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MM203 Mechanics of Machines: Part 2

MM203 Mechanics of Machines: Part 2. Kinetics of systems of particles. Extension of basic principles to general systems of particles Particles with light links Rigid bodies Rigid bodies with flexible links Non-rigid bodies Masses of fluid. Newton’s second law. G – centre of mass

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MM203 Mechanics of Machines: Part 2

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  1. MM203Mechanics of Machines: Part 2 Dr. Alan Kennedy

  2. Kinetics of systems of particles • Extension of basic principles to general systems of particles • Particles with light links • Rigid bodies • Rigid bodies with flexible links • Non-rigid bodies • Masses of fluid Dr. Alan Kennedy

  3. Newton’s second law • G– centre of mass • Fi – external force, fi – internal force • ri – position of mi relative to G Dr. Alan Kennedy

  4. Newton’s second law • By definition • For particle i • Adding equations for all particles Dr. Alan Kennedy

  5. Newton’s second law • Differentiating w.r.t. time • gives • Also • so • (principle of motion of the mass centre) Dr. Alan Kennedy

  6. Newton’s second law • Note that is the acceleration of the instantaneous mass centre – which may vary over time if body not rigid. • Note that the sum of forces is in the same direction as the acceleration of the mass centre but does not necessarily pass through the mass centre Dr. Alan Kennedy

  7. Example • Three people (A, 60 kg, B, 90 kg, and C, 80 kg) are in a boat which glides through the water with negligible resistance with a speed of 1 knot. If the people change position as shown in the second figure, find the position of the boat relative to where it would be if they had not moved. Does the sequence or timing of the change in positions affect the final result? (Answer: x = 0.0947 m).(Problem 4/15, M&K) Dr. Alan Kennedy

  8. Example • The 1650 kg car has its mass centre at G. Calculate the normal forces at A and B between the road and the front and rear pairs of wheels under the conditions of maximum acceleration. The mass of the wheels is small compared with the total mass of the car. The coefficient of static friction between the road and the rear driving wheels is 0.8. (Answer: NA = 6.85 kN, NB = 9.34 kN).(Problem 6/5, M&K) Dr. Alan Kennedy

  9. Work-energy • Work-energy relationship for mass i is • where (U1-2)i is the work done on mi during a period of motion by the external and internal forces acting on it. • Kinetic energy of mass i is Dr. Alan Kennedy

  10. Work-energy • For entire system Dr. Alan Kennedy

  11. Work-energy • Note that no net work is done by internal forces. • If changes in potential energy possible (gravitational and elastic) then • as for single particle Dr. Alan Kennedy

  12. Work-energy • For system • Now • and note that • so Dr. Alan Kennedy

  13. Work-energy • Since ri is measured from G, • Now Dr. Alan Kennedy

  14. Work-energy • Therefore • i.e. energy is that of translation of mass-centre and that of translation of particles relative to mass-centre Dr. Alan Kennedy

  15. Example • The two small spheres, each of mass m, are rigidly connected by a rod of negligible mass and are released in the position shown and slide down the smooth circular guide in the vertical plane. Determine their common velocity v as they reach the horizontal dashed position. Also find the force R between sphere 1 and the guide the instant before the sphere reaches position A. (Answer: v = 1.137(gr)½, R = 2.29mg). (Problem 4/9, M&K) Dr. Alan Kennedy

  16. Rigid body • Motion of particles relative to mass-centre can only be due to rotation of body • Velocity of particles due to rotation depends on angular velocity and the distance to centre of rotation. Where is centre of rotation? • Need to examine kinematics of rotation Dr. Alan Kennedy

  17. Plane kinematics of rigid bodies • Rigid body • distances between points remains unchanged • position vectors, as measured relative to coordinate system fixed to body, remain constant • Plane motion • motion of all points is on parallel planes • Plane of motion taken as plane containing mass centre • Body treated as thin slab in plane of motion – all points on body projected onto plane Dr. Alan Kennedy

  18. Kinematics of rigid bodies Dr. Alan Kennedy

  19. Translation • All points move in parallel lines or along congruent curves. • Motion is completely specified by motion of any point – therefore can be treated as particle • Analysis as developed for particle motion Dr. Alan Kennedy

  20. Kinematics of rigid bodies Dr. Alan Kennedy

  21. Rotation about fixed axis • All particles move in circular paths about axis of rotation • All lines on body (in plane of motion) rotate through the same angle in the same time • Similar to circular motion of a particle • where riO is distance to O, the centre of rotation, and IO is massmoment of inertia about O Dr. Alan Kennedy

  22. Mass moment of inertia • Mass moment of inertia in rotation is equivalent to mass in translation • Rotation and translation are analogous Dr. Alan Kennedy

  23. General plane motion • Combination of translation and rotation • Principles of relative motion used Dr. Alan Kennedy

  24. Rotation • Angular positions of two lines on body are measured from any fixed reference direction Dr. Alan Kennedy

  25. Rotation • All lines on a rigid body in its plane of motion have the same angular displacement, the same angular velocity, and the same angular acceleration • Angular motion does not require the presence of a fixed axis about which the body rotates Dr. Alan Kennedy

  26. Angular motion relations • Angular position, angular velocity, and angular acceleration • Similar to relationships between s, v, and a. • Also, combining relationships and cancelling out dt Dr. Alan Kennedy

  27. Angular motion relations • If constant angular acceleration • Direction of +ve sense must be consistent • Analogous to rectilinear motion with constant a • Same procedures used in analysis Dr. Alan Kennedy

  28. Example • The angular velocity of a gear is controlled according to w = 12 – 3t2 where w, in radians per second, is positive in the clockwise sense and where t is the time in seconds. Find the net angular displacement Dq from the time t= 0 to t= 3 s. Also find the total number of revolutions N through which the gear turns during the 3 seconds. (Answer: Dq= 9rad, N = 3.66 rev). (Problem 5/5, M&K) Dr. Alan Kennedy

  29. Kinetic energy of rigid body • If rotation about O Dr. Alan Kennedy

  30. Parallel axis theorem • Now (P.A.T.) • and so Dr. Alan Kennedy

  31. Radius of gyration • Radius of gyration • Mass moment of inertia of point mass m at radius of gyration is the same as that for body • P.A.T. Dr. Alan Kennedy

  32. Work done on rigid body Dr. Alan Kennedy

  33. Work done by couple • Couple is system of forces that causes rotation but no translation • Moment about G • Moment about O Dr. Alan Kennedy

  34. Work done by couple • Moment vector is a free vector • Forces have turning effect or torque • Torque is force by perpendicular distance between forces • Work done • positive or negative Dr. Alan Kennedy

  35. Forces and couples • Torque is • Also unbalanced force Dr. Alan Kennedy

  36. Work-energy principle • When applied to system of connected bodies only consider forces/moments of system – ignore internal forces/moments. • If there is significant friction between components then system must be dismembered Dr. Alan Kennedy

  37. Example • A steady 22 N force is applied normal to the handle of the hand-operated grinder. The gear inside the housing with its shaft and attached handle have a combined mass of 1.8 kg and a radius of gyration about their axis of 72 mm. The grinding wheel with its attached shaft and pinion (inside housing) have a combined mass of 0.55 kg and a radius of gyration of 54 mm. If the gear ratio between gear and pinion is 4:1, calculate the speed of the grinding wheel after 6 complete revolutions of the handle starting from rest. (Answer: N = 3320 rev/min).(Problem 6/119, M&K) Dr. Alan Kennedy

  38. Rotation about fixed axis • Motion of point on rigid body Dr. Alan Kennedy

  39. Vector notation • Angular velocity vector, w, for body has sense governed by right-hand rule • free vector Dr. Alan Kennedy

  40. Vector notation • Velocity vector of point A • What are magnitude and direction of this vector? • Note that Dr. Alan Kennedy

  41. Vector notation • Acceleration of point Dr. Alan Kennedy

  42. Vector notation • Vector equivalents • Can be applied in 3D except then angular velocity can change direction and magnitude Dr. Alan Kennedy

  43. Example • The T-shaped body rotates about a horizontal axis through O. At the instant represented, its angular velocity is w = 3 rad/s and its angular acceleration is a = 14 rad/s2. Determine the velocity and acceleration of (a) point A and (b) point B. Express your results in terms of components along the n- and t- axes shown. (Answer: vA = 1.2etm/s, aA = −5.6et + 3.6enm/s2, vB = 1.2et + 0.3en m/s, aB = −6.5et + 2.2enm/s2). (Problem 5/2, M&K) Dr. Alan Kennedy

  44. Example • The two V-belt pulleys form an integral unit and rotate about the fixed axis at O. At a certain instant, point A on the belt of the smaller pulley has a velocity vA = 1.5 m/s, and the point B on the belt of the larger pulley has an acceleration aB = 45 m/s2 as shown. For this instant determine the magnitude of the acceleration aC of the point C and sketch the vector in your solution. (Answer: aC = 149.6 m/s2). (Problem 5/16, M&K) Dr. Alan Kennedy

  45. Linear impulse and momentum • Returning to general system: Linear momentum of mass i is • For system (assuming m does not change with time) Dr. Alan Kennedy

  46. Linear impulse and momentum • Differentiating w.r.t. time • Same as for single particle – only applies if mass constant • Same for rigid body Dr. Alan Kennedy

  47. Example • The 300 kg and 400 kg mine cars are rolling in opposite directions along a horizontal track with the speeds shown. Upon impact the cars become coupled together. Just prior to impact, a 100 kg boulder leaves the delivery chute and lands in the 300 kg car. Calculate the velocity v of the system after the boulder has come to a rest relative to the car. Would the final velocity be the same if the cars were coupled before the boulder dropped? (Answer: v = 0.205 m/s).(Problem 4/11, M&K) Dr. Alan Kennedy

  48. Angular impulse and momentum • Considered about a fixed point O and about the mass centre. Dr. Alan Kennedy

  49. Angular impulse and momentum • About O • First term is zero since vi×vi=0 so • - sum of all external moments (net moment of internal forces is zero) Dr. Alan Kennedy

  50. Angular impulse and momentum • About O: same as for single particle. As before, does not apply if mass is changing. • About G Dr. Alan Kennedy

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