Physics For Engineers 1 Notes PDF

lOMoARcPSD|46385866 Physics for engineers - Lecture notes 1-10 Mechatronics engineering (Jomo Kenyatta University of Agriculture and Technology) Scan to open on Studocu Studocu is not sponsored or endorsed by any college...

lOMoARcPSD|46385866 Physics for engineers - Lecture notes 1-10 Mechatronics engineering (Jomo Kenyatta University of Agriculture and Technology) Scan to open on Studocu Studocu is not sponsored or endorsed by any college or university Downloaded by Nate Player ([email protected])  lOMoARcPSD|46385866 1 SPH 110: FUNDAMENTALS OF PHYSICS I 1.0 UNITS AND DIMENSIONS 1.1 Introduction Measurable quantities in physics are assigned units of measurements. Quantities are divided into 2 namely:- 1) Basic / fundamental quantities 2) Derived quantities 1.2 Basic /Fundamental quantities They don’t depend on other quantities. These quantities are used to fully describe other physical quantities. They include:- Basic Quantity S. I. Unit Symbol Length Metre m Mass Kilogramme kg Time second s Amount of substance mole mol Electric current Ampere A Thermodynamic temp. Kelvin K Luminous intensity candela cd 1.3 Derived quantities They are described in terms of basic or fundamental quantities e.g. volume, area, pressure, density etc. Metre: It’s the distance between two points. The standard of a metre is marked on a bar of platinum (90%) – Iridium (10%) alloy kept at 0oc. Second: It’s the duration of 9, 192, 631, 770 periods of certain microwave radiation emitted by the ceasium atom. The atomic clock is the most accurate and other clocks (secondary) are set compared to it. Kilogramme: The standard mass is the platinum. Iridium cylinder whose mass is exactly one kilogramme Downloaded by Nate Player ([email protected])  lOMoARcPSD|46385866 2 Note: The physical quantities, time, mass and length are fundamental quantities we use in our study of mechanics. 1.4 Dimension and dimension Analysis Dimension: It is a physical property described by the words time, length or mass. (This property is the same no matter what units it is expressed. Dimension: Symbol Length: L Time: T Mass: M Dimension Analysis: It is a technique of establishing the validity of a solution to a problem, a unit or an equation by checking for dimensional consistency. Dimensional units must have the following properties:- 1) For addition and subtraction, quantities must have the same dimensional units. 2) For division and multiplication they may have different units 3) For equations to hold they must have the same dimensional units on both sides. Note: Constants and angles have their dimensional units as 1 e.g 𝜋, Cos θ, Sin θ, Tan θ, 1, 2…., ½ , 4/3, exp, ln, log, etc. Examples 1. Define and give the dimensional units for the following:- a) Speed: It is the rate of change of displacement 𝐿 Dimension of velocity = = 𝐿𝑇 −1 (S.I unit is metre per second) 𝑇 b) Acceleration: It is the rate of change of velocity 𝐿𝑇 −1 Dimensions of acceleration = = LT-2 𝑇 Density: It is the mass per unit volume M Dimensions of density 3 = 𝑀𝐿−3 (S.I unit kg/m3) 𝐿 Downloaded by Nate Player ([email protected])  lOMoARcPSD|46385866 3 c) Dimension of energy ; Energy = Force x distance E = Ma.S E = MLT-2.L E = ML2T-2 2. Prove if the following dimensions are correct. 𝑙 a) T =2π√ , 2π = 1 𝑔 𝑙 L T =√ , T = √(𝐿𝑇 −2 ) = T 𝑔 Both sides have dimensions of time. b) V = πr2h π= 1 V = r2h = L2. L = L3 Both sides have dimensions of volume c) V = u + a t V = u + a t = LT-1 + LT-2 x T = LT-1 + LT-1 = LT-1 Both sides have dimensions of velocity. d) E = mc2 (C is speed of light) E = M(LT-1)2 = ML2T-2 Both sides have dimensions of energy e) d Sin θ = n, n and Sin θ are dimensionless L = L. Both have dimensions of length. 3. Find the units of constants below a) F = -kx where k is spring constant 𝐹 𝑀𝑎 𝑀𝑇 −2 k =- =- 𝑥 =- = -ML-1T-2 𝐿 𝐿 b) N(t) = N exp (-kt) exponential is dimensionless kt = 1 kT = 1 1 k = 𝑇 k = T-1 Downloaded by Nate Player ([email protected])  lOMoARcPSD|46385866 4 4. The expression for kinetic energy is K.E = ½ mv2 and potential energy is P.E = mgh. Show that both expressions have the same dimensions, hence can be subtracted or added from each other. ½ mv2 = mgh M(LT-1)2 = MLT-2 x L ML2T-2 = ML2T-2 5. The period T of a pendulum is given by the dimension equation T = kmxlygz, where m is mass of the bob, l is the length of the string, g is acceleration due to gravity and k, x, y, z, are constants. Calculate the values of x, y, and z. T =MxLy (LT-2)z = MxLy + z T-2z - 2z = 1, z = -1/2 y + z = 0, y = 1/2 x = 0. Downloaded by Nate Player ([email protected])  lOMoARcPSD|46385866 5 2.0 VECTORS 2.1 Introduction If a sack of flour has a mass of 10kg, that mass is not dependent on where the flour, whether it at rest in a storeroom on land or in motion on a ship in sea. The above statement describes only the magnitude / size (10kg) but not the position. This shows that mass is a scalar quantity. For a quantity like velocity it is quite different. To a passenger in Mombasa desiring to go to Nairobi city on a bus moving at 20m/s, it obviously makes a big difference whether the bus is moving towards Nairobi city or Malindi town. Here both direction and size/magnitude are vitally important. Such a quantity like velocity is a vector quantity. 2.2 Scalar and Vector quantities Scalar quantity: It is a physical quantity that has no direction and it is completely specified by its magnitude / size alone, e.g. mass, energy, time, etc. Vector quantity: It is a physical quantity that is completely specified only when both its magnitude / size and direction are given, e.g. velocity, displacement, force, momentum, acceleration etc. 2.3 Representing vectors A vector quantity is represented in many ways. Pictorial representation: A vector is represented by a directed line segment (arrow). Where length of the line is the size/magnitude while the arrow shows direction. Symbol representation: Vector A can be represented as in:- A - Arrow on top A - wavy line below A - Bold face Downloaded by Nate Player ([email protected])  lOMoARcPSD|46385866 6 Vectors can be analyzed when represented on a coordinate system. i) Cartesian or rectangular co-ordinate. (xy plane/2 dimension) ii) xyz plane (3 dimension) Downloaded by Nate Player ([email protected])  lOMoARcPSD|46385866 7 Vectors can also be represented in terms of i, j and k. 3 i.e OA = [ ] ; OA = 3i + 4j 4 3 OA = ; OA = 3i + 4j + 6k 6 Position vector: It is a vector drawn from the origin of some coordinate system to a point in space to indicate position of object with respect to origin, i.e 3 3 OA = [ ] or OA = 4 6 Displacement vector: It is a directed line segment (arrow) whose length indicates the magnitude of the displacement and whose direction is the direction of displacement. 2.4 Operation on vectors 2.4.1 Vector addition In addition it means two vectors are added to get another vector, i.e A + B = C There are two ways of doing this:- Triangle method: If A and B are drawn to scale with tail of B at the tip of A, then C is a vector from the tail of A to the tip of B. Downloaded by Nate Player ([email protected])  lOMoARcPSD|46385866 8 Tip – to – tip Method (polygon): It is an extension of the triangle method to two or more than two vectors. 2.4.2: Vector subtraction The negative of a vector of equal magnitude but different direction. A = -B Vector subtraction is vector addition of opposite vectors. A - B = A + (-B) Example: 1. Given that A = 5i + 3j and B = 2i - 4j Find: a) A + B b) A - B a) A + B = 7i - j b) A - B = 3i - 7j Downloaded by Nate Player ([email protected])  lOMoARcPSD|46385866 9 2.4.3: Multiplication of Vectors We have two ways of a vector multiplication  Dot / scalar product  Cross/ vector product Dot / Scalar product Means the result is a scalar. If there are two vectors A and B then the dot product of the two vectors is defined as A.B = |A||B| Cos θ Where θ is the angle between the two vectors. Note: dot product commute, i.e A.B = B.A If the vectors are perpendicular to each other then the angle between them is 90o. A.B = |A||B| Cos 90o = 0 If we express in terms of i, j and k then k.j = i.k = 0, k.j = j.k = 0 and i.j = j.i = 0 Also i.i = j.j = k.k = 1 Consider vectors A = a1i + a2j + a3k and B = b1i + b2j + b3k then, A.B = (a1i + a2j + a3k) (b1i + b2j + b3k) = a1b1i.i + a1b2i.j + a1b3i.k + a2b1j.i + a2b2j.j + a2b3j.k + a3b1k.i + a3b2k.j + a3b3k.k = a1b1 + 0 + 0 + 0 + a2b2 +0 + 0 + 0 + a3b3 = a1b1 + a2b2 + a3b3 Downloaded by Nate Player ([email protected])  lOMoARcPSD|46385866 10 Example 1) Find A.B and B.A given that A = 2i + 3j + 4k and B = -j – 2j + k A.B = (2x – 1) + (3 x –2) + (4 x 1) = -4 B.A = (-1 x 2) + (-2 x 3) + (1 x 4) = -4 2) Given that, |A| = √14, |B| = √16 and the angle between A and B is 30o, find A.B A. B = |A||B| Cos θ = √14 x √16 Cos 300 = 12.9514 Cross Product Given that two vectors A and B the cross product of A and B is defined as A x B = |A||B| Sin θ Consider two vectors A = a1i + a2j + a3k and B = b1i + b2j + b3k 𝒊 𝒋 𝒌 Represent in matrix form A x B = |𝑎1 𝑎2 𝑎3 | 𝑏1 𝑏2 𝑏3 A x B = i [(a2 b3) – (b2 a3)] + j [(b1 a3) – (a1 b3)] +k [(a1b2) – (b1a2)] Example. Given that A = 2i + 3j - k and B = -i + j + 2k Find A x B 𝒊 𝒋 𝒌 AxB =| 2 3 −1| −1 1 2 A x B = i [(3 x 2) – (1 x -1)] + j [(-1 x -1) – (2 x 2)] +k [(2 x 1) – (3 x -1)] = 7i – 3j + 6k 2.4.4: Multiplication with scalars Consider vectors A, B and scalar S then S (A + B) = SA + SB Downloaded by Nate Player ([email protected])  lOMoARcPSD|46385866 11 2.4.5: Magnitude and Direction of a vector Given that A = a1i + a2 j + a3k, then |A| = √[(a1)2 + (a2)2 + (a3)2] Example: 1. Given that A = 2i + 3j + 4k, find |A| |A| = √[(2)2 + (3)2 + (4)2] = 5.39 units 2. If A + B + C = 0 and A = 2i + 3j + 4k, B = 5j + 6j + 7k. What is C, |C| and angle between C and x axis. C = -A –B = -7i - 9j - 11k |𝐂| = √[(-7)2 + (-9)2 + (-11)2] = 15.84 units θ= Tan (-9/-7) =52.13o 2.4.6: Angle between vectors We find angles between vectors by using the dot product. This is because dot product gives the result of a scalar. 𝑨.𝑩 A.B = |A||B| Cos θ, θ = Cos-1 ( ) |𝑨||𝑩| 2.4.7. Angle between vector and axes Consider vector A as shown:- Downloaded by Nate Player ([email protected])  lOMoARcPSD|46385866 12 Note: The negative sign shows the angle is below x – axis Example: 2 5 Given that OA = and OB = [−2], find the angle between them. 1 2 |𝐀| = √[(2)2 + (3)2 + (1)2] = √14, |𝐁| = √[(5)2 + (-2)2 + (2)2] = √33, 𝑨.𝑩 𝟔 θ = Cos-1 ( )= Cos-1 ( ) = 73.790 |𝑨||𝑩| √(𝟏𝟒 𝐱 𝟑𝟑) Magnitude and Direction in Two Dimension The rectangular co-ordinates (x,y) and polar co-ordinates (r,θ) are related by x = r Cos θ, y = r Sin θ, r = √𝑥 2 + 𝑦 2 and Tan θ = y/x Downloaded by Nate Player ([email protected])  lOMoARcPSD|46385866 13 Three dimension (x,y,z) co-ordinate and spherical co-ordinate. The rectangular co-ordinate (x, y, z) and spherical co-ordinates (r, θ,𝜙) are related by: 𝑥2 + 𝑦2 x =r Sin θ Cos𝜙, y = r Sin θ Sin𝜙, z = r Cos θ, r = √𝑥2 + 𝑦2 + 𝑧2 , Tan θ =√ 𝑍2 and Tan 𝜙 =y/x Examples. 1) Find the magnitude and direction of the following vectors. a) A = 5i + 3j b) B = 10i – 7j c) C = -2i - 3j + 4k Solution a) |A| = r = √52 + 32 = 5.83 5 θ = Cos -1 (5.83) = 30.960 b) |B| = r = √102 + −72 = 12.21 10 θ = Cos -1 ( ) = 35.020 12.21 c) |C| = r = √−22 + −32 + 42 = 5.39 √−22 + −32 θ = Tan -1 ( )= 42.030 √4 −3 ϕ = Tan -1 (−2) =56.310 2) The rectangular components of the vectors which lie in x – y plane have their magnitudes and directions given below. Find the x and y components of the vectors. a) r = 10 and θ = 300 b) r = 7 and θ = 600 Downloaded by Nate Player ([email protected])  lOMoARcPSD|46385866 14 Solution a) x = r Cos θ = 10 Cos 300 = 8.66, y = r Sin θ = 10 Sin 300 = 5 b) x = r Cos θ = 7 Cos 600 = 3.5, y = r Sin θ = 7 Sin 600 = 6.06 3. a) Find the magnitude and direction of the resultant vector A = 5i + 3j and B = 2i – 4j Solution R = A + B = 7i – j |R| = √(72 + −12 ) = 7.07 𝑦 −1 Tan θ = = 𝑥 7 θ = -8.13 0 2.4.8 Resolution of Vectors A component of a vector is the effective part of a vector in that direction. Consider a Force F pulling in the direction as shown. X component of F is F Cos θ Y component of F is F Sin θ Example Consider two forces F1 and F2 pulling as shown below. Find the X and Y components of the forces given that |F1| = 2.88 and |F2| = 3.44 Downloaded by Nate Player ([email protected])  lOMoARcPSD|46385866 15 F1x = |F1| Cos 33.7o = 2.88 Cos 33.7o = 2.40, F1y = |F1| Sin 33.7o = 2.88 Sin 33.7o = 1.60 F1 = 2.4i + 1.6j Similarly F2x = |F2| Cos 35.5O = 3.44 Cos 35.50 = 2.80, F2y =|F2| Sin (-35.5) = 3.44 Sin (-35.5o) = 2.00 F2 = 2.80i – 2.00j Downloaded by Nate Player ([email protected])  lOMoARcPSD|46385866 16 3.0: FORCE Introduction Force is defined as pull or push in on a body, it is a vector quantity it is measured in Newtons. A Newton is the force that gives a mass of 1kg an acceleration of 1m/s2 Task: (1) Give five effects of force. (2) Name and explain at least 10 different types of force. 3.1: Resolution of forces Forces is a vector quantity which can also be expressed in x and y on rectangular coordinates Consider a force F pulling a load along a surface at an angle θ The horizontal component of the force is F Cos θ while the vertical component is F Sin θ. Also consider a load sliding along an inclined plane at a constant acceleration Downloaded by Nate Player ([email protected])  lOMoARcPSD|46385866 17 Example Find the tension in each cord if the weight of suspended object is 490N Solution ∑Fy = 0 T3 – 490 = 0 ∑Fx = 0 T2 Cos400 – T1 Cos 600 = 0 T2 = 0.653T1 ∑Fy = 0, T2Sin400 + T1 Sin 600 – 490N = 0 (0.653T1) Sin 400 +T1 Sin 600 = 490 T1 = 381N T2 = 0.653 (381) = 249N Downloaded by Nate Player ([email protected])  lOMoARcPSD|46385866 18 3.2: Friction Force Friction force is a force that opposes relative motion between two surfaces. Friction is an example of dissipative or resistive force this force convert mechanical energy (kinetic energy) to other energies e.g. sound and heat by resisting motion of bodies. Other examples are air resistance and viscosity of fluids. Task: 1. Friction is a nuisance, explain? 2. Friction is vital explain? 3. How can you reduce friction? For a body resting on a rough surface, when external force (F) is applied on it (pushing /pulling) Then this force has to overcome friction (Fr) before the body moves. The body is moved by a net force (F net) i.e. F net = F – Fr Frictional force Fr depends on the normal reaction (R = mg) Fr ⋉ R Fr = μR Where μ is the coefficient of friction, μ depends on the nature of two surfaces that are in relative motion. Static friction:- This is the frictional force exerted by one surface on another when there is no relative motion of the two surfaces Downloaded by Nate Player ([email protected])  lOMoARcPSD|46385866 19 Sliding (Kinetic friction):- The frictional force exerted by one surface on another when one surface slides over another surface. Fsliding/kinetic=Fk= μkR- μk - coefficient of kinetic/sliding friction Fstatic=Fs≤ μsR – μs coefficient of static friction Task: (1) State the Laws of friction Examples 1. A block of wood of mass 20kg requires a horizontal force of 50N to pull it with a uniform velocity along a horizontal surface. Calculate the coefficient of friction between the block and the surface. R = mg = 20 x 10 =200N, Fr = μR μ = 50/200 = 0.25 2. A mass of 5kg is placed on a plane inclined at an angle of 300 to be horizontal. Calculate the force required to pull the mass up the plane at uniform velocity if μ = 0.5 Downloaded by Nate Player ([email protected])  lOMoARcPSD|46385866 20 R = mgCos300 = 50Cos300 = 43.3N Fr = μR = μmgCosθ = 0.5 x 50Cos300 = 21.65N There are two forces opposing motion, Fr and mgSinθ Fnet = F – (Fr + mgSinθ) We have uniform motions, net force is zero F = Fr + mgSinθ F = 21.65 + 50Sin300 = 46.65N 3.3: Newton Laws First Law (law of inertia):- every body continues to be in state of rest or to move with uniform velocity unless a resultant force acts on it. Implication of the 1st law – causes inertia Inertia:- is the property of an object that resists change of motion Second law:– The rate of change of momentum is directly proportional to the change causing it (resultant force) and takes place in the direction of force. Momentum:- – is defined as the product of mass of a body and its velocity. p =mv………………………………..(1) Consider a force F, acting on a body of mass, m for a time t, causing a change in velocity from u to v, then:- ∆p = mv – mu………………………. (2) ∆𝑷 𝑚𝒗−𝑚𝒖 From 2nd law = ∆𝑡 ∆𝑡 ∆𝑷 F∝ ∆𝑡 ∆𝑷 F=𝐾 however, K = 1 ∆𝑡 ∆𝑷 𝑚𝒗−𝑚𝒖 𝑚(𝒗−𝒖) F=𝐾 = = ∆𝑡 ∆𝑡 ∆𝑡 (𝒗−𝒖) But a = ∆𝑡 Therefore F =ma………………………….. (3) 𝑝2 Task: Kinetic energy, K = ½mv2 and momentum, p = mv, show that, K = 2𝑚 Downloaded by Nate Player ([email protected])  lOMoARcPSD|46385866 21 Examples 1) A car of mass 1200kg travelling at 45m/s is brought to rest in 9 seconds. Calculate the average retardation of the car and the average force applied by the brakes. 𝑣−𝑢 0−45 a= = = -5ms-2 𝑡 9 F = ma = (1200 x -5)N = -6000N 2) A truck weighs 1.0 x 1.05N and free to move. What force will give it an acceleration of 1.5ms-2 1.0 𝑥 105 F = ma = x 1.5N = 1.5 x 104N 10 Third Law: For Every Action, There is An Equal and Opposite Reaction. Note: A body moving in a straight line has linear momentum. 3.4: Impulse – momentum theorem States that the change in momentum of a body is equal to the sum of the force acting on the body with respect to time. mv-mu = ∫ Fdt Impulse change the velocity of a body of mass m from u to v. 3.4.1: Conservation of linear momentum Momentum of an isolated system is always conserved, An isolated system is one that has zero interaction with its environment e.g, pressure, temperature. e.t.c. In practice it is easy to isolate a system. Consider two bodies of mass m1 and m2 moving in the same direction on a smooth horizontal surface at velocities u1 and u2 respectively. Initial momentum will be p1 = m1u1 and p2 = m2 u2 Total momentum before collision p = p1 + p2 = m1u1 + m2 u2 After collision, their velocities are v1 and v2 respectively. Total momentum after collision, p = m1v1 + m2v2 By conservation of linear momentum therefore, Downloaded by Nate Player ([email protected])  lOMoARcPSD|46385866 22 Total initial momentum = Total final momentum i.e, m1u1 + m2 u2 = m1v1 + m2v2 (isolated system) 3.4.2: Elastic Collision. It’s a case where bodies separate after collision 3.4.3: Inelastic Collision Colliding bodies coalesce/combine after collision. During perfect elastic collision, the momentum p of the bodies is conserved and also the kinetic energy possessed by the body is conserved. i.e, ½m1u12 + ½m2 u22 = ½m1v12 + ½m2v22. The bodies combine and move with a common final velocity v i.e, ½m1u12 + ½m2 u22 = ½(m1 +m2)v2 During imperfect elastic collision, momentum is conserved but energy is not conserved. 3.4.4 Collision in Two Dimensions. In this kind of collision the bodies collide at angles. This is represented in a rectangular coordinate. Downloaded by Nate Player ([email protected])  lOMoARcPSD|46385866 23 The law that controls the above scattering starter:- “If no external force acts on a system, the components of the total linear momentum of the system along any chosen axis are the same after a collision as before the collision.” On resolution:- X - Components: m1u1 + 0 = m1v1Cosθ1 + m2v2Cosθ2 Y - Components: 0 + 0 = m1 v1 sinθ1 – m2 v2Sinθ2 Or m1 v1 Sinθ1 = m2 v2 Sinθ2 If the collision is elastic then kinetic energy is conserved. ½m1u12 + 0 = ½m1v12 + ½m2v22. 3.4.5 Co-efficient of resolution, e Is the ratio of the velocity of separation to the velocity of approach. 𝑣 −𝑣 e =𝑢2 −𝑢1 2 1 e >1 - explosive / super elastic collision e = 1 - elastic collision e < 1 - inelastic collision e = 0 –completely inelastic. Examples 1) An object A of mass 2kg is moving with velocity of 3ms -1 and collides head-on with an object B of mass 1kg moving in the opposite direction with a velocity of 4ms-1. After collision both objects stick so that they move with a common velocity v, calculate v. Solution. Total momentum before collision is equal to total momentum after collision. mAvA – mB vB = (mA +mB )v ( 3 ×2) – (1×4) = (2+ 1) v v = 2/3 = 0.67 ms-1 Downloaded by Nate Player ([email protected])  lOMoARcPSD|46385866 24 2) A bullet of mass 20g, traveling horizontally at 100ms-1 embeds itself in the centre of the block of wood of mass 1kg which is suspended by a light vertical string 1m long. Calculate maximum inclination to the string to the vertical. (take g = 9.8ms-2) Solution. Using the principle of conservation of momentum then M1 u1+ m2 u2 = (m1 +m2) v v = 1.96ms-1 From the law of conservation of energy then ½ mv 2 = mg h v2 =gh h = l – l Cos θ h = l (1 – Cos θ) v2 = 2gl (1 – Cos θ) (1.96)2 = 2×9.8 ×1 ( 1- Cos θ) 1.962 1 –Cos θ = –1 19.6 θ = Cos-1(0.8038) θ = 370 3) A pool ball x of mass 0.3kg moving with velocity 5 ms-1, hits a stationary ball y of mass 0.4kg. Y moves off with a velocity of 2ms-1 at 300 to the initial direction of x. Find the velocity of x and its direction after hitting y Downloaded by Nate Player ([email protected])  lOMoARcPSD|46385866 25 In initial direction of x, from conservation of momentum 0.3v Cosθ + 0.4 x 2 Cos 300 = 0.3 x 5 0.3v Cosθ = 1.5 – 0.8 Cos 300 = 0.8……………………………….. (1) Along y, 900 to initial x direction, initial momentum = 0, so in this direction, 0.4 x 2 Sin300 – 0.3 x v Sinθ = 0 0.3 v Sinθ = 0.4 ………………………………………………………..……… (2) Dividing equation (2) by (1) 𝑆𝑖𝑛𝜃 0.4 = Tanθ = = 0.5 𝐶𝑜𝑠𝜃 0.8 𝜃 = 270 0.4 From (2) v = 0.3 𝑆𝑖𝑛 270 = 3ms -1 4) A trolley of mass 8kg is held at rest on a smooth inclined plane as shown. When released, it moves down through a vertical height of 2.5m while accelerating. It then collides with a second trolley of mass 6kg which is at rest on a smooth horizontal plane. After collision, the two trolleys coalesce and move forward. Calculate: a. The velocity of A just before collision. b. The common velocity of the two trolleys after the collision. c. The kinetic energy just before and after the collision. Account for the difference in kinetic energy. Solution: a) From the principle of conservation of energy 1 2 mv2 = mgh, v = √2𝑔ℎ v = √2 𝑥 10 𝑥 2.5 = 7.07m/s b) m1u1 + m2u2 = (m1 + m1)v 8 x 7.07 + 0 = (8 +6)v v = 4.04m/. Downloaded by Nate Player ([email protected])  lOMoARcPSD|46385866 26 1 1 c) 𝐾𝑒 = m1u12 = x 8 x (7.07)2 = 200J 2 2 1 1 𝐾𝑒 = (m1 + m2)v2 = x (8 +6) (4.04)2 = 114.25J 2 2 3.4.6 Variation of weight in a lift Weight of a body inside a lift increases when the lift accelerates upwards and decreases when the lift accelerates downwards. Suppose a person is standing on a weighing balance the gravitation force acts vertically downwards. A weight balance exerts reaction force (R ) in the upward direction hence the net force on the person is F = mg - R But the person is stationary, hence the net force acting on the person is considered to be zero, if the lift is moving at a constant velocity (a = 0) Mg –R = 0 R = mg The weighing balance reaches the actual weight of person When the lift moves up at acceleration, a then; F = R – mg = ma R = mg + ma R = m (g+a) Hence the weighing balance reads slightly higher value than actual weight of the person. Thus person feels his/her weight has increased slightly above the normal value. When a lift is moving downwards. F = mg - R Ma = mg - R R = mg - ma R = m (g-a) The weight of the body will be slightly less than the weight in normal state. Example 1) A man of mass 75kg stands on a weighing machine in a lift. Determine the readings in the weighing machine whom the lift moves. a) Upward with acceleration if 2ms-2 b) Downward with acceleration of 2ms-2 Downloaded by Nate Player ([email protected])  lOMoARcPSD|46385866 27 Solution. a) R= m(g+a) = 75 (10 +2) = 900 N b) R= m(g-a) = 75 (10 -2) = 600 N Downloaded by Nate Player ([email protected])  lOMoARcPSD|46385866 28 4.0: RECTILINEAR MOTION Introduction Motion: is continuous change of position of a body. We have three types of motion namely: 1) Rectilinear (linear) motion 2) Circular motion 3) Vibrational motion Rectilinear (linear) motion: This is motion on a straight line. Mechanics: This is the study of motion of objects and the causes of that motion. Kinematics: This is the study of motion without considering its causes. Dynamics: This is the study of motion considering its causes. 4.1: Linear motion It is the study of motion on a straight line. 4.1.1: Definition of terms Distance (D): It is the measure of length between two points, it is a scalar quantity and it is measured in metres. Displacement (S): It is the measure of length between two points in a specified direction, it is a vector quantity and it is measured in meters. Speed (s): It is the rate of change of distance, it is a scalar quantity and it is measured in metres per second, (m/s or ms-1). Velocity (v): It is the rate of change of displacement, it is a vector quantity and it is measured in meters per second, (m/s or ms-1). 𝑑𝑆 v= 𝑑𝑡 Acceleration (a): It is the rate of change of velocity, it is a vector quantity measured in meters per second squared, (m/s 2 or ms – 2) 𝑑𝑣 𝑑2 𝑣 a= = 𝑑𝑡 𝑑𝑡 2 Average velocity (vav): This is the ratio of average velocity to time interval. 𝑣+𝑢 vav = 2 Downloaded by Nate Player ([email protected])  lOMoARcPSD|46385866 29 Displacement - time graph. (i) The gradient at any given instant of this graph for a body that is changing position represents the instantaneous velocity. (ii) If the displacement – time graph is a straight line then the body is undergoing uniform velocity. Downloaded by Nate Player ([email protected])  lOMoARcPSD|46385866 30 (iii) S Negative gradient means the body is moving in the opposite direction to the original direction. Note: The area under the graph has no meaning Velocity – time graphs (i) The gradient at any instant of this graph for a body that is changing position represents the instantaneous velocity Downloaded by Nate Player ([email protected])  lOMoARcPSD|46385866 31 (ii) If the velocity – graph is a straight line then the body undergoing uniform acceleration. (iii) Downloaded by Nate Player ([email protected])  lOMoARcPSD|46385866 32 Negative gradient shows that the body is decelerating or retarding. The area under the graph is the displacement. Task: Describe the motion in the graphs below. (i) (ii) 4.2: Equations of linear motion 4.2.1: First equation of linear motion Consider a body starting with an initial velocity u and it increases or (accelerates) to a final velocity v in time t then the:- Downloaded by Nate Player ([email protected])  lOMoARcPSD|46385866 33 first equation: 𝑣−𝑢 a= 𝑡 v=u+at 4.2.2: Second equation of a linear motion Suppose a body accelerates with uniform acceleration a, in a time t and reaches a velocity v. The distance S traveled by the object in time t is given by S = average velocity x time 𝑣+𝑢 Where average velocity =, vav = 2 𝑣+𝑢 s=( )xt 2 𝑢+𝑎𝑡+𝑢 s=( )t 2 2𝑢 𝑎𝑡 s=( ) +( )t 2 2 𝑎𝑡 2 𝑠 = 𝑢𝑡 + 2 4.2.3: Third Equation of linear motion 𝑣−𝑢 Also t =( ) from 1st equation a 𝑢+𝑣 𝑣−𝑢 𝑣 2 − 𝑢2 s=( )( )=( ) 2 a 2a v2 = u2 + 2as 4.3: Velocity vectors for motion in a plane and acceleration 4.3.1: Velocity vectors Downloaded by Nate Player ([email protected])  lOMoARcPSD|46385866 34 ∆r is the displacement vector i.e the vector difference between position vectors r1 and r2 and ∆r = r2 - r1 If an object undergoes a displacement ∆r in time ∆t then its average velocity vector (v) is defined as ∆𝒓 𝒓2 −𝒓1 𝑣= = ∆𝑡 𝑡2 −𝑡1 The bar means average value of velocity. Instantaneous velocity in vector notation. 𝐿𝑖𝑚 ∆𝒓 𝒗= ∆𝑡 → 0 ∆𝑡 The limit of the average velocity over a time interval that approaches zero but always includes the desired instant of time. 4.3.2: Acceleration vectors Suppose that at time t, the velocity is specified by vector v1 where as at sometime later t2, the velocity vector has a new value v2 then we define acceleration vector as the change of velocity of an object during a given time interval divided by that time interval 𝒗2 −𝒗1 ∆𝒗 ā= = 𝑡2 −𝑡1 ∆𝑡 Instantaneous acceleration 𝐿𝑖𝑚 ∆𝒗 ā= ∆𝑡→0 ∆𝑡 The acceleration at a particular instant of time, the limit of the average acceleration over a time interval that approaches zero but always includes the desired instant time. The displacement vector ∆r = r2 – r1 could be replaced by two vector displacements along X and Y axes of the rectangular coordinate system. These components are the magnitude ∆x and ∆y. The scalar magnitudes of ∆x and ∆y are called rectangular components of ∆r. The same approach is taken by velocity and accelerations A velocity v in x-y plane can be replaced by, one along x axis and another along y axis of magnitude vx and vy respectively. Downloaded by Nate Player ([email protected])  lOMoARcPSD|46385866 35 The vector equations introduced for displacement, velocity and acceleration can be summarized in table below. Equations for Physical quantity Vector notation Rectangular components X - axis y - axis Displacement ∆r = r2 – r1 ∆x = x2-x1 ∆y = y2-y1 Average velocity ∆𝒓 ∆𝑥 𝑥2 − 𝑥1 ∆𝑦 𝑦2 − 𝑦1 𝒗= 𝒗𝑥 = = 𝒗𝑦 = = ∆𝑡 ∆𝑡 𝑡2 − 𝑡1 ∆𝑡 𝑡2 − 𝑡1 Instantaneous velocity 𝐿𝑖𝑚 ∆𝒗 𝐿𝑖𝑚 ∆𝑥 𝐿𝑖𝑚 ∆𝑦 𝒗= 𝑣𝑥 = 𝑣𝑦 = ∆𝑡 → 0 ∆𝑡 ∆𝑡 → 0 ∆𝑡 ∆𝑡 → 0 ∆𝑥 Average velocity ā= ∆𝒗 = 𝒗2 −𝒗1 ∆ 𝑣𝑥 ∆ 𝑣𝑦 ∆𝑡 𝑡2 −𝑡1 ā𝑥 = ā𝑦 = ∆𝑡 ∆𝑡 𝑣2𝑥 − 𝑣1𝑥 𝑣2𝑦 − 𝑣1𝑦 = = 𝑡2 − 𝑡1 𝑡2 − 𝑡1 Instantaneous 𝐿𝑖𝑚 ∆𝒗 𝐿𝑖𝑚 ∆𝑣𝑥 𝐿𝑖𝑚 ∆𝑣𝑦 ā= 𝑎𝑥 = 𝑎𝑦 = acceleration ∆𝑡 → 0 ∆𝑡 ∆𝑡 → 0 ∆𝑡 ∆𝑡 → 0 ∆𝑡 Downloaded by Nate Player ([email protected])  lOMoARcPSD|46385866 36 4.3.3: Uniformly accelerated motion on a straight line Uniformly accelerated motion is motion on a straight line with an acceleration of constant magnitude. We modify the notation we have used so far i.e t1 = 0 and t2 = t. The initial position be x0 and initial velocity be v0 while final position be x and final velocity be v. If we consider motion in x direction. ∆𝑥 𝑥 − 𝑥0 𝑥 − 𝑥0 𝑣= = = ………………… (1) ∆𝑡 𝑡−0 𝑡 On solving we have x=x0+vt…………………………….…….(2) The average velocity 𝑣0+𝑣 𝑣= …………………………………. (3) 2 Substituting equation (3) into (2) 𝑣0+𝑣 𝑥 = 𝑥0 + ( ) 𝑡 … … … … ….…………….(4) 2 The constant acceleration is given by ∆𝑣 𝑣−𝑣0 𝑎= = ………………………….….(5) ∆𝑡 𝑡−0 Therefore v = v0 + at…………………………..……. (6) If we substitute equation (4) into 6) 𝑣0 +𝑣0 +at 𝑎𝑡 2 𝑥 = 𝑥0 + ( ) 𝑡 = 𝑥0 + 𝑣0 + ………(7) 2 2 This is a perfect general equation of a body moving in a straight line with uniform acceleration a and initial position x0 and initial velocity v0 at t = 0. We can also develop an equation governing motion but time does not explicitly appear. We use equations and (1) in (4). 𝑣−𝑣0 𝑣−𝑣0 I.e x = x0 + v( 𝑎 )i.e. t=( ) 𝑎 𝑣−𝑣0 From (6) t=( ) 𝑎 Substituting equation (8) into (4) 𝑣0 + 𝑣 𝑣 − 𝑣0 𝑥 = 𝑥0 + ( )( ) 2 𝑎 𝑣−𝑣0 𝑣−𝑣0 𝑣 2 −𝑣0 2 𝑥 − 𝑥0 = ( )( )= 2 𝑎 2𝑎 v2=v02+2a (x-x0) Downloaded by Nate Player ([email protected])  lOMoARcPSD|46385866 37 The equation of uniformly accelerated motion in one dimension. x = x0 +vt 𝑎𝑡 2 x = x0 +vot+ 2 v2 = v02 +2a (x-x0) v = v0 +at 𝑣0 +𝑣 v= 2 Example 1) A force of 140N acts on a body rest, the body moves a distance in 10 seconds. Calculate: a) Acceleration of the body b) The distance moved by the body c) The velocity of the body Solution a) Fx = max, =>140 = (32.5kg)a ax = 4.31ms-2 b) x = x0 +v0xt + ½ axt2 x = ½ (4.31)1 (10.0)2 = 216m c) Vx = v0x +axt Vx = O+(4.31) (10) = 43.1 m/s 4.4: Motion Under Gravity We have three types of motion under gravity, namely: 1) Free fall 2) Vertical projection 3) Horizontal projection 4.4.1: Vertical Projection The Earth pulls all bodies towards its centre. A pull of gravity produces a constant acceleration on a body moving vertically upwards. This is called gravitational acceleration (g). Equations of linear motion can be modified by replacing a with –g, since the body is moving away from the centre of the Earth and S is replaced by H. Downloaded by Nate Player ([email protected])  lOMoARcPSD|46385866 38 v = u – gt………………………………….….(1) H = ut –1/2gt2……………………………………. (2) v2 = u2 – 2gH ………………………….…….(3) Time taken to reach maximum height (t) From equation (1) v = u-gt………………………………………..… (4) At maximum height v = 0, the body becomes momentarily at rest, equation (4) becomes u = gt, => t = u/g……………………………... (5) Time of flight (T) Time taken by body to reach maximum height and back to point of projection. T=2t = 2u/g………………………………………….. (6) Maximum Height Reached (Hmax) At maximum height v = 0 using equation (3) v2 = u2- 2gh 2g Hmax = u2 u 2 H max = 2𝑔 ……………………………………………….(7) 4.4.2: Free Fall In free fall or downward motion in equations of linear motion, a is replaced by +g while S is replaced by H v=u+gt………………(1) H=ut + ½ gt2………..(2) v2 = u2 + 2gH………….(3) 4.4.3: Horizontal projection This occurs when a body is projected horizontally from a height, h. The body has not continued moving but lands after having same horizontal displacement (range) R. Downloaded by Nate Player ([email protected])  lOMoARcPSD|46385866 39 We consider the vertical and horizontal motions from 0 independently consider vertical motion from 0, the vertical velocity u=0 from. h=ut +1/2 gt2 h = 1/2gt2………………………………………………(1) For horizontal projection acceleration is zero. From S = ut + 1/2at2 a = 0 and S = R R = ut…………………………………….………(2) 4.4.4: Projectile Motion Consider a body thrown with a velocity u at an angle θ to the horizontal. We consider the vertical and horizontal motion separately in motion of this kind and use components. Downloaded by Nate Player ([email protected])  lOMoARcPSD|46385866 40 Vertical motion component y = u sin θ and a = -g, When the project reaches the ground at B, the vertical distance covered will be zero from S = ut +1/2at 2 O= u Sin θ t - ½ gt2 2𝑢 𝑠𝑖𝑛𝜃 𝑡 = 𝑔 ……………………………………………..(1) Where t, is the time of flight. Horizontal motion since g acts vertically, it has no component in a horizontal direction. From S= ut +1/2 at2, a=0 and S=R 2𝑢 𝑠𝑖𝑛𝜃 2𝑢2 sin 𝜃 𝐶𝑜𝑠 𝜃 R = OB = ut = ucos θ x = 𝑔 𝑔 From 2 Sin θ Cos θ = Sin 2 θ – Trigonometric identity u2 (𝑆𝑖𝑛 2 𝜃) Therefore R = 𝑔 Note: For a given velocity of projection, the range is maximum when Sin 2 θ=1, i.e 𝑢2 the range has a maximum value At θ = 450 which is R= 𝑔 Trajectory The horizontal component of motion (x) is given by ux = u Cos θ. At any given time the horizontal component is given by horizontal velocity x time. x =uxt = ut cos θ………………………………………… (1) Vertical Displacement, y y = uyt – 1/2gt2…………………………………………….. (2) But uy = u Sin θ…………………………………………… (3) From (1), making t the subject then 𝑥 t= …………………………………………………. (4) 𝑢 𝐶𝑜𝑠 𝜃 Substituting (3) and (4) into (2), we get 𝑥 𝑥 y = u sin θ - 1/2g ( )2 𝑢 𝐶𝑜𝑠 𝜃 𝑢 𝐶𝑜𝑠 𝜃 𝑆𝑖𝑛 𝜃 For = tan θ – Trigonometric identity 𝐶𝑜𝑠 𝜃 Downloaded by Nate Player ([email protected])  lOMoARcPSD|46385866 41 Therefore, y = x tan θ – 1/2gx2 u cos2 θ Compare y = ax – bx2, by taking Tan θ and ½gu Cos2θ as constants a and b respectively. This is an equation of a parabola described. Example: 1) A golfer strikes a golf ball on level ground, the ball leaves the ground with initial velocity of 30.0m/s at an angle of 450 above the horizontal. Calculate the ball’s: a) Horizontal and vertical components of initial velocity b) Time taken to reach maximum height c) Maximum height kicked d) Range of the ball on the ground e) Velocity with which it strikes Solution. a) vx = u Cosθ = 30 Cos 450 = 25.98m/s , vy = u Sinθ = 30 Sin 450 = 15ms 𝑢 𝑆𝑖𝑛𝜃 30 𝑆𝑖𝑛 450 b) t = = = 2.16 sec. 𝑔 9.8 c) y = u Sinθ.t - ½gt = 30 Sin 450 x 2.16 - ½ x 9.8 x (2.16)2 = 22.9m 2 𝑢2 302 d) R = = = 91.8 m 𝑔 9.8 e) v =√𝑣𝑥 2 + 𝑣𝑦 2 But vx = u Cosθ = 21.2m/s, & vy = u Sinθ – gt = -21.2 m/s v = √(21.2)2 + (−21.2)2 = 30.0 m/s 4.4.5: Motion of an inclined plane Body moving down a rough inclined plane. Downloaded by Nate Player ([email protected])  lOMoARcPSD|46385866 42 Fr = mg Sin θ……………………………………… (1) R = mg Cos θ……………………………………… (2) 𝐹𝑟 𝑚𝑔 𝑠𝑖𝑛 𝜃 𝑆𝑖𝑛 𝜃 μ= = Using = tan θ - Trigonometric identity 𝑅 𝑚𝑔 𝐶𝑜𝑠 𝜃 𝐶𝑜𝑠 𝜃 Then μ=tan θ ……………..(3) Body moving up a rough inclined plane R – mg Cosθ = 0, R = mg Cosθ………………………………………. (1) F – mg Sin θ - Fr = 0, F – mg Sin θ – μR = 0 ……………….……….(2) Substituting equation (1) into (2) for R F – mg Sin θ –μmgCosθ = 0, F = mg Sin θ + μmgCosθ F = mg (Sinθ + μ Cosθ) ……………………………………………….(3) Examples 1) A man pulls a sledge of mass 16kg up an inclined plane of slope 220 to the horizontal. If he uses an effort of 1400N and coefficient of friction of the slope is 0.130. What extra load of wood can he place on the sledge? (Use g = 9.8ms-2). Solution m =msledge + m wood F = mg (Sinθ + μ Cosθ) 1400N = m x 9.8 (Sin 220 + 0.13 Cos 220) m = 289kg m wood = 289kg – 16kg = 273kg. Downloaded by Nate Player ([email protected])  lOMoARcPSD|46385866 43 4.5: Motion of connected bodies Case I: One body on a horizontal plane and other hanging freely The string is inextensible, the pulley and the horizontal surface are frictionless. For m2>m1 and ignoring friction forces, the equations of motion are: T = m1a…………………………………………….…… (1) m2g – T = m2a…………………………………………. (2) Adding (1) and (2) m2g = m1a = m2a= a(m1+m2) 𝑚2𝑔 Therefore a = 𝑚1 +𝑚2 … … … … … … … … … … ….. (3) From Equation (1) 𝑚2𝑔 𝑚1 𝑚2𝑔 T = m1a = m1[ (𝑚1 +𝑚2 ) ] =𝑚 ………………....(4) 1 +𝑚2 Suppose the horizontal surface is rough with coefficient of friction, μ. Then T = m2g – m2a ………………………………..… (1) R – m1g = 0 =>R = m1g T – Fr = m1a, But Fr = μR T – μR = m1a, But R = m1g T – μ m1g = m1a………………………………. (2) Downloaded by Nate Player ([email protected])  lOMoARcPSD|46385866 44 Subtracting (2) from (1) T - m2g = – m2a T – μ m1g = m1a μ m1g - m2g =– m2a - m1a Solving for a we get (𝑚2 −𝜇𝑚1 )g a = ……………………………………..….. (3) (𝑚2 + 𝑚1 ) Substituting (3) into (2) for a we get (𝑚1 𝑚2 g) (1+ μ) T= (𝑚2 + 𝑚1 ) …………………………………..… (4) Case II: Two bodies having freely and supported by a frictionless pulley. If m2 >m1 then the equations are: T- m1g = m1a……………………………..……………. (1) m2 g-T = m2a……………………………………..…….. (2) Adding (1) and (2) m2g –m1g = m1a+m2a and solving for a Downloaded by Nate Player ([email protected])  lOMoARcPSD|46385866 45 (m2-m1) g = a (m1 + m2) (𝑚2 −𝑚1 )g a= (𝑚1 +𝑚2 ) ………………………………….. (3) Substituting (3) into (1) and solving for T (𝑚2 −𝑚1 )g T= m1 𝑚1 +𝑚2 + m1g (2𝑚1 𝑚2 )g T= 𝑚1 +𝑚2 ……………………………………(4) Case III: One body lying on an inclined plane and the other hanging freely. For m2 > m1 T – m1 g sin θ = m1 a……………………………………..……. (1) M2 g – T = m2 a…………….…………………………………. (2) Adding (1) and (2) m2g – m1 sin θ = a (m1 + m2) Solving for a we get (𝑚2 −𝑚1 Sin θ)g a= ………………………………………..………(3) 𝑚1 +𝑚2 Substituting (3) into (1) and solving for T, we get (𝑚2 +𝑚1 𝑆𝑖𝑛𝜃)𝑔 T = m1[ (𝑚1 +𝑚2 ) ] + m1g sin θ 𝑚1 𝑚2 𝑔(1+𝑆𝑖𝑛 𝜃) T= ………………………………………. (4) 𝑚1 +𝑚2 Suppose the inclined plane is rough with coefficient of friction, μ, then T - m1gSinθ - Fr = m1a, But R = m1g Cos θ and Fr = μR T - m1gSinθ – μ m1g Cos θ = m1a,…………………………….. (5) Downloaded by Nate Player ([email protected])  lOMoARcPSD|46385866 46 Subtracting (5) from (2) T – m2g = -m2a T - m1gSinθ – μ m1g Cos θ = m1a m1gSinθ – m2g + μ m1g Cos θ = -m2a - m1a Solving for a we get (𝑚2 −𝑚1 𝑆𝑖𝑛 𝜃− 𝜇𝑚1 𝐶𝑜𝑠 𝜃)g a= …………………………………….. (6) 𝑚2 +𝑚1 Substituting (6) into (1) and solving for T, we get (1+ 𝑆𝑖𝑛 𝜃+ 𝜇 𝐶𝑜𝑠 𝜃)𝑚1 𝑚2 g T= 𝑚2 +𝑚1 ………………………………….. (7) Case IV: Two Bodies on two inclined planes for m1 >m2 ϕ ϕ ϕ We have T – m2g Sin θ = m2a……………………………………………………….. (1) m1g Sin 𝜙 – T = m1a……………………………………….…………..… (2) Adding (1) and (2) m1g Sin 𝜙 – m2g Sin θ = (m1+m2)a (𝑚1 Sin 𝜙−𝑚2 Sin θ )g a= …………………………………………..….(3) 𝑚1 +𝑚2 From (1) T=m2 a +m2 g Sin θ…………………………………………………….……(4) Substituting (4) into (3) (𝑚1 Sin 𝜙−𝑚2 Sin θ )g T = m2[ ]+m2 g Sin θ 𝑚1 +𝑚2 𝑚1 𝑚2 g (Sin 𝜙+Sin θ) T= 𝑚1 +𝑚2 ………………………………………………………. (5) Downloaded by Nate Player ([email protected])  lOMoARcPSD|46385866 47 Examples. 1) A wooden block of mass 2kg is pulled along a horizontal surface by light inextensible spring attached over frictionless pulley to a mass of 3kg hanging vertically. Calculate the acceleration of the system and tension in the string. Solution: GRAVITATION Keplers Law Kepless laws deal with a motion of planets 30 – T = 3a ………………………………………………….….. (1) T = 2a………………………………………………………….….(2) Adding (1) and (2) 3 0 =5a a= 6m/s2 3x10 T = 2x6 = 12N or a = =6m/s2 2+3 =2x6=12N 2) A light inextensible ship passes over a smooth fixed pulley. A mass of 0.3kg is attached to one of the spring while a mass of 0.2kg is attached to the other end. The system is held at rest then released so that the 0.3kg is descends. Calculate the acceleration of the two masses and the term in the spring (g = 10m/s2). Solution. (𝑚2 −𝑚1 )𝑔 0.1 𝑥 10 a= = = 2 m/s2 𝑚2 +𝑚1 0.3+0.2 2𝑚2 𝑚1 𝑔 2 𝑥 0.2 𝑥 0.3 𝑥 10 T= 𝑚2 +𝑚1 = 0.3+0.2 = 2.4N Downloaded by Nate Player ([email protected])  lOMoARcPSD|46385866 48 3) A mass of 3kg has on earth inclined plane and connected by a light inextensible string passing over a frictionless pulley at the top of the inclined plane to a mass of 5kg as shown. Calculate the acceleration and tension of the system. Take g = 10m/s2 Solution. (𝑚2 − 𝑚1 𝑆𝑖𝑛 𝜃)𝑔 (5−3 𝑆𝑖𝑛 30)10 a= = = 4.38m/s2 𝑚2 +𝑚1 5+3 𝑚1 𝑚2 𝑔(1 + 𝑆𝑖𝑛 𝜃) 5 𝑥 3 𝑥 10 (1−𝑆𝑖𝑛 30) T= = = 28.125N 𝑚2 +𝑚1 5+3 4.6: Uniform Circular Motion Introduction When a particle moves on a circular part with a constant speed, then its motion is said to be uniform circular motion in a plane. The magnitude of speed remains constant but direction changes. Downloaded by Nate Player ([email protected])  lOMoARcPSD|46385866 49 Angular Displacement (θ) Suppose a particle covers a distance ∆S along the circular path in the time interval ∆t = t2 – t1 It revolves through an angle ∆θ = θ2 – θ1 During the interval, the angle of revolution ∆θ is called angular displacement of the particle if ∆𝑠 r is radius of the circle then angular displacement ∆θ = 𝑟 𝐴𝑟𝑐−𝑙𝑒𝑛𝑔𝑡ℎ Angle = SI unit is the radian. 𝑅𝑎𝑑𝑖𝑢𝑠 If the arc-length = Radius of Circle = 1 radian Angular Velocity (ω) Angular velocity of a particle is given by the displacement per unit time 𝐴𝑛𝑔𝑢𝑎𝑟 𝑑𝑒𝑠𝑝𝑙𝑎𝑐𝑒𝑚𝑒𝑛𝑡 ω= 𝑇𝑖𝑚𝑒 𝑡𝑎𝑘𝑒 ∆𝜃 ∆𝜃 𝑑𝜃 ω= → ω= Lim ∆t →0 = SI unit radian per second. ∆𝑡 ∆𝑡 𝑑𝑡 If T is the time taken for one revolution 2𝜋 1 then, ω = , But f = 𝑇 𝑇 ω = 2πf Relationship between Angular Velocity and linear velocity The particle covers an arc-length ∆s in time ∆t, hence, ∆𝑠 Angular displacement ∆θ = , dividing through by ∆t 𝑟 ∆𝜃 ∆𝑠 ∆𝜃 1 ∆𝑠 𝑣 = , Lim ∆t →0 = (Lim ∆t →0 )= ∆𝑡 𝑟∆𝑡 ∆𝑡 𝑟 ∆𝑡 𝑟 Therefore v = r ω which is instantaneous linear velocity. Centripetal Acceleration (a). A particle is uniform circular motion changes direction hence velocity changes thus the body accelerates. The direction of acceleration is towards the centre thus centripetal petal acceleration. ∆𝑠 ∆𝑣 𝑣 𝑟 = → ∆v = ∆s, dividing on both sides by ∆t we have 𝑣 𝑟 ∆𝑣 𝑣∆𝑆 = , if ∆t is infitesmally small ∆t → 0, then. ∆𝑡 𝑟∆𝑡 ∆𝑣 𝑣 ∆𝑆 𝑣 𝑣2 Lim ∆t →0 ∆𝑡 = 𝑟 (Lim ∆t → 0 ∆𝑡 ) = 𝑟 𝑥𝑣 = 𝑟 Downloaded by Nate Player ([email protected])  lOMoARcPSD|46385866 50 𝑣2 a= But v = rω 𝑟 a = ω2 r Centripetal force (Fc) From Newton’s second law, acceleration is always produced by a force whose direction is the same as that of the acceleration. A body performing circular motion is acted upon by a force, which is always directed towards the centre of the circle. 𝑣2 Fc = F = ma = m = mω2r 𝑟 Examples of center petal forces 1) A car taking a turn requires a centripetal force. This is provided by the frictional force between the tyres and the road. 2) When a stone is tied on the end of a string is whirled on a circular path, the centripetal force provided by the tension in the string created by drawing the string inward. Cases of Circular Motion. Case I: Motion of a bicycle rider round circular track. When a person on bicycle rides around a circular racing track, the necessary centripetal 𝑚𝑣2 force is actually provided by the frictional force Fr at the ground. Fr has amoment about 𝑟 the c.o.g, G equals to F.h which tends to turn the rider outwards. When the rider leans inwards as shown, this is counter balanced by the moment R.a about G which is equal to mg.a thus provided no slipping occurs. Downloaded by Nate Player ([email protected])  lOMoARcPSD|46385866 51 F.h = mg.a 𝐹 a /h = tan θ = , where θ is the angle of inclination to the vertical, now, 𝑚𝑔 𝑚𝑣 2 F= 𝑟 =mg tanθ 𝑣2 Tanθ =𝑟𝑔 When F is greater than the limiting friction, skidding occurs. Case II: Motion of a Car Around a Circular Track Suppose a car is moving with velocity v around a horizontal circular track of radius r, and let R1 and R2 be respective normal re actions at the wheels A, B and F1, F2 the corresponding frictional forces. Then for circular motion we have F1 + F2 = mv2 and vertically R1 + R2 = mg, also taking moments at 𝐺, (F1 + F2 ) h + R1a – R2a = 0, where 2a is the distance between the wheels. G is midway between the wheels and h is the height of G above the ground. From the three equations we find, 𝑣2 ℎ 𝑣2 ℎ 𝑎𝑟𝑔 R2=1/2m (g + 𝑟𝑎 ) and vertically, R1=1/2m (g - ). If v2 = , R1 = 0 and the car 𝑟𝑎 ℎ1 𝑎𝑟g turning left is about to overturn outwards. R1 will be positive if v2< ℎ Case III: Motion of a Car Around A Banked Track. Centripetal force is required for a car to go round a bend on a level surface. The force is provided by the frictional force exerted on the tyres by the road. A suitable banked road removes the need to rely on friction. This provided, no slipping occurs. Downloaded by Nate Player ([email protected])  lOMoARcPSD|46385866 52 The normal reaction R, of the road on the car acquires a horizontal component, R Sin θ as a result of banking. Using Newton’s 2nd law of motion, F = ma, we have, R Sin θ = ma m𝑣 2 R Sin θ = …………………………………………..…….. (1) 𝑟 There is no vertical acceleration hence R Cos θ = mg ………………………….. (2) Dividing (1) by (2) 𝑣2 𝑣2 Tan θ = , The angle of banking θ is given by θ = tan -1 ( ) 𝑟𝑔 𝑟𝑔 Case IV: Conical Pendulum. If a pendulum bob moves in such a way that the string sweeps out a cone, then the bob describes a horizontal circle. Downloaded by Nate Player ([email protected])  lOMoARcPSD|46385866 53 There are two forces acting on the bob, its weight (mg) and tension in string (F) centripetal force is provided by the horizontal component of the tension. F Sin θ, hence from Newton’s 2nd law 𝑚𝑣 2 F = ma = 𝑟 𝑚𝑣 2 F Sinθ = ………………………………………………... (1) 𝑟 Since there is no vertical acceleration, F Cos θ = mg ……………………………………………………….. (2) Dividing (1) by (2) 𝐹𝑆𝑖𝑛𝜃 𝑚𝑣 2 /𝑟 Sin θ = , Tan θ = , Trigonometric identity. 𝐹𝐶𝑜𝑠𝜃 𝑚𝑔 𝐶𝑜𝑠 𝜃 𝑣2 Tan θ =𝑟𝑔 Also v = ωr 𝜔2 𝑟2 𝜔2𝑟 Tanθ = 𝑟𝑔 = 𝑔 𝜔2 𝑟 𝜔 2 𝑙 𝑆𝑖𝑛 𝜃 Also r = l Sin θ, hence Tan θ = = 𝑔 𝑔 𝑆𝑖𝑛𝜃 𝜔 2 𝑙 𝑆𝑖𝑛 𝜃 = 𝐶𝑜𝑠𝜃 𝑔 𝑔 2π Therefore ω = √ , Using equation T = 𝑙 𝐶𝑜𝑠 𝜃 𝜔 𝑙 𝐶𝑜𝑠 𝜃 T = 2π √ 𝑔 Case V: Motion on a Vertical Circle. Downloaded by Nate Player ([email protected])  lOMoARcPSD|46385866 54 The string experiences the highest tension when at C 𝑚𝑣 2 Tc = 𝑟 + mg …………………………………………………. (1) The string has the lowest tension at A 𝑚𝑣 2 TA = 𝑟 - mg …………………………………………………. (2) The tension at B and D is equal to centripetal force 𝑚𝑣 2 TB = TD = 𝑟 ………………………………………………….. (3) Summary ∆𝑠 1) ∆θ = 𝑟 2𝜋 2) ω = 𝑇 = 2πf = v/r 3) v = rω 𝑣2 4) a = 𝑟 5) a = ω2 r 𝑣2 6) F = m = mω2r 𝑟 𝑣2 7) Tan θ =𝑟𝑔 𝑣2 ℎ 8) R2=1/2m (g + ) 𝑟𝑎 𝑣2 ℎ 9) R1=1/2m (g - ). 𝑟𝑎 √𝑙𝐶𝑜𝑠𝜃 10) T = 2π=. 𝑔 𝑚𝑣 2 11) Tbottom = + mg. 𝑟 𝑚𝑣 2 12) Ttop = - mg. 𝑟 Downloaded by Nate Player ([email protected])  lOMoARcPSD|46385866 55 5.0: SIMPLE HAMORNIC MOTION Simple Harmonic motion (S.H.M) is defined as a periodic motion in which the restore force is proportional to the displacement and in the opposite direction F = - kx or S.H.M is motion change of a particle when directed towards a fixed point and its directly proportional to distance from fixed point. Fixed point is equilibrium position or rest position. Periodic motion is motion that repeats itself after a fixed time interval. Period (T) is the time during which spherical system completes one full cycle of motion. Frequency (f) is the number of complete cycles per second T = 1/f Displacement (x) is the distance of a moving object from its equilibrium position at any instant of time. Amplitude (x0) is the maximum value of displacement. A system vibrates /oscillates if it satisfies the following conditions; 1. Should be able to store P.E (should have elastic properties ) 2. Should have some inertia which enables it to possess K.E. The motion of a body is described by the equation of the form 𝑑2 𝑥 𝑑2 𝑥 = -ω2x where = 𝑎 where a is acceleration of the body. 𝑑𝑡 2 𝑑𝑡 2 The negative sign indicates that the acceleration is always directed towards the equilibrium position. For a body moving in a circular path, the linear displacement x is displaced by an angular displacement θ and linear acceleration: 𝑑2 𝑥 𝑑2 𝜃 𝑑𝑡 2 is replaced by 𝑑𝑡 2 𝑑2 𝑥 𝑑2 𝜃 = = -ω2θ 𝑑𝑡 2 𝑑𝑡 2 2𝜋 Period T = , where ω is angular velocity. 𝜔 N/B: Period is independent of amplitude 1 𝜔 Frequency (f), f = = 𝑇 2𝜋 Downloaded by Nate Player ([email protected])  lOMoARcPSD|46385866 56 Phase It expresses position and direction of motion of the vibrating particle at any instant. Displacement is given by y = y0 Sin (ω t + 𝜙 ) 𝜙 – Phase angle Time Example of S.H.M 1. Simple pendulum 2. Mass on a helical spring 3. Vibration of the prongs of the tuning fork. 4. Liquid in a U –tube 5. Oscillation of atoms in a molecule 5.1: Relationship between S.H.M and circular motion Case I: Consider a small ball rotating at constant speed in a vertical circle of radius r. we are interested in the x component of this circular motion. To find this, we might imagine a large spotlight shining down on the rotating ball from the top of the page and projecting on the x axis a shadow of the ball. As the ball rotates through 3600 or 2π radians, its shadow goes from +x0 to –x0 and back again to +x0 as shown below. If we plot the position of the ball’s shadow as a function of angle θ swept out by the rotating ball, we find that x executes a perfect cosine curve. This has to be the case because the cosine of θ is the ratio of the adjacent side (x) to the hypotenuse r. Since r is constant and numerically equal to x0, as the ball rotates its shadow along x axis is always proportional to Cosine θ hence, 𝑥 Cos θ = or x = r Cos θ = x0 Cos θ, 𝑟 But θ = ωt = 2π ft. Hence x = x0 Cos θ = x0 Cos ωt = x0 Cos 2π ft. The period of the projected motion along the x axis is the same as the period of the circular motion and both complete one cycle in the same time, i.e. 1 2𝜋 T= = 𝑓 𝜔 Downloaded by Nate Player ([email protected])  lOMoARcPSD|46385866 57 Case II: Consider a particle moving with a uniform speed along a circumference of a circle as shown below. When particle is at point P, the foot of the perpendicular drawn from P to the diameter AA1 of the circle is at point N. As the particle moves around the circle, the foot N moves on a straight line to and fro about point O. The motion of N can be projected as shown above. The to and fro motion of N along AOA1 is called S.H.M Note: The period and hence frequency of rotation of P is equal to the period and 2𝜋 frequency of oscillation of N which is 𝜔 Displacement Equation of S.H.M. 𝜃 At P, ω= , the displacement of the foot of N from O is ON = Y = OP Sin θ t But OP which is the radius of circular path y = r Sin θ, θ = ωt y = r Sin ωt, y = S S = r Sin ωt ……………………………………………….(1) Downloaded by Nate Player ([email protected])  lOMoARcPSD|46385866 58 Velocity Equation for S.H.M. 𝑑𝑆 𝑑(𝑟 𝑆𝑖𝑛𝜔𝑡) 𝑑(𝑆𝑖𝑛 𝑎𝑥) v= , = rω Cos ωt Recall = a Cos ax Standard derivative 𝑑𝑡 𝑑𝑡 𝑑𝑥 v = rω Cos ωt…………………………………………………. (2) But NP = r Cos θ = r Cos ωt = √𝑟 2 – 𝑦 2 v = ω√𝑟 2 – 𝑦 2………………………………………………… (3) at y = O, v = ±ωr, (velocity in circular path) At y = r, v = ω x 0 Hence v = 0 Acceleration Equation in S. H. M. 𝑑𝑣 𝑑2 S a= = 𝑑𝑡 𝑑𝑡 2 𝑑(𝐶𝑜𝑠𝜔𝑡) 𝑑(𝐶𝑜𝑠 𝑎𝑥) a = rω = -rω2 Sinωt, Recall = -a Sin ax Standard derivative. 𝑑𝑡 𝑑𝑥 Since y = r Sin ωt, Then a = -ω2y…………………………………….. (4) Using (4), acceleration is proportional to displacement and in opposite direction. At y = 0, a = 0 and At y = r, a = ω2r 5.2: Determination of g Using a Simple pendulum Suppose the mass m is displaced slightly through an angle θ and l is the length of the string at A, the forces acting on the mass m are i) Weight, mg ii) Tension in the string, T = mg Cos θ Components of weight along x – direction = mg sin θ and is the force that tends to restore the bob to the equilibrium point O Thus the restoring force is F = - mg Sin θ The negative sign is due to the fact that the force is opposite to the angular displacement. Downloaded by Nate Player ([email protected])  lOMoARcPSD|46385866 59 For smaller angles of θ, where θ is in radians Sin θ ≃ θ = arc length/radius = r/l. Hence F = -mg Sin θ = -mg r/l If a is the acceleration of m towards O, then by Newton’s second law motion. 𝑟 𝑟 F = ma= -mg , a= - g also a = -ω2y where 𝑙 𝑙 𝑔 ω2y = g r/l for y = r, ω2 = 𝑙 𝑔 2𝜋 2𝜋 Hence ω2 =√ But T = ;w= 𝑙 𝜔 𝑇 2𝜋 𝑔 4𝜋 2 𝑔 4𝜋 2 =√ , = , T2 = xl 𝑇 𝑙 𝑇2 𝑙 𝑔 𝑙 T = 2π√ , Equation for period of a pendulum in S. H. M. 𝑔 By measuring l and determining corresponding values of T at small angles of 𝑙 displacement, g can be evaluated by using T = 2π√. A graph of T2 against l has the 𝑔 gradient given by 4𝜋 2 4𝜋 2 Gradient/slope =. Hence g = , where s is slope/gradient. 𝑔 𝑆 Note: If we have a spring of mass ms with a mass of mass m suspended on it then the body oscillates and the period of the system is given by (𝑚𝑠 +m) 1/2 T = 2π [ ] K Downloaded by Nate Player ([email protected])  lOMoARcPSD|46385866 60 5.3: Motion of a body suspended on a spring. Oscillating system –spring and mass. When the spring is stretched through x and released then it oscillates as shown above. Suppose the extension x of the spring is stretching proportional to F in spring (Hookes law). F acts in opposite direction to x so F= - kx where F is force constant force per unit extension called spring stiffness. If m is the mass of the body then acceleration a is given by F = ma ma = - kx −𝑘 a= 𝑥 But a = - ω2x 𝑚 −𝑘 𝑎= 𝑥 = - ω2 x 𝑚 𝑘 2𝜋 𝑚 = ω2 But ω= 𝑇 𝑘 4𝜋 2 4𝜋 2 𝑚 𝑚 = 2 , T2= , 𝑇 = 2𝜋√ , 𝑚 𝑇 𝑘 𝑘 𝑚 𝑇 = 2𝜋√ Equation for period of oscillating spring in S. H. M. 𝑘 𝑚𝑔 𝑒 Also k = , Therefore 𝑇 = 2𝜋√ 𝑒 𝑔 Downloaded by Nate Player ([email protected])  lOMoARcPSD|46385866 61 5.4: Oscillations of a liquid in a U-tube If the liquid on one side of a U-tube T is depressed by blowing gently down that side, the levels of the liquid will oscillate for short time about their respective initial positions O, before finally coming to rest. At some instant, suppose that the level of the liquid on the left side of T is at D, at a height x above its original (undisturbed) position O. The level B of the liquid on the other side is then at a depth x below its original position C. So excess pressure on the whole liquid is = excess height x liquid density x g = 2xρg Since pressure = force per unit area Force on liquid = pressure x area of cross section of the tube = 2xρg x A Mass of a liquid in U-tube =Volume of liquid x density = 2h Aρ where 2h is total length of liquid at T From F= ma where a is acceleration towards O or C then -2xρgA=2hAρa 𝑔𝑥 a= - = -ω2x ℎ 𝑔 2𝜋 ℎ = ω 2 for ω = 𝑇 ℎ Then T2=4π2 𝑔 Downloaded by Nate Player ([email protected])  lOMoARcPSD|46385866 62 ℎ T= 2π√𝑔 Equation for period of oscillating liquid in U-Tube in S. H. M. 5.5: Conservation of mechanical energy in S.H.M If E is the total energy of S.H.M, the we have E=K.E +P.E is constant. Consider a mass m executing S.H.M K.E=1/2 mv2=1/2mω2 [√(𝑟 2 − 𝑦 2 ) ] 2 =1/2 mω2(r2-y2)…………………………….… (1) P.E for S.H.M. 𝑑2 y a= = -ω2y………………………………………..… (2) 𝑑𝑡 2 Drawing graphs of x vsy, the area under curve is equal to P.E of m y A O B x Area = ½ (AB x OB) = 1/2 Fy = ½ mω2y2 Total energy = 1/2mω2y2 + mω2(r2-y2) = 1/2mω2r2 Total energy, E is independent of time When y = r, E is the form of P.E When y = o, E is the form of K. E The variation of P.E with K.E with displacement is shown below, Downloaded by Nate Player ([email protected])  lOMoARcPSD|46385866 63 Summary 1. a = -ω2x the minus sign shows that a is always directed towards a fixed point when at distance x from centre of oscillation. 2. T= 2π/ ω, ω =2πf 3. v = ω√(r2-y2) 4. vmax = ±ωr 5. amax = -ω2r 6. K.Emax = ½mr2ω2 ℎ 7. T= 2π√ 𝑔 𝑚 𝑚𝑔 𝑒 8. 𝑇 = 2𝜋√ Also k = , Therefore ?