PHY 102 Introductory Mechanics and Properties of Matter PDF

PHY 102 Introductory Mechanics and Properties of Matter (3 units) Lecturers: Prof. O.E.Awe Dr. Titus Ogunseye Recommended Texts: Fundamentals of Physics: Resnick and Halliday Advanced Level Physics: Nelkon and Parker Colle...

PHY 102 Introductory Mechanics and Properties of Matter (3 units) Lecturers: Prof. O.E.Awe Dr. Titus Ogunseye Recommended Texts: Fundamentals of Physics: Resnick and Halliday Advanced Level Physics: Nelkon and Parker College Physics: Any Good Author College Physics: Schaum Series (Worked Examples) Mechanics and Properties of Matter Idowu Farai 1  Course content Useful Mathematics. Quantities, Units and Their Dimensions. Introduction to Vectors. Kinematics. Newton’s laws of motion. Gravitational Force of Attraction. Work and Energy. Dynamics of Rotating Rigid Bodies. Simple Harmonic Motion. Elastic Properties. Some Properties of Liquids. 2 Useful Mathematics ❖ There is so much order in the physical world. ❖ The orderliness in the physical world has allowed for accurate or near accurate prediction of the behaviours of many physical systems using laws or theorems. ❖ Laws and theorems are often expressed in the form of mathematical functions. ❖ Mathematical functions can be represented in a variety of ways:Table, Formula and Graph. 3 a) Common functions include Periodic and Sinusoidal functions. A periodic function repeats its value at regular intervals. Examples of periodic functions are trigonometric functions, inverse trigonometric functions and hyperbolic functions. A periodic wave repeats its shape at regular intervals. A function that is based on the sine function which oscillates between a low (minimum) value and a high (maximum) value at regular intervals is known as Sinusoidal function. Since the cosine function (y2 = cos B) is a shifted sine function (y1 = sin A), cosine function is an example of sinusoidal function. Thus, both sine and cosine functions are examples of Sinusoidal functions. Sinusoidal functions are necessarily periodic functions but not all periodic functions are sinusoidal functions. Note that A and B as used above may contain more than one term. 4 ❖The graph of either Sine or Cosine function is known as Sinusoidal wave. Sine wave 5 Sine and Cosine waves: The cosine wave leads the sine wave 0 𝜋 by 90 or 𝑟𝑎𝑑𝑖𝑎𝑛𝑠 as seen below. 2 6 Examples of Sinusoidal wave or Sinusoidal function include AC Voltage, AC Current, Simple harmonic motion (SHM) and Displacement in a wave motion. 7 8 b.) Logarithm Functions and Indices We can recall that if x is related to y by: y = logb x, then, y is related to x by: x = b y. e.g. 23 = 8 or log2 8 = 3 The letter b is called the base of the logarithm and it is usually omitted when it is 10. e.g. 1000 = 103 or log 1000 = 3 For the exponential function, ex = N, then x = loge N, where e ≈ 2.7183 and loge e = 1. This logarithm is called a Naperian or natural log because many natural systems are described by exponential functions. 9 10 Example For a mass of a gas at pressure P and volume V, PVγ = C, where γ and C are constants. From the values of P and V obtained in an experiment, determine the values of γ and C and estimate P when V = 100 cm3 Take log of both sides: log P + γ logV = log C Rearrange: log P = - γ logV + log C Compare with straight line graph: y = mx + c Plot a graph of log P on y-axis against log V on x-axis 1. Slope = - γ and intercept on y-axis = log C 2. Read log P corresponding to log 100, hence find P 11 Derived Functions For linear equations like: y = 4x +2, the rate of change of y with x is constant and it is equal to the slope of the graph of y against x For functions like: y = 2x2 or y = 5x3 + 2x, the slope is different at different values of x. A new equation, called the derivative of the original equation is required to get the slope at any value of x 12 Differentiation Operation: It can be shown that if: y = kxn, then its first derivative is 𝒅𝒚 = 𝒏𝒌𝒙𝒏−𝟏. 𝒅𝒙 That is, if 𝒚 = 𝟖𝒙𝟓 + 𝟒𝒙𝟑 + 𝟐𝒙 + 𝟕. 𝒅𝒚 The first derivative is = 𝟒𝟎𝒙𝟒 + 𝟏𝟐𝒙𝟐 + 𝟐 𝒅𝒙 𝒅𝟐 𝒚 The second derivative is: = 𝟏𝟔𝟎𝒙𝟑 + 𝟐𝟒𝒙 and so on 𝒅𝒙𝟐 Exercise: Find the first and the second derivatives 𝟏 of 𝒙 = 𝒖𝒕 + 𝒂𝒕𝟐 , if u and a are constants. 𝟐 13 14 For example, 15 3  = + = +c 2 3 15 x dx x c 5 x 3 1 A special integration is that of  dx = log e x + c x From Summation to Integration The summation of all integer numbers between 1 and 10 is 10  x =  (1 + 2 + 3 + 4 +...... + 10) = 55 1 When the quantity x can assume any value (e.g. 3.6434..) and not only integers, we say it is continuous. The summation operation is replaced by integration. 15 Application If we divide a circle into an infinite number of tiny rings such that the radius of a ring of thickness ∆x varies continuously between point 0(i.e. O is the centre of the circle) and point r (i.e. r is the distance between point O and the circumference of the circle) then , the area of the circle is the summation of the areas of all the tiny rings. Area of each ring, ∆A = 2π x ∆x, where ∆x is the thickness of the ring. 16  𝒓 Total area, A = ‫∆ 𝟎׬‬A Therefore, 𝒓 𝒙𝟐 𝑨 = න 𝟐𝝅𝒙𝒅𝒙 = 𝟐𝝅 − 𝟎 = 𝝅𝒓𝟐 𝟎 𝟐 We are all familiar with the formula A = πr2 for a circle. Homework The surface area of a sphere is 4πr2. By dividing the sphere into an infinite number of spheres, each of very small thickness ∆x when radius is x, show that the volume of the whole sphere of radius r is 𝟒 𝟑 𝑨 = 𝝅𝒓 𝟑 17 Quantities, Units and Dimensions There are two groups of physical or measurable quantities. These are fundamental and derived quantities. A Fundamental quantity: Does not depend on prior knowledge of any other quantity. A derived quantity: It is derived from the knowledge of some other quantities. The seven fundamental quantities are Mass, Time, Length, Temperature, Electric current, Amount of substance and Luminous intensity. 1 All other quantities such as volume, density, force, energy, resistance, specific heat, current etc. are derived from these fundamental quantities. Unit of a Quantities The unit of a quantity tells us the scale in which its measurement is made. A value without its unit of measurement has incomplete meaning. In Systeme Internationale d’Unites (SI) units, the fundamental units are: metre (m) for length, kilogram (kg) for mass, seconds (s) for time, Kelvin (K) for temperature, mole (mol) for amount of substance, Ampere (A) for electric current, and candela (cd) for luminous intensity. 2 Derived units: Force in Newton (N) (i.e kg m/𝑠 2 ),Energy in Joule (J) (i.e. kg 𝑚2 /𝑠 2 ), Power in Watts (W), (i.e. J/s or kg 𝑚2 /𝑠 3 ). SI units (Since 1973): Super Units Sub Units 103 = kilo (k) 10-3 = milli (m), 106 = Mega (M) 10-6 = micro (μ) 109 = Giga (G) 10-9 = nano(n) 1012 = Tera (T) 10-12 = pico (p) 1015 = Peta (P) 10-15 = femto (f) Examples of some quantities and their units Diameter of a nucleus is of the order of 1 fm. Thickness of human hair is about 100 μm Electric Power generation in Nigeria is 7,000 MW 3 Radius of orbit of the sun is 4.5 Tm Dimensions The dimension of a physical quantity is an expression which shows how the physical quantity is related to the fundamental quantities from which it has been derived. All quantities in the mechanics branch of physics are derived by the combination of the three fundamental(or basic) units of length L, mass M and time T. Examples For velocity [v] = LT -1 For acceleration [a ] = LT -2 For force [f] = MLT -2 For density [d] = ML-3 For energy [e] = ML2T -2, For pressure [p] = ML-1T -2 4 Applications of Dimensional Analysis 1. Checking the validity or otherwise of an equation. The terms of a correct physical equation must have the same dimensions For example: v 2 = u 2 + 2as [v 2] = [u2 ]+ [a][s] (LT -1)2 = (LT -1) 2 + (LT -2)(L) L2T -2 = L2T -2 + L2T -2 Since each term of the equation has the same dimension of L2T -2, the equation is homogenous and hence, valid. 5  Typical MCQ 1. The displacement of a particle in an x-y plane is given by 𝑥 = 𝐴𝑡 2 + 𝐵 and 𝑦 = 𝐷𝑡 3 − 𝐸 where t is time from rest and A, B, D and E are constants. The respective dimensions of A, B and D are: A. (𝐿𝑇 −2 , 𝐿2 , 𝐿𝑇 3 ) B. (𝐿𝑇 −2 , 𝐿𝑇 3 , 𝐿2 ) C. 𝐿𝑇 −2 , 𝐿, 𝐿𝑇 −3 D. (𝐿𝑇 −3 , 𝐿, 𝐿𝑇 3 ) 6 2. Deriving the exact dependence of one quantity on other quantities in an equation. Example: The variables in a simple pendulum on which the period T of the simple pendulum may depend are: 1.)the length l of the pendulum 2.)the mass m of the bob. 3.)the acceleration due to gravity g at the location. On the basis of the information given above, we can assume that 𝑇 𝛼 𝑙 𝑥 𝑚𝑦 𝑔 𝑧 𝑜𝑟 𝑇 = 𝑘𝑙 𝑥 𝑚𝑦 𝑔 𝑧 …...(**) where the k in eqn.(**) is a dimensionless constant. 7 Dimensionally, we can rewrite eqn.(**) as 𝑇 1 = 𝐿𝑥 𝑀 𝑦 𝐿𝑇 −2 𝑧 or 𝑇 1 = 𝐿𝑥+𝑧 𝑀 𝑦 𝑇 −2𝑧 …. (∗∗∗) Equating the indices on both sides of eqn.(***), for L : x+z = 0 which implies x = - z ….(+) for T: -2z = 1 which implies z = - ½.Hence, from eqn.(+) x = ½ for M: y = 0 (no dependence on mass) If we substitute the values of x, y and z we have just obtained in eqn.(**), 𝑙 we shall have 𝑻 = 𝒌 𝒍𝟏/𝟐 𝒎𝟎 𝒈−𝟏/𝟐 or 𝑇 = 𝑘 𝑔 The value of the constant k will be shown later to be 2π so that the 𝑙 actual relationship is 𝑇 = 2𝜋 𝑔 8  Generally, the dimension of any physical quantity can be written as: 𝐿𝑎 𝑀𝑏 𝑇 𝑐 𝐼𝑑 θ𝑒 𝑁 𝑓 𝐽 𝑔 ---------------------------------------------------------(*) Where L, M, T, I, θ, N and J are the symbols for the dimensions of Length, Mass, Time, Electric current, Thermodynamic temperature, Amount of substance, and Luminous intensity, respectively; a, b,c, d, e, f, and g are the assumed powers or exponents of the dimensions of the fundamental or basic quantities. The value of each of the exponents has to be carefully determined to validate the correctness of the physical quantity of interest. Dimensionless quantities are also known as PURE NUMBERS. 9  Some dimensionless quantities have units. E.g,Angles(radians/degrees). Cycle in the unit of frequency(Hertz(Hz) = cycle/second), is dimensionless. The dimension of length is L because 𝐿1 = 𝐿1 𝑀0 𝑇 0 𝐼 0 θ0 𝑁 0 𝐽0 = 𝐿1 = 𝐿 The dimension of mass is M because 𝑀1 = 𝐿0 𝑀1 𝑇 0 𝐼 0 θ0 𝑁 0 𝐽0 = 𝑀1 = 𝑀 Any quantity whose dimension is such that all the powers or indices or exponents in eqn(*) is zero(i.e. a=b=c=d=e=f=g=0) is said to be DIMENSIONLESS or has a dimension of UNITY. Homework A. Find the dimension and the SI unit of η which stands for viscosity in the following equation if F is force, r is radius, 𝑙 is length, v is speed and R is distance? 𝑣 𝐹 = −2 𝜋 𝑟 𝑙 𝜂 𝑅 B. What is the dimension of the derivative of the physical quantity v(velocity) with 𝑑𝑣 respect to the physical quantity t(time)?(i.e. Find ?) 𝑑𝑡 C. What is the dimension of the integral of v(velocity) with respect to t(time)? (i.e. Find ‫)? 𝑡𝑑𝑣 ׬‬ 11  Scalars and Vectors Scalars: Are completely described by their magnitudes, expressed in the appropriate units. Examples: A mass of 20 kg, A time of 2 hours, well understood A work (energy) of 100 J etc. Addition of scalars: 20 kg +40 kg = 60 kg Vectors: Are described by their magnitudes and directions A displacement of 1,000 km from Ibadan is not well understood until the direction is stated. 1 Other quantities in the category of vectors include 1. Force, 2. Velocity, 3. Acceleration, 4. Electric field intensity etc. Vector Addition The sum of 20N, 25N and 40N along east, west and north respectively is not equal to 85N. The computation has to take their direction into account. 2 Geometrical Method: Representing the two vectors A and B say, by two adjacent sides of a triangle inclined to each other at the angle between the two vectors. The third side R is the resultant vector Use of Cosine Rule R2 = A2 + B2 – 2AB cos (1800 – θ) = A2 + B2 + 2AB cos θ Exercise Two boys pull a toy with forces 10 N and 15 N inclined at 600 to each other. Find their resultant in magnitude and direction. (R = 21.8N at 36.60 to the 10 N force). We can use the sine rule if we want to state R relative to vectors A or B 𝑨 𝑩 𝑪 = = 𝑺𝒊𝒏𝒂 𝑺𝒊𝒏𝒃 𝑺𝒊𝒏𝒄 where A, B and C are the lengths opposite the angles a, b and c, respectively. Subtraction of Vectors The same process as addition if we note that A – B = A + (- B) where –B is a vector equal in magnitude to B but oppositely directed. The vector diagram is the form: A -B R2=A2 + B2 + 2AB cos θ R The cosine rule becomes: R2 = A2 + B2 - 2AB cos θ Scale Drawing: If convenient scales are used and the lengths and angles are very carefully measured, the geometrical method can be carried out by scale drawing  Scalars and Vectors Scalars: Are completely described by their magnitudes, expressed in the appropriate units. Examples: A mass of 20 kg, A time of 2 hours, well understood A work (energy) of 100 J etc. Addition of scalars: 20 kg +40 kg = 60 kg Vectors: Are described by their magnitudes and directions A displacement of 1,000 km from Ibadan is not well understood until the direction is stated. 1 Other quantities in the category of vectors include 1. Force, 2. Velocity, 3. Acceleration, 4. Electric field intensity etc. Vector Addition The sum of 20N, 25N and 40N along east, west and north respectively is not equal to 85N. The computation has to take their direction into account. 2 Geometrical Method: Representing the two vectors A and B say, by two adjacent sides of a triangle inclined to each other at the angle between the two vectors. The third side R is the resultant vector Use of Cosine Rule R2 = A2 + B2 – 2AB cos (1800 – θ) = A2 + B2 + 2AB cos θ Exercise Two boys pull a toy with forces 10 N and 15 N inclined at 600 to each other. Find their resultant in magnitude and direction. (R = 21.8N at 36.60 to the 10 N force). We can use the sine rule if we want to state R relative to vectors A or B 𝑨 𝑩 𝑪 = = 𝑺𝒊𝒏𝒂 𝑺𝒊𝒏𝒃 𝑺𝒊𝒏𝒄 where A, B and C are the lengths opposite the angles a, b and c, respectively. Subtraction of Vectors The same process as addition if we note that A – B = A + (- B) where –B is a vector equal in magnitude to B but oppositely directed. The vector diagram is the form: A -B R2=A2 + B2 + 2AB cos θ R The cosine rule becomes: R2 = A2 + B2 - 2AB cos θ Scale Drawing: If convenient scales are used and the lengths and angles are very carefully measured, the geometrical method can be carried out by scale drawing That is, V = iVx + jVy + kVz The magnitude, 𝑉 = 𝑽 = 𝑉𝑥2 + 𝑉𝑦2 + 𝑉𝑧2 −1 𝑉𝑦 Direction to x-axis, 𝜃 = 𝑡𝑎𝑛 ( ) 𝑉𝑥 Example 1 If three forces, F1 =( 4i – j)N, F2 = (-3i + 2j) N and F3 = -3j N act on a body in a plane. Find the resultant  F = i(4 – 3) + j(-1 + 2 – 3)= i – 2j F = 1 + 2 = 1 + 4 = 5 = 2.23 N 2 2 y −1 − 2  = tan −1 = tan = −63.430 x 1 That is, the resultant is 2.23 N at angle 63.430 to the x-axis in the 4th quadrant. Example 2 If three forces F1 = 20N, 300NE, F2 = 50N along W, F3 = 40N 500NW act on a body. Find the resultant in magnitude and direction. Vector Multiplication There are generally three types of multiplication operations: 1. a vector by a scalar which always yields a vector, such as F = ma 2. vector by a vector to yield a scalar, such as W = F.S (Scalar or Dot product product) 3. a vector by a vector to yield a vector, such as τ = F × S (Vector or Cross product) Operation (i) The product is always a vector in the direction of the vector being multiplied with the scalar. For example: (m × a = F) in direction of a (Newton second law) Operation (ii) Dot product of two vectors and it is defined by: 𝑨. 𝑩 = 𝐴𝐵𝑐𝑜𝑠𝜃 where θ is the angle between them. Analytically, 𝑨. 𝑩 = 𝒊𝐴𝑥 + 𝒋𝐴𝑦 + 𝒌𝐴𝑧. (𝒊𝐵𝑥 + 𝒋𝐵𝑦 + 𝒌𝐵𝑧 ) = 𝒊. 𝒊 𝐴𝑥 𝐵𝑥 + ⋯ + 𝒋. 𝒋 𝐴𝑦 𝐵𝑦 + ⋯ + 𝒌. 𝒌 𝐴𝑧 𝐵𝑧 = 𝑨𝒙 𝑩𝒙 + 𝑨𝒚 𝑩𝒚 + 𝑨𝒛 𝑩𝒛 Note that: i. i = j. j = k. k = 1×1×Cos0 = 1 and i. j = i. k = j. k = 1×1×Cos90 = 0 so that terms like (i. j)(Ax.By) and so on are zero. Example A force F = 2i + 4j (in N) acts on a body and causes a displacement S = i + 5j (in m). Find the work done. W = F.S = (2 × 1) + (4 ×5) = 22 J. Operation (iii) Vector (or cross) product and it is defined as A × B = n AB sin θ where θ is the angle between them and n is a unit vector perpendicular to both A and B A  B = ABSin It has the property that A × B = - B × A A×B B ×A A × B and B × A are in opposite directions  Use of a Matrix The product vector is equal to the determinant of the matrix formed by the two vectors. A × B = (iAx + jAy + kAz) × (iBx + jBy + kBz) i j k = Ax Ay Az Bx By Bz = (Ay Bz – ByAz)i – (AxBz –Bx Az)j +(AxBy – BxAy)k  Example If A = 5i – 2j + k and B = 2i + 4j – 3k i j k A×B = 5 −2 1 2 4 −3 = (6-4)i – (-15 – 2)j + (20+4)k = 2i + 17j + 24k A B = 2 2 + 17 2 + 24 2 = 29.5 in appropriate unit Exercise Find B × A and show that it is equal in magnitude and opposite in direction to A × B. Exercise Two insects A and B fly in space with uniform velocities VA = i + 4j + 3k and VB = 4i + 2j – 4k in m/s with respect to a stationary observer at the origin (0,0,0). Show that the insects fly at right angle to each other and determine their distance apart after 5s. (39.37 m).  Kinematics This is the study of motions without considering the forces causing the motions. We can have motion along a line (1D), on a surface or plane (2D) or in space (3D). Displacement. Displacement is the effective distance between two points. A motion from position P1(2, 2) to P2(6,4) is of the form: Velocity and Acceleration Velocity is the rate at which displacement is changing with time. We may be interested in its 1. Average value over a period of time 2. Instantaneous value at a point of interest. Example A particle moving in a plane (or surface) has its motion described by: x = 20t + 10 and y = 6t2 + 4t where distances are in m and time is in seconds. Find the magnitude and direction of its velocity at the instant 5s from rest Acceleration When the velocity of a body is changing either in magnitude or direction (or both) the body is said to be undergoing acceleration, which is the rate of change of velocity with time. Uniform Motion A motion for which the acceleration is constant is referred to as uniform motion. For all uniform motions, some useful equations can be derived, which are generally applicable.  𝑢+𝑣 𝑠 = 𝑣𝑡lj = ( )𝑡. The distance 2  Example 2 A proton is accelerated in a particle accelerator in the horizontal x – y plane. If its velocity changes from 3000 m/s along y-axis to 4000 m/s along x-axis in 20 μs, what is its average acceleration in magnitude and direction? From eqn i, the acceleration is v−u a= t An approach is to find v – u with the geometrical method of vector addition, recalling that the change in velocity, D = v – u = v + (-u)  𝑅= 40002 + 30002 = 2.5 × 107 = 5.0 × 103 5.0×103 Acceleration = Δv/Δt = = 2.5 × 108 m/s2 20×10−6 −3  = tan −1 = −36.87o to x-axis 4 Exercise: Solve the problem by analytical method Relative Motion Every motion is observed (or measured) relative to an observer and the observer is always considered to be at rest in the frame of reference in which the motion is being measured. The velocity of A as measured by B as VAB and that of B as measured by (relative to) A is VBA. Example The velocity of a boat B can be measured relative to another boat A (VBA), relative to water W (VBW) or relative to the shore S (VBS). Note that: VBA≠VBW ≠VBS Rule 1: If VBA is the velocity of B relative to A and VAB is the velocity of A relative to B, then: VBA = - VAB Rule 2: If VBA is the velocity of B relative to A and VCB is the velocity of C relative to B, then the velocity of C relative to A is: VCA = VCB + VBA If you are in car A, which a stationary roadside observer O sees as moving with VAO = 100 km/hr and another car B overtakes your car with a velocity which you reckon as VBA = 20 km/hr, the roadside observer will see car B as moving with the velocity VBO = VBA + VAO = 120 km/hr. Example 1 A boy runs inside a train with a velocity VBT = 4i – 2j relative to a passenger sitting in the train while the train is moving with a velocity VTR = 50i + 30j relative to a roadside observer. Find the velocity of boy relative to roadside observer, VBR If you are in car A, which a stationary roadside observer O sees as moving with VAO = 100 km/hr and another car B overtakes your car with a velocity which you reckon as VBA = 20 km/hr, the roadside observer will see car B as moving with the velocity VBO = VBA + VAO = 120 km/hr. Example 1 A boy runs inside a train with a velocity VBT = 4i – 2j relative to a passenger sitting in the train while the train is moving with a velocity VTR = 50i + 30j relative to a roadside observer. Find the velocity of boy relative to roadside observer, VBR The velocity of the boy relative to the roadside observer VBR = VBT + VTR = (4i – 2j)+(50i + 30j) = 54i + 28j VRB = - (54i + 28j) (this is not required in the given question; it is just to remind you that VBR is different from VRB ) = - 54i - 28j (backwards) Example 2 A boy can swim 1.2 m/s in still water. Find total time to swim a distance 1 km upstream and return to the starting point 1. When the water is still 2. When the water has a steady speed of 0.5 m/s There are three velocities to consider Velocity of boy relative to water, VBW Velocity of water relative to the ground, VWG Velocity of boy relative to the ground, VBG From Rule 2: VBG = VBW + VWG 1. When the water is still, VWG = 0 or VBG = VBW = 1.2 m/s 𝑆 1000 Time to go upstream, 𝑡1 = = = 833 𝑠 𝑣 1.2 Time to go up and return = 2t = 1.67 × 103 s 2. When the water is not still, VBG = VBW + VWG Upstream, VWG = - 0.5 m/s or VBG = 1.2 – 0.5 = 0.7 m/s 𝑆 1000 Time to go upstream, 𝑡1 = = = 1.43 × 103 𝒔 𝑣 0.7 Downstream, VWG = 0.5 m/s or VBG = 1.2 + 0.5 = 1.7 m/s 𝑆 1000 Time to go Downstream, 𝑡2 = = = 588 𝐬 𝑣 1.7 Time to go up and return = t1 + t2 = 2.02 × 103 s Projectiles If the velocity of projection of an object is U at an angle θ (different from 90o) to the horizontal, then its motion becomes 2-D. That is, its velocity has both x and y components. The initial components are: Ux = U cos θ and Uy = U sin θ The resultant path is parabolic and such a motion is called projectile motion.  Along x-direction, no force and hence, no acceleration. Horizontal displacement, x = Ut cos θ i Along y-direction, we have negative acceleration (g) due to gravitational force. The vertical displacement, 1 2 y = Ut sin  − gt ii 2 x From eqn. i, t= , so that ii becomes U cos g y = (tan  ) x − ( 2 2 ) x 2 iii 2U cos  Since the terms in brackets are constants, we can write y = ax – bx2 iv This is a quadratic equation for which there are two values of x for the same height y. This explains the parabolic nature of projectile motions. 𝟏 If we use the trigonometric relation: = 𝟏 + 𝒕𝒂𝒏𝟐 𝜽, 𝑪𝒐𝒔𝟐 𝜽 𝑔𝑥 2 eqn. iii becomes: 𝑦 = 𝑡𝑎𝑛 𝜃 𝑥 − 2 (1 + 𝑡𝑎𝑛2 𝜃) v 2𝑈 𝑔𝑥 2 𝑔𝑥 2 vi Rearranging: 𝑦=− 𝑡𝑎𝑛2 𝜃 + 𝑥 𝑡𝑎𝑛𝜃 − 2𝑈 2 2𝑈 2 When θ is the only variable in the equation, we can write eqn.vi as: y = a tan2 θ + b tan θ + c vii with two values of θ for the same height y. Exercise A ball is thrown towards a wall 10 m away with a velocity of 40 m/s. At what angle must it be thrown if it is to enter a hole on the wall at a height of 7m from the ground, neglecting wind effects (88.20, 40.10). The Range R of the projectile is the horizontal distance covered before the object hits the ground and it is given as: R = UT cos θ viii where U cos θ is the horizontal component of initial velocity and T is the total time of the flight. To determine the time of the flight, we us the vertical component of the velocity: V y = U sin  − gt At the maximum height H, Vy = 0 and hence U sin θ = gt. U sin  That is, time at this point is t= g  2U sin  The total time of flight, T = 2t = g 2U 2 sin  cos  U 2 sin 2 From eqn viii R = = g g Note that: Sin2θ = 2Sinθ Cosθ. The maximum range is achieved when θ = 450 Example Find the minimum velocity with which a missile can be projected from a military base to hit a target 500 km away, assuming only gravitational force.  U 2 sin 2 Rg R= That is,. U 2 = g sin 2 The minimum value of U is when sin 2θ = 1 or when θ = 45o At this point U 2 = Rg or 𝑈= 500,000 × 9.8 = 2213.6𝑚/𝑠 at 450 to the horizontal. At any other angle, U will be more than this to achieve the same range.  Note that: The maximum height H is at a time, t = U sin . g Substituting for t in the second equation of motion 1 𝐻= 𝑈𝑦 𝑡 + 𝑔𝑡 2 , noting that 𝑈𝑦 = 𝑈𝑆𝑖𝑛𝜃, 2 U 2 sin 2  U 2 sin 2  U 2 sin 2  H= − That is H= g 2g 2g Hmax is achieved when θ = 900 (i.e. vertical throw) Uniform Circular Motion Some bodies are not free to move along a line but are constrained to circular paths. Common examples: 1. the earth around the sun 2. a satellite (the moon) around the earth, 3. an electron around the nucleus 4. a stone whirled round a circle by means of a string attached to it. Speed is constant in the circle but the direction of motion is constantly changing and as such, the velocity vector is changing. That implies acceleration. We talk of linear velocity in linear motion and angular velocity in circular motion.  𝑑𝜃 Angular velocity is a vector quantity with magnitude and 𝑑𝑡 direction at right angle to both the tangential velocity and the radius of the circle. A useful relationship which exists between angular velocity ω and the linear or instantaneous velocity v can be derived by considering a motion in a circle as shown below. For a very small change in instantaneous position, the distance ΔS can be taken as a straight line so that: 1  This is the mathematical relationship Therefore, 𝑣 = 𝑟𝜔 between instantaneous/linear velocity and angular velocity. Just as we have angular velocity, we also have angular acceleration which come is a reality when the rotating body slows down or speeds up. Angular acceleration is defined mathematically as ∆𝜔 𝑑𝜔 α= =. ∆𝑡 𝑑𝑡 There is also radial or centripetal acceleration as a result of tangential velocity constantly changing in direction, although the magnitude is 𝑉2 −𝑉1 ∆𝑉 constant. This is defined as 𝑎 = =. Its value in terms of the velocity ∆𝑡 ∆𝑡 can be derived by considering the triangle AOB (shown below) when the velocity changes from 𝑉1 to 𝑉2 as illustrated in the figure. 4 The two triangles AOB and MNP in the two are both isosceles with ∆θ equal, hence they are similar.  This is the centripetal or radial acceleration. This acceleration is at right angle to v and it is directed towards the centre of the circle and it is not constant, since the direction is not constant. There are corresponding equations of motion for constant angular motion as in the case of uniform linear motion. Example The moon revolves round the earth, making a complete revolution in about 27.3 days. If the radius of its orbit is 3.85 x 108 m, calculate the 1. angular speed 2. tangential velocity of the motion. 3. angular acceleration 4. centripetal accelerations? Angular speed  Linear speed Tangential velocity is the 𝑣 = 𝜔𝑟 = 2.664 × 10−6 × 3.85 x 108 = 1025.57 m/s linear speed. Angular acceleration The angular velocity of the moon’s motion round the earth is constant (unless the motion) slows and hence, 𝑑𝜔 angular acceleration is zero ie 𝛼 = =0 𝑑𝑡 Centripetal acceleration Dynamics The study of motion and the way in which forces produce motion. The solid foundation for dynamics was laid by Isaac Newton in his three legendary laws of motion. Inertia A body at rest will remain at rest, while a body in motion will maintain the motion with a constant speed in a straight line for as long as no net force acts on it. This natural tendency of a body is called its inertia and it is determined by the mass of the object. Newton’s First law of Motion A body at rest will remain at rest and a body in motion will maintain the motion with a constant speed in a straight line, as long as no net force acts on it. That is, inertia must be overcome (by an applied force) before we can have motion or before we can bring motion to rest. Force The action on a body which changes or tends to change the state of rest or of uniform motion of a body. Momentum When a body is at rest, its inertia is completely determined by its mass but when in motion, both mass and velocity determine its inertia. The product of mass m and velocity v is called momentum. That is, momentum p = mv. It is a vector quantity with magnitude (mv) and direction that of the velocity A body at rest has zero momentum and the one moving with constant velocity has a constant momentum. Newton’s First Law Restated: A net force is necessary to produce a change in momentum.  Newton’s second law of motion: Relationship between the net force and the change in momentum produced. It states that the magnitude of the rate of change of momentum is proportional to the magnitude of the force producing it and its direction is the same as that of the force. 𝒅𝒑 𝒅(𝒎𝒗) 𝒅𝒗 𝑭𝜶 𝒐𝒓 𝑭 = 𝒌 = 𝒌𝒎 = 𝒌𝒎𝒂 𝒅𝒕 𝒅𝒕 𝒅𝒕 where a is the acceleration produced. In S.I. units, 1 Newton (N) is defined as the net force that produces an acceleration of 1 m/s2 on a body of mass 1 kg. That is the constant k above has been conveniently set to unity so that: F = ma Impulsive Force: An impulsive force F is a force produced as a result of a change in momentum within a time interval ∆t. Converse of 2nd Law The smaller the value of ∆t, the larger the force produced(recall that 𝑑𝑃 : F α ). 𝑑𝑡 This is what we have when ✓ a ball hits a wall and its motion reverses in a fraction of a second ✓ a man jumps from height and his momentum becomes zero in a fraction of a second Impulse The product of the force and the time interval ∆t for which it acts is called the impulse, denoted by ∆j (Unit: Ns). That is: ∆𝑝 ∆𝑗 = 𝐹∆𝑡 =. ∆𝑡 = ∆𝑝 ∆𝑡 where ∆p is change in momentum, which can be due to either a change in velocity or a change in mass. The total impulse when momentum changes from the initial value pi to the final value pf is obtained by integration: 𝑝 ‫𝑝𝑑 𝑓 𝑝׬ = 𝑗𝑑 ׬‬ 𝑜𝑟 𝐽 = 𝑝𝑓 − 𝑝𝑖 𝑖 Example A man of mass 80 kg jumps from a height of 4 m unto the ground. Compare the force that act on him if 1. he lands barefooted and the landing takes place in 0.01s 2. he lands with a cushioned (padded) shoe such that the landing is delayed to 0.08s. If the initial velocity, u = 0 on top of the height, the velocity on landing, 𝑣 = 2𝑔ℎ = 8.85 𝑚/𝑠 That is, momentum on landing, 𝑝𝑖 = 80 × 8.85 𝑘𝑔𝑚/𝑠. After landing, momentum 𝑝𝑓 = 0 Impulse, 𝐽 = 𝑝𝑓 − 𝑝𝑖 = −708.35 Ns  J 708.35 1 F1 = =− = − 70835 N t1 0.01 J − 708.35 2 F2 = = = −8854.4 N t 2 0.08 The minus sign implies that the impulse opposes a change in momentum caused by the upward reaction of the ground. Increase in the time of landing (gentle landing) reduces the force of impact. Remember: speed kills Newton’s third law of motion Whenever a body exerts a force (action) on another body, the second body exerts a force (reaction) equal in magnitude and opposite in direction on the first body 1. the operation of a lawn sprayer in which the hose pushes water out (action) and water pushes back the hose (reaction). The reaction results in the rotation of the sprayer. 2. The rocket expels a large amount of products of combustion from its rear as the action while the reaction from the expelled products pushes the rocket upward. Atwood Machine This consists of two bodies suspended by a rope passing over a frictionless fixed pulley. We can apply the second law to determine both the acceleration and the tension in the ropes. From the net force on each mass, we can write, T1 – m1g = m1 a1 - T2 + m2g = m2a2 But the string is continuous, so that T1 = T2 = T a1 = -a2 = a  T – m1g = m1a i T – m2g = - m2a ii From where we get that m2g – m1g = m2a – m1a  m 2 − m1  a = g   That is,  m 2 + m1  Exercise Show from eqns i and ii that the tension  2m1 m2  T =   g  m1 + m2   This is Atwood machine on an inclined plane. T – m1g sin30 = m1a - T + m2g = m2a g (m2 − m1 sin 30) (2 −1.5) a= = 9.8  = 0.98m / s 2 m1 + m2 5 A Passenger in a Lift Another application of the second law of motion. R Two forces on the passenger 1. His weight W directed vertically downward 2. The reaction R from the floor, which is vertically upward. W The weight W = mg is always constant while the reaction R (what the man feels), depends on the motion of the lift. 1. When the lift is at rest, R – W = 0 or R = W. The man feels his real weight 2. When the lift is ascending with an acceleration a, R – W = ma or R = mg + ma The man feels heavier than his real weight and the more the acceleration, the heavier he feels. 3. When the lift is descending with an acceleration a, the force of motion, mg – R = ma or R = mg – ma. The man feels lighter than his real weight and the more the acceleration, the lighter he feels. 4. When the lift descends with an acceleration a = g, R = 0. The man feels weightless. Example In the diagram, a painter is in a crate which hangs alongside a building. When the painter who weighs 1000 N pulls on the rope, the force he exerts on the floor is 450N. If the crate weighs 250 N, find the acceleration. Take the acceleration a to be upward