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Kinematics of Particles

Kinematics of Particles 27
Kinematics of Particles Space Curvilinear Motion Threedimensional motion of a particle along a space curve. Three commonly used coordinate systems to describe this motion: 1. Rectangular Coordinate System (xyz) 2. Cylindrical Coordinate System (rθz) 3. Spherical Coordinate System (RθΦ) ME101 Division III Kaustubh Dasgupta 1Kinematics of particles :: motion in space • Example 2 ME101 Division III Kaustubh Dasgupta 2Kinematics of particles :: motion in space • Example 3 ME101 Division III Kaustubh Dasgupta 3Kinematics of particles :: motion in space n and tcoordinates for plane curvilinear motion can also be used for space curvilinear motion of a particle :: Considering a plane containing the curve and the n and taxes at a particular location (instance) •This plane will continuously shift its location and orientation in case of space curvilinear motion  difficult to use. ME101 Division III Kaustubh Dasgupta 4Kinematics of particles :: motion in space Rectangular Coordinates (xyz) •Simply extend the previously derived equations to include third dimension. Plane Curvilinear Motion (2D) Space Curvilinear Motion (3D) In three dimensions, R is used in place of r for the position vector ME101 Division III Kaustubh Dasgupta 5Kinematics of Particles Cylindrical Coordinates (rθz) •Extension of the Polar coordinate system. •Addition of zcoordinate and its two time derivatives Position vector R to the particle for cylindrical coordinates: R = r e + zk r Velocity: Acceleration: Polar Cylindrical Polar Cylindrical 2    vrerezk arrer2re  vrere r r r 2   v r  r arrer2rezk r  v r r  v r 2 2     a rr a rr  r r v r   v z  z a r2r a r2r   2 2 v vv 2 2 2 r  a z 2 2 z v vvv a aa r z r 2 2 2 a aaa r z Unit vector k remains fixed in direction  has a zero time derivative ME101 Division III Kaustubh Dasgupta 6Kinematics of Particles Spherical Coordinates (RθΦ) •Utilized when a radial distance and two angles are utilized to specify the position of a particle. •The unit vector e is in the direction in which the particle P R would move if R increases keeping θ and Φ constant. •The unit vector e is in the direction in which the particle P θ would move if θ increases keeping R and Φ constant. •The unit vector e is in the direction in which the particle P Φ would move if Φ increases keeping R and θ constant. Resulting expressions for v and a: vv ev ev e aa ea ea e R R R R 2 2 2   a RRR cos v R R R cos d 2  aR2Rsin v R cos   R dt  1 d v R 2 2  aRR sincos  R dt ME101 Division III Kaustubh Dasgupta 7Example (1) on cylindrical/spherical coordinate ME101 Division III Kaustubh Dasgupta 8Example (1) on cylindrical/spherical coordinate ME101 Division III Kaustubh Dasgupta 9Example (1) on cylindrical/spherical coordinate ME101 Division III Kaustubh Dasgupta 10Example (1) on cylindrical/spherical coordinate ME101 Division III Kaustubh Dasgupta 11Kinematics of Particles Relative Motion (Translating Axes) • Till now particle motion described using fixed reference axes  Absolute Displacements, Velocities, and Accelerations • Relative motion analysis is extremely important for some cases  measurements made wrt a moving reference system Motion of a moving coordinate system is specified wrt a fixed coordinate system (whose absolute motion is negligible for the problem at hand). Current Discussion: • Moving reference systems that translate but do not rotate • Relative motion analysis for plane motion Relative Motion Analysis is critical even if aircrafts ME101 Division III Kaustubh Dasgupta 12 are not rotatingKinematics of Particles Relative Motion (Translating Axes) Vector Representation Two particles A and B have separate curvilinear motions in a given plane or in parallel planes. • Attaching the origin of translating (nonrotating) axes xy to B. • Observing the motion of A from moving position on B. • Position vector of A measured relative to the frame xy is r = xi + yj. Here x and y are the coordinates of A A/B measured in the xy frame. (A/B  A relative to B) • Absolute position of B is defined by vector r B measured from the origin of the fixed axes XY. • Absolute position of A  r = r + r A B A/B • Differentiating wrt time  Velocity of A wrt B: Acceleration of A wrt B: Unit vector i and j have constant direction  zero derivatives ME101 Division III Kaustubh Dasgupta 13Kinematics of Particles Relative Motion (Translating Axes) Vector Representation Velocity of A wrt B: Acceleration of A wrt B: Velocity or Absolute Velocity or Absolute Velocity  = + Acceleration of Acceleration of A or Acceleration of B A relative to B. • The relative motion terms can be expressed in any convenient coordinate system (rectangular, normaltangential, or polar) • Already derived formulations can be used.  The appropriate fixed systems of the previous discussions becomes the moving system in this case. ME101 Division III Kaustubh Dasgupta 14Kinematics of Particles Relative Motion (Translating Axes) Selection of Translating Axes Instead of B, if A is used for the attachment of the moving system:  r = r + r B A B/A v = v + v B A B/A a = a + a B A B/A  r = r ; v = v ; a = a B/A A/B B/A A/B B/A A/B Relative Motion Analysis: • Acceleration of a particle in translating axes (xy) will be the same as that observed in a fixed system (XY) if the moving system has a constant velocity  A set of axes which have a constant absolute velocity may be used in place of a fixed system for the determination of accelerations  Interesting applications of Newton’s Second law of motion in Kinetics A translating reference system that has no acceleration  Inertial System ME101 Division III Kaustubh Dasgupta 15Kinematics of Particles Relative Motion (Translating Axes) Inertial Reference Frame Or Newtonian Reference Frame • When applying the eqn of motion (Newton’s Second Law of Motion), it is important that the acceleration of the particle be measured wrt a reference frame that is either fixed or translates with a constant velocity. • The reference frame should not rotate and should not accelerate. • In this way, the observer will not accelerate and measurements of particle’s acceleration will be the same from any reference of this type.  Inertial or Newtonian Reference Frame Study of motion of rockets and satellites: inertial reference frame may be considered to be fixed to the stars. Motion of bodies near the surface of the earth: inertial reference frame may be considered to be fixed to the earth. Though the earth rotates its own axis and revolves around the sun, the accelerations created by these motions of the earth are relatively small and can be neglected. ME101 Division III Kaustubh Dasgupta 16Example on relative motion ME101 Division III Kaustubh Dasgupta 17Example on relative motion Graphical method Trigonometric method Vector algebra ME101 Division III Kaustubh Dasgupta 18Kinematics of Particles Constrained Motion of Connected Particles • Interrelated motion of particles One Degree of Freedom System Establishing the position coordinates x and y measured from a convenient fixed datum.  We know that horz motion of A is twice the vertical motion of B. Total length of the cable: L, r , r and b are constant. First and second time derivatives: 2 1  Signs of velocity and acceleration of A and B must be opposite  v is positive to the left. v is positive to the down A B Lower Pulley  Equations do not depend on lengths or pulley radii Alternatively, the velocity and acceleration magnitudes can be determined by inspection of lower pulley. SDOF: since only one variable (x or y) is needed to specify the positions of all parts of the system ME101 Division III Kaustubh Dasgupta 19Kinematics of Particles Constrained Motion of Connected Particles One Degree of Freedom System Applying an infinitesimal motion of A’ in lower pulley. • A’ and A will have same motion magnitudes • B’ and B will have same motion magnitudes • From the triangle shown in lower figure, it is clear that B’ moves half as far as A’ because point C has no motion momentarily since it is on the fixed portion on the cable. • Using these observations, we can obtain the velocity and acceleration magnitude relationships by inspection. • The pulley is actually a wheel which rolls on the fixed cable. Lower Pulley ME101 Division III Kaustubh Dasgupta 20Kinematics of Particles Constrained Motion of Connected Particles Two Degrees of Freedom System Two separate coordinates are required to specify the position of lower cylinder and pulley C y and y . A B Lengths of the cables attached to cylinders A and B: Their time derivatives: Eliminating the terms in  It is impossible for the signs of all three terms to be positive simultaneously.  If A and B have downward (+ve) velocity, C will have an upward (ve) velocity. ME101 Division III Kaustubh Dasgupta 21Kinematics of Particles Constrained Motion of Connected Particles Two Degrees of Freedom System Same results can be obtained by observing the motions of the two pulleys at C and D. • Apply increment dy (keeping y fixed) A B  D moves up an amount dy /2 A  this causes an upward movement dy /4 of C A • Similarly, for an increment dy (keeping y fixed) B A  C moves up an amount dy /2 B • A combination of the two movements gives an upward movement:  v = v /4 + v /2 as obtained earlier. c A B ME101 Division III Kaustubh Dasgupta 22Kinematics of Particles Constrained Motion of Connected Particles Example Determine the velocity of B if the cylinder A has a downward velocity of 0.3 m/s. Use two different methods. Solution Method I: Centers of pulleys at A and B are located by the coordinates y and y measured from fixed positions. A B Total constant length of the cable in the system: L = 3 y + 2 y + constants B A Differentiating wrt time: Substituting  v = 0.2 m/s B ME101 Division III Kaustubh Dasgupta 23Kinematics of Particles Constrained Motion of Connected Particles Example C Solution Method II: Graphical method: A Enlarged views of the pulleys at A, B, and C are shown. B • Apply a differential movement ds at center of pulley A A  no motion at left end of its horz diameter since it is attached to the fixed part of the cable  right end will move by 2ds A  this movement will be transmitted to the left end of horz diameter of pulley B but in the upward direction. • Pulley C has a fixed center  disp on each side are equal and opposite (ds ) B  Right end of pulley B will also have a downward displacement equal to the upward displacement of the center of the pulley B (both ds ) B  2ds = 3 ds ds = 2/3 ds A B B A Dividing by dt │v │ = 2/3 v = 0.2 m/s (upwards) B A ME101 Division III Kaustubh Dasgupta 24Kinematics of Particles Summary ME101 Division III Kaustubh Dasgupta 25Kinetics of Particles Kinetics: Study of the relations between unbalanced forces and the resulting changes in motion. Newton’s Second Law of Motion : The acceleration of a particle is proportional to the resultant force acting on it and is in the direction of this force.  A particle will accelerate when it is subjected to unbalanced forces Three approaches to solution of Kinetics problems: 1. ForceMassAcceleration method (direct application of Newton’s Second Law) 2. Use of Work and Energy principles 3. Impulse and Momentum methods Limitations of this chapter: • Motion of bodies that can be treated as particles (motion of the mass centre of the body) • Forces are concurrent through the mass center (action of nonconcurrent forces on the motion of bodies will be discussed in chapter on Kinetics of rigid bodies). ME101 Division III Kaustubh Dasgupta 26Kinetics of Particles ForceMassAcceleration method Newton’s Second Law of Motion Subject a mass particle to a force F and measure accln of the particle a . 1 1 Similarly, F and a …  The ratio of magnitudes of force and resulting 2 2 acceleration will remain constant. Constant C is a measure of some invariable property of the particle  Inertia of the particle Inertia = Resistance to rate of change of velocity Mass m is used as a quantitative measure of Inertia. C = km where k is a constant introduced to account for the units used. F = magnitude of the resultant force acting on the particle of mass m  F = kma a = magnitude of the resulting acceleration of the particle. Accln is always in the direction of the applied force  Vector Relation: F = kma In Kinetic System of units, k is taken as unity  F = ma  units of force, mass and acceleration are not independent  Absolute System since the force units depend on the absolute value of mass. 0 Values of g at Sea level and 45 latitude: 2 For measurements relative to rotating earth: Relative g 9.80665 9.81 m/s 2 For measurements relative to nonrotating earth: Absolute g 9.8236 m/s ME101 Division III Kaustubh Dasgupta 27Kinetics of Particles ForceMassAcceleration method Equation of Motion Particle of mass m subjected to the action of concurrent forces F , F ,… whose 1 2 vector sum is ∑F:  Equation of motion: ∑F = ma  ForceMassAcceleration equation Equation of Motion gives the instantaneous value of the acceleration corresponding to the instantaneous value of the forces. • The equation of motion can be used in scalar component form in any coordinate system. • For a 3 DOF problem, all three scalar components of equation of motion will be required to be integrated to obtain the space coordinates as a function of time. • All forces, both applied or reactive, which act on the particle must be accounted for while using the equation of motion. Free Body Diagrams: In Statics: Resultant of all forces acting on the body = 0 In Dynamics: Resultant of all forces acting on the body = ma Motion of body ME101 Division III Kaustubh Dasgupta 28Kinetics of Particles: ForceMassAcceleration method Rectilinear Motion Motion of a particle along a straight line For motion along xdirection, accelerations along y and zdirection will be zero  ∑F = ma x x ∑F = 0 y ∑F = 0 z For a general case:  ∑F = ma x x ∑F = ma y y ∑F = ma z z The acceleration and resultant force are given by: ME101 Division III Kaustubh Dasgupta 29Kinetics of Particles: ForceMassAcceleration method Rectilinear Motion Example A 75 kg man stands on a spring scale in an elevator. During the first 3 seconds of motion from rest, the tension T in the hoisting cable is 8300 N. Find the reading R of the scale in Newton during this interval and the upward velocity v of the elevator at the end of the 3 seconds. Total mass of elevator, man, and scale is 750 kg. Solution motion Draw the FBD of the elevator and the man alone ME101 Division III Kaustubh Dasgupta 30Kinetics of Particles: ForceMassAcceleration method Rectilinear Motion Example Solution During first 3 seconds, the forces acting on the elevator are constant. Therefore, the acceleration a will also y remain constant during this time. Force registered by the scale and the velocity of the elevator depend on the acceleration a y From FBD of the elevator, scale, and man taken together: 2 ∑F = ma 83007360 = 750a a = 1.257 m/s y y y y From FBD of the man alone: ∑F = ma R736 = 75a R = 830 N y y y Velocity reached at the end of the 3 sec: Δv = ∫a dt v = 3.77 m/s ME101 Division III Kaustubh Dasgupta 31
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