Kinetics of motion

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 24 Theory of Machines 3 Features Kinetics of 1. Introduction. 2. Newton's Laws of Motion. 3. Mass and Weight. 4. Momentum. Motion 5. Force. 6. Absolute and Gravitational Units of Force. 3.1. Introduction 7. Moment of a Force. 8. Couple. In the previous chapter we have discussed the 9. Centripetal and Centrifugal kinematics of motion, i.e. the motion without considering Force. the forces causing the motion. Here we shall discuss the 10. Mass Moment of Inertia. kinetics of motion, i.e. the motion which takes into 11. Angular Momentum or consideration the forces or other factors, e.g. mass or weight Moment of Momentum. of the bodies. The force and motion is governed by the three 12. Torque. laws of motion. 13. Work. 3.2. Newton’s Laws of Motion 14. Power. 15. Energy. Newton has formulated three laws of motion, which 16. Principle of Conservation of are the basic postulates or assumptions on which the whole Energy. system of kinetics is based. Like other scientific laws, these 17. Impulse and Impulsive Force. are also justified as the results, so obtained, agree with the 18. Principle of Conservation of actual observations. These three laws of motion are as Momentum. follows: 19. Energy Lost by Friction 1. Newton’s First Law of Motion. It states, “Every Clutch During Engagement. body continues in its state of rest or of uniform motion in 20. Torque Required to a straight line, unless acted upon by some external force.” Accelerate a Geared System. This is also known as Law of Inertia. 21. Collision of Two Bodies. The inertia is that property of a matter, by virtue of 22. Collision of Inelastic Bodies. which a body cannot move of itself, nor change the motion 23. Collision of Elastic Bodies. imparted to it. 24. Loss of Kinetic Energy During Elastic Impact. 24 Chapter 3 : Kinetics of Motion 25 2. Newton’s Second Law of Motion. It states, “The rate of change of momentum is directly proportional to the impressed force and takes place in the same direction in which the force acts.” 3. Newton’s Third Law of Motion. It states, “To every action, there is always an equal and opposite reaction.” 3.3. Mass and Weight Sometimes much confu-sion and misunder- standing is created, while using the various systems of units in the measurements of force and mass. This happens because of the lack of clear understanding of the difference between the mass and the weight. The following definitions of mass and weight should be clearly understood : The above picture shows space shuttle. 1. Mass. It is the amount of matter contained in a All space vehicles move based on given body, and does not vary with the change in its Newton’s third laws. position on the earth's surface. The mass of a body is measured by direct comparison with a standard mass by using a lever balance. 2. Weight. It is the amount of pull, which the earth exerts upon a given body. Since the pull varies with distance of the body from the centre of the earth, therefore the weight of the body will vary with its position on the earth’s surface (say latitude and elevation). It is thus obvious, that the weight is a force. The earth’s pull in metric units at sea level and 45° latitude has been adopted as one force unit and named as one kilogram of force. Thus, it is a definite amount of force. But, unfor- tunately, it has the same name as the unit of mass. The weight of a body is measured by the use of a spring balance which indicates the varying ten- sion in the spring as the body is moved from place to place. Note: The confusion in the units of mass and weight is eliminated, to a great extent, in S.I. units. In this system, the mass is taken in kg and force in newtons. The relation between the mass (m) and the weight (W) of a body is W = m.g or m = W/g where W is in newtons, m is in kg and g is acceleration due to gravity. 3.4. Momentum It is the total motion possessed by a body. Mathematically, Momentum = Mass × Velocity Let m = Mass of the body, u = Initial velocity of the body, v = Final velocity of the body, a = Constant acceleration, and t = Time required (in seconds) to change the velocity from u to v. 26 Theory of Machines Now, initial momentum = m.u and final momentum = m.v ∴ Change of momentum = m.v – m.u mv..−− mu m(v u) vu −  and rate of change of momentum = == ma . ... ∵ = a  tt t  3.5. Force It is an important factor in the field of Engineering-science, which may be defined as an agent, which produces or tends to produce, destroy or tends to destroy motion. W, weight (force) applied force, F f, friction force N, normal force According to Newton’s Second Law of Motion, the applied force or impressed force is directly proportional to the rate of change of momentum. We have discussed in Art. 3.4, that the rate of change of momentum = m.a where m = Mass of the body, and a = Acceleration of the body. ∴ Force , F ∝ m.a or F = k.m.a where k is a constant of proportionality. For the sake of convenience, the unit of force adopted is such that it produces a unit acceleration to a body of unit mass. ∴ F = m.a = Mass × Acceleration In S.I. system of units, the unit of force is called newton (briefly written as N). A newton may be defined as the force while acting upon a mass of one kg produces an acceleration of 2 1 m/s in the direction of which it acts. Thus 2 2 1 N = 1 kg × 1 m/s = 1 kg-m/s Note: A force equal in magnitude but opposite in direction and collinear with the impressed force producing the acceleration, is known as inertia force. Mathematically, Inertia force = – m.a 3.6. Absolute and Gravitational Units of Force We have already discussed, that when a body of mass 1 kg is moving with an acceleration of 2 1 m/s , the force acting on the body is one newton (briefly written as N). Therefore, when the same 2 body is moving with an acceleration of 9.81 m/s , the force acting on the body is 9.81 newtons. But 2 we denote 1 kg mass, attracted towards the earth with an acceleration of 9.81 m/s as 1 kilogram- force (briefly written as kgf) or 1 kilogram-weight (briefly written as kg-wt). It is thus obvious that 2 2 2 1 kgf = 1 kg × 9.81 m/s = 9.81 kg-m/s = 9.81 N ... (∵ 1 N = 1 kg-m/s ) The above unit of force i.e. kilogram-force (kgf ) is called gravitational or engineer's unit Chapter 3 : Kinetics of Motion 27 of force, whereas newton is the absolute or scientific or S.I. unit of force. It is thus obvious, that the gravitational units are ‘g’ times the unit of force in the absolute or S.I. units. It will be interesting to know that the mass of a body in absolute units is numerically equal to the weight of the same body in gravitational units. For example, consider a body whose mass, m = 100 kg. ∴ The force, with which it will be attracted towards the centre of the earth, F = m.a = m.g = 100 × 9.81 = 981 N Now, as per definition, we know that the weight of a body is the force, by which it is attracted towards the centre of the earth. Therefore, weight of the body, W = 981 N = 981 / 9.81 = 100 kgf ... (∵ 1 kgf = 9.81 N) In brief, the weight of a body of mass m kg at a place where gravitational acceleration is ‘g’ 2 m/s is m.g newtons. 3.7. Moment of a Force It is the turning effect produced by a force, on the body, on which it acts. The moment of a force is equal to the product of the force and the perpendicular distance of the point about which the moment is required, and the line of action of the force. Mathematically, Moment of a force = F × l where F = Force acting on the body, and l = Perpendicular distance of the point and the line of action of the force, as shown in Fig. 3.1. Fig. 3.1. Moment of a force. 3.8. Couple The two equal and opposite parallel forces, whose lines of action are different, form a couple, as shown in Fig. 3.2. The perpendicular distance (x) between the lines of action of two equal and opposite parallel forces (F) is known as arm of the couple. The magnitude of the couple (i.e. moment of a couple) is the product of one of the forces and the arm of the couple. Fig. 3.2. Couple. Mathematically, Moment of a couple = F × x A little consideration will show, that a couple does not produce any translatory motion (i.e. motion in a straight line). But, a couple produces a motion of rota- tion of the body, on which it acts. 3.9. Centripetal and Centrifugal Force Consider a particle of mass m moving with a linear velocity v in a circular path of radius r. We have seen in Art. 2.19 that the centripetal acceleration, 2 2 a = v /r = ω .r c and Force = Mass × Acceleration ∴ Centripetal force = Mass × Centripetal acceleration 2 2 or F = m.v /r = m.ω .r c 28 Theory of Machines This force acts radially inwards and is essential for circular motion. We have discussed above that the centripetal force acts radially inwards. According to Newton's Third Law of Motion, action and reaction are equal and opposite. Therefore, the particle must exert a force radially outwards of equal magnitude. This force is known as centrifugal force whose magnitude is given by 2 2 F = m.v /r = m.ω r c 3.10. Mass Moment of Inertia It has been established since long that a rigid body is composed of small particles. If the mass of every particle of a body is multiplied by the square of its perpendicular distance from a fixed line, then the sum of these quantities(for the whole body) is known as mass moment of inertia of the body. It is denoted by I. Consider a body of total mass m. Let it is composed of small particles of masses m , m , m , m etc. If k , k , k , k are 1 2 3 4 1 2 3 4 the distances of these masses from a fixed line, as shown in Fig. 3.3, then the mass moment of inertia of the whole body is given Fig. 3.3. Mass moment of inertia. by 2 2 2 2 I = m (k ) + m (k ) + m (k ) + m (k ) +.... 1 1 2 2 3 3 4 4 If the total mass of body may be assumed to concentrate at one point (known as centre of mass or centre of gravity), at a distance k from the given axis, such that 2 2 2 2 2 m.k = m (k ) + m (k ) + m (k ) + m (k ) +... 1 1 2 2 3 3 4 4 2 then I = m.k The distance k is called the radius of gyration. It may be defined as the distance, from a given reference, where the whole mass of body is assumed to be concentrated to give the same value of I. 2 The unit of mass moment of inertia in S.I. units is kg-m . Notes : 1. If the moment of inertia of a body about an axis through its centre of gravity is known, then the moment of inertia about any other parallel axis may be obtained by using a parallel axis theorem i.e. moment of inertia about a parallel axis, 2 I = I + m.h p G where I = Moment of inertia of a body about an axis through its centre of gravity, and G h = Distance between two parallel axes. 2. The following are the values of I for simple cases : (a) The moment of inertia of a thin disc of radius r, about an axis through its centre of gravity and perpendicular to the plane of the disc is 2 I = m.r 2 / and moment of inertia about a diameter, 2 I = m.r /4 (b) The moment of inertia of a thin rod of length l, about an axis through its centre of gravity and perpendicular to its length, 2 I = m.l /12 G and moment of inertia about a parallel axis through one end of a rod, 2 I = m.l /3 p Chapter 3 : Kinetics of Motion 29 3. The moment of inertia of a solid cylinder of radius r and length l, about the longitudinal axis or polar axis 2 = m.r /2 and moment of inertia through its centre perpendicular to longitudinal axis 22  rl =+   412  3.11. Angular Momentum or Moment of Momentum Consider a body of total mass m rotating with an angular velocity of ω rad/s, about the fixed axis O as shown in Fig. 3.4. Since the body is composed of numerous small particles, therefore let us take one of these small particles having a mass dm and at a distance r from the axis Fig. 3.4. Angular of rotation. Let v is its linear velocity acting tangentially at any instant. momentum. We know that momentum is the product of mass and velocity, therefore momentum of mass dm = dm × v = dm × ω × r ... ( v = ω.r)  and moment of momentum of mass dm about O 2 = dm × ω × r × r = dm × r × ω = I × ω m 2 where I = Mass moment of inertia of mass dm about O = dm × r m ∴ Moment of momentum or angular momentum of the whole body about O =ωII..=ω ∫ m I = where Mass moment of inertia of the m ∫ whole body about O. Thus we see that the angular momentum or the moment of momentum is the product of mass moment of inertia ( I ) and the angular velocity (ω) of the body. 3.12. Torque It may be defined as the product of force and the perpendicular distance of its line of action from the given point or axis. A little consideration will show that the torque is Torque equivalent to a couple acting upon a body. The Newton’s Second Law of Double Motion, when applied to rotating bodies, torque states that the torque is directly proportional to the rate of change of angular momentum. Mathematically, Torque, Same d (. ω) force T ∝ dt applied Double length spanner Since I is constant, therefore dω dω  TI=× =I.α ...  =α  dt dt  30 Theory of Machines 2 2 The unit of torque (T ) in S.I. units is N-m when I is in kg-m and α in rad/s . 3.13. Work Whenever a force acts on a body and the body undergoes a displacement in the direction of the force, then work is said to be done. For example, if a force F acting on a body causes a displacement x of the body in the direction of the force, then Work done = Force × Displacement = F × x If the force varies linearly from zero to a maximum value of F, then 01 + F Work done =×xF =××x 22 When a couple or torque ( T ) acting on a body causes the angular displacement (θ) about an axis perpendicular to the plane of the couple, then Work done = Torque × Angular displacement = T.θ The unit of work depends upon the unit of force and displacement. In S.I. system of units, the practical unit of work is N-m. It is the work done by a force of 1 newton, when it displaces a body through 1 metre. The work of 1 N-m is known as joule (briefly written as J ) such that 1 N-m = 1 J. Note: While writing the unit of work, it is general practice to put the unit of force first followed by the unit of displacement (e.g. N-m). 3.14. Power It may be defined as the rate of doing work or work done per unit time. Mathematically, Work done Power = Time taken In S.I. system of units, the unit of power is watt (briefly written as W) which is equal to 1 J/s or 1 N-m/s. Thus, the power developed by a force of F (in newtons) moving with a velocity v m/s is F.v watt. Generally a bigger unit of power called kilowatt (briefly written as kW) is used which is equal to 1000 W. Notes: 1. If T is the torque transmitted in N-m or J and ω is the angular speed in rad/s, then Power, P = T.ω = T × 2 π N/60 watts ... (∵ ω = 2 π N/60) where N is the speed in r.p.m. 2. The ratio of power output to power input is known as efficiency of a machine. It is always less than unity and is represented as percentage. It is denoted by a Greek letter eta (η). Mathematically, Power output Efficiency, = η Powerinput 3.15. Energy It may be defined as the capacity to do work. The energy exists in many forms e.g. mechanical, electrical, chemical, heat, light etc. But we are mainly concerned with mechanical energy. The mechanical energy is equal to the work done on a body in altering either its position or its velocity. The following three types of mechanical energies are important from the subject point of view. Chapter 3 : Kinetics of Motion 31 1. Potential energy. It is the energy possessed by a body for doing work, by virtue of its position. For example, a body raised to some height above the ground level possesses potential energy because it can do some work by falling on earth’s surface. Let W = Weight of the body, m = Mass of the body, and h = Distance through which the body falls. Then potential energy, P.E. = W.h = m.g.h ... (∵ W = m.g) It may be noted that (a) When W is in newtons and h in metres, then potential energy will be in N-m. (b) When m is in kg and h in metres, then the potential energy will also be in N-m as discussed below : We know that potential energy, m 1kg–m  P.E.== mg . .h kg×× m= N− m 1N =  2 2  s s 2. Strain energy. It is the potential energy stored by an elastic body when deformed. A com- pressed spring possesses this type of energy, be- cause it can do some work in recovering its original shape. Thus if a compressed spring of stiffness s newton per unit deformation (i.e. extension or com- pression) is deformed through a distance x by a load W, then 1 Strain energy = Work done = Wx . 2 1 2 = sx . ...(∵ W = s × x) 2 In case of a torsional spring of stiffness q N-m per unit angular deformation when twisted through as angle θ radians, then 1 2 =θ q. Strain energy = Work done 2 3. Kinetic energy. It is the energy possessed by a body, for doing work, by virtue of its mass and velocity of motion. If a body of mass m attains a velocity v from rest in time t, under the influence of a force F and moves a distance s, then Work done = F.s = m.a.s ... (∵ F = m.a) ∴ Kinetic energy of the body or the kinetic energy of translation, 2 v 1 2 = mv . K.E. = m.a.s = m × a × 22 a 2 2 We know that, v – u = 2 a.s Since u = 0 because the body starts from rest, therefore, 2 2 v = 2 a.s or s = v /2a 32 Theory of Machines It may be noted that when m is in kg and v in m/s, then kinetic energy will be in N-m as discussed below: We know that kinetic energy, 2 1mkg-m  1kg-m 2  = K.E. = m.v = kg × = × m = N-m ... 1N  2 22  s 2 ss Notes : 1. When a body of mass moment of inertia I (about a given axis) is rotated about that axis, with an angular velocity ω, then it possesses some kinetic energy. In this case, 1 2 Kinetic energy of rotation = I.ω 2 2. When a body has both linear and angular motions e.g. in the locomotive driving wheels and wheels of a moving car, then the total kinetic energy of the body is equal to the sum of kinetic energies of translation and rotation. 11 22 mv.. +ω I ∴ Total kinetic energy = 22 Example 3.1. The flywheel of a steam engine has a radius of gyration of 1 m and mass 2500 kg. The starting torque of the steam engine is 1500 N-m and may be assumed constant. Determine : 1. Angular acceleration of the flywheel, and 2. Kinetic energy of the flywheel after 10 seconds from the start. Solution. Given : k = 1 m ; m = 2500 kg ; T = 1500 N-m 1. Angular acceleration of the flywheel Let α = Angular acceleration of the flywheel. Flywheel We know that mass moment of inertia of the flywheel, 22 2 Im== .k 2500×1= 2500 kg-m We also know that torque ( T ), 2 or Ans. 1500=α . = 2500×α α= 1500 / 2500= 0.6 rad/s I 2. Kinetic energy of the flywheel after 10 seconds from start First of all, let us find the angular speed of the flywheel (ω ) after t = 10 seconds from the 2 start (i.e. ω = 0 ). 1 We know that ω = ω + α.t = 0 + 0.6 × 10 = 6 rad/s 2 1 ∴ Kinetic energy of the flywheel, 11 22 EI=ω ( )=× 2500×6= 45 000 J= 45 kJ Ans. 2 22 Example 3.2. A winding drum raises a cage of mass 500 kg through a height of 100 metres. The mass of the winding drum is 250 kg and has an effective radius of 0.5 m and radius of gyration is 0.35 m. The mass of the rope is 3 kg/m. 2 The cage has, at first, an acceleration of 1.5 m/s until a velocity of 10 m/s is reached, after 2 which the velocity is constant until the cage nears the top and the final retardation is 6 m/s . Find 1. The time taken for the cage to reach the top, 2. The torque which must be applied to the drum at starting; and 3. The power at the end of acceleration period. Solution. Given : m = 500 kg ; s = 100 m ; m = 250 kg ; r = 0.5 m ; k = 0.35 m, C D m = 3 kg/m Chapter 3 : Kinetics of Motion 33 Fig. 3.5 Fig. 3.5 shows the acceleration-time and velocity-time graph for the cage. 1. Time taken for the cage to reach the top Let t = Time taken for the cage to reach the top = t + t + t 1 2 3 where t = Time taken for the cage from initial velocity of u = 0 to final 1 1 2 velocity of v = 10 m/s with an acceleration of a = 1.5 m/s , 1 1 t = Time taken for the cage during constant velocity of v = 10 m/s until the 2 2 cage nears the top, and t = Time taken for the cage from initial velocity of u = 10 m/s to final velocity 3 3 2 of v = 0 with a retardation of a = 6 m/s . 3 3 We know that v = u + a .t 1 1 1 1 10 = 0 + 1.5 t or t = 10/1.5 = 6.67 s 1 1 and distance moved by the cage during time t , 1 vu++ 10 0 11 st =×= × 6.67= 33.35 m 11 22 Similarly, v = u + a .t 3 3 3 3 0 = 10 – 6 × t or t = 10/6 = 1.67 s 3 3 vu++ 010 33 st =×= × 1.67= 8.35 m and 33 22 Now, distance travelled during constant velocity of v = 10 m/s, 2 ss =−s −s = 100− 33.35− 8.35= 58.3m 213 We know that s = v .t or t = s /v = 58.3/10 = 5.83 s 2 2 2 2 2 2 ∴ Time taken for the cage to reach the top, t = t + t + t = 6.67 + 5.83 + 1.67 = 14.17 s Ans. 1 2 3 2. Torque which must be applied to the drum at starting Let T = Torque which must be applied to the drum at starting = T + T + T , 1 2 3 where T = Torque to raise the cage and rope at uniform speed, 1 T = Torque to accelerate the cage and rope, and 2 T = Torque to accelerate the drum. 3 34 Theory of Machines Since the mass of rope, m = 3 kg/m, therefore total mass of the rope for 100 metres, m = m.s = 3 × 100 = 300 kg R We know that the force to raise cage and rope at uniform speed, F = (m + m ) g = (500 + 300) 9.81 = 7850 N 1 C R ∴ Torque to raise cage and rope at uniform speed, T = F .r = 7850 × 0.5 = 3925 N-m 1 1 Force to accelerate cage and rope, F = (m + m ) a = (500 + 300) 1.5 = 1200 N 2 C R 1 ∴ Torque to accelerate the cage and rope, T = F .r = 1200 × 0.5 = 600 N-m 2 2 We know that mass moment of inertia of the drum, 2 2 2 I = m .k = 250 (0.35) = 30.6 kg-m D and angular acceleration of the drum, a 1.5 2 1 α= = =3rad/s r 0.5 ∴ Torque to accelerate the drum, T = I.α = 30.6 × 3 = 91.8 N-m 3 and total torque which must be applied to the drum at starting, T = T + T + T = 3925 + 600 + 91.8 = 4616.8 N-m Ans. 1 2 3 3. Power at the end of acceleration period When the acceleration period is just finishing, the drum torque will be reduced because there will be s = 33.35 m of rope less for lifting. Since the mass of rope is 3 kg/m, therefore mass of 1 33.35 m rope, m = 3 × 33.35 = 100.05 kg 1 ∴ Reduction of torque, T = (m .g + m .a ) r = (100.05 × 9.81 + 100.05 × 1.5) 0.5 4 1 1 1 = 565.8 N-m and angular velocity of drum, ω = v /2πr = 10 / 2π × 0.5 = 3.18 rad/s We know that power = T .ω = 565.8 × 3.18 = 1799 W = 1.799 kW Ans. 4 Example 3.3. A riveting machine is driven by a 4 kW motor. The moment of inertia of the 2 rotating parts of the machine is equivalent to 140 kg-m at the shaft on which the flywheel is mounted. At the commencement of operation, the flywheel is making 240 r.p.m. If closing a rivet occupies 1 second and consumes 10 kN-m of energy, find the reduction of speed of the flywheel. What is the maximum rate at which the rivets can be closed ? 2 Solution : Given : P = 4 kW = 4000 W ; I = 140 kg-m ; N = 240 r.p.m. or ω = 2π × 1 1 240/60 = 25.14 rad/s Reduction of speed of the flywheel Let ω = Angular speed of the flywheel immediately after closing a rivet. 2  Chapter 3 : Kinetics of Motion 35 Since the power of motor is 4000 W, therefore energy supplied by motor in 1 second, E = 4000 N-m ... (∵ 1 W = 1 N-m/s) 1 We know that energy consumed in closing a rivet in 1 second, E = 10 kN-m = 10 000 N-m 2 ∴ Loss of kinetic energy of the flywheel during the operation, E = E – E = 10 000 – 4000 = 6000 N-m 2 1 We know that kinetic energy of the flywheel at the commencement of operation 1 1 2 2 = I (ω ) = × 140 (25.14) = 44 240 N-m 1 2 2 ∴ Kinetic energy of the flywheel at the end of operation = 44 240 – 6000 = 38 240 N-m ... (i) We also know that kinetic energy of the flywheel at the end of operation 1 1 2 2 2 = I (ω ) = × 140 (ω ) = 70 (ω ) ... (ii) 2 2 2 2 2 Equating equations (i) and (ii), 2 2 70 (ω ) = 38 240 or (ω ) = 38 240/70 = 546.3 and ω = 23.4 rad/s 2 2 ∴ Reduction of speed = ω – ω = 25.14 – 23.4 = 1.74 rad/s 1 2 = 1.74 × 60/2π = 16.6 r.p.m. Ans. ... (∵ ω = 2π N/60) Maximum rate at which the rivets can be closed Maximum rate at which the rivets can be closed per minute Energy supplied by motor per min 4000 × 60 =2 ==4 Ans. Energyconsumed to close a rivet 10000 Example 3.4. A wagon of mass 14 tonnes is hauled up an incline of 1 in 20 by a rope which is parallel to the incline and is being wound round a drum of 1 m diameter. The drum, in turn, is driven through a 40 to 1 reduction gear by an electric motor. The frictional resistance to the move- ment of the wagon is 1.2 kN, and the efficiency of the gear drive is 85 per cent. The bearing friction at the drum and motor shafts may be neglected. The rotating parts of the drum have a mass of 1.25 tonnes with a radius of gyration of 450 mm and the rotating parts on the armature shaft have a mass of 110 kg with a radius of gyration of 125 mm. At a certain instant the wagon is moving up the slope with a velocity of 1.8 m/s and an 2 acceleration of 0.1 m/s . Find the torque on the motor shaft and the power being developed. Solution. Given : m = 14 t = 14 000 kg ; Slope = 1 in 20 ; d = 1m or r = 0.5 m ; F = 1.2 kN = 1200 N ; η = 85% = 0.85 ; m = 1.25 t = 1250 kg ; k = 450 mm = 0.45 m ; m = 110 kg; 1 1 2 2 k = 125 mm = 0.125 m; v = 1.8 m/s ; a = 0.1 m/s 2 Torque on the motor shaft We know that tension in the rope, P = Forces opposing the motion as shown in 1 Fig. 3.6 Fig. 3.6. 36 Theory of Machines = Component of the weight down the slope + Inertia force + Frictional resistance 1 =× mg.. +m.a+ F 20 14 000 × 9.81 =+ 14 000× 0.1+ 1200= 9467 N 20 ∴ Torque on the drum shaft to accelerate load, T = P .r = 9467 × 0.5 = 4733.5 N-m 1 1 We know that mass moment of inertia of the drum, 2 2 2 I = m (k ) = 1250 (0.45) = 253 kg-m 1 1 1 and angular acceleration of the drum, α = a/r = 0.1/0.5 = 0.2 rad/s ∴ Torque on the drum to accelerate drum shaft, T = I .α = 253 × 0.2 = 50.6 N-m 2 1 1 Since the drum is driven through a 40 to 1 reduction gear and the efficiency of the gear drive is 85%, therefore Torque on the armature to accelerate drum and load, 11 11 TT=+ ( T ) × = (4733.5+ 50.6) × = 140.7 N-m 31 2 40 0.85 40 0.85 We know that mass moment of inertia of the armature, 2 2 2 I = m (k ) = 110 (0.125) = 1.72 kg-m 2 2 2 and angular acceleration of the armature, a 0.1 2 α= × 40= × 40= 8 rad / s 2 r 0.5 ... (∵ Armature rotates 40 times that of drum) ∴ Torque on the armature to accelerate armature shaft, T = I .α = 1.72 × 8 = 13.76 N-m 4 2 2 and torque on the motor shaft T = T + T = 140.7 + 13.76 = 154.46 N-m Ans. 3 4 Power developed by the motor We know that angular speed of the motor, v 1.8 ω= ×= 40 ×= 40 144 rad/s r 0.5 ∴ Power developed by the motor = T.ω = 154.46 × 144 = 22 240 W = 22.24 kW Ans. Inertia force is equal and opposite to the accelerating force. Chapter 3 : Kinetics of Motion 37 Example 3.5. A road roller has a total mass of 12 tonnes. The front roller has a mass of 2 tonnes, a radius of gyration of 0.4 m and a diameter of 1.2 m. The rear axle, together with its wheels, has a mass of 2.5 tonnes, a radius of gyration of 0.6 m and a diameter of 1.5 m. Calculate : 1. Kinetic energy of rotation of the wheels and axles at a speed of 9 km/h, 2. Total kinetic energy of road roller, and 3. Braking force required to bring the roller to rest from 9 km/h in 6 m on the level. Solution. Given : m = 12 t = 12 000 kg ; m = 2 t = 2000 kg ; k = 0.4 m ; d = 1.2 m or r = 0.6 m ; m = 2.5 t = 2500 kg ; k = 0.6 m ; d = 1.5 1 1 1 1 2 2 2 m or r = 0.75 m ; v = 9 km/h = 2.5 m/s; s = 6 m 2 1. Kinetic energy of rotation of the wheels and axles We know that mass moment of inertia of the front roller, 2 2 2 I = m (k ) = 2000 (0.4) = 320 kg-m 1 1 1 and mass moment of inertia of the rear axle together with its wheels, 2 2 2 I = m (k ) = 2500 (0.6) = 900 kg -m 2 2 2 Angular speed of the front roller, ω = v/r = 2.5/0.6 = 4.16 rad/s 1 1 and angular speed of rear wheels, ω = v/r = 2.5/0.75 = 3.3 rad/s 2 2 We know that kinetic energy of rotation of the front roller, 11 22 EI=ω ( )=× 320(4.16)= 2770 N-m 11 1 22 and kinetic energy of rotation of the rear axle together with its wheels, 11 22 EI=ω ( )=× 900(3.3)= 4900 N-m 222 22 ∴ Total kinetic energy of rotation of the wheels, EE= +E=+= 2770 4900 7670 N-m Ans. 12 2. Total kinetic energy of road roller We know that the kinetic energy of motion (i.e. kinetic energy of translation) of the road roller, 11 22 Em== .v ×12000(2.5)=37500N-m 3 22 This energy includes the kinetic energy of translation of the wheels also, because the total mass (m) has been considered. ∴ Total kinetic energy of road roller, E = Kinetic energy of translation + Kinetic energy of rotation 4 = E + E = 37 500 + 7670 = 45 170 N-m Ans. 3 38 Theory of Machines 3. Braking force required to bring the roller to rest Let F = Braking force required to bring the roller to rest, in newtons. We know that the distance travelled by the road roller, s = 6 m ... (Given) ∴ Work done by the braking force = F × s = 6 F N-m This work done must be equal to the total kinetic energy of road roller to bring the roller to rest, i.e. 6 F = 45 170 or F = 45 170/6 = 7528.3 N Ans. Example 3.6. A steam engine drop-valve is closed by a spring after the operation of a trip gear. The stiffness of the spring is such that a force of 4 N is required per mm of compression. The valve is lifted against the spring, and when fully open the compression is 75 mm. When closed the compression is 30 mm. The mass of the valve is 5 kg and the resistance may be taken as constant and equal to 70 N. Find the time taken to close the valve after the operation of the trip. Solution. Given : s = 4 N/mm = 4000 N/m ; x = 75 mm = 0.075 m ; x = 30 mm = 0.03 m; 1 2 m = 5 kg ; R = 70 N Let x = Displacement of the valve (in metres) from its highest position in time t seconds. When the valve is closed, then the value of x = x – x = 0.075 – 0.03 = 0.045 m 1 2 Since the stiffness of the spring is 4000 N/m ; therefore in any position, the push of the spring Q = 4000 (0.075 – x ) N If P is the downward force on the valve, then P = Q + m.g – R = 4000 (0.075 – x) + 5 × 9.81 – 70 = 279 – 4000 x Also Force, P = Mass × Acceleration 2 dx 279 – 4000 x = 5 × 2 dt 2 dx 279 − 4000x == 56− 800xx=− 800(− 0.07) or 2 5 dt Let y = x – 0.07 22 2 dy d x dy ==− 800y += 800y 0 ∴ or 22 2 dt dt dt The solution of this differential equation is ya=+ cos 800t b sin 800t xa −= 0.07 cos 800t+b sin 800t ... (i) where a and b are constants to be determined. Now when t = 0, x = 0, therefore from equation (i), a = – 0.07 Differentiating equation (i), dx =− 800at sin 800 + 800bcos 800t ... (ii) dt Chapter 3 : Kinetics of Motion 39 dx Now when t = 0, = 0, therefore from equation (ii), b = 0 dt Substituting the values of a and b in equation (i), x – 0.07 = – 0.07 or cos 800 txt =− 0.07 (1 cos 800 ) When x = 0.045 m, then 0.045=− 0.07 (1 cos 800 t) or or 1−= cos 800 t 0.045 / 0.07= 0.642 cos 800 t=− 1 0.642= 0.358 π −1 800 t== cos (0.358) 69°= 69×= 1.2 rad 180 ∴ Ans. t== 1.2 / 800 1.2 / 28.3= 0.0424 s 3.16. Principle of Conservation of Energy It states “The energy can neither be created nor destroyed, though it can be transformed from one form into any of the forms, in which the energy can exist.” Note : The loss of energy in any one form is always accompanied by an equivalent increase in another form. When work is done on a rigid body, the work is converted into kinetic or potential energy or is used in overcom- ing friction. If the body is elastic, some of the work will also be stored as strain energy. Thus we say that the total energy possessed by a system of moving bodies is constant at every instant, provided that no energy is rejected to or received from an external source to the system. 3.17. Impulse and Impulsive Force The impulse is the product of force and time. Mathematically, Impulse = F × t where F = Force, and t = Time. Now consider a body of mass m. Let a force F changes its velocity from an initial velocity v 1 to a final velocity v . 2 We know that the force is equal to the rate of change of linear momentum, therefore mv() −v 21 () or Ft ×=mv −v F = 21 t i.e. Impulse = Change of linear momentum If a force acts for a very short time, it is then known as impulsive force or blow. The impulsive force occurs in collisions, in explosions, in the striking of a nail or a pile by a hammer. Note: When the two rotating gears with angular velocities ω and ω mesh each other, then an impulsive 1 2 torque acts on the two gears, until they are both rotating at speeds corresponding to their velocity ratio. The impulsive torque, T.t = I (ω – ω ) 2 1 3.18. Principle of Conservation of Momentum It states “The total momentum of a system of masses (i.e. moving bodies) in any one direc- tion remains constant, unless acted upon by an external force in that direction.” This principle is applied to problems on impact, i.e. collision of two bodies. In other words, if two bodies of masses m and m with linear velocities v and v are moving in the same straight line, and they collide and 1 2 1 2 begin to move together with a common velocity v, then 40 Theory of Machines Momentum before impact = Momentum after impact i.e. mv±= m v() m+m v 11 2 2 1 2 Notes : 1. The positive sign is used when the two bodies move in the same direction after collision. The negative sign is used when they move in the opposite direction after collision. 2. Consider two rotating bodies of mass moment of inertia I 1 and I are initially apart from each other and are made to engage as in 2 the case of a clutch. If they reach a common angular velocity ω, after slipping has ceased, then I .ω ± I .ω = (I + I ) ω 1 1 2 2 1 2 The ± sign depends upon the direction of rotation. 3.19. Energy Lost by Friction Clutch During Engagement Consider two collinear shafts A and B connected by a friction clutch (plate or disc clutch) as shown in Fig. 3.7. Let I and I = Mass moment of inertias of the A B rotors attached to shafts A and B respectively. ω and ω = Angular speeds of shafts A and B A B respectively before engagement of clutch, and ω = Common angular speed of shafts A and B after engagement of clutch. By the principle of conservation of momentum, Fig. 3.7. Friction clutch. I .ω + I .ω = (I + I ) ω A A B B A B II.. ω+ ω AA B B ∴ ... (i) ω= II + AB Total kinetic energy of the system before engagement, 22 11 II ()ω+ (ω) 22 AA B B EI=ω()+I(ω)= 1AA BB 22 2 Kinetic energy of the system after engagement, 2 11  II.. ω+ ω 2 AA B B EI=+()Iω=() I+I  2AB AB 22 II +  AB 2 (.II ω+ .ω) AA B B = 2(II + ) AB ∴ Loss of kinetic energy during engagement, 22 2 II ()ω+ (ω) (I.ω+I.ω) AA B B A A B B EE=−E= − 12 22(II +) AB Please refer Chapter 10 (Art. 10.32) on Friction. Chapter 3 : Kinetics of Motion 41 2 II.(ω−ω) AB A B = ... (ii) 2(II + ) AB Notes: 1. If the rotor attached to shaft B is at rest, then ω = 0. Therefore, common angular speed after engagement, B I ω . AA ω= ... Substituting ω = 0 in equation (i) ... (iii) B II + AB 2 II.(ω) AB A and loss of kinetic energy, E = ... Substituting ω = 0 in equation (ii) ... (iv) B II + 2( ) AB 2. If I is very small as compared to I and the rotor B is at rest, then B A I .ω AA ω= = ω A ... (Neglecting I ) B II + AB 11 2 EI=ω ..ω=I .ω and ... From equations (iii) and (iv) BA B 22 = Energy given to rotor B Example 3.7. A haulage rope winds on a drum of radius 500 mm, the free end being attached to a truck. The truck has a mass of 500 kg and is initially at rest. The drum is equivalent to a mass of 1250 kg with radius of gyration 450 mm. The rim speed of the drum is 0.75 m/s before the rope tightens. By considering the change in linear momentum of the truck and in the angular mo- mentum of the drum, find the speed of the truck when the motion becomes steady. Find also the energy lost to the system. Solution. Given : r = 500 mm = 0.5 m ; m = 500 kg ; m = 1250 kg ; k = 450 mm = 0.45 m ; 1 2 u = 0.75 m/s We know that mass moment of inertia of drum, 2 2 2 I = m .k = 1250 (0.45) = 253 kg-m 2 2 Speed of the truck Let v = Speed of the truck in m/s, and F = Impulse in rope in N-s. We know that the impulse is equal to the change of linear momentum of the truck. Therefore F = m .v = 500 v N-s 1 and moment of impulse = Change in angular momentum of drum uv −   uv u −v i.e. Fr×=I() ω −ω =I ... ω−ω = − =  21  22 1 2  r rr r  0.75 − v  500v×= 0.5 253 250vv =− 380 506  or 0.5  ∴ 250 v + 506 v = 380 or v = 380/756 = 0.502 m/s Ans. Energy lost to the system We know that energy lost to the system = Loss in K.E. of drum – Gain in K.E. of truck 11 22 2  ×ωIm ()−(ω) − × .v = 22 1 1  22 42 Theory of Machines 22  11 uv − 2 =×Im −× .v 21  2 22 r  22  1 (0.75) − (0.502) 1 2 =× 253 −× 500(0.502) N-m  2 22 (0.5)  = 94 N-m Ans. Example 3.8. The two buffers at one end of a truck each require a force of 0.7 MN/m of compression and engage with similar buffers on a truck which it overtakes on a straight horizontal track. The truck has a mass of 10 tonnes and its initial speed is 1.8 m/s, while the second truck has mass of 15 tonnes with initial speed 0.6 m/s, in the same direction. Find : 1. the common velocity when moving together during impact, 2. the kinetic energy lost to the system, 3. the compression of each buffer to store the kinetic energy lost, and 4. the velocity of each truck on separation if only half of the energy offered in the springs is returned. 6 3 Solution. Given : s = 0.7 MN/m = 0.7 × 10 N/m ; m = 10 t = 10 × 10 kg ; v = 1.8 m/s; 1 3 m = 15 t = 15 × 10 kg ; v = 0.6 m/s 2 2 1. Common velocity when moving together during impact Let v = Common velocity. We know that momentum before impact = Momentum after impact i.e. m . v + m .v = (m + m ) v 1 1 2 2 1 2 3 3 3 3 10 × 10 × 1.8 + 15 × 10 × 0.6 = (10 × 10 + 15 + 10 ) v 3 3 3 3 27 × 10 = 25 × 10 v or v = 27 × 10 /25 × 10 = 1.08 m/s Ans. 2. Kinetic energy lost to the system Since the kinetic energy lost to the system is the kinetic energy before impact minus the kinetic energy after impact, therefore Kinetic energy lost to the system 11 1  22 2 =+ mv m v− m+ m v ()  11 2 2 1 2 22 2  11  32 32 ×× 10 10 (1.8)+ × 15× 10 (0.6) =  22  1 33 2 −× 10 10+ 15× 10 (1.08) () 2 3 = 4.35 × 10 N-m = 4.35 kN-m Ans. 3. Compression of each buffer spring to store kinetic energy lost Let x = Compression of each buffer spring in metre, and s = Force required by each buffer spring or stiffness of each spring 6 = 0.7 MN/m = 0.7 × 10 N/m ... (Given) Since the strain energy stored in the springs (four in number) is equal to kinetic energy lost in impact, therefore 1 23 4×= sx . 4.35× 10 2 Chapter 3 : Kinetics of Motion 43 1 62 3 4×× 0.7× 10 x = 4.35× 10 2 62 3 or 1.4×= 10 x 4.35× 10 2 3 6 –3 ∴ x = 4.35 × 10 /1.4 × 10 = 3.11 × 10 or x = 0.056 m = 56 mm Ans. 4. Velocity of each truck on separation Let v = Velocity of separation for 10 tonnes truck, and 3 v = Velocity of separation for 15 tonnes truck. 4 The final kinetic energy after separation is equal to the kinetic energy at the instant of com- mon velocity plus strain energy stored in the springs. Since it is given that only half of the energy stored in the springs is returned, therefore Final kinetic energy after separation 1 = Kinetic energy at common velocity + Energy stored in springs 2 11 1 1 1  22 2 2 mv()+= m(v) (m+m)v+ 4× s.x or 13 2 4 1 2  22 2 2 2  11 1 1 32 3 2 3 3 2 3 ×× 10 10 (vv )+ × 15× 10 ( )= (10× 10+ 15× 10 ) (1.08)+ (4.35× 10 ) 34 22 2 2 1  23 ×= sx × ... 4 . 4.35 10  2  22 ... (i) 10(vv )+= 15( ) 33.51 34 We know that initial momentum and final momentum must be equal, i.e. m .v + m .v = (m + m ) v 1 3 2 4 1 2 3 3 3 3 10 × 10 × v + 15 × 10 × v = (10 × 10 + 15 × 10 ) 1.08 3 4 10v + 15 v = 27 ... (ii) 3 4 From equations (i) and (ii), v = 0.6 m/s, and v = 1.4 m/s Ans. 3 4 Example 3.9. A mass of 300 kg is allowed to fall vertically through 1 metre on to the top of a pile of mass 500 kg. Assume that the falling mass and pile remain in contact after impact and that the pile is moved 150 mm at each blow. Find, allowing for the action of gravity after impact 1. The energy lost in the blow, and 2. The average resistance against the pile. Solution. Given : m = 300 kg ; s = 1 m ; m = 500 kg ; x = 150 mm = 0.15 m 1 2 1. Energy lost in the blow First of all, let us find the velocity of mass m with which it hits the pile. 1 Let v = Velocity with which mass m hits the pile. 1 1 22 We know that vu−=2.gs 1 2 vv −= 0 2× 9.81× 1= 19.62 or = 4.43 m/s ... (∵ u = 0 ) 11

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