Ten: Electromagnetic Induction and AC current (2 hrs.) 10.1 Magnetic flux and Gauss law 10.2 Faradays Law 10.3 AC current 10.4 Transformer References 1. High school textbooks 2. Raymond A. Serway, John W. Jewett - Physics for scientists and engineers (2004, Thomson-Brooks_Cole) 3. Robert Resnick and David Halliday, Fundamentals of Physics Extended, HRW 8th ed., 2008 4. Douglas C. Giancoli, Physics for scientists and engineers, Printice Hall, 4th, 2005 Mode of delivery Activity based interactive teaching approach will be applied. WACHEMO UNIVERSITY COLLEGE OF NATURAL AND COMPUTATIONAL SCIENCE Department of Physics Physics Module for Pre-University Remedial Program for ESSLCE Examinees March 6, 2023 WACHEMO UNIVERSITY COLLEGE OF NATURAL AND COMPUTATIONAL SCIENCE Department of Physics Physics Module for Pre-University Remedial Program for ESSLCE Examinees Prepared By Teshome Gerbaba (PhD, Wachemo University) Tesfaye Tadele (MSc, Wachemo University) Mathewos Tulore (PhD Candidate, Wachemo University) Edited by Adane Tadese (MSc, Wachemo University) March, 2023 Hossana, Ethiopia Contents Contents ii 1 Vectors 1 1.1 Vectors.................................... 1 1.1.1 Vector and Scalar quantities.................... 1 1.1.2 Vector Representation....................... 2 1.1.3 Vector Addition and subsection.................. 2 1.1.4 Vector Components......................... 5 1.1.5 Unit Vector............................. 5 1.2 Multiplication of vectors.......................... 6 1.2.1 Scalar-vector multiplication.................... 7 1.2.2 Dot product............................. 7 1.2.3 Cross Product............................ 9 2 Kinematics 13 2.1 Kinematics of the particle......................... 13 2.1.1 One or two dimensional (2D) motion............... 14 2.1.2 Motion in 1D............................ 14 2.1.3 Two dimension (2D) motion.................... 21 3 Angular Motion 29 3.1 Angular Motion............................... 29 3.1.1 Rotational Kinematics....................... 30 4 Dynamics 35 4.1 Dynamics.................................. 36 4.2 linear momentum............................. 42 4.2.1 Conservation of Momentum.................... 43 4.2.2 Collision............................... 43 4.3 Center of Mass and Moment of Inertia.................. 45 4.3.1 Center of Mass........................... 45 4.3.2 Moment of Inertia......................... 46 4.4 Torque and angular momentum...................... 48 4.5 Conditions of Equilibrium (First and second)............... 49 5 Work, Energy and Power 51 5.1 Work done by constant and variable forces................ 51 5.1.1 Work done by a variable force................... 53 5.2 Conservation of energy........................... 55 5.3 Work energy theorem........................... 57 ii CONTENTS 1 5.4 Conservative forces............................. 58 5.5 Power.................................... 58 6 Oscillation and Waves 61 6.1 Oscillatory motion............................. 61 6.1.1 Harmonic Motion.......................... 62 6.1.2 Damped and Forced Oscillation.................. 66 6.2 Properties of wave (frequency, wave length, period)........... 69 6.3 Types of Waves............................... 70 6.3.1 Transverse and longitudinal.................... 70 6.3.2 Mechanical and Electromagnetic wave.............. 71 6.4 Wave behavior (reflection, refraction, interference, diffraction)..... 71 6.5 Wave equation................................ 73 7 Heat and thermodynamics 77 7.1 Temperature and Heat........................... 77 7.2 The effect of heat on materials (change of Temperature, expansion, change of phase, heat capacity....................... 80 7.2.1 Specific Heat and Latent Heat................... 82 7.3 Laws of thermodynamics (zeros, first and second Laws)......... 85 7.3.1 Zeros Laws of thermodynamics.................. 85 7.3.2 First Laws of thermodynamics................... 86 8 Electrostatics 89 8.1 Coulomb’s law................................ 89 8.2 Electric Field (E).............................. 91 8.2.1 Electric Field Intensity....................... 92 8.3 Electric Field Lines............................. 95 8.4 Electric potential of a point charge.................... 97 8.4.1 Motion of charged particles in an electric field.......... 99 8.5 Capacitance and Capacitor networks................... 101 8.6 The Parallel Plate Capacitor........................ 103 8.6.1 Energy Stored in a Capacitor................... 105 8.7 Capacitance net work............................ 107 Chapter 1 Vectors 1.1 Vectors Learning competencies Demonstrate an understanding of the difference between scalars and vectors and give common examples. Explain what a position vector is. Use vector notation and arrow representation of a vector. Specify the unit vector in the direction of a given vector. Determine the magnitude and direction of the resolution of two or more vectors using Pythagoras’s theorem and trigonometry. Add vectors by graphical representation to determine a resultant. Add/subtract two or more vectors by the vector addition rule. Use the geometric definition of the scalar product to calculate the scalar product of two given vectors. Use the scalar product to determine projection of a vector onto another vector. Use the vector product to test for collinear and orthogonality vectors. Explain the use of knowledge of vectors in understanding natural phenomena. 1.1.1 Vector and Scalar quantities Scalars are the physical quantities that have the only magnitude. Examples of scalars are electric charge, density, mass etc. Vectors are physical Quatities that must be described by both magnitude and direction. Example: Velocity, Force, Torque, Electric field etc. 1 2 CHAPTER 1. VECTORS 1.1.2 Vector Representation Vectors are represented in two methods (Analytical/Algebric) and Graphical/Geometrical) 1. Analytical methods: Vectors are representated analytically by a letter with an arrow over its head or with bold face letter. → − → − → − Example: Force =⇒ F or F, Momentum =⇒ P or P, Vector A =⇒ A or A 2. Graphical/Geometrical methods: Graphically vectors are representated by a straight line and arrow drown to the scale. The length of the line is the magnitude of the vector and arrow tells us the direction. 1.1.3 Vector Addition and subsection ~ Note that subtraction The sum of two or more vectos is called resultant vector (R). is addition of the negative ie ~ =A R ~−B ~ =A ~ + (−B) ~ Vector addition is not simple algebraic addition of numbers that is handled with the normal rules of arithmetic. It Obeys the laws of vector addition as follows The resultant of two vectos having the same direction is algebraic sum of the two vectors with the same direction as both. ~ = 8m East and B Example A ~ = 6m East then R ~ =A ~+B ~ = 14m East The resultant of two vectors having opposite direction has magnitude equal to the difference of magnitudes of the vectors and the resultant has the same direction ~ = 8m East and B as the larger vector. Example: A ~ = 6m West then R ~ = ~ + (−B) A ~ = 2m East 1.1. VECTORS 3 The resultant of two vectors acting at right angle with each other is obtained ~ = 8m East and B using Pythagorus theorem. Example: A ~ = 6m North then, ~ obtained using Pythagorus theorem as And then the magnitude of R R2 = A2 + B 2 + 2ABcosθ = A2 + B 2 + 2ABcos900 = A2 + B 2 p R= (8m)2 + (6m)2 = 10m ~ obtained by trigonometery Direction of R Opp. |B| tanθ = = adj. |A| B 6 θ = tan−1 ( ) = tan−1 ( ) = 36.87o A 8 If the two vectors inclined at a certain anle θ to each other. ~ =A The magnitude of the vector R ~+B ~ is given by √ ~ = |R| A2 + B 2 + 2ABcosθ ~ =A The magnitude of the vector R ~−B ~ is given by √ ~ = |R| A2 + B 2 − 2ABcosθ And its direction is given by α = tan−1 ( A+Bcosθ Bsinθ ) 4 CHAPTER 1. VECTORS sin(1800 −θ) sinα sin(1800 −θ) Or using sin law R = B or sinα = R B sin(1800 − θ) α = sin−1 (B ) R NB: If the vectors form a closed polygon when joined head to tail in a certain order, their resultant is zero or null vector Two or more vectors are equal if and only if they are – the same physical quantities – have the same magnitude and – have the direction Class work ~ and B. 1.Given Vector A ~ Find the resultant vector R= ~ A+ ~ B ~ ~ = 4bunit East and B a) If the A ~ = 3unit East ~ = 4bunit East and B b) If the A ~ = 3unit West ~ = 4bunit East and B c) If the A ~ = 3unit north ~ = 4bunit East and B d) If the A ~ = 3unit at 60o north of east 1.1. VECTORS 5 2. A car travels 20.0km due north and then 35.0km in direction 600 west of north. Find ~ = 48.2km at the magnitude and direction of the car’s resultant displacement. (ans S 39.0o west of north). 1.1.4 Vector Components Components of vectors are projection of vectors along coordinate axis (x, y, z-axis). This meanse splitting vector into its Components. Consider the following figures. From the figures we can see that Ax and By forms two sides of right angle triangle with hypotenuse of length A. Using simple trigonometery (definition of sin and cosin) we see that Figure 1.1: Vector Components Ax cosθ = =⇒ Ax = Acosθ ~ |A| Ay sinθ = =⇒ Ay = Asinθ ~ |A| ~ = A~x + A~y for three dimension A Thus A ~ = A~x + A~y + A~z 1.1.5 Unit Vector Unit vector is dimensionless vector with unit magnitude. ~ A Â = ~ |A| Â read as A hat or caret is a unit vector that points in the direction of vector A. We shall use the symbols î, ĵ and k̂ to representat a unit vector pointing in the positive x, y and z direction respectively as we can see from the figure above. The unit vectors î, ĵ and k̂ in rectangular coordinate System. î, ĵ and k̂ are mutually perpendiculr axes. 6 CHAPTER 1. VECTORS Figure 1.2: Unit Vector in Rectangular Coordinat Axis In general Vector A in rectangular coordinate system can be written as the sum ~ = Ax î + Ay ĵ + Az k̂ of three vectors each which is parallel to a coordinate axes A ~ and B Addition and Subtraction of two vectors A ~ can be written intermes of unit vector as ~+B A ~ = (Ax î + Ay ĵ + Az k̂) + (Bx î + By ĵ + Bz k̂) = (Ax î + Bx î) + (Ay ĵ + By ĵ) + (Az k̂ + Bz k̂) = (Ax + Bx )î + (Ay + By )ĵ + (Az + Bz )k̂ Class work ~ = 4mî + 3mĵ, B 1. Given vectors A ~ = 2mî − 3mĵ, C ~ = 2mî + 3mĵ − 2mk̂ and ~ = 1mî − 2mĵ + 2mk̂. Find D ~ a) |A| ~+B b) 2A ~ −C ~ ~ +B c) Unit vector in the direction of vector R such that 2C ~ −R ~ =0 2. A particle undergoes three consecutive displacements d~1 = (15î + 30ĵ + 12k̂)cm, d~2 = (23î − 14ĵ − 5.0k̂)cm, d~3 = (−13î + 15ĵ)cm. Find a) The components of the resultant displacement and its magnitude b) Unit vector in the direction of resultant displacement 1.2 Multiplication of vectors Vector multiplication refer to several operations between two (or more) vectors. It may concern any of the following articles: Scalar-vector multiplication Dot product Cross product 1.2. MULTIPLICATION OF VECTORS 7 1.2.1 Scalar-vector multiplication Multiplication of a vector by a scalar changes the magnitude of the vector, but leaves its direction unchanged. The scalar changes the size of the vector. The scalar ”scales” the vector. For example, If ~ = ax î + ay ĵ + az k̂ A ~ by the scalar b is Multiplied A ~ = b(ax î + ay ĵ + az k̂) = bax î + bay ĵ + baz k̂ bA Scalar multiplication obeys the following rules: Additivity in the scalar: (c + d)~v = c~v + d~v ; Additivity in the vector: c(~v + w) ~ = c~v + cw; ~ Compatibility of product of scalars with scalar multiplication: (cd)~v = c(d~v ); Multiplying by 1 does not change a vector: 1~v = ~v ; Multiplying by 0 gives the zero vector: 0~v = 0; Multiplying by -1 gives the additive inverse: (−1)~v = −~v. 1.2.2 Dot product The dot product of two vectors is the magnitude of one times the projection of the second onto the first. The symbol used to represent this operation is a small dot at middle height (·), which is where the name ”dot product” comes from. Since this product has magnitude only, it is also known as the scalar product. Mathematically defined as ~·B A ~ = ABcos(θ) ~ and B where θ the angle btween A ~ Let ~ = ax î + ay ĵ + az k̂ A 8 CHAPTER 1. VECTORS and ~ = bx î + by ĵ + bz k̂ B Figure 1.3: Dot Product ~·B A ~ = (ax î + ay ĵ + az k̂) · (bx î + by ĵ + bz k̂) = ax bx î · î + ay by ĵ · ĵ + az bz k̂ · k̂ = ax b x + ay b y + az b z since î · î = (1)(1)cos(0) = 1, ĵ · ĵ = (1)(1)cos(0) = 1 , k̂ · k̂ = (1)(1)cos(0) = 1 but î · ĵ = ĵ · î = (1)(1)cos(900 ) = 0, ĵ · k̂ = k̂ · ĵ = (1)(1)cos(900 ) = 0 , k̂ · î = î · k̂ = (1)(1)cos(900 ) = 0 Dot Product Properties of Vector: ~·B Dot product of two vectors is commutative i.e. A ~ =B ~ ·A ~ ~·B If A ~ = 0 then it can be clearly seen that either A ~ or B ~ is zero or cos(θ) = 0. ~ · (q B) Also we know that using scalar product of vectors (pA) ~ = (pB) ~ · (q A) ~ = ~ · B) pq(A ~ The dot product of a vector to itself is the magnitude squared of the vector i.e. ~ · A) A ~ = AAcos(0) = A2 ~ · (B Distributive Property: A ~ + C) ~ =A ~·B ~ +A ~·C ~ ~ · (B Non-Associative Property: A ~ · C) ~ 6= (A ~ · B) ~ · (A ~ · C), ~ because the dot product between a scalar and a vector is not allowed. 1.2. MULTIPLICATION OF VECTORS 9 1.2.3 Cross Product The cross product of two vectors ~a and ~b is vector ~c which is perpendicular to both ~a and ~b and equal magnitude to the area of the parallelogram between ~a and ~b. The symbol used to represent this operation is a large diagonal cross (×), which is where the name ”cross product” comes from. Since this product has magnitude and direction, it is also known as the vector product. ~a × ~b = absin(θ)n̂ Figure 1.4: Cross Product The vector n̂ (n hat) is a unit vector perpendicular to the plane formed by the two vectors and θ is the angle between ~a and ~b. The direction of n̂ is determined by the right hand rule. Cross Product Properties : the cross product is distributive: ~a × (~b + ~c) = (~a × ~b) + (~a × ~c) the cross product is not commutative: ~a × ~b 6= ~b × ~a but ~a × ~b = −~b × ~a the cross product of any vector with itself is zero: ~a × ~a = ~b × ~b = 0 cross product of any unit vector with itself is zero: î × î = (1)(1)sin(0) = 0, ĵ × j = (1)(1)cos(0) = 0 , k̂ × k̂ = (1)(1)cos(0) = 0 any cyclic product of the three coordinate axes is positive and any anticyclic product is negative as shown bellow. 10 CHAPTER 1. VECTORS Let ~a = ax î + ay ĵ + az k̂ and ~b = bx î + by ĵ + bz k̂ ~a × ~b = (ax î + ay ĵ + az k̂) × (bx î + by ĵ + bz k̂) ~a×~b = ax î×bx î+ax î×by ĵ+ax î×bz k̂+ay ĵ×bx î+ay ĵ×by ĵ+ay ĵ×bz k̂+az k̂×bx î+az k̂×by ĵ+az k̂×bz k̂ ~a × ~b = 0 + (ax by )k̂ − (ax bz )ĵ − (ay bx )k̂ + 0 + (ay bz )î + (az bx )ĵ − (az by )î + 0 ~a × ~b = (ay bz − az by ) î + (az bx − ax bz )ĵ + (ax by − ay bx )k̂ Or using determinat form î ĵ k̂ ay az az ax ax ay ~a × ~b = ax ay az = î + ĵ + k̂ = (ay bz − az by ) î + (az bx − by bz bz bx bx b y bx by b z ax bz )ĵ + (ax by − ay bx )k̂ Class work 1. Which of the following statements is true about the relation-ship between the dot product of two vectors and the product of the magnitudes of the vectors? (a) ~·B A ~ is larger than AB; (b) A ~·B ~ is smaller than AB; (c) A ~·B ~ could be larger ~ ·B or smaller than AB, depending on the angle between the vectors; (d) A ~ could be equal to AB. ~ × B) 2. Which of the following is equivalent to the following scalar product: (A ~ · ~ × A)? (B ~ ~·B (a) A ~ +B ~ ·A~ (b) (A~ × A) ~ · (B ~ × B) ~ (c) (A ~ × B) ~ × B) ~ · (A ~ (d) ~ × B) −(A ~ · (A ~ × B) ~ 1.2. MULTIPLICATION OF VECTORS 11 3. Which of the following statements is true about the relationship between the magnitude of the cross product of two vectors and the product of the magnitudes ~ × B| of the vectors? (a) |A ~ is larger than AB; (b) |A ~ × B| ~ is smaller than AB; ~ × B| (c) |A ~ could be larger or smaller than AB, depending on the angle between ~ × B| the vectors; (d) |A ~ could be equal to AB. 4. Is the triple product defined by A · (B × C) a scalar or a vector quantity? Explain why the operation A · (B × C) has no meaning. 5. Vector A is in the negative y direction, and vector B is in the negative x direction. What are the directions of (a) A × B (b)B × A? 6. Given M = 6î + 2ĵ − k̂ and N = 2î − ĵ − 3k̂ calculate the M · N, M × N and the angle between M&N and. 12 CHAPTER 1. VECTORS Chapter 2 Kinematics Learning competencies Describe Kinematical terms such as distance, Displacement, average speed (velocities) and instantaneous speed (velocity). Solve numerical problems involving average velocity and instantaneous velocity. Derive equations of motion for uniformly accelerated motion. Apply equations of uniformly accelerated motion in solving problems. Relate scientific concepts to issues in everyday life. Explain the science of kinematics underlying familiar facts, observations and related phenomena. Describe the conditions at which freely falling bodies attain their terminal velocity. Define, Analyse and predict, terms in 2D motion Apply equations to solve problems related 2D motion. Distinguish between uniform and non-uniform circular motion. Analyse and predict, in quantitative terms, and explain uniform circular motion in the horizontal and vertical planes with reference to the forces involved. 2.1 Kinematics of the particle The word Kinematics comes from Greek word ”kinesis” meaning motion, thus Kinematics: is a branch of mechanics that describes the motion of an object without refer- 13 14 CHAPTER 2. KINEMATICS ence to couse of motion (force). It does not give any information about force that couses it to move. 2.1.1 One or two dimensional (2D) motion What is motion? Motion is continuous change of position with time. Position is location of an object with respect to a choosen reference frame or point. Reference frame, also called frame of reference, in dynamics, system of graduated lines symbolically attached to a body that serve to describe the position of points relative to the body. In physics we are considered three type of motion 1. Translational motion: is type motion in which all points (parts) of an object move the same distance in a given a given time. Example: A car moving in a straight line, a bullet which gets fired moves in rectilinear motion, child going down, a bird flying in the sky. In the above example, all the points of the body/object in motion are in the same direction. Translational motion can be of two types, rectilinear and curvilinear. Rectilinear motion is when an object in translational motion moves in a straight line motion. When an object in translational motion moves along a curved path, it is said to be in curvilinear motion 2 Rotational motion: is when an object moves about an axis and different parts of it move by different distances in a given interval of time. Examples: blades of a rotat- ing fan, merry-go-round, blades of a windmill. When an object undergoes rotational motion, all its parts do not move the same distance in a given interval of time. For example, the outer portion of the blades of a windmill moves much more than the portion closer to the centre. 3 Vibrational motion: is when a body moves to and fro about its mean position is called vibratory motion. Vibratory motion can be described as any object mov- ing/swinging back and forth, moving up and down, pulsating, etc. Examples Pen- dulums, swings, tuning forks, etc are of vibratory motion. Vibrational motion can be periodic or non-priodic 2.1.2 Motion in 1D This is motion of a particle along straight line in fixed direction (or motion along one coor- dinate axis). Example: a car moving along a flat straight narrow road. 2.1. KINEMATICS OF THE PARTICLE 15 p p x xi xf ∆x = xf − xi Definition of Kinematical terms Distance and Displacement Distance: is the total path length covered by the moving object. Displacement: is change of position i.e the shortest distance between start and end of motion. For example: a particle moving from point A to B as shown in figure below. Figure 2.1: Comparison of Distance and Displacement Speed and Velocity Speed (v): is the rate of change of distance in a unit time. Average Speed (vav ): is total distance traveled by the total time required to cover the distance. total..distance vav = total..time Velocity(~v ): the rate of change of displacement as a function of time. Average Velocity (~vav ): is change of displacement ∆x divided by the time interval ∆t during which the displacement occure. ~xf − ~xi ∆~x ~vav = = tf − ti ∆t Instantaneous Velocity and Speed Instantaneous Velocity v(t): is the Velocity of the particle at a given instant of time. It is the limit of average velocity as ∆t approaches to zero. ∆x ~x(t + ∆t) − ~x(t) v(t) = lim = lim ∆t→0 ∆t ∆t→0 ∆t 16 CHAPTER 2. KINEMATICS This can be rewritten as frist derivatives of displacement with respect time. dx v(t) = dt The magnitude of Instantaneous velocity is instantaneous speed Class work 1. Which of the following is true for displacement? (a) It cannot be zero. (b) Its magnitude is greater than the distance travelled by the object. (c) displacement may or may not be equal to distance 2. If the displacement of the body is zero, the distance covered by it may not be zero. 3. In which of the following cases of motions, the distance moved and the magnitude of displacement are equal ? (a) If the object is moving along straight road (b) If the object is moving along staight path (c) The pendulum is moving back and fro (d) The earth revolving around the sun 4. A particle moves along the x-axis according to the equation given below. ~x(t) = (4 + 2t − t2 )mî where t is in Second. a) Determine the displacement of this particle between the time interval t = 0 and t = 1s b) Determine the average velocity during those two time intervals c) Dirive a general expression for the instantaneous velocity as a function of time. d) calculate instantaneous velocity at t = 2s. 5. A boy walk from his home to school at constant speed of 5m/s along straight line and then back along the same line (road) from school to his home at constant speed of 6m/s. 2.1. KINEMATICS OF THE PARTICLE 17 a) What is his average speed? b) What is his average velocity? Acceleration (~a): is the rate of change of velocity. Average acceleration (~aav ): is the change in velocity divided by time interval during which it occure. xf − xi ∆x ~aav = = tf − ti ∆t Instantaneous acceleration ~a(t): is the acceleration of the particle at a given instant of time. It is the limit of average acceleration as ∆t approaches to zero. ∆v ~v (t + ∆t) − ~v (t) ~a(t) = lim = lim ∆t→0 ∆t ∆t→0 ∆t This can be rewritten as Second derivatives of displacement or first derivatives of velocity with respect time. dv d2 x ~a(t) = = 2 dt dt Class work 1. A particle moves along the x-axis component varies with time according to equation given below. ~x(t) = (20 − 2t + t3 )mî where t is in Second. a) Determine initial position of the particle. b) Determine the displacement of this particle between the time interval t = 0 and t = 1s; t = 1 and t = 4s c) Determine the average velocity during those two time intervals d) Dirive a general expression for the instantaneous velocity as a function of time. e) calculate instantaneous velocity at t = 3s. f) average acceleration between t = 2s to t = 3s. g) Dirive a general expression for the instantaneous acceleration as a function of time. h) calculate instantaneous acceleration at t = 2s. 18 CHAPTER 2. KINEMATICS Uniform Motion in 1D Uniform Motion: A body is said to be in a state of uniform motion if it travels equal distances in equal intervals of time. If the time distance graph is a straight line the motion is said to be uniform motion. This meanse that the velocity of the body remain constant as it cover equal distance in equal interval of time, in case of uniform rectilinear motion acceleration of the body will be zero. Here, the avrage speed and instantaneous speed will be equal to the actual speed; avrage velocity and instantaneous velocity will be equal to the actual velocity and the magintude of velocity is equal speed. ~vav = ~v = ~v (t) |~v | = v ∆x = ~v (t)t Uniformely accelerated Motion in 1D This is motion with constant acceleration ie velocity change with uniform rate. dv v − vi ~a(t) = = = constant dt t v(t) = vi + at (2.1) Average velocity for uniformely accelerated motion is given by ~v + v~i ~vav = 2 Thus ~v + v~i ~x − x~i = ∆~xt = ~vav t = ( )t (2.2) 2 Using equation (2.1) into (2.2) v~i + at + v~i ∆~x = ( )t 2 1 ∆~x = vi t + at2 (2.3) 2 From equation (2.1) v − vi t= (2.4) a 2.1. KINEMATICS OF THE PARTICLE 19 Using equation (2.4) into (2.2) v + vi v − vi v 2 − vi2 ∆x = ( )( )= 2 a 2a v 2 = vi2 + 2a∆x (2.5) Class work 1. An electron in a cathod ray tube accelerate uniformely from 2.0 × 104 m/s to 6.0 × 106 m/s over 1.5cm. a) How long does the electron take the to travel this 1.5cm? b) What is its acceleration? 2. A track covers 40m in 8.5s while smoothely slowing down to a final speed of 2.8m/s. Find a) its original velocity b) its acceleration 3. A jet lands on an air craft at 140mi/hr and stops in 2s due to an arresting cable that snags the air plane. a) What is its acceleration? b) If the plane touches down at position xi = 0 what is the final position of the plane? 4. A car traveling at constant speed of 45m/s passes a tropper hidden behind a billboard. One second after the speeding car passes the billboard, the tropper sets out from the billboard to catch it, accelerating at constant rate of 3.0m/s2. How long does it take her to over take the car? 5. A jet plane lands with a speed of 100m/s and slow down at rate of 5m/s2 as it comes to rest. a) What is the time interval needed by the jet to come to rest? b) Can this jet land on an airport where the runway is 0.8km long? 20 CHAPTER 2. KINEMATICS Free falling bodies A freely falling object is any object moving freely under the influence of gravity alone, re- gardless of its initial motion. Example: object thrown upward or down ward and object released from rest. Free fall is motion with constant gravitational acceleration g = 9.81m/s2 toward the center of the earth. So we can use equation of uniformely accelerated as in table below When released from rest When thrown up When thrown down vy = gt vy = voy − gt vy = voy + gt ∆y = 12 gt2 ∆y = voy − 21 gt2 ∆y = voy + 12 gt2 vy2 = 2g∆y vy2 = voy 2 − 2g∆y vy2 = voy 2 + 2g∆y Class work 1. A girl thows a ball upwards, moving it an initial speed u = 15m/s. Neglect air resistance a) How long does the ball take to return to the girl’s hand? b) What will be its velocity then? 2. A ball is thrown upward. While the ball is in free fall, does its acceleration (a) increase (b) decrease (c) increase and then decrease (d) decrease and then increase (e) remain constant? 3. After a ball is thrown upward and is in the air, its speed (a) increases (b) decreases (c) increases and then decreases (d) decreases and then increases (e) remains the same. 2.1. KINEMATICS OF THE PARTICLE 21 2.1.3 Two dimension (2D) motion 2D motion is motion in a plane (This meanse object moving along two coordinate axis simul- taneously, and its position can be described by two coordinate). Example: Projectile motion, Circular motion Figure 2.2: Motion in a plane If a particle move from point A to point B in figure 2.2 its displacement is given by ∆~r = ~rB − ~rA ∆~r = (xB î + yB ĵ) − (xA î + yA ĵ) = ∆xî + ∆y ĵ For infintesmal change d~r = dxî + dy ĵ Average velocity (vav ) is given by ∆~r ∆x ∆y ~vav = = î + ĵ = ~vx î + ~vy ĵ ∆t ∆t ∆t Instantaneous velocity is given by ∆r ~r(t + ∆t) − ~r(t) v(t) = lim = lim ∆t→0 ∆t ∆t→0 ∆t v(t) = ~vx (t)î + ~vy (t)ĵ Average acceleration is given by ∆v ∆vx ∆vy aav = = î + ĵ ∆t ∆t ∆t 22 CHAPTER 2. KINEMATICS Instantaneous acceleration ∆v ~v (t + ∆t) − ~v (t) a(t) = lim = lim ∆t→0 ∆t ∆t→0 ∆t a(t) = ~ax (t)î + ~ay (t)ĵ Class work 1. A bird flies in xy plane with velocity vector given by ~v = (α − βt2 )î + γtĵ where α = 2.1m/s and γ = 2.8m/s2 and the positive y direction is vertically upward at t = 0, the bird is at the origin. a) Determine average acceleration between time interval t = 0 to 1s b) calculate the general expression for instantaneous acceleration at any time t c) What is the birds altitude (y cordinate) as it flies over x = 0 for the first time after t = 0. 2.1.3.1 Projectile Motion Projectile motion is motion of an object in a plan under the infuelence of gravity alone, regardless of its initial motion(neglacting air resistance). Examples: A ball kicked from the horizontal ground. The path followed by projectile motion is trajectory and downward Figure 2.3: Projectil Motion parabola due to gravitational acceleration and combination of horizontal and vertical velocity as we can see in figure 2.3. As projectile motion is 2D motion we can regard it as two separate and independent horizontal (x-component) and vertical (y-component) motion. 2.1. KINEMATICS OF THE PARTICLE 23 Horizontal motion of projectile Horizontal motion of projectile motion is uniform motion (velocity constant, ax = 0). Because no net force act on horizontal motion of projectile motion. ie v0x = v0 cosθ this is horizontal component of initial velocity. vx = v0x = v0 cosθ = constant and x(t) = v0x t = vx t = v0 cosθt Vertical motion of projectile Vertical motion of projectile motion is uniformely accelerated motion. It is motion with constant gravitational acceleration of g = 9.8m/s2 to ward the center of the earth. At the origin v0y = v0 sinθ this is vertical component of initial velocity. Now we can use equation uniformely accelerated motion as vy = v0y − gt (v0y + vy ) ∆y = t 2 1 ∆y = v0y t − at2 2 vy2 = v0y 2 − 2g∆y But at the maximum height vy = 0 so 2 0 = v0y − 2gymax 2 v0y v 2 sin2 θ ymax = = 0 2g 2g Total time (t) is time of flight is given by vy = v0y − gt 24 CHAPTER 2. KINEMATICS but at ymax , vy = 0 0 = v0y − gta v0y v0 sinθ ta = = g g But total time is t = ta + td and ta = td thus 2v0y 2v0 sinθ t= = g g Range (R) is maximum horizontal displacement (xmax ) 2v0 sinθ R = vox t = v0 cosθ( ) g v02 2cosθsinθ R= g v02 sin2θ R= g The maximum range is reached at an angle of projection θ = 450 v02 sin(2 × 450 ) v 2 sin(900 ) v2 R= = 0 = 0 g g g Figure 2.4: A projectile launched from the origin with an initial speed of 50 m/s at various angles of projection. Note that complementary values of θi result in the same value of R (range of the projectile). Class work 1. A ball is kicked with an initial velocity if 40m/s from the ground at an angle of 300 to the horizontal. (Use g = 10m/s2 ) Calculate a) Horizontal and vertical component of initial velocity b) The vertical velocity after t = 1s, 2s, 3s, and 4s 2.1. KINEMATICS OF THE PARTICLE 25 c) Position (~r = xî + y ĵ) after t = 1s, 2s, 3s, and 4s d) Time of flight (total time) e) maximum height f) Range of projectile 2. An air plane moving horizontally with velocity of 500km/hr at a height of 2km above the ground dropped a bomb when it directly above the target. By how much distance will the bomb miss the target? 3. An astronaut on a strange planet finds that she can jump a maximum horizontal distance of 15.0m if her initial speed is 3.00m/s. What is the free-fall acceleration on the planet? 26 CHAPTER 2. KINEMATICS 2.1.3.2 Circular Motion Uniform Circular Motion Uniform circular motion: is a type of motion in which an object moves in a circular path at a constant speed. The direction of motion is constantly changing as the object moves around the circle. For example, imagine a car moving around a circular racetrack at a constant speed of 100 km/h. The car is always moving in a circle, and the direction of the car is constantly changing as it goes around the track. However, the car’s speed is always the same, so the car is said to be in uniform circular motion. Another example of uniform circular motion is a planet orbiting around a star. The planet is constantly moving in a circular path around the star, and its speed is constant as it moves around the orbit. Figure 2.5: Uniform Circular Motion. In uniform circular motion velocity is not constant because continuous variation of direction. So the two triangle are similar by SÂS. For similar triangle the ratio of their side is equal |∆~r| |~r| |∆~v | |~v | =..or.. = |∆~v | |~v | |∆~r| |~r| v ∆v = × ∆r r 2.1. KINEMATICS OF THE PARTICLE 27 Dividing bothside by ∆t ∆v v ∆r = × ∆t r ∆t v ac = × v r v2 ac = r Thus ac is radial or centerpital accelaration and it is always toward the center of the circle perpendiculr to velocity. Its magnitude is constant but its direction change continuous perpendiculr to velocity. This accelaration is due to a centripetal force. A centripetal force is a net force that acts on an object to keep it moving along a circular path and its direction is toward the center the circle. Example: The tension force in the string of a swinging tethered ball and the gravitational force keeping a satellite in orbit are both examples of centripetal forces. Multiple individual forces can even be involved as long as they add up (by vector addition) to give a net force towards the center of the circular path. Period (T): it is time taken for one complete rotation. 2πr T = v Non-uniform Circular Motion Non-uniform circular motion is a type of circular motion in which the speed of an object moving in a circular path changes at different points along the path. In other words, the magnitude of the velocity vector of the object is not constant, meaning that the object is accelerating even though it is moving in a circle. An example of non-uniform circular motion is a car driving around a curved road. The car’s speed may change as it navigates the curve, depending on factors such as the car’s position on the curve and the road conditions. The car’s direction is constantly changing as it moves around the curve, but the speed is not constant. As a result, the car is undergoing non-uniform circular motion. Another example of non-uniform circular motion is a planet orbiting a star, as the planet speeds up and slows down in its elliptical orbit. In this case there are two type of acceleration:- 1. Radial or centerpital accelaration:- due to change of direction of motion v2 ~ac = r 2. Tangential accelaration:- due to change magnitude of velocity. Its magintude is 28 CHAPTER 2. KINEMATICS change of velocity over change of time. ~vf − ~vi ∆v ~aT = = tf − ti ∆t Its direction is in the direction of Velocity which is perpendicular to centerpital acce- laration. Thus total accelaration ~a = ~ac + ~aT Its magintude p ~a = (~ac )2 + (~aT )2 Its direction −1 aT θ = tan ac Class work 1. A ball tied to the end of a string 1m in length swings in a vertical circle under the infuelence of gravity. When the string makesan angle of 200 its speed was 2m/s. Calculate a) magintude of centerpital accelaration b magintude of centerpital Tangential accelaration c) magintude and direction of of total accelaration Chapter 3 Angular Motion Learning competencies Describe the rotational kinematical quantities. Give the angular speed and angular velocity of a rotating body. Determine the velocity of a point in a rotating body. Derive equations of motion with constant angular acceleration. Use equations of motion with constant angular acceleration to solve related problems. State the law of conservation of angular momentum. Apply the law of conservation of angular momentum in Understanding various natural phenomena, and solving problems. Express angular momentum as a cross product of r and p. Derive an expression for angular momentum in terms of I and ω. Use the relationship between torque and angular momentum, according to Newton’s second law. Apply the relationship between torque and angular momentum to solve problems in- volving rigid bodies. 3.1 Angular Motion Angular motion is a type of motion that occurs when an object moves along a circular path or rotates around a fixed axis. Angular motion is characterized by two main quantities: angular displacement and angular velocity. Angular motion is important in many areas of physics, including mechanics, electromagnetism, and quantum mechanics. It is also used in many practical applications, such as in the design of engines, turbines, and other rotating machinery. 29 30 CHAPTER 3. ANGULAR MOTION 3.1.1 Rotational Kinematics Angular displacement: is the change in the angle (θ) of rotation of an object with respect to a fixed axis as we can see in figurebellow. Radian (rad) is SI unit of angular displacement, one radian is angle sutended by an arc length equal to radius of the arc. The relation between revolutio(rev), degree(deg or 0 ) and radian (rad) 2πrad = 3600 = 1rev (3.1) Average Angular velocity: is the rate of change of the angular displacement of an object with respect to time. Its represented by Greek letter ω. It is measured in radians per second or (degrees per second or revolation per second) and is equal to the ratio of the change in the angular displacement of the object to the time interval over which the change occurred. θ − θ0 ∆θ ω ~ = = t − t0 ∆t Instantaneous Angular Velocity: It is average angular velocity as ∆t o. This meanse angular velocity at instant of time (for infintesmal change) dθ(t) ω(t) = dt Angular acceleration: the rate of change of the angular velocity of an object with respect to time. It is denoted by Greek letter α. Average angular acceleration (ωav ): is the ratio of andgular velocity to time interval ∆t during which the change occure. ~ −ω ω ~0 ∆~ ω α ~= = t − t0 ∆t 3.1. ANGULAR MOTION 31 Instantaneous Angular acceleration: It is average angular acceleration as ∆t o. This meanse angular acceleration at instant of time (for infintesmal change). Its SI unit is rad/s2 dω(t) d2 θ(t) α(t) = = dt dt2 Uniformley Accelarated Rotational Motion Uniformly accelerated angular motion refers to the motion of an object rotating around an axis at a constant rate of acceleration (α). This means that the angular velocity of the object is changing at a constant rate over time. In this type of motion, the angular acceleration of the object is constant, which means that the rate of change of the angular velocity is also constant. The equation that describes this relationship is: ω = ωo + αt (3.2) For uniformly accelerated angular motion the average angular velocity is given by ω + ω0 ωav = 2 Therefore ω + ω0 θ − θ0 = ( )t (3.3) 2 Using all those together 1 θ = ω0 t + α 2 t 2 (3.4) 2 and ω 2 = ω02 + 2αθ (3.5) Class work 1. The angular position of a fly wheel of car engines is given by θ = (2rad/s3 )t3 the diametre of the flywheel is 0.36m. a) Find the angle θ in radian, degree and revolution at time t1 2secandt2 = 5sec. b) Find the distance that the particle of the rim moves during that time intervaly. c) Find the average angular velocity in rad/s, and rev./m 32 CHAPTER 3. ANGULAR MOTION d) Find the general expression for the angular velocity at any time t. e) Find the general expression for the angular accelaration at any time t. 2. A wheel rotates with angular acceleration of 3.5 rad/s2. If the angular speed of the wheel is 2.0m/s at t=0. a) through what angle doese the wheel rotate in 2.0s b) what is the angular speed at t = 2.0s Angular Momentum ~ of a moving particle with respect to a given axis is given by The angular momentum (L) ~ = ~r × p~ L Where ~r is didtance from axis of rotation and p~ = m~v is linear momentum. ~ = ~r × p~ = ~r × m~v = m~r × ~v L and we know that v = rω ~ = m~r × r~ L ω = mr2 ω ~ And we know that I = mr2 ~ = I~ L ω (3.6) Eqn. 3.6 is angular momentum. Taking time derivatives of equation 3.6 ~ dL d~v = m~r × = m~r × ~a = ~r × F~net = ~τnet dt dt Law of Conservation of Angular Momentum This states that if the net external torque acting on the system is zero (the system is isolated ) then the angular momentum of the system is conserved (remain unchenged) ~i = L L ~f Ii ωi = If ωf (3.7) Class work 3.1. ANGULAR MOTION 33 1. The position vector of a particle of mass 2kg is given as a function of time by ~r = (6î + 5tĵ. Determine the angular momentum of the particle about the origin as a function of time. 2. A large circular disk of mass 2kg and radius 0.2m initially rotating at 50rad/s is coupled a smaller circular disk of mass 4kg and radius 0.1m initially rotating at 20rad/s in the same direction as large disk. a) Find the commen angular velocity after the disk are coupled. b) Calculate the loss of kinetic energy during this collision. 34 CHAPTER 3. ANGULAR MOTION Chapter 4 Dynamics Learning competencies Identify the four basic forces in nature. Define and describe the concepts and units related to force. Define the term dynamics. Define and describe the concepts and units related to coefficients of friction. Use the laws of dynamics in solving problems. Interpret Newton’s laws and apply these to moving objects. Explain the conditions associated with the movement of objects at constant velocity. Solve dynamics problems involving friction. Analyse, in qualitative and quantitative terms, the various forces acting on an object in a variety of situations, and describe the resulting motion of the object. Describe the terms momentum and impulse. State the law of conservation of linear momentum. Discover the relationship between impulse and momentum, according to Newton’s sec- ond law. Apply quantitatively the law of conservation of linear momentum. Distinguish between elastic and inelastic collisions. Describe head-on collisions. Describe glancing collisions. Define and describe the concepts and units related to torque. Describe centre of mass of a body. Determine the position of centre of mass of a body. Interpret Newton’s laws and apply these to objects undergoing uniform circular motion. Solve dynamics problems involving friction. 35 36 CHAPTER 4. DYNAMICS 4.1 Dynamics Dynamics: In physics, dynamics is the branch of mechanics that deals with the study of motion and the forces that cause or affect that motion. It involves the analysis of how an object moves and the forces that cause it to move, including the study of the forces that cause changes in the motion of an object, such as acceleration, deceleration, and changes in direction. The fundamental concepts in dynamics are force, mass, and acceleration, as described by Newton’s laws of motion. Dynamics is used to describe a wide range of physical phenomena, from the motion of particles at the subatomic level to the motion of planets in the solar system. It is used in many fields, including engineering, physics, and applied mathematics, to understand and predict the behavior of physical systems. Force is a physical quantity that describes an interaction between two objects that can cause a change in motion of one or both objects. It is defined as the product of mass and acceleration, or more formally as the rate of change of momentum with respect to time. It is a push or a pull of an object (Intraction that change state of motion). We can’t see force with our necked-eye but, in everyday life, we experience the following effects of force all the time. Force set or tends to set an object to motion Force stop or tends to stop motion Force change direction of motion Force accelerate or decelerate motion Force change shape and size of materials Type of force Force usually catagorized into two 1. Contact Force: This is a force that requires physical contact between two objects in order for the force to be applied. Examples: Frictional force, Tension force, Normal force, Air resistance force, and Applied force. 2. Non-contact Force: This is a force that can act over a distance without any phys- ical contact between the objects. Examples: Gravitational force, Magnetic force, Electrostatic force, Electromagnetic force, Nuclear force. 4.1. DYNAMICS 37 Newton’s Law of Motion Newton’s laws of motion are three fundamental principles that describe the behavior of objects in motion. They were first introduced by Sir Isaac Newton in his 1687 work ”Philosophiæ Naturalis Principia Mathematica”. The laws are:, 1. Newton’s frist Law of motion (Law of inertia), 2. Newton’s second law (Law of acceleration) and 3. Newton’s third law (action and reaction force) Newton’s first law: This states that ”Unless an external force exerted on the body the state of motion the body remain as it is”. This is called law of inertia. Inertia: is the tedencey of the body to resist its change of state of motion. Newton’s second law: This states that ”accelaration of an object is directely proportional to the net force acting on it and inversely proportional to is mass”. X F~ = m~a X X X X F~ = F~x + F~y + F~z = m(~ax + ~ay + ~az ) Newton’s third law: This states that ”if object A exert force on object B, then object B exert a force on object A that is equal in magnitude and opposite in direction”. Thus force always occure in pair. This pair of force are called action and reaction force. For every action force there is reaction force. Action and reaction force are always: the same in magnitude opposite in direction act on different bodies the same type Class work 1. Object of mass 10kg is exerted on by a force of F~1 = (2î + 3ĵ)N , F~2 = (4î − 3ĵ)N and F~3 = (−î + 3ĵ)N , calculate a) net force on this object b) its accelaration 2. A 2kg object undergoes an accelaration of given by ~a = (3î + 4ĵ)m/s2. Find the magnitude resultant of force. 38 CHAPTER 4. DYNAMICS 3. A particle of mass 2unit moves along space curve defined by ~r(t) = (4t2 − t3 )î − 5tĵ + (t4 − 2)k̂. Find the force acting on it at any time t. 4. Find the force needed to accelarate a mass of 400kg from velocity ~v0 = (4î−5ĵ +3k̂)m/s to ~vf = (8î + 3ĵ − 5k̂)m/s in 10s. Friction Force Friction force: is a force generated in opposite direction to the motion when solid object slide or attempt to slid over each other. Its magnitude is given by f = FN µ Where µ is coefficient of friction (constant that depend on the nature of the surface in contact), FN is normal force. There are two type of frictional force Static friction:- friction occure when object attempt to slid over each other but not yet slid over each other. Its magnitude given fs = FN µs µs is coefficient of static friction Kinetic friction:- friction force occure when object sliding over eachother fs = FN µk µk is coefficient of kinetic friction NB µs > µk thus fs > fk Normal Force (FN ):- Is a force or component of force that is perpendicular to the surface in contact and equal in magnitude to the force that holds the surface press together. Normal force equal to mg when the sliding object is on horizontal surface and acted on by horizontal force as shown in figure below. But if the force acted on the object is at a certain angle to the horizontal the normal force is different as shown in figure below 4.1. DYNAMICS 39 Class work 1. A 20kg block is initially at rest on a horizontal surface. A horizontal force of 75N is a required to set the block in motion. After it is in motion a horizontal force of 60N is required to keep the block moving with constant speed. Find the coefficient of static and kinetic friction. Applied Force:Applied force is a physical force that is applied to an object by a person or another object. It is a force that causes an object to move, accelerate, or change direction. Applied force is an important concept in physics and is used to describe many physical phenomena, including the motion of objects, the behavior of fluids, and the behavior of electromagnetic fields. Gravitational force is the force by which a planet or other body draws objects toward its center. The force is always attractive and acts along the line connecting the two bodies. The force is proportional to the product of the two masses and inversely proportional to the square of the distance between them. The proportionality constant is known as the gravitational constant. The gravitational force is responsible for keeping the planets in orbit around the sun and for keeping the moon in orbit around the Earth. The mathematical formula for the gravitational force between two objects can be expressed using Newton’s law of gravitation: m1 m2 F =G r2 where: - F is the gravitational force between the two objects, measured in Newtons (N), - G is the gravitational constant, which has a value of approximately 6.674 × 10−11 N m2 /kg 2 , - m1 and m2 are the masses of the two objects in kilograms (kg), - r is the distance between the centers of mass of the two objects, measured in meters (m) A restoring force is a force that acts to bring an object back to its original position after it has been displaced. In other words, it is a force that opposes displacement. For example, 40 CHAPTER 4. DYNAMICS the force exerted by a spring when it is stretched or compressed is a restoring force. In a simple harmonic motion, restoring force is directly proportional to the displacement and acts in the direction opposite to the displacement.The mathematical expression of the restoring force for a spring is: F = −kx where F is the restoring force, x is the displacement from the equilibrium position, and k is the spring constant, which is a measure of the stiffness of the spring. Another example of a restoring force is the force of gravity acting on a pendulum. The restoring force of a pendulum is given by: F = −mgsin(θ) where F is the restoring force, m is the mass of the pendulum, g is the acceleration due to gravity, and θ is the angle between the pendulum and the vertical. In general, the mathemat- ical expression of the restoring force depends on the specific physical system being considered, and can be derived from the laws of physics governing that system. Application and Newton’s law of motion In this case we apply Newton’s law to objects either in equilibrium (~a = 0) or accelarating along straight line under action of constant force. The following procedure is recommanded when dealing problems with involving Newton’s law. 1. Identify the object or particle on which force are exerted. 2. Identify the force exerted on the object (Draw free body diagram) 3. Decompose each force into their x,y and z-components. 4. Calculate net force, accelaration, velocity and so’on. 4.1. DYNAMICS 41 Class work 1. A block of mass 10kg hungs from three cords as shown below Figure 4.1: 10kg hungs from three cords 2. A block of mass m slids down an inclined plane as shown in the figure below. Figure 4.2: mass m slids down an inclined plane 3. The block of mass m sliding horizontally as shown in figure below. Figure 4.3: block of mass m sliding horizontally 4. Two object of unequal mass are hung vertically over a frictionless pully of negligible mass as in figure below 42 CHAPTER 4. DYNAMICS Figure 4.4: unequal mass are hung vertically over a frictionless pully 4.2 linear momentum Linear momentum (~ p) is defined as quality of an object to exert a force on any thing that tries to change its state of motion. Linear momentum is an important concept in physics because it is a measure of an object’s ability to cause change through its motion. For example, a moving car has a lot of linear momentum and is able to do a lot of damage in a collision because it is difficult to stop. On the other hand, a stationary car has no linear momentum and is not able to cause much change through its motion. Its magnitude is the product of mass of the system with its velocity. p~ = m~v (4.1) For an objet’s in three dimension p~ = m(vx î + vy ĵ + vz k̂) (4.2) Linear momentum is a vector quantity, meaning it has both magnitude and direction. The direction of an object’s linear momentum is the same as the direction of its velocity. Its SI unit is kgm/s. ~ Impulse is defined as the product of the force acting on an object and the Impulse (J): time for which the force acts. Mathematically, impulse can be expressed as: From Newton’s second law ∆v ∆ ∆~ p F~net = m = (m~v ) = (4.3) ∆t ∆t ∆t This can be rewritten as p = F~net ∆t = J~ ∆~ (4.4) Or ∆~ p F~net = (4.5) ∆t 4.2. LINEAR MOMENTUM 43 This is the relation between p~ and resultant force acting on it. 4.2.1 Conservation of Momentum When ever two or more particles in an isolated system (in which net external force acting on the system is zero) intract, the total momentum of the system remain constant (conserved) ie X X p~i = p~f (4.6) m1 u~1 + m2 u~2 = m1 v~1 + m2 v~2 (4.7) Where u1 & u2 are initial velocity of m1 and m2 respectively, and v1 & v2 are final velocity of m1 and m2 respectively Thus p~xi = p~xf , p~yi = p~yf , p~zi = p~zf Class work 1. A 60 kg archer stands at rest on a frictionless ice and fires a 0.5 kg arrow horizontally at 50m/s. With what velocity doese archer move across the ice after firing the arrow? 4.2.2 Collision Collision: is the event of two particles comming together for short time and thereby pro- ducing impulsive force on each other. Collisions are an important topic in physics because they can be used to understand a wide range of phenomena, from the behavior of subatomic particles to the motion of celestial bodies in the universe. Depending on kinetic energy, the Q-value and coefficient of restitution Collision grouped in two (1) Elastic Collision, and (2) Inelastic Collision. Elastic Collision It is type of collision in which both kinetic energy and momentum are conserved. m1 u1 + m2 u2 = m1 v1 + m2 v2 (4.8) 1 1 1 1 m1 u21 + m2 u22 = m1 v12 + m2 v22 (4.9) 2 2 2 2 44 CHAPTER 4. DYNAMICS The collision in which kinetic energy is fully conserved is called perfectely elastic collision. Figure 4.5: perfectely elastic collision Inelastic Collision It is a type of collision in which only momentum is conserved but kinetic energy is not conserved. m1 u1 + m2 u2 = m1 v1 + m2 v2 (4.10) 1 1 1 1 m1 u21 + m2 u22 6= m1 v12 + m2 v22 (4.11) 2 2 2 2 A collision in which a colliding object stick together after collision is called perfectely in- elastic collision In this collision kinetic energy lost as a form of heat and sound during Figure 4.6: perfectely inelastic collision collision. This lost of kinetic enegy represented by Q-value. The Q-value is equal zero for elastic collision and less than zero (Q < 0) for inelastic collision. Coefficient of restitution (e) u2 −u1 = v2 −v1 is equal 1 for elastic collision and zero for inelastic collision. Head-on Collisions: Collision, when objects rebound on straight line paths that co-incide with original direction of motion. These collisions can be treated one dimen- sionally. Glancing Collisions: When Object do not collide on the same path line they make glancing collision. To solve this problem, break it into components as shown in figure 4.3. CENTER OF MASS AND MOMENT OF INERTIA 45 bellow. Figure 4.7: Glancing Collision Class work 1. A ball of mass 2 kg is moving with a velocity of 12m/s collides with a stationary ball of mass 6 kg and comes to rest. Calculate velocity of ball of mass 6kg after collision. 2. A 10.0g bullet is fired into a stationary block of wood (m = 5.00kg). The bullet sticks into the block, and the speed of the bullet-plus combination immedately after collision is 0.600m/s. What was the original speed of the bullet? 3. A block of mass m1 = 1.6kg initially moving to the right with a speed of 4m/s on a horizontal frictionless track collides with a block of mass m2 = 2.1kg initially moving to the left with speed of 2.5m/s. If the collision is elastic, find the velocities of the two block after collision? 4. A partcle of mass 4.0kg initially moving with velocity of 2.0m/s collides with a partcle of mass 6.0kg, initially moving velocity of -4m/s. What are the velocity of the two particle after collision? 5. A 4kg block moving right at 6m/s collides elastically with a 2kg moving at 3m/s left, find final velocities the blocks. 4.3 Center of Mass and Moment of Inertia 4.3.1 Center of Mass The center of mass of an object or system is the unique point at which the entire mass of the object or system can be considered to be concentrated.It is the point about which the object 46 CHAPTER 4. DYNAMICS or system will balance if it is supported at that point, and it is the point around which the object or system will rotate if it is free to do so. In a system of particles, the center of mass is the average position of all the particles in the system, weighted according to their masses. It is a useful concept in mechanics because it allows the analysis of the motion of an object or system as if all of its mass were concentrated at a single point. It is located somwher on the line joining the partcle and closser to the larger mass. The center of mass of an object or system can be found by taking the sum of the positions of all the particles in the system multiplied by their masses, and then dividing by the total mass of the system. Center of mass of several partcle with mass m1 , m2 , · · · , mn at a distance ~r1 , ~r2 , · · · , ~rn from each other is given by Pn Pn i=1 mi~ri mi~ri m1~r1 + m2~r2 + · · · + mn~rn ~rcm = Pn = i=1 = (4.12) i=1 mi M m1 + m2 + · · · + mn Where n X M= mi = total.mass i=1 For coordinate x, y and z, center of mass given by Pn Pn Pn i=1 mi xi i=1 mi yi i=1 mi zi ~xcm = î,..~ycm = ĵ..and..~zcm = k̂ (4.13) M M M This is known as the center of mass formula.The concept of center of mass is closely related to other important concepts in mechanics, such as the center of gravity, which is the point at which the gravitational force acting on an object or system can be considered to be concentrated. In many cases, the center of mass and the center of gravity of an object or system are at the same location, but this is not always the case, especially for objects or systems that are not symmetrical. 4.3.2 Moment of Inertia Moment of Inertia: is the a measure of body’s resistance to rotational motion about about a particular axis. It is typically denoted by the symbol I and is measured in kg ∗ m2. Its magnitude is affected by distribution of mass of the body in relation to its axis of rotation. Thus there is no single value of moment of inertia of an object. But for a point mass moment of inertia is given by I = mr2 4.3. CENTER OF MASS AND MOMENT OF INERTIA 47 Moment of inertia of a rigid object, made up of a particles of mass m1 , m1 ,... at respective distance r1 , r2 ,... from its axis of rotation, its moment of inertial about that axis is given by X I = m1 r12 + m2 r2 +...mi ri2 = mi ri2 i For continuous mass distribution ˆ m I= r2 dm m0 In general moment of inertia of a body depend on size of the body Shape of the body For example consider disk and sphare of the same mass and the same radius. I = 21 mr2 for uniform disk I = 52 mr2 for uniform sphare It also depend on a point of axis of rotation Example: For uniform rod axis of rotation through its center 1 I= M L2 12 For uniform rod axis of rotation through its one end 2 I = M L2 3 Class work 1. A system consists three partcles of m1 = 1kg, m2 = 1kg and m3 = 2kg located as in figure below. Find the center of mass of the system. 48 CHAPTER 4. DYNAMICS 2. A partcles are connected by a light rod as shown bellow 3. Calculate moment of inertia of 4 equal masses be each having a mass of 50g and situated at the corner of the square of side 30cm. when axis of rotation passes through a) the center of the square perpendicular to the plane of the square b) the center of the square perpendicular to sides of the square. c) along the side of the square. 4.4 Torque and angular momentum Torque: is defined as rotational effect of force ie (it is measure of force that couse an object to rotate around an axis. Torque is represented by symbol (greek letter) 0 τ 0. Torque is vector quality with both magintude and direction. It is calculated by the product of force and perpendicular distance from its axis of rotation. ~τ = F rsin(θ)n̂ = F~ × ~r Where θ is the direction of line of action of force n̂ is unit vector in the direction of torque (clockwise or anticlockwise), r is moment arm of force (point of application of force from axis of rotation) F is applied. Magintude of torque is depend on – Size of force – moment arm of force (radius or point of application of force) – line of action of force (direction of line of act

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