J3010   Unit 1
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J3010 - Mechanics of Machines 1

J3010 - Mechanics of Machines 1

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  • 1. GENERAL DYNAMICS J3010/1/1 UNIT 1 GENERAL DYNAMICS OBJECTIVES General Objective : To understand the concept of general dynamics Specific Objectives : At the end of this unit you should be able to :  relate linear and angular velocity, linear and angular acceleration.  solve problem using equation uniformly accelerated and angular motion.  describe tangent acceleration and centripetal acceleration, centripetal and centrifugal force, work and power.  explain the principle conservation of energy and momentum. .
  • 2. GENERAL DYNAMICS J3010/1/2 INPUT 1.0 INTRODUCTION. A vector quantity requires a number and a direction to specify it completely; that is, a vector has magnitude and direction. Examples of vectors are velocity, acceleration and force. Mechanics is the study of object or bodies, as we shall call them, when subjected to force. 1.1 VELOCITY AND ACCELERATION Velocity The velocity of a body may be defined as its rate of change of displacement, with respect to its surroundings, in a particular direction. As the velocity is always expressed in a particular direction, it is also a vector quantity. Acceleration The acceleration of a body may be defined as the rate of change of its velocity. It is said to be positive, when the change in velocity of a body increases with time, and negative when the velocity decreases with time. The negative acceleration is also called retardation.
  • 3. GENERAL DYNAMICS J3010/1/3 In general, the term acceleration is used to denote the rate at which the velocity is changing. It may be uniform or variable. 1.2 EQUATION FOR LINEAR, UNIFORMLY ACCELERATED MOTION Suppose a body moving in a straight line has an initial speed u and that it undergoes uniform acceleration a for time t considering, let the final speed be v and the distance traveled in the time t be s. The speed –time curve will be show in fig.1.1 Fig. 1.1 Uniform – accelerated linear motion. Acceleration a is uniform, its magnitude is change in speed a = change in time = v–u/t or at = v – u or v = u + at (1.1) In this case, the average speed will be the speed at time t/2. Hence average speed = 1 ( u + v) 2 Further, since distance travel = average speed x t then s = 1 ( u + v)t (1.2) 2 Substituting for v from (1.1) into (1.2) gives s = 1 ( u + u + at )/ t 2 Or s = u t + 1 at2 2
  • 4. GENERAL DYNAMICS J3010/1/4 Substituting for t from (1.1) into (1.2) gives s = 1 ( u + v ) (v - u )/ a 2 or 2as = v - u2 2 or v2 = u2 + 2as Example 1.1 A workman drops a hammer from the top of scaffolding. If the speed of sound in air is 340 m/s, how long does the workman have before shouting to another workman 60 m vertically below him if his warning is to arrive before the hammer. Neglect air resistance. Solution 1.1 For hammer Inertial speed = 0 m/s. Acceleration = 9.81 m/s2. Distance = 60 m s = ut + 1 at2 2 60 = 0 + ½ (9.81)2 t = 3.50s 60 The hammer takes 3.50 s to fall 60 m. The sound takes = 0.18 to travel the same 340 distance so the workman has (3.50 – 0.18) = 3.32 s before shouting if the sound is to arrive before the hammer.
  • 5. GENERAL DYNAMICS J3010/1/5 1.3 RELATIONSHIP BETWEEN LINEAR SPEED AND ANGULAR SPEED If a point P moves round in a circle with a of radius r with constant linear speed v then the angular speed  will be constant and   = t (1.3) where t is the time to move from Q to P along the arc QP of the curve. Fig 1.2 Fig. 1.2 Circular motion However, arc length QP is r  when  is measured in radians and hence linear speed v is arc QP r V= = (1.4) t t Using (1.3) and (1.4) leads to V = r  for circular motion. (1.5) Or linear speed = radius x angular speed. Example 1.2 What is the peripheral speed of the tread on a tire of a motor car if the wheel spins about the axle with an angular velocity of 6 radian/ second. Diameter of tires is 0.7 m. Solution 1.2 V = r where r = 0.35 m and  = 6 rad/s = 0.35 x 6 V = 2.1 m/s
  • 6. GENERAL DYNAMICS J3010/1/6 1.4 RELATIONSHIP BETWEEN LINEAR ACCELERATION AND ANGULAR ACCELERATION d By equation   and V = r  dt d 1 dv Hence    since r is constant  v  r dt dt   r dv However is linear acceleration a dt a Hence  = r Or a = r  Or linear acceleration = radius x angular acceleration Example 1.3 A grinding wheel is accelerated uniformly from rest to 3000 rev/min in 3 seconds. Find it angular and linear acceleration. If the wheel diameter is 200 mm, find the final linear speed of a point on its rim. Solution 1.3 t = 3s  1 = 0 rad/s  2 = 3000 rev/min = 2πN/60 2 x  x 3000 = 60 = 314.16 rad/s  2  1  = t 314.16  0 = 3  = 104.72 rad/s2
  • 7. GENERAL DYNAMICS J3010/1/7 a = r = 0.1 x 104.72 a = 10.47 m/s2 v = r 2 = 0.1 x 314.16 = 31.42 m/s
  • 8. GENERAL DYNAMICS J3010/1/8 Activity 1A TEST YOUR UNDERSTANDING BEFORE YOU CONTINUE WITH THE NEXT INPUT…! 1.1 Which of the following is a vector quantity?. A. density B. speed C. area D. acceleration 1.2 Velocity is the rate of change with time of A. displacement B. acceleration C. speed D. distance 1.3 When a body moves round a circle with radius r at uniform speed v, the angular speed ω is A. vr B. v/r C. v2/r D. 2πv 1.4 A 5 kg block, at rest on a smooth horizontal surface, is acted on by a resultant force of 2.5 N parallel to the surface. The acceleration, in m/s2 is A. 0.5 B. 2 C. 12.5 D. 2000 1.5 A car travel along a straight road at a steady speed of 13 m/s, accelerates uniformly for 15 s until it is moving at 25 m/s. Find its acceleration. 1.6 A particle moves from rest with an acceleration of 2 m/s2. Determine the velocity and displacement of the particle after 20 seconds. 1.7 A parcel, starting at rest, is moved by conveyor belt with an acceleration of 1.5 m/s2. What will its velocity be after it has moved 3 meters?
  • 9. GENERAL DYNAMICS J3010/1/9 S = ½ ( u + v) t V = u + at V2 = u2 + 2as S = ut + ½ at. Feedback to Activity 1A Have you tried the questions????? If “YES”, check your answers now 1.1 D. acceleration 1.2 A. displacement 1.3 B. v/r 1.4 A. 0.5 1.5 0.8 m/s2 1.6 40 m/s; 400 m. 1.7 3 m/s.
  • 10. GENERAL DYNAMICS J3010/1/10 INPUT 1.5 WORK DONE BY A CONSTANT FORCE Work done = Force x distance. Unit of work is joule (J) or Kilojoule (kJ) When the point at which a force acts moves, the force is said to have done work. When the force is constant, the work done is defined as work done = force x distance moved in the direction of the force. It is a scalar quantity. If a constant force F moves a body from A to B then distance moved in the direction of F is s cos  fig.1.3. The work done by a constant force is thus: Work done = F s cos  fig. 1.3 Notation for the work done
  • 11. GENERAL DYNAMICS J3010/1/11 If the body moves in the same direction as the force, where by  = 00 and work done is Fs. Work done is zero if direction force  = 900. If F is in Newton and s is in meters, the work done will be measured in joules (J) Example 1.4 How much work is done when a force of 5 KN moves its point of application 600 mm in the direction of the force. Solution. 1.4 Work done = force x distance = 5 X 103 X 600 X10-3 = 3000 J = 3 KJ. 1.6 POWER Power is the rate of doing work, i.e. the work done in unit time. The SI unit of power is the watt; it is 1 joule per second and is written 1 W. The British unit of power use earlier was the Horse- Power, and is equivalent to about 746 watts . If a force of F Newton keeps its point of application moving in the direction of the force with uniform speed v meters per second, the work done per second is Fv joules, and is the power is Fv watts. Example 1.5 The total mass of an engine and train is 200 Mg; what is the power of the engine if it can just keep the train moving at a uniform speed of 100 km/h1 on the level, the 1 resistances to friction, amounting to of the weight of the train. 200
  • 12. GENERAL DYNAMICS J3010/1/12 Solution 1.5 Since the speed is uniform, the pull of the engine is equal to the total resistance, i.e. 1000g N.= 1000 x 9.81 = 9810 N. The speed is 100 km/h1 = 1000/36 m/s The work per second = 1000 x 9.8 x 1000/36 J Power = 105 x 2.72 W = 272 kW 1.7 ENERGY The energy may be defined as the capacity to do work. It exists in many form e.g. Mechanical, electrical, chemical, heat, light etc. But in applied Mechanics, we shall deal in Mechanical Energy only. The unit for energy is the same as those of work i.e. example joules. 1.7.1 CONSERVATION OF ENERGY Energy cannot be created or destroyed but can be transformed from one to another form of energy. For instance water stored in a dam possesses potential energy which changes to kinetic energy as it flows downwards through a tunnel to turn turbines, which in turn changes to electric energy which can be used to produce heat energy. 1.7.2 POTENTIAL ENERGY The potential energy of a body may be defined as the amount of work it can do when it moves from its actual position to the standard position chosen. The work done lifting a load of mass M and weight W = Mg through a height h is Wh. This is known as the potential energy of the load referred to its original position and its unit in that energy, i.e. the basic unit is the joule (J). Potential energy = Wh = Mgh (zero at earth’s surface)
  • 13. GENERAL DYNAMICS J3010/1/13 Example 1.6 What is the potential energy of a 10 kg mass? (a) 100 m above the surface of the earth. (b) at the bottom of a vertical mine shaft 1000 m deep. Solution 1.6 (a) Potential energy = mgh = 10 x 9.81 x 100 J = 9.81 KJ. (b) Potential Energy = -10 x 9.81 x 1000 J = -9.81 x 104 = -98.1 KJ. 1.7.3 KINETIC ENERGY A body may possess energy due to its motion as well as due to its position. For example, when a hammer is used to drive in a nail, work is done on the nail by the hammer, hence it must have possessed energy. Also a rotating flywheel possess energy due its motion. These are example of the form of energy call kinetic energy. Kinetic energy may be described as energy due to motion. Only linear motion will be considered. The kinetic energy of a body may be defined as the amount of work it can do before being brought to rest. 1.7.4 FORMULA FOR KINETICS ENERGY Let a body of mass m moving with a speed v be brought to rest with a uniform retardation by constant force P in a distance s.
  • 14. GENERAL DYNAMICS J3010/1/14 v2 = u2 + 2 as 0 = v2 - 2 as since a is negative 2 or s = v 2a work done = force x distance = Ps 2 = Ps 2a However P = ma And Hence Work done = mav2/2a = ½ mv2 The kinetic energy is thus given by Kinetic energy = ½ mv2 1.7.5 STRAIN ENERGY The work done in compressing or stretching a spring is stored as strain energy in the spring provided that there is no permanent deformation (over stretching). The stiffness of a spring is the load per unit extension and is approximately constant within the working range of the spring; thus if S is the stiffness, the load P required to produce an extension x is given by P=Sx Suppose a load is gradually applied to a spring so that it varies from zero to maximum value P and produce a maximum extension x. Then Work done = average load x extension =½Pxx = ½ Sx x x = ½ Sx2 since strain energy U = Work done Thus U= ½Px = ½ Sx2 The units of strain energy are same as those of work, i.e. joules (J)
  • 15. GENERAL DYNAMICS J3010/1/15 Example 1.7 A wagon of mass 12 tone traveling at 16 km/h strikes a pair of parallel spring-loaded stops. If the stiffness of each spring is 600 KN/m, calculate the maximum compression in bringing the wagon to rest. Solution 1.7 16 m V = 16 km/h = 3.6 s Kinetic energy of wagon = ½ Mv2 2  16  = ½ x 12 x 1000 x    3. 6  = 118,500 J This kinetic energy may be assumed to be absorbed equally by the two springs. Strain energy stored per spring is ½ x 118,500 = 59,250 J Thus x is the maximum compression of the springs, ½ Sx2 = 59,250 Or ½ x 600 x 1,000 x2 = 59,250 x = 0.446 m = 446 mm 1.8 MOMENTUM AND CONSERVATION OF MOMENTUM 1.8.1 MOMENTUM The momentum of a particle is the product of the mass of the particle and its velocity. If m is the mass of the particle and v its velocity the momentum is m v. The unit of momentum is equivalent, i.e. Ns = kg m/s.
  • 16. GENERAL DYNAMICS J3010/1/16 1.8.2 CONSERVATION OF MOMENTUM If two bodies collide then the sum of the momentum before the collision is equal to the sum of the momentum after collision measured in the same direction. m1u1 + m2u2 = m1v1 + m2v2 Where m1 = mass of the first body m2 = mass of the second body u1 = initial velocity of the first body u2 = initial velocity of the second body v1 = final velocity of the first body v2 = final velocity of the second body Example 1.8 A 750 kg car collided head on with a 1 tone car. If both cars are travel at 16 km/h at the time of impact and after impact the second car rebounds at 3 km/h, find the velocity of the first car after collision (assume perfect elastic collision) Solution 1.8 By the conservation of momentum and assuming that the first car also rebound. + m1u1 + m2u2 = m1v1 + m2v2 ﴾750 x (+16)﴿ + ﴾ 100 x (-16) ) = ﴾ 750 x (-v1) ﴿ + ( 1000 + ( +3) ) 12 x 103 – 16 x 103 = -750 v1 + 3 x 103 (12 – 16 – 3) x 103 = -750 v1 -7 x 103 = -750 v1 7000 v1 = 750 v1 = 9.333 km/h Where m1= 750 kg ; m2 = 1 tone = 1000 kg ; u1= + 16 km/h ; u2 = -16 km/h; v2 = + 3km/h
  • 17. GENERAL DYNAMICS J3010/1/17 Activity 1B TEST YOUR UNDERSTANDING BEFORE YOU CONTINUE WITH THE NEXT INPUT…! 1.8 A flywheel rotating at 1200 rev/min slow down at a constant rate of 900 rev/min in 30 seconds. Find: a. the initial angular speed b. the final angular speed c. the angular acceleration d. the initial speed of a point on the rim of the flywheel if its diameter is 1.1 m. 1.9 A constant force of 2 kN pulls a crate along a level floor for a distance of 10 m in 50 seconds. What power was used?. 1.10 A car of mass 1000 kg traveling at 30 m/s has its speed reduced to 10 m/s by constant breaking force over a distance of 75 meter. Find the initial and final kinetic energy and the breaking force.
  • 18. GENERAL DYNAMICS J3010/1/18 work done Power = = Fv time taken Feedback to Activity 1B Have you tried the questions????? If “YES”, check your answers now 1.8 a. 125.7 rad/s b. 94.2 rad/s c. -1.05 rad/s2 d. 69.1 m/s 1.9 400 W 1.10 0.5 x 105 J, 5333 N
  • 19. GENERAL DYNAMICS J3010/1/19 SELF-ASSESSMENT 1 You are approaching success. Try all the questions in this self-assessment section and check your answers with those given in the Feedback on Self-Assessment 1 given on the next page. If you face any problems, discuss it with your lecturer. Good luck. 1. The spin drier in a washing machine is a cylinder with a diameter of 500 mm. It spins at 900 rev/min. Find the speed and acceleration of a point on the side of the drum. 2. Find the work done in raising 100 kg of water through a vertical distance of 3 m. 3. A cyclist, with his bicycle, has a total mass 80 kg. He reaches the top of the hill, with a slope 1 in 2 measured along the slope, at a speed of 2 m/s. He then free-wheels to the bottom of the hill where his speed has increased to 9 m/s. How much energy has been lost on the hill which is100 m long? 4. An electric motor is rated at 400 W. If its efficiency is 80%, find the maximum torque which it can exert when running at 2850 rev/min. 5. The engine of a car has a power output of 42 KW. It can achieve a maximum speed of 120 km/h along the level. Find the resistance to motion. If the power output and resistance remained the same, what would be the maximum speed a car could achieve up an incline of 1 in 40 along the slope if the car mass is 900 kg?
  • 20. GENERAL DYNAMICS J3010/1/20 Feedback to Self-Assessment 1 Have you tried the questions????? If “YES”, check your answers now. 1. 23.6 m/s; 2230 m/s2 2. 2943 J 3. 844 J CONGRATULATIONS!!!!….. May success be with you 4. 1.07 Nm always…. 5. 1260 N; 102 km/h