General Physics I, Tech. Uni., Chapter 1 & 2(1) PDF

Physics Mechanics and Heat Presented by Dr. Hadeer H. El-Hawary Ph.D. Theoretical Physics  To Contact Me Dr. Hadeer Hassan Elhawary – PhD. Theoretical Physics Email: [email protected] [email protected] Office hours: 2 hou...

Physics Mechanics and Heat Presented by Dr. Hadeer H. El-Hawary Ph.D. Theoretical Physics  To Contact Me Dr. Hadeer Hassan Elhawary – PhD. Theoretical Physics Email: [email protected] [email protected] Office hours: 2 hours/week – Tuesdays from 11 am : 1 pm Don’t hesitate to contact me @ General Physics I Dr. Hadeer Hassan El-Hawary 2  Course Assessments 10% Attendance- 15 points 20% Class work/Quizzes/Assignments-30 points 20% Practical/Laboratory Exam- 30 points 20% Midterm Exam- 30 points 30% Final Exam- 45 points You have to exceed 50% of the total course points to pass this course General Physics I Dr. Hadeer Hassan El-Hawary 3  Course Content Chapter 1: Units, Dimensions and Measurements Chapter 2: The Laws of Motion Chapter 3: Properties of Matter Chapter 4: Temperature and Heat Chapter 5: Internal Energy and Heat Transfer General Physics I Dr. Hadeer Hassan El-Hawary 4  Course Timetable 1st week: Ice breaking – Introduction in chapter 1 2nd week: Chapter 1. 3rd & 4th weeks: Chapter 2. 5th & 6th weeks: Chapter 3. 7th week: Midterm Exam. 8th & 9th weeks: Chapter 4. 10th : 12th weeks: Chapter 5. 13th & 14th weeks: Problem Solving and Revision. General Physics I Dr. Hadeer Hassan El-Hawary 5  Chapter 1 Units, Dimensions and Measurements General Physics I Dr. Hadeer Hassan El-Hawary 6  Chapter 1 Units, Dimensions and Measurements 1-1 Basic and derived quantities 1-2 Systems of units 1-3 Density and atomic mass 1-4 Dimensional analysis 1-5 Conversion of units 1-6 Vector and scalar quantities 1-7 Vectors properties General Physics I Dr. Hadeer Hassan El-Hawary 7  1-1 Basic and Derived Quantities The laws of physics are expressed in terms of basic quantities that require a clear definition and cannot be explained in terms of other physical quantities. Basic Quantities Length Mass Time L M T Any other mechanical quantities can be expressed in terms of the basic quantities and are called derived quantities Derived Quantities Force Velocity Volume ……What else?. General Physics I Dr. Hadeer Hassan El-Hawary 8  1-2 Systems of Units 1- CGS System: The CGS system of units is a French system. This system deals with only three fundamental units – Centimeter, Gram and Second for length, mass and time respectively. 2- MKS Units The MKS system of units was set up by France. This system also deals with three fundamental units – Meter, Kilogram and Second for length, mass and time respectively. This system is also called the metric system of units and is closely related to C.G.S system of units. General Physics I Dr. Hadeer Hassan El-Hawary 9  1-2 Systems of Units 3- FPS system The FPS system of units is a British system. This system deals with only three fundamental units – the Foot, Pound and the Second for length, mass and time respectively. Although the FPS system is often associated with the British Imperial system, it has become more prevalent in American engineering and everyday measurements. General Physics I Dr. Hadeer Hassan El-Hawary 10  1-2 Systems of Units 4- SI system The measurement system which is internationally accepted now is suggested by the Eleventh general conference of weights and Measures in 1960 – France and is known as SI units of measurement. There are seven basic or fundamental units and three supplementary units. General Physics I Dr. Hadeer Hassan El-Hawary 11  1-2 Systems of Units 4- SI system The basic units are: The supplementary units are: Meter (m) for length, Radian (rad) for angle, Mechanical Kilogram (kg) for mass, Steradian (sr) for solid angle, Quantities Second (s) for time, Becquerel (Bq) for radioactivity. Kelvin (K) for temperature, Non-Mechanical Quantities Ampere (A) for electric current, Candela (cd) for luminous intensity, Mole (mol) for the amount of substance. General Physics I Dr. Hadeer Hassan El-Hawary 12  1-2 Systems of Units The International committee established a set of standards for the three basic quantities as follows: 1- Length: Standard of length is based on speed of light in vacuum. Meter (m): distance travelled by light in vacuum during a time of 𝑳 (𝒎) 𝟏 𝒎 𝟏 𝑳𝒊𝒈𝒕𝒉 𝑺𝒑𝒆𝒆𝒅 = = 𝟐𝟗𝟗 𝟕𝟗𝟐 𝟒𝟓𝟖 (𝒎/𝒔) 𝒕 𝒔 = 𝒎 ≅ 𝟖 𝒔 𝒕 (𝒔) 𝟐𝟗𝟗 𝟕𝟗𝟐 𝟒𝟓𝟖 𝒔 𝟑 × 𝟏𝟎 General Physics I Dr. Hadeer Hassan El-Hawary 13  1-2 Systems of Units The International committee established a set of standards for the three basic quantities as follows: 2- Mass: Standard of mass is based on the mass of an alloy. Kilogram (kg): mass of a specific Pt-Ir alloy cylinder. (Platinum–iridium alloy is very stable ) General Physics I Dr. Hadeer Hassan El-Hawary 14  1-2 Systems of Units The International committee established a set of standards for the three basic quantities as follows: 3- Time: Standards of time is based on the atomic clock of Cs133 atom Second (s): obtained by the characteristic frequency (period of vibration) of radiation from the Cs133 atom. (Cs: Cesium) Periodic time is 𝟏 𝑻 = 𝒔 𝟏 𝒔 = 𝟗 𝟏𝟗𝟐 𝟔𝟑𝟏 𝟕𝟕𝟎 × 𝑻 𝟗𝟏𝟗𝟐𝟔𝟑𝟏𝟕𝟕𝟎 General Physics I Dr. Hadeer Hassan El-Hawary 15  1-3 Density and Atomic Mass Density “𝜌”: is defined as the mass per unit volume 𝜌 = 𝑚/𝑉 For Aluminum (Al) ρ = 2.7 gm/cm3 A piece of Al of volume 10 cm3 has a mass 𝑚 = 𝜌𝑉 = 2.7 × 10 = 27𝑔𝑚 For Lead (Pb) ρ = 11.3 gm/cm3 Why they are A piece of Pb of volume 10 cm3 has a mass different for 𝑚 = 𝜌𝑉 = 113 × 10 = 113 𝑔𝑚 same volume? General Physics I Dr. Hadeer Hassan El-Hawary 16  1-3 Density and Atomic Mass Why they are different? Because the atomic masses are different  Al AAl = 27 Note : the atomic mass  Pb APb = 207  Pb 11.3 APb 207  Pb APb = = 4.19  = = 7.67   Al 2.7 AAl 27  Al AAl The ratios of densities and atomic masses are not the same because of atomic spacing and atomic arrangements in both materials (Al & Pb) are different. General Physics I Dr. Hadeer Hassan El-Hawary 17  1-3 Density and Atomic Mass Mole: One mole of a substance is that a mount of it that consists of Avogadro’s number of atoms or molecules. Avogadro’s Number (Na): The volume of a gas is directly proportional to the amount of the gas at constant pressure and temperature. The proportional constant is Avogadro's number. 𝑁𝑎 = 6.02 × 1023 𝑎𝑡𝑜𝑚 /𝑚𝑜𝑙 One mole of Al has a mass of AAL=27 gm One mole of Pb has a mass of APb=207 gm General Physics I Dr. Hadeer Hassan El-Hawary 18  1-3 Density and Atomic Mass A (gm/mol) Mass of any atom = Na (atom/mol) where A is the atomic mass in gm (mass of one mole) Example: A solid cube of Aluminum of 0.2 cm3.How many Aluminum atoms contained in the cube?(ρ=2.7 gm/cm3,A=27 gm) m  =  m =   V  m = 2.7  0.2 = 0.54gm V A 27 − 23 matom = = = 4.5  10 gm N A 6.02 1023 mtotal 0.54 gm Number of atoms = = = 1.2  1022 atoms matom 4.5 10 atom/gm - 23 General Physics I Dr. Hadeer Hassan El-Hawary 19  1-4 Dimensional Analysis Dimension denotes the physical nature of a quantity and is described using the basic quantities: Length (L) - Mass (M) - Time (T) Dimension of some physics quantities: Area [A] =L.L = L2 Volume [V] =L.L.L = L3 Velocity [ν] =L / T = LT-1 Acceleration [a] =L / T2 = LT-2 Momentum [P] = [m]. [ν] = MLT-1 Kinetic Energy [K.E] = [1/2].[m]. [ν2] = M((LT-1)2) = ML2T-2 Potential Energy [P.E] = [m]. [g]. [h] = M(LT-2) L = ML2T-2 General Physics I Dr. Hadeer Hassan El-Hawary 20  1-4 Dimensional Analysis Dimensional analysis can be used to assist in checking the validity of a specific formula and deriving some formulas. Example: A car started from rest and moves with a constant acceleration a. What is the distance x travelled in a time t. 1) If we know the formula and want to check it. x=½ a t2 [L.H.S] = [x ] = L [R.H.S] = [ ½] [a]. [t2] = LT-2. T2 = L [R.H.S]= [L.H.S] Since dimensions of both side are the same: The formula is correct. General Physics I Dr. Hadeer Hassan El-Hawary 21  1-4 Dimensional Analysis 2) If we wish to derive the formula We assume the factors that affect the displacement (x) are the acceleration (a) and the time (t). 𝑥 ∝ 𝑡 𝑛 , 𝑥 ∝ 𝑎𝑚 𝑥 ∝ 𝑡 𝑛. 𝑎𝑚 ⇒ 𝐿 = 𝑇 𝑛. 𝐿. 𝑇 −2 𝑚 𝐿1. 𝑇 0 = 𝑇 𝑛. 𝐿𝑚. 𝑇 −2𝑚 = 𝐿𝑚. 𝑇 𝑛−2𝑚 ⇒ 𝑚 = 1, 𝑛 − 2𝑚 = 0 ⇒ 𝑛 = 2 ⇒ 𝑥 ∝ 𝑡 2. 𝑎1 ⇒ 𝑥 = 𝑘 𝑎𝑡 2 While k is a numerical constant. Dimensional analysis can tell you what units you need and lead you to the general form but cannot tell you what the numerical value of constants must be. General Physics I Dr. Hadeer Hassan El-Hawary 22  1-4 Dimensional Analysis Example: Suppose we are told that the acceleration a of a particle moving with a uniform speed v in a circular orbit of radius r is proportional to some power of r say rn and some power of v say vm. How can we determine the power of r and v ? a  rn ( LT -2 ) = (L) n.(LT -1 )m a  vm LT -2 =L nL m T -m a  r n vm LT - 2 = L n + m T -m a = k r n vm 1= n+ m  -2 = -m m = 2, n = 1 − 2 = −1 2 a=k a=kr −1 2 r General Physics I Dr. Hadeer Hassan El-Hawary 23  1-5 Conversion of Units Sometimes it is necessary to convert units from one system to another. Quantities FPS Unit SI Unit Conversion Factor Length Foot (ft) Meter (m) 1 ft = 0.3048 m Length Yard (yd) Meter (m) 1 yd = 0.9144 m Length Inch (in) Meter (m) 1 in = 0.0254 m Mass Pound (lb) Kilogram (kg) 1 lb = 0.453592 kg Mass Slug (slug) Kilogram (kg) 1 slug = 14.5939 kg Velocity Foot per second (ft/s) Meter per second (m/s) 1 ft/s = 0.3048 m/s Force Pound-force (lbf) Newton (N) 1 lbf = 4.44822 N Energy Foot-pound (ft.lb) Joule (J) 1 ft·lb = 1.35582 J Power Horsepower (hp) Watt (W) 1 hp = 745.7 W General Physics I Dr. Hadeer Hassan El-Hawary 24  1-5 Conversion of Units Sometimes it is necessary to convert units from one system to another. SI Unit Abbreviation Equivalent in Meters (m) Kilometer km 1 km = 1000 m =103 m Meter m 1m Centimeter cm 1 cm = 0.01 m =10-2 m Millimeter mm 1 mm = 0.001 m =10-3 m Micrometer µm 1 µm = 0.000001 m =10-6 m Nanometer nm 1 nm = 0.000000001 m =10-9 m Picometer pm 1 pm = 0.000000000001 m =10-12 m General Physics I Dr. Hadeer Hassan El-Hawary 25  1-5 Conversion of Units Example 1- Length Conversion: Convert 150 centimeters to meters. 150 × 0.01 = 1.5 𝑚 Example 2- Weight Conversion: Convert 200 pounds to kilograms. 200 × 0.453592 ≈ 90.72 𝑘𝑔 Example 3- Velocity Conversion: Convert 60 miles per hour to meters per second. 1 mile = 1609.34 meters, and 1 hour = 3600 seconds. 60 × 1609.34 /3600 ≈ 26.82 𝑚/𝑠 General Physics I Dr. Hadeer Hassan El-Hawary 26  1-6 Vector and Scalar Quantities A scalar quantity is completely specified by a single value with an appropriate unit and has no direction. A vector quantity is completely described by a number and appropriate units plus a direction. A particle travels from A to B along the path shown by the broken line. This is the distance traveled and is a scalar. The displacement is the solid line from A to B and is independent of the path taken between the two points. The displacement is a vector. General Physics I Dr. Hadeer Hassan El-Hawary 27  1-6 Vector and Scalar Quantities Displacement is the difference Distance adds Up between start and end points. General Physics I Dr. Hadeer Hassan El-Hawary 28  1-7 Vector Properties The vector magnitude has physical units (m/sec., N, m/sec.2,…) The magnitude is always a positive number. The direction is an angle relative to a reference line (positive x-axis direction). Vectors could be written as 𝐴Ԧ or in a bold print A While the magnitude of a vector, is written in italic letter as A or 𝐴Ԧ General Physics I Dr. Hadeer Hassan El-Hawary 29  1-7 Vector Properties Two vectors are equal if they have the same magnitude and the same direction (i.e. A = B and they point along parallel lines). All of the vectors shown are equal. Vector can be moved to a position parallel to itself General Physics I Dr. Hadeer Hassan El-Hawary 30  1-7 Vector Properties Vector components are the projections of a vector along the x- and y-axes. The magnitude of the x-component of a vector is the projection along the x-axis. 𝐴𝑥 = 𝐴 cos 𝜃 The magnitude of the y-component of a vector is the projection along the y-axis. 𝐴𝑦 = 𝐴 𝑠𝑖𝑛 𝜃 This assumes the angle θ is measured with respect to the x-axis. General Physics I Dr. Hadeer Hassan El-Hawary 31  1-7 Vector Properties The magnitude A and the direction θ of any vector 𝐴Ԧ could be found using the magnitude of the components Ax and Ay. 𝐴= 𝐴 𝑥 2 + 𝐴𝑦 2 𝐴𝑦 −1 𝜃 = 𝑡𝑎𝑛 𝐴𝑥 The components could be positive or negative, have the same units as the original vector and their signs depend on the angle. General Physics I Dr. Hadeer Hassan El-Hawary 32  1-7 Vector Properties A unit vector is a dimensionless vector with a magnitude of exactly 1 and are only used to specify the direction. 𝑖Ƹ = 𝑗Ƹ = 𝑘෠ = 1 The vector form in two dimensions: 𝐴Ԧ = 𝐴𝑥 𝑖Ƹ + 𝐴𝑦 𝑗Ƹ The complete vector (in three dimensions) can be expressed as: 𝐴Ԧ = 𝐴𝑥 𝑖Ƹ + 𝐴𝑦 𝑗Ƹ + 𝐴𝑧 𝑘෠ General Physics I Dr. Hadeer Hassan El-Hawary 33  1-7 Vector Properties Adding Vectors We could add or subtract two or more vectors algebraically by using the complete vector form described by the unit vectors 𝑅 = 𝐴Ԧ + 𝐵 𝑅 = 𝐴𝑥 𝑖Ƹ + 𝐴𝑦 𝑗Ƹ + 𝐵𝑥 𝑖Ƹ + 𝐵𝑦 𝑗Ƹ 𝑅 = 𝐴𝑥 + 𝐵𝑥 𝑖Ƹ + 𝐴𝑦 + 𝐵𝑦 𝑗Ƹ 𝑅 = 𝑅𝑥 𝑖Ƹ + 𝑅𝑦 𝑗Ƹ Where 𝑅𝑥 = 𝐴𝑥 + 𝐵𝑥 and 𝑅𝑦 = 𝐴𝑦 + 𝐵𝑦 are the components of 𝑅. While the magnitude R and the direction θ are given by 𝑅= 𝑅𝑥 2 + 𝑅𝑦 2 −1 𝑅𝑦 𝜃 = 𝑡𝑎𝑛 𝑅𝑥 General Physics I Dr. Hadeer Hassan El-Hawary 34  1-7 Vector Properties Example: Given the vectors A = 2.00 i +6.00 j and B = 3.00 i - 2.00 j, C = A + B and the vector difference D = A - B. Calculate C and D, first in terms of unit vectors and then in terms of magnitudes and angles 𝑪 = 𝑨 + 𝑩 𝑪 = 2.00 𝒊 + 6.00 𝒋 + 3.00 𝒊 − 2.00 𝒋 𝑪=5𝒊 + 4𝒋 Then the magnitude is 𝑪 = 52 + 42 = 25 + 16 = 6.4 In addition, the direction is −1 4 𝜃𝐶 = 𝑡𝑎𝑛 = 38.7° 5 General Physics I Dr. Hadeer Hassan El-Hawary 35  1-7 Vector Properties 𝑫 = 𝑨− 𝑩 𝑫 = 2.00 𝒊 + 6.00 𝒋 − 3.00 𝒊 − 2.00 𝒋 𝑫=−𝒊 + 8𝒋 Then the magnitude is 𝑫 = −1 2 + 82 = 1 + 64 = 8.06 In addition, the direction is −1 8 𝜃𝐷 = 𝑡𝑎𝑛 = 180° − 82.875° = 97.125° −1 Note while calculations: If the x-component of any vector is a negative value, then add 180° to the angle calculated using the calculator to correct the sign error. General Physics I Dr. Hadeer Hassan El-Hawary 36  Problems 1- What mass of a material with density is required to make a hollow spherical shell having inner radius r1 and outer radius r2? 2- Calculate the mass of an atom of (a) helium, (b) iron, and (c) lead. Give your answers in grams. The atomic masses of these atoms are 4.00 g/mol, 55.9 g/mol, and 207 g/mol, respectively. 3- Which of the following equations are dimensionally correct? (a) vf = vi + ax (b) y = (2 m) cos(kx), where k = 2 m–1. General Physics I Dr. Hadeer Hassan El-Hawary 37  Problems 𝑀𝑚 4- Newton’s law of universal gravitation is represented by 𝐹 = 𝐺 2 𝑟 Here F is the gravitational force exerted by one small object on another, M and m are the masses of the objects, and r is a distance. Force has the SI units kg·m/s2. What are the SI units of the proportionality constant G. 5- The period of a pendulum t is measured in time units and is  t = 2 g  = length of the pendelium g = free - fall acceleration measured in length divided by square of time Show that the equation is dimensionally correct. General Physics I Dr. Hadeer Hassan El-Hawary 38  Chapter 2 The Laws of Motion General Physics I Dr. Hadeer Hassan El-Hawary 39  Chapter 2 The Laws of Motion 2-1 Position, velocity and speed 2-2 Instantaneous velocity and speed 2-3 Acceleration 2-4 One-Dimensional motion with constant acceleration 2-5 Freely falling objects 2-6 Two-Dimensional motion with constant acceleration 2-7 Projectile motion 2-8 Uniform circular motion General Physics I Dr. Hadeer Hassan El-Hawary 40  Chapter 2 The Laws of Motion 2-9 The concept of force 2-10 Newton’s first law 2-11 Mass 2-12 Newton’s second law 2-13 The gravitational force and weight 2-14 Newton’s third law 2-15 Some applications of Newton’s law 2-16 Forces of friction General Physics I Dr. Hadeer Hassan El-Hawary 41  2-1 Position, Velocity and Speed Displacement is the change of the position of a particle. Displacement is a vector quantity. Displacement ∆𝑥Ԧ = 𝑥Ԧ𝑓 − 𝑥Ԧ𝑖 If 𝑥Ԧ𝑓 > 𝑥Ԧ𝑖 then ∆𝑥Ԧ = +ve, If 𝑥Ԧ𝑓 < 𝑥Ԧ𝑖 then ∆𝑥Ԧ = −ve Displacement ≠ Distance travelled Displacement ∆𝑥Ԧ = 𝑥Ԧ𝑓 − 𝑥Ԧ𝑖 = 𝐴𝐵 Distance= 𝐴𝐶 + 𝐶𝐵 General Physics I Dr. Hadeer Hassan El-Hawary 42  2-1 Position, Velocity and Speed Average velocity is the ratio of its displacement Δ𝑥 and the time interval Δ𝑡. A particle is moved from point A to point B Point A : position= 𝑥Ԧ𝑖 , time= 𝑡𝑖 Point B : position= 𝑥Ԧ𝑓 , time= 𝑡𝑓 Displacement is ∆𝑥Ԧ = 𝑥Ԧ𝑓 − 𝑥Ԧ𝑖 and time interval is ∆t = 𝑡𝑓 − 𝑡𝑖 ∆𝑥Ԧ 𝑥Ԧ𝑓 −𝑥Ԧ𝑖 The average velocity is 𝑣𝑥 = = ∆t 𝑡𝑓 −𝑡𝑖 General Physics I Dr. Hadeer Hassan El-Hawary 43  2-1 Position, Velocity and Speed Geometrical meaning of 𝒗 ഥ? 𝑣ҧ is the slope of the straight-line joining initial (A) and final (B) points on the position-time graph. ∆𝑥Ԧ Slope of 𝐴𝐵 = = 𝑣𝑥 ∆t Total distance traveled Average Speed = Total time Average speed is a scalar quantity (has no direction ). General Physics I Dr. Hadeer Hassan El-Hawary 44  2-1 Position, Velocity and Speed Example: A car is moving along the x-axis from point A to point F. Point A : 𝑥Ԧ𝑖 = 30 𝑚, 𝑡𝑖 = 0 𝑠 Point B : 𝑥Ԧ = 52 𝑚, 𝑡𝑖 = 10 𝑠 Point F : 𝑥Ԧ𝑓 = −53 𝑚, 𝑡𝑓 = 50 𝑠 Find: (i) Displacement (ii) Average velocity (iii) Average speed General Physics I Dr. Hadeer Hassan El-Hawary 45  2-1 Position, Velocity and Speed Answer: (i) Displacement ∆𝑥Ԧ = 𝑥Ԧ𝑓 − 𝑥Ԧ𝑖 = −53 − 30 = −83 m (ii) Average velocity ∆𝑥Ԧ 𝑥Ԧ𝑓 − 𝑥Ԧ𝑖 −83 𝑣𝑥 = = = = −1.7 𝑚/𝑠 ∆t 𝑡𝑓 − 𝑡𝑖 50 − 0 (iii) Average speed Total distance traveled 𝐴𝐵 + 𝐵𝐹 22 + 52 + 53 𝑣ҧ = = = = 2.5 𝑚/𝑠 Total time ∆t 50 General Physics I Dr. Hadeer Hassan El-Hawary 46  2-2 Instantaneous Velocity and Speed Instantaneous velocity is the limit of the average velocity as the time interval becomes infinitesimally short, or as the time interval approaches zero. If B approach A then ∆𝑡 → 0. ∆𝑥Ԧ 𝑑𝑥 𝑣𝑥 = lim = ∆𝑡→0 ∆t 𝑑𝑡 The instantaneous velocity can be positive, negative or zero. The instantaneous speed of a particle is defined to be equal to the magnitude of its the instantaneous velocity. The instantaneous velocity is also called “velocity”. General Physics I Dr. Hadeer Hassan El-Hawary 47  2-2 Instantaneous Velocity and Speed Example: A particle moves along the x-axis. Its x-coordinates varies with time according to the expression 𝑥 = −4𝑡 + 2𝑡 2 where 𝑥 is in meters and 𝑡 is in seconds. The position–time graph for this motion is shown below. a) Determine the displacement of the particle in the time intervals i) t = 0 to t = 1s and ii) t = 1s to t = 3s b) Calculate the average velocity during these two time intervals i) t = 0 to t = 1s and ii) t = 1s to t = 3s c) Find the instantaneous velocity of the particle at t = 2.5s General Physics I Dr. Hadeer Hassan El-Hawary 48  2-2 Instantaneous Velocity and Speed Answer: 𝑥 = −4𝑡 + 2𝑡 2 a) Determine the displacement of the particle in the time intervals i) t = 0 to t = 1s and ii) t = 1s to t = 3s 𝑡(𝑠) 𝑥(𝑚) i) t = 0 to t = 1s from A to B 0 0 ∆𝑥Ԧ = 𝑥Ԧ𝐵 − 𝑥Ԧ𝐴 = −2 − 0 = −2 𝑚 1 -2 ii) t = 1 to t = 3s from B to D 2 0 3 6 ∆𝑥Ԧ = 𝑥Ԧ𝐷 − 𝑥Ԧ𝐵 = 6 − −2 = 8 𝑚 4 16 General Physics I Dr. Hadeer Hassan El-Hawary 49  2-2 Instantaneous Velocity and Speed Answer: b) Calculate the average velocity during these two time intervals i) t = 0 to t = 1s and ii) t = 1s to t = 3s 𝑡(𝑠) 𝑥(𝑚) i) t = 0 to t = 1s, ∆𝑥Ԧ = −2 𝑚 0 0 ∆𝑥Ԧ𝐴𝐵 −2 𝑣𝐴𝐵 = = = −2 𝑚/𝑠 1 -2 ∆t 1−0 2 0 ii) t = 1 to t = 3s, ∆𝑥Ԧ = 8 𝑚 3 6 ∆𝑥Ԧ𝐵𝐷 8 𝑣𝐵𝐷 = = = 4 𝑚/𝑠 4 16 ∆t 3−1 General Physics I Dr. Hadeer Hassan El-Hawary 50  2-2 Instantaneous Velocity and Speed Answer: 𝑥 = −4𝑡 + 2𝑡 2 c) Find the instantaneous velocity of the particle at t = 2.5s 𝑑𝑥 𝑑 𝑣Ԧ = = −4𝑡 + 2𝑡 2 = −4 + 4𝑡 𝑑𝑡 𝑑𝑡 𝑣Ԧ 𝑡 = 2.5 𝑠 = −4 + 4 × 2.5 = 6 𝑚/𝑠 General Physics I Dr. Hadeer Hassan El-Hawary 51  2-3 Acceleration Average acceleration: When the velocity of a particle changes with time, the particle is said to be accelerating. The average acceleration of the particle in the time interval Δ𝑡 = 𝑡𝑓 – 𝑡𝑖 ∆𝑣𝑥 is defined as the ratio , where ∆𝑣𝑥 = 𝑣𝑥𝑓 − ∆𝑣𝑥𝑖 ∆t ∆𝑣𝑥 𝑣𝑥𝑓 − ∆𝑣𝑥𝑖 𝑎𝑥 = = ∆t 𝑡𝑓 – 𝑡𝑖 The instantaneous acceleration It is the limit of the average acceleration as ∆t approaches zero. ∆𝑣𝑥 𝑑𝑣𝑥 𝑑 2 𝑥 𝑎𝑥 = lim = = 2 ∆𝑡→0 ∆t 𝑑𝑡 𝑑𝑡 General Physics I Dr. Hadeer Hassan El-Hawary 52  2-3 Acceleration Example: The velocity of a particle moving along the x-axis varies in time according to the expression 𝑣𝑥 = 40 − 5𝑡 2 𝑚/𝑠, where t is in seconds. a) Find the average acceleration in the time interval t = 0 to t = 2.0 s. ∆𝑣𝑥 𝑣𝑥𝑓 − ∆𝑣𝑥𝑖 20 − 40 𝑎𝑥 = = = = −10 𝑚/𝑠 2 𝑡(𝑠) 𝒗𝒙 (𝑚/𝒔) ∆t 𝑡𝑓 – 𝑡𝑖 2−0 0 40 (b) Determine the acceleration at t = 2.0 s. 1 35 𝑑𝑣𝑥 𝑑(40 − 5𝑡 2 ) 2 20 𝑎𝑥 = = = 0 − 10𝑡 𝑑𝑡 𝑑𝑡 3 -5 𝑎𝑥 𝑡 = 2 𝑠 = −10 × 2 = −20 𝑚/𝑠 2 4 -40 General Physics I Dr. Hadeer Hassan El-Hawary 53 2-4 One-Dimensional Motion with Constant Acceleration Position- time graph Velocity- time graph Acceleration- time graph General Physics I Dr. Hadeer Hassan El-Hawary 54 2-4 One-Dimensional Motion with Constant Acceleration When the velocity of an object changes at the same rate throughout the motion the acceleration is said to be constant. Thus, the velocity-time graph becomes a straight line. 1) Velocity as a function of time 𝑣𝑥𝑓 − 𝑣𝑥𝑖 𝑎𝑥 = = const. 𝑡𝑓 – 𝑡𝑖 𝑣𝑥𝑓 −𝑣𝑥𝑖 At 𝑡𝑖 = 0, 𝑡𝑓 = 𝑡, Thus 𝑎𝑥 = ⇒ 𝑣𝑥𝑓 − 𝑣𝑥𝑖 = 𝑎𝑥 𝑡 𝑡 𝑣𝑥𝑓 = 𝑣𝑥𝑖 + 𝑎𝑥 𝑡 General Physics I Dr. Hadeer Hassan El-Hawary 55 2-4 One-Dimensional Motion with Constant Acceleration 2) Displacement as a function of velocity and time Since 𝑣𝑥 varies linearly with time, we have 𝑣𝑥𝑓 + 𝑣𝑥𝑖 ∆𝑥 𝑥𝑓 − 𝑥𝑖 𝑣𝑥 = = =. 2 ∆𝑡 𝑡 Thus, 1 𝑥𝑓 − 𝑥𝑖 = 𝑣𝑥𝑓 + 𝑣𝑥𝑖 𝑡 2 General Physics I Dr. Hadeer Hassan El-Hawary 56 2-4 One-Dimensional Motion with Constant Acceleration 3) Displacement as a function of time 1 𝑣𝑥𝑓 = 𝑣𝑥𝑖 + 𝑎𝑥 𝑡 and 𝑥𝑓 − 𝑥𝑖 = 𝑣𝑥𝑓 + 𝑣𝑥𝑖 𝑡 2 Thus, 1 1 𝑥𝑓 − 𝑥𝑖 = 𝑣𝑥𝑖 + 𝑎𝑥 𝑡 + 𝑣𝑥𝑖 𝑡 = 𝑣𝑥𝑖 + 𝑎𝑥 𝑡 2 2 2 1 𝑥𝑓 − 𝑥𝑖 = 𝑣𝑥𝑖 𝑡 + 𝑎𝑥 𝑡 2 2 General Physics I Dr. Hadeer Hassan El-Hawary 57 2-4 One-Dimensional Motion with Constant Acceleration 4) Velocity as a function of displacement 𝑣𝑥𝑓 − 𝑣𝑥𝑖 𝑣𝑥𝑓 = 𝑣𝑥𝑖 + 𝑎𝑥 𝑡 ⇒ 𝑡 = 𝑎𝑥 1 1 𝑣𝑥𝑓 − 𝑣𝑥𝑖 𝑥𝑓 − 𝑥𝑖 = 𝑣𝑥𝑓 + 𝑣𝑥𝑖 𝑡 ⇒ 𝑥𝑓 − 𝑥𝑖 = 𝑣𝑥𝑓 + 𝑣𝑥𝑖 2 2 𝑎𝑥 2 2 𝑣 1 𝑥𝑓 − 𝑣𝑥𝑖 𝑥𝑓 − 𝑥𝑖 = 2 𝑎𝑥 2 2 𝑣𝑥𝑓 − 𝑣𝑥𝑖 = 2𝑎𝑥 𝑥𝑓 − 𝑥𝑖 2 2 𝑣𝑥𝑓 = 𝑣𝑥𝑖 + 2𝑎𝑥 𝑥𝑓 − 𝑥𝑖 General Physics I Dr. Hadeer Hassan El-Hawary 58 2-4 One-Dimensional Motion with Constant Acceleration Kinematic Equations of motion in a straight line with constant acceleration General Physics I Dr. Hadeer Hassan El-Hawary 59 2-4 One-Dimensional Motion with Constant Acceleration Example: A jet lands on an aircraft carrier at 63 m/s. (a) What is its acceleration (assumed constant) if it stops in 2.0 s? 𝑚 𝑣𝑥𝑓 = 0, 𝑣𝑥𝑖 = 63 , 𝑡 = 2 𝑠 𝑠 𝑣𝑥𝑓 − 𝑣𝑥𝑖 63 𝑣𝑥𝑓 = 𝑣𝑥𝑖 + 𝑎𝑥 𝑡 ⇒ 𝑎𝑥 = = = 31.5 𝑚/𝑠 2 𝑡 2 (b) What is the displacement of the plane while it is stopping ? 1 1 𝑥𝑓 − 𝑥𝑖 = 𝑣𝑥𝑓 + 𝑣𝑥𝑖 𝑡 ⇒ ∆𝑥 = 63 + 0 × 2 = 63 𝑚 2 2 General Physics I Dr. Hadeer Hassan El-Hawary 60  2-5 Freely Falling Objects A freely falling object is any object moving freely under the influence of gravity alone, regardless of its initial motion. Acceleration: Directed downward, regardless of the initial motion of the object. Kinematic equations are valid with change: 𝑥 → 𝑦, 𝑎𝑥 → 𝑎𝑦 = −𝑔 = −9.8 𝑚/𝑠 2 General Physics I Dr. Hadeer Hassan El-Hawary 61  2-5 Freely Falling Objects General Physics I Dr. Hadeer Hassan El-Hawary 62  2-5 Freely Falling Objects Example: A stone thrown from the top of a building is given an initial velocity of 20.0 m/s straight upward. The building is 50.0 m high, and the stone just misses the edge of the roof on its way down. Using tA = 0, find (a) The time at which the stone reaches its maximum height? At maximum height 𝑣𝑦𝐵 = 0, 𝑣𝑦𝐵 = 𝑣𝑦𝐴 − 𝑔𝑡 𝑣𝑦𝐴 − 𝑣𝑦𝐵 20 − 0 𝑡= = = 2.04 𝑠 𝑔 9.8 General Physics I Dr. Hadeer Hassan El-Hawary 63  2-5 Freely Falling Objects (b) The maximum height At maximum height 𝑣𝑦𝐵 = 0, 𝑣𝑦𝐵 = 𝑣𝑦𝐴 − 𝑔𝑡 𝑣𝑦𝐴 − 𝑣𝑦𝐵 20 − 0 𝑡= = = 2.04 𝑠 𝑔 9.8 1 2 𝑦𝑓 − 𝑦𝑖 = 𝑣𝑦𝑖 𝑡 − 𝑔𝑡 2 1 𝑦𝑓 − 𝑦𝑖 = 20 × 2.04 − × 9.8 × 2.042 = 20.4 𝑚 2 General Physics I Dr. Hadeer Hassan El-Hawary 64  2-5 Freely Falling Objects (c) The time at which the stone returns to the height from which it was thrown and the velocity of the stone at this instant? 1 2 𝑦𝑓 − 𝑦𝑖 = 𝑣𝑦𝑖 𝑡 − 𝑔𝑡 2 𝑦𝑓 − 𝑦𝑖 = 0, 𝑣𝑦𝑖 = 20 𝑚/𝑠 1 0 = 20 × 𝑡 − × 9.8 × 𝑡 2 2 𝑡 20 − 4.9𝑡 = 0 ⇒ 𝑡 = 4.08 𝑠 𝑣𝑓 = 𝑣𝑖 − 𝑔𝑡 ⇒ 𝑣𝑓 = 20 − 9.8 × 4.08 = −20 𝑚/𝑠 General Physics I Dr. Hadeer Hassan El-Hawary 65  2-5 Freely Falling Objects (d) The velocity and position of the stone at t = 5 s.? 𝑣𝑓 = 𝑣𝑖 − 𝑔𝑡 ⇒ 𝑣𝑓 = 20 − 9.8 × 5 = −29 𝑚/𝑠 1 𝑦𝑓 − 𝑦𝑖 = 20 × 5 − × 9.8 × 52 = −22.5 𝑚 2 𝑦𝑖 = 0 ⇒ 𝑦𝑓 = −22.5 𝑚 Exercise: Find the velocity of the stone just before it hits ground and the total time the stone in air ? (-37.1 m/s, 5.83 s) General Physics I Dr. Hadeer Hassan El-Hawary 66 2-6 Two-Dimensional Motion with Constant Acceleration The position vector of a particle in the x‐y plane is a vector drawn from the origin (0,0) to the location of the particle. A particle moves from point Ⓐ to point Ⓑ in the x‐y plane The displacement vector is defined as the difference between the final and the initial position vectors. ∆𝑟Ԧ = 𝑟Ԧ𝑓 − 𝑟Ԧ𝑖 The average velocity, 𝑣ҧ of a particle during the time interval Δt is defined as: 𝑣ҧ = ∆𝑟/∆𝑡 Ԧ General Physics I Dr. Hadeer Hassan El-Hawary 67 2-6 Two-Dimensional Motion with Constant Acceleration The instantaneous velocity 𝑣Ԧ is defined as ∆𝑟Ԧ 𝑑 𝑟Ԧ 𝑟Ԧ = 𝑥 𝑖Ƹ + 𝑦 𝑗Ƹ 𝑣Ԧ = lim = 𝑣Ԧ = 𝑣𝑥 𝑖Ƹ + 𝑣𝑦 𝑗Ƹ 𝑡→0 ∆𝑡 𝑑𝑡 The direction of at any point of the path is along the tangent to the path at this point and the magnitude of the velocity is called the speed. General Physics I Dr. Hadeer Hassan El-Hawary 68 2-6 Two-Dimensional Motion with Constant Acceleration The average acceleration of a particle is defined as 𝑣Ԧ𝑓 − 𝑣Ԧ𝑖 𝑎Ԧ = 𝑡𝑓 − 𝑡𝑖 𝑎Ԧ is a vector a quantity directed along the direction of 𝑣Ԧ. The instantaneous acceleration of a particle is defined as ∆𝑣Ԧ 𝑑𝑣Ԧ 𝑎Ԧ = lim = 𝑎Ԧ = 𝑎𝑥 𝑖Ƹ + 𝑎𝑦 𝑗Ƹ 𝑡→0 ∆𝑡 𝑑𝑡 General Physics I Dr. Hadeer Hassan El-Hawary 69 2-6 Two-Dimensional Motion with Constant Acceleration The motion in two dimensions with constant acceleration can be considered as two independent motions: One along x‐direction with constant acceleration (𝑎𝑥 ) One along y‐direction with constant acceleration (𝑎𝑦 ) General Physics I Dr. Hadeer Hassan El-Hawary 70 2-6 Two-Dimensional Motion with Constant Acceleration Example: A particle starts from the origin at t = 0 with 𝑣𝑥𝑖 = 20 𝑚 /𝑠, 𝑣𝑦𝑖 = −15 𝑚/𝑠. The particle moves in the x-y plane with an x- component of acceleration only, 𝑎𝑥 = 4.0 𝑚/𝑠 2. a) Determine the velocity vector at any time? 𝑣𝑖 = 𝑣𝑥𝑖 𝑖Ƹ + 𝑣𝑦𝑖 𝑗Ƹ = 20 𝑖Ƹ − 15 𝑗Ƹ 𝑎Ԧ = 𝑎𝑥 𝑖Ƹ + 𝑎𝑦 𝑗Ƹ = 4 𝑖Ƹ − 0 𝑗Ƹ 𝑣𝑓 = 𝑣𝑖 + 𝑎𝑡 Ԧ 𝑣𝑓 = 20 𝑖Ƹ − 15 𝑗Ƹ + 4𝑡 𝑖Ƹ = (20 + 4𝑡) 𝑖Ƹ − 15 𝑗Ƹ General Physics I Dr. Hadeer Hassan El-Hawary 71 2-6 Two-Dimensional Motion with Constant Acceleration b) Calculate the magnitude of 𝑣Ԧ (speed) and its direction after 5s? 𝑣Ԧ = 20 + 4𝑡 𝑖Ƹ − 15 𝑗Ƹ = 20 + 4 × 5 𝑖Ƹ − 15 𝑗Ƹ = 40𝑖Ƹ − 15 𝑗Ƹ 2 2 2 2 𝑚 𝑣Ԧ = 𝑣𝑥 + 𝑣𝑦 = 40 + (−15) = 43 𝑠 −1 𝑣𝑦 −15 𝜃 = 𝑡𝑎𝑛 = = −21° = 339° 𝑣𝑥 40 General Physics I Dr. Hadeer Hassan El-Hawary 72  2-7 Projectile Motion The motion of the projectiles near the surface of the earth is an example of two dimensional motion with constant acceleration. Assumptions: 1. The constant acceleration is directed vertically downward and equal 𝑔. ⟹ 𝑎𝑦 = −𝑔 2. The effect of air resistance is negligible. ⟹ 𝑎𝑥 = 0 3. The particle start motion from the origin. at 𝑡 = 0 ⇒ 𝑥𝑖 = 𝑦𝑖 = 0 4. The initial velocity vector makes an angle 𝜃𝑖 with the horizontal. 𝑣𝑥𝑖 = 𝑣𝑖 cos 𝜃𝑖 , 𝑣𝑦𝑖 = 𝑣𝑖 sin 𝜃𝑖 General Physics I Dr. Hadeer Hassan El-Hawary 73  2-7 Projectile Motion Trajectory Equation: Being a motion with constant acceleration in two dimensions, the position components become: 1 2 1 𝑥𝑓 = 𝑥𝑖 + 𝑣𝑥𝑖 𝑡 + 𝑎𝑥 𝑡 , 𝑦𝑓 = 𝑦𝑖 + 𝑣𝑦𝑖 𝑡 + 𝑎𝑦 𝑡 2 2 2 𝑥𝑖 = 0, 𝑦𝑖 = 0, 𝑎𝑥 = 0, 𝑎𝑦 = −𝑔, 𝑣𝑥𝑖 = 𝑣𝑖 cos 𝜃𝑖 , 𝑣𝑦𝑖 = 𝑣𝑖 sin 𝜃𝑖 1 2 𝑥𝑓 = 𝑣𝑖 cos 𝜃𝑖 𝑡 → (1), 𝑦𝑓 = 𝑣𝑖 sin 𝜃𝑖 𝑡 − 𝑔𝑡 → (2) 2 2 𝑥𝑓 𝑥𝑓 1 𝑥𝑓 𝑡= , 𝑦𝑓 = 𝑣𝑖 sin 𝜃𝑖 − 𝑔 𝑣𝑖 cos 𝜃𝑖 𝑣𝑖 cos 𝜃𝑖 2 𝑣𝑖 cos 𝜃𝑖 General Physics I Dr. Hadeer Hassan El-Hawary 74  2-7 Projectile Motion Trajectory Equation: Being a motion with constant acceleration in two dimensions, the position components become: 1 2 1 𝑥𝑓 = 𝑥𝑖 + 𝑣𝑥𝑖 𝑡 + 𝑎𝑥 𝑡 , 𝑦𝑓 = 𝑦𝑖 + 𝑣𝑦𝑖 𝑡 + 𝑎𝑦 𝑡 2 2 2 𝑥𝑖 = 0, 𝑦𝑖 = 0, 𝑎𝑥 = 0, 𝑎𝑦 = −𝑔, 𝑣𝑥𝑖 = 𝑣𝑖 cos 𝜃𝑖 , 𝑣𝑦𝑖 = 𝑣𝑖 sin 𝜃𝑖 1 2 𝑥𝑓 = 𝑣𝑖 cos 𝜃𝑖 𝑡 → (1), 𝑦𝑓 = 𝑣𝑖 sin 𝜃𝑖 𝑡 − 𝑔𝑡 → (2) 2 2 𝑥𝑓 𝑥𝑓 1 𝑥𝑓 𝑡= ⇒ 𝑦𝑓 = 𝑣𝑖 sin 𝜃𝑖 − 𝑔 𝑣𝑖 cos 𝜃𝑖 𝑣𝑖 cos 𝜃𝑖 2 𝑣𝑖 cos 𝜃𝑖 General Physics I Dr. Hadeer Hassan El-Hawary 75  2-7 Projectile Motion Trajectory Equation: 2 1 1 𝑦𝑓 = t𝑎𝑛 𝜃𝑖 𝑥𝑓 − 𝑔 𝑥𝑓2 2 𝑣𝑖 cos 𝜃𝑖 This expression is of the form: 𝑦𝑓 = 𝑎 𝑥𝑓 − 𝑏 𝑥𝑓2 Then, the trajectory is a parabola. General Physics I Dr. Hadeer Hassan El-Hawary 76  2-7 Projectile Motion The parabolic path of a projectile that leaves the origin with a velocity 𝑣𝑖. No acceleration in x-direction 𝑎𝑥 = 𝑧𝑒𝑟𝑜 ⇒ 𝑣𝑥 = 𝑣𝑥𝑖 The x-component of the velocity remains constant. Acceleration due to gravity 𝑎𝑦 = −𝑔 ⇒ 𝑣𝑦 = 𝑣𝑦𝑖 − 𝑔𝑡 The y-component of the velocity changes its magnitude and direction with time. General Physics I Dr. Hadeer Hassan El-Hawary 77  2-7 Projectile Motion R: is called the Range of the projectile. h: is called the Maximum height of the projectile. General Physics I Dr. Hadeer Hassan El-Hawary 78  2-7 Projectile Motion At the maximum height of a projectile (point (A)) 𝑣𝑦 = 0. 𝑣𝑦𝑓 = 𝑣𝑦𝑖 + 𝑎𝑦 𝑡 ⇒ 0 = 𝑣𝑖 sin 𝜃𝑖 − 𝑔𝑡𝐴 𝑣𝑖 sin 𝜃𝑖 𝑡𝐴 = → 1 𝑔 1 𝑦𝑓 = 𝑦𝑖 + 𝑣𝑦𝑖 𝑡 + 𝑎𝑦 𝑡 2 At point A ⇒ 𝑦𝑓 = ℎ 2 2 2 1 2 𝑓𝑟𝑜𝑚 1 𝑣𝑖 sin 𝜃𝑖 1 𝑣𝑖2 sin2 𝜃𝑖 ℎ = 0 + 𝑣𝑖 sin 𝜃𝑖 𝑡 − 𝑔 𝑡 ℎ= − 𝑔 2 𝑔 2 𝑔2 𝑣𝑖2 sin2 𝜃𝑖 ⇒ℎ= 𝑔 General Physics I Dr. Hadeer Hassan El-Hawary 79  2-7 Projectile Motion The range is the distance travelled in the x-direction in twice the time it takes to reach maximum height. 𝑡𝐵 = 2𝑡𝐴 2𝑣𝑖 sin 𝜃𝑖 𝑡𝐵 = 2𝑡𝐴 = → 2 𝑔 1 𝑥𝑓 = 𝑥𝑖 + 𝑣𝑥𝑖 𝑡 + 𝑎𝑥 𝑡 2 At point B ⇒ 𝑥𝑓 = 𝑅 2 𝑓𝑟𝑜𝑚 2 2𝑣𝑖 sin 𝜃𝑖 𝑅 = 0 + 𝑣𝑖 cos 𝜃𝑖 𝑡 + 0 𝑅 = 𝑣𝑖 cos 𝜃𝑖 𝑔 Remember: 2 cos 𝜃 sin 𝜃 = sin 2𝜃 𝑣𝑖2 sin(2𝜃𝑖 ) ⇒𝑅= 𝑔 General Physics I Dr. Hadeer Hassan El-Hawary 80  2-7 Projectile Motion Maximum height at 𝜃𝑖 = 90° , which not projectile motion 𝑣𝑖2 sin2 90 𝑣𝑖2 ℎ𝑚𝑎𝑥 = = 𝑔 𝑔 Maximum Range at 𝜃𝑖 = 45° 𝑣𝑖2 sin(2 × 45) 𝑣𝑖2 𝑅𝑚𝑎𝑥 = = 𝑔 𝑔 General Physics I Dr. Hadeer Hassan El-Hawary 81  2-7 Projectile Motion Example: A long‐jumper leaves the ground at an angle of 20.0° above the horizontal and at a speed of 11.0m/s. a) How far does he jumps in the horizontal direction? 𝑣𝑖2 sin(2𝜃𝑖 ) 112 sin(2 × 20) 𝑅= = = 7.93 𝑚 𝑔 9.8 b) What is the maximum height reached? 𝑣𝑖2 sin2 𝜃𝑖 112 sin2 (20) ℎ= = = 0.722 𝑚 𝑔 9.8 General Physics I Dr. Hadeer Hassan El-Hawary 82  2-7 Projectile Motion Example: A plane drops a package of supplies to a party of explorers, as shown. If the plane is traveling horizontally at 40.0 m/s and is 100 m above the ground, where does the package strike the ground relative to the point at which it is released? 𝑥𝑖 = 𝑦𝑖 = 0 , at origin 𝑣𝑥𝑓 = 𝑣𝑥𝑖 = 40 𝑚Τ𝑠 , 𝑎𝑥 = 0, 𝑣𝑦𝑓 = 𝑣𝑦𝑖 = 0, 𝑎𝑦 = −𝑔 1 𝑥𝑓 = 𝑥𝑖 + 𝑣𝑥𝑖 𝑡 + 𝑎𝑥 𝑡 2 ⇒ 𝑥𝑓 = 40𝑡 2 1 1 𝑦𝑓 = 𝑦𝑖 + 𝑣𝑦𝑖 𝑡 + 𝑎𝑦 𝑡 ⇒ −100 = − × 9.8𝑡 2 2 2 2 𝑡 = 200/9.8 = 4.52 𝑠 ⇒ 𝑥𝑓 = 40 × 4.52 = 181 𝑚 General Physics I Dr. Hadeer Hassan El-Hawary 83  2-8 Uniform Circular Motion Motion of an object in a circular path with constant speed, is called uniform circular motion. Even though the object moves at a constant speed in a circular path, it still has an acceleration due to change of the direction of the velocity. This acceleration is called centripetal acceleration and points toward the center. General Physics I Dr. Hadeer Hassan El-Hawary 84  2-8 Uniform Circular Motion Centripetal Acceleration: By definition ,the average acceleration is given by 𝑣𝑓 − 𝑣𝑖 ∆𝑣Ԧ 𝑎Ԧ = = ∆𝑡 ∆𝑡 Δ𝜃 is small, and so Δ𝑟 and Δ𝑡 , 𝑣𝑓 = 𝑣𝑖 = 𝑣 and 𝑟𝑓 = 𝑟𝑖 = 𝑟, 𝑣 = ∆𝑟/ ∆𝑡 ∆𝑣 ∆𝑟 = 𝑣 𝑟 ∆𝑟 𝑣 𝑣 2 ⇒ 𝑎𝑐 = 𝑟 ⇒ 𝑎𝑐 = ∆𝑡 𝑟 General Physics I Dr. Hadeer Hassan El-Hawary 85  2-8 Uniform Circular Motion Period of circular motion: In many situations, it is convenient to describe the motion of a particle moving with constant speed in a circle of radius r in terms of the period T ,which is defined as the time required to complete one revolution. 𝑐𝑖𝑟𝑐𝑙𝑒 𝑝𝑒𝑟𝑖𝑝ℎ𝑒𝑟𝑦 (2𝜋𝑟) 𝑣= 𝑝𝑒𝑟𝑖𝑜𝑑 𝑡𝑖𝑚𝑒 𝑇 2𝜋𝑟 𝑇= 𝑣 General Physics I Dr. Hadeer Hassan El-Hawary 86  2-8 Uniform Circular Motion Example: What is the centripetal acceleration of the Earth as it moves in its orbit around the Sun? (r =1.496x1011 m, T=1Year) 2𝜋𝑟 2 × 3.14 × 1.496 × 1011 𝑣= = = 2.977 × 104 𝑚/𝑠 𝑇 1 × 365.25 × 24 × 60 × 60 𝑣2 2.977 × 104 2 −3 𝑚/𝑠 2 𝑎𝑐 = = = 5.924 × 10 𝑟 1.496 × 1011 General Physics I Dr. Hadeer Hassan El-Hawary 87  2-9 The Concept of Force Classical Mechanics: describes the relation between motion of a body and the forces acting on it. Classical Mechanics deals with objects that satisfy the following conditions: 1) Objects are large compared with the dimension of the atom (10-10m). 2) Objects move at speeds that are much less than speed of light (c = 3x108m/s). General Physics I Dr. Hadeer Hassan El-Hawary 88  2-9 The Concept of Force When you push or pull an object, you exert a force on it. In order to change the velocity of an object, we have to exert a force on it. An object accelerates due to an external force. The object accelerates only if the net force acting on it is not equal to zero. Several forces act simultaneously on an object General Physics I Dr. Hadeer Hassan El-Hawary 89  2-9 The Concept of Force Definition of Equilibrium: If the net force acting on a body is zero, the body is said to be in equilibrium. At equilibrium: ෍ 𝐹Ԧ = 0 ⇒ 𝐹1 + 𝐹2 + 𝐹3 + ⋯ = 0 ⇒ 𝐹𝑥 = 0, 𝐹𝑦 = 0 ⇒ 𝑎𝑥 = 0, 𝑎𝑦 = 0 ⇒ 𝑣𝑥 = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡, 𝑣𝑦 = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 General Physics I Dr. Hadeer Hassan El-Hawary 90  2-10 Newton’s First Law Newton’s First law: “An object at rest remains at rest and an object in motion will continue in motion with a constant velocity, unless it experiences a net external force.” When no force acts on an object, the acceleration of the object is zero. General Physics I Dr. Hadeer Hassan El-Hawary 91  2-11 Mass If you attempt to change the velocity of an object, the object resists this change. Mass is that property of an object that specifies how much resistance an object exhibits to changes in its velocity. General Physics I Dr. Hadeer Hassan El-Hawary 92  2-12 Newton’s Second Law Mass = Constant The acceleration of an object is directly proportional to the resultant force acting on it. 𝑎Ԧ ∝ ෍ 𝐹Ԧ General Physics I Dr. Hadeer Hassan El-Hawary 93  2-12 Newton’s Second Law Force = Constant The acceleration of an object is inversely proportional to its mass. 1 𝑎Ԧ ∝ 𝑚 General Physics I Dr. Hadeer Hassan El-Hawary 94  2-12 Newton’s Second Law Newton 2nd Law: The acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass. ෍ 𝐹Ԧ = 𝑚𝑎Ԧ Components Form: ෍ 𝐹𝑥 = 𝑚𝑎𝑥 , ෍ 𝐹𝑦 = 𝑚𝑎𝑦 , ෍ 𝐹𝑧 = 𝑚𝑎𝑧 , The force unit is Newton (N) which equals (kg.m/t2). General Physics I Dr. Hadeer Hassan El-Hawary 95  2-12 Newton’s Second Law Example: A hockey puck of mass 0.3 kg slides on a horizontal frictionless surface of an ice rink. Two forces act on the puck as shown. The force F1 has a magnitude = 5.0 N and F2 has a magnitude of 8.0 N. Determine the magnitude and the direction of the puck’s acceleration? 𝑚 = 0.3 𝑘𝑔 𝐹1 = 5.0 cos 20 𝑖Ƹ − 5.0 sin 20 𝑗,Ƹ 𝐹2 = 8.0 cos 60 𝑖Ƹ + 8.0 sin 60 𝑗Ƹ General Physics I Dr. Hadeer Hassan El-Hawary 96  2-12 Newton’s Second Law ෍ 𝐹𝑥 = 5.0 cos 20 + 8.0 cos 60 = 8.7 𝑁 ෍ 𝐹𝑦 = −5.0 sin 20 + 8.0 sin 60 = 5.2 𝑁 Applying Newton's 2nd law, we get ෍ 𝐹𝑥 = 𝑚𝑎𝑥 , ෍ 𝐹𝑦 = 𝑚𝑎𝑦 σ 𝐹𝑥 8.7 σ 𝐹𝑦 5.2 𝑎𝑥 = = = 29 𝑚/𝑠 2 , 𝑎 𝑦 = = = 17 𝑚/𝑠 2 𝑚 0.3 𝑚 0.3 General Physics I Dr. Hadeer Hassan El-Hawary 97  2-12 Newton’s Second Law σ 𝐹𝑥 8.7 𝑎𝑥 = = = 29 𝑚/𝑠 2 𝑚 0.3 σ 𝐹𝑦 5.2 𝑎𝑦 = = = 17 𝑚/𝑠 2 𝑚 0.3 𝑎Ԧ = 29 𝑖Ƹ + 17 𝑗Ƹ 𝑎Ԧ = 292 + 172 = 34 𝑚/𝑠 2 The direction of acceleration is 𝑎 −1 𝑦 −1 17 𝜃 = tan = tan = 30° 𝑎𝑥 29 General Physics I Dr. Hadeer Hassan El-Hawary 98 2-13 The Gravitational Force and Weight The attractive force exerted by the Earth on an object is called the gravitational force Fg. This force is directed toward the center of the Earth. ෍ 𝐹Ԧ = 𝑚𝑎, Ԧ ෍ 𝐹Ԧ = ෍ 𝐹Ԧ𝑔 𝑎Ԧ = 𝑔, Ԧ ෍ 𝐹𝑔 = 𝑚𝑔Ԧ The weight of an object is defined as the magnitude of its gravitational force. 𝑊 = 𝑚𝑔 General Physics I Dr. Hadeer Hassan El-Hawary 99  2-14 Newton’s Third Law Newton 3rd Law: If two objects interact, the force F12 exerted on object 1 by object 2 is equal in magnitude and opposite in direction to the force F21 exerted on object 2 by object 1. 𝐹12 = −𝐹21 The force F12 that object 1 exerts on object 2 may be called the action force while the force F21 of object 2 on object 1 called the reaction force. “The action force is equal in magnitude to the reaction force and opposite in direction.” General Physics I Dr. Hadeer Hassan El-Hawary 100 2-15 Some Applications of Newton’s Law When we apply Newton’s Laws to an object, we are interested only in external forces that act on the object. We have to draw Free Body Diagrams for all objects. In the free body diagram, the only forces acting on a stationary object are and 𝐹𝑔 and 𝑛. ෍ 𝐹𝑦 = 𝑛 − 𝐹𝑔 Monitor is in equilibrium ෍ 𝐹𝑦 = 0 ⇒ 𝑛 − 𝐹𝑔 = 0 ⇒ 𝐹𝑔 = 𝑛 = 𝑚𝑔 General Physics I Dr. Hadeer Hassan El-Hawary 101 2-15 Some Applications of Newton’s Law When a rope attached to an object is pulling on the object, the rope exerts a force T on the object, and the magnitude T of that force is called the tension in the rope. We usually neglect the mass of any ropes, strings, or cables involved. The suspended lamp is in equilibrium σ 𝐹𝑦 = 0 ෍ 𝐹𝑦 = 𝑇 − 𝐹𝑔 ⇒ 𝑇 − 𝐹𝑔 = 0 ⇒ 𝐹𝑔 = T = 𝑚𝑔 General Physics I Dr. Hadeer Hassan El-Hawary 102 2-15 Some Applications of Newton’s Law Objects experiences a net force: A crate being pulled on a frictionless surface. If an object that can be modeled as a particle experiences an acceleration, then there must be a nonzero net force acting on the object. The acceleration is in x-direction only then 𝑎𝑦 = 0 ෍ 𝐹𝑥 = 𝑚𝑎𝑥 , ෍ 𝐹𝑦 = 0 General Physics I Dr. Hadeer Hassan El-Hawary 103 2-15 Some Applications of Newton’s Law ෍ 𝐹𝑥 = 𝑚𝑎𝑥 , ෍ 𝐹𝑦 = 0 From the free body diagram: ෍ 𝐹𝑥 = 𝑇 ⇒ 𝑚𝑎𝑥 = 𝑇 ⇒ 𝑎𝑥 = 𝑇/𝑚 ෍ 𝐹𝑦 = 𝑛 − 𝐹𝑔 ⇒ 0 = 𝑛 − 𝐹𝑔 ⇒ 𝑛 = 𝐹𝑔 = 𝑚𝑔 General Physics I Dr. Hadeer Hassan El-Hawary 104 2-15 Some Applications of Newton’s Law Example: A traffic light at rest. A traffic light weighting 122 N hangs from a cable tied to two other cables fastened to a support. The upper cables make angles of 37° and 53° with the horizontal. Find the tension in the three cables? General Physics I Dr. Hadeer Hassan El-Hawary 105 2-15 Some Applications of Newton’s Law Answer: At first, we must draw the free body diagram. Traffic Light: in equilibrium ෍ 𝐹𝑥 = 0, ෍ 𝐹𝑦 = 0 ෍ 𝐹𝑦 = 𝑇3 − 𝐹𝑔 = 0 ⇒ 𝑇3 = 𝐹𝑔 = 122 𝑁 ⟶ (1) General Physics I Dr. Hadeer Hassan El-Hawary 106 2-15 Some Applications of Newton’s Law Answer: At first, we must draw the free body diagram. Kont: in equilibrium ෍ 𝐹𝑥 = 0, ෍ 𝐹𝑦 = 0 ෍ 𝐹𝑥 = −𝑇1 cos 37 + 𝑇2 cos 53 = 0 ⟶ (2) ෍ 𝐹𝑦 = 𝑇1 sin 37 + 𝑇2 sin 53 − 𝑇3 = 0 ⟶ 3 General Physics I Dr. Hadeer Hassan El-Hawary 107 2-15 Some Applications of Newton’s Law Answer: Now, Solve the three equations to determine T1, T2 and T3. From Eq (1) 𝑻𝟑 = 𝟏𝟐𝟐 𝑵 cos 37 From Eq (2) 𝑇1 cos 37 = 𝑇2 cos 53 ⇒ 𝑇2 = 𝑇1 ⇒ 𝑇2 = 1.33 𝑇1 cos 53 From Eq (3), using (1) & (2) 𝑇1 sin 37 + 𝑇2 sin 53 − 𝑇3 = 0 𝑇1 sin 37 + 1.33 × sin 53 𝑇1 − 122 = 0 0.602 𝑇1 + 1. 062 𝑇1 − 122 = 0 ⇒ 𝑻𝟏 = 𝟕𝟑. 𝟒 𝑵 𝑇2 = 1.33 𝑇1 ⇒ 1.33 × 73.4 ⇒ 𝑻𝟐 = 𝟗𝟕. 𝟒 𝑵 General Physics I Dr. Hadeer Hassan El-Hawary 108 2-15 Some Applications of Newton’s Law Answer: Now, Solve the three equations to determine T1, T2 and T3. From Eq (1) 𝑻𝟑 = 𝟏𝟐𝟐 𝑵 cos 37 From Eq (2) 𝑇1 cos 37 = 𝑇2 cos 53 ⇒ 𝑇2 = 𝑇1 ⇒ 𝑇2 = 1.33 𝑇1 cos 53 From Eq (3), using (1) & (2) 𝑇1 sin 37 + 𝑇2 sin 53 − 𝑇3 = 0 𝑇1 sin 37 + 1.33 × sin 53 𝑇1 − 122 = 0 0.602 𝑇1 + 1. 062 𝑇1 − 122 = 0 ⇒ 𝑻𝟏 = 𝟕𝟑. 𝟒 𝑵 𝑇2 = 1.33 𝑇1 ⇒ 1.33 × 73.4 ⇒ 𝑻𝟐 = 𝟗𝟕. 𝟒 𝑵 General Physics I Dr. Hadeer Hassan El-Hawary 109 2-15 Some Applications of Newton’s Law Atwood's Machine: When two unequal masses are hung vertically over a frictionless pulley of negligible mass, the arrangement is called Atwood's machine. Example: Determine the magnitude of the acceleration of the two masses m1, m2 and the tension in the string of the Atwood's machine? General Physics I Dr. Hadeer Hassan El-Hawary 110 2-15 Some Applications of Newton’s Law Answer: Draw the Free Body Diagram for each mass. General Physics I Dr. Hadeer Hassan El-Hawary 111 2-15 Some Applications of Newton’s Law Answer: Applying Newton's 2nd Law to both masses in motion we get: ෍ 𝐹𝑦 = 𝑚1 𝑎𝑦 = 𝑇 − 𝑚1 𝑔 ⟶ (1) ෍ 𝐹𝑦 = −𝑚2 𝑎𝑦 = 𝑇 − 𝑚2 𝑔 𝑚2 𝑎𝑦 = 𝑚2 𝑔 − 𝑇 ⟶ (2) Add (1) to (2) 𝑚1 + 𝑚2 𝑎𝑦 = 𝑚2 − 𝑚1 𝑔 General Physics I Dr. Hadeer Hassan El-Hawary 112 2-15 Some Applications of Newton’s Law Answer: 𝑚1 + 𝑚2 𝑎𝑦 = 𝑚2 − 𝑚1 𝑔 𝒎𝟐 − 𝒎𝟏 𝒂𝒚 = 𝒈 ⟶ (3) 𝒎𝟏 + 𝒎𝟐 Substituting (3) into (1), 𝑚1 𝑚2 − 𝑚1 𝑚1 𝑎𝑦 = 𝑇 − 𝑚1 𝑔 ⇒ 𝑔 = 𝑇 − 𝑚1 𝑔 𝑚1 + 𝑚2 𝑚1 𝑚2 − 𝑚1 𝑚1 𝑚1 + 𝑚2 𝟐𝒎𝟏 𝒎𝟐 ⇒𝑇= 𝑔+ 𝑔⇒𝑻= 𝒈 𝑚1 + 𝑚2 𝑚1 + 𝑚2 𝒎𝟏 + 𝒎𝟐 General Physics I Dr. Hadeer Hassan El-Hawary 113 2-15 Some Applications of Newton’s Law Example: Find the acceleration & tension of Atwood's machine in which m1= 2kg, m2=4 kg. Answer: 𝑚2 − 𝑚1 4−2 𝑎𝑦 = 𝑔= × 9.8 = 3.27 𝑚/𝑠 2 𝑚1 + 𝑚2 2+4 2𝑚1 𝑚2 2×2×4 𝑇= 𝑔= × 9.8 = 26.1 𝑁 𝑚1 + 𝑚2 2+4 General Physics I Dr. Hadeer Hassan El-Hawary 114  2-16 Forces of Friction When a body is in motion either on a surface or through a viscous medium such as water, there is a resistance to its motion. We call such resistance a force of friction. General Physics I Dr. Hadeer Hassan El-Hawary 115  2-16 Forces of Friction General Physics I Dr. Hadeer Hassan El-Hawary 116  2-16 Forces of Friction The force of static friction 𝑓𝑠 is equal to the applied external force F as far as there is no motion. If F is increased and the object start slipping, 𝑓𝑠 becomes maximum (𝑓𝑠, max ) For objects in motion, the friction forces is called force of kinetic friction 𝑓𝑘. 𝑓𝑘 < 𝑓𝑠, max General Physics I Dr. Hadeer Hassan El-Hawary 117  2-16 Forces of Friction For static friction: ෍ 𝐹𝑥 = 𝐹 − 𝑓𝑠 = 0 ⇒ 𝐹 = 𝑓𝑠 For kinetic friction: ෍ 𝐹𝑥 = 𝐹 − 𝑓𝑘 For 𝐹 > 𝑓𝑘 ⇒ σ 𝐹𝑥 > 𝑍𝑒𝑟𝑜 ⟹ The object accelerates. For 𝐹 = 𝑓𝑘 ⇒ σ 𝐹𝑥 = 𝑍𝑒𝑟𝑜 ⟹The object moves with constant velocity. General Physics I Dr. Hadeer Hassan El-Hawary 118  2-16 Forces of Friction Coefficients of Friction (Static 𝜇𝑠 ) and (Kinetic 𝜇𝑘 ): If the body is in the verge of slipping, we may write 𝑓𝑠,𝑚𝑎𝑥 = 𝜇𝑠 𝑛 𝜇𝑠 : coefficient of static friction. 𝑛: magnitude of the normal force. Note: if 𝐹 < 𝑓𝑠,𝑚𝑎𝑥 we may write 𝑓𝑠 < 𝜇𝑠 𝑛 If the body is in motion, we may write 𝑓𝑘 = 𝜇𝑘 𝑛 𝜇𝑘 : coefficient of kinetic friction General Physics I Dr. Hadeer Hassan El-Hawary 119  2-16 Forces of Friction The direction of 𝑓𝑠 is always opposite to the direction of the external force 𝐹. The direction of 𝑓𝑘 is opposite to the direction of motion. 𝜇𝑠 and 𝜇𝑘 depend on the nature of the two surfaces and are independent on the area of contact between the surfaces. General Physics I Dr. Hadeer Hassan El-Hawary 120  2-16 Forces of Friction Example: Suppose a block of mass m is placed on a rough surface inclined relative to the horizontal. The angle of the incline is increased until the block slips at critical angle 𝜃𝑐. Derive an expression of the relation between 𝜃𝑐 and 𝑓𝑘 ? Answer: While the block is not moving: ෍ 𝐹𝑥 = −𝑓𝑠 + 𝑚𝑔 sin 𝜃 = 0 ⇒ 𝑚𝑔 = 𝑓𝑠 / sin 𝜃 ෍ 𝐹𝑦 = −𝑚𝑔 cos 𝜃 + 𝑛 = 0 ⇒ 𝑚𝑔 = 𝑛/ cos 𝜃 sin 𝜃 𝑓𝑠 = 𝑛 ⇒ 𝒇𝒔 = 𝒏 𝐭𝐚𝐧 𝜽 cos 𝜃 When the inclined plane is at the critical angle 𝜃𝑐 𝑓𝑠 = 𝑓𝑠,𝑚𝑎𝑥 = 𝜇𝑠 𝑛 ⇒ 𝜇𝑠 𝑛 = 𝑛 𝑡𝑎𝑛 𝜃𝑐 ⇒ 𝝁𝒔 = 𝒕𝒂𝒏 𝜽𝒄 General Physics I Dr. Hadeer Hassan El-Hawary 121  2-16 Forces of Friction Example: A hockey puck on a frozen bond is hit and given an initial speed of 20.0 m/s. If the puck always remains on the ice and slides 115 m before coming to rest, determine the coefficient of kinetic friction between the puck and the ice. Answer: ෍ 𝐹𝑥 = −𝑓𝑘 = 𝑚𝑎𝑥 ෍ 𝐹𝑦 = 𝑛 − 𝑚𝑔 = 0 ⇒ 𝑛 = 𝑚𝑔 𝑓𝑘 = 𝜇𝑘 𝑛 General Physics I Dr. Hadeer Hassan El-Hawary 122  2-16 Forces of Friction Answer: 𝑓𝑘 = −𝑚𝑎𝑥 = 𝜇𝑘 𝑛 𝑛 = 𝑚𝑔 −𝑚𝑎𝑥 = 𝜇𝑘 𝑚𝑔 𝑎𝑥 = −𝜇𝑘 𝑔 From kinematics equations 𝑣𝑓2 = 𝑣𝑖2 + 2𝑎𝑥 (𝑥𝑓 − 𝑥𝑖 ) 02 = 202 + 2𝑎𝑥 (115) 400 𝑎𝑥 = − = −1.74 𝑚/𝑠 2 230 17.4 −𝜇𝑘 𝑔 = −17.4 ⇒ 𝜇𝑘 = = 0.177 9.8 General Physics I Dr. Hadeer Hassan El-Hawary 123