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Mechanics

This chapter discusses the dynamics of particle systems, specifically the impact and collision between particles. It explores the general concepts, direct central impact, oblique central impact, and provides sample examples. The methods used are the impulse-momentum method and the work-energy method.

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Mechanics

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  1. Mechanics Chapter 6. Dynamics of Particle system 6.6 Impact Section 15.8,15.9,15.10 Chapter 6. Dynamics of particle systems

  2. 6.6 Impact As a example of system of particles, we’ll discuss the dynamics of the impact, or collision, between particles • The forces during an impact are usually unknown  can not use FMA method • The mechanical energy is generally lost during the impact (converted into sound or heat)  the work-energy method can not be used directly • Can only use impulse-momentum method Chapter 6. Dynamics of particle systems

  3. Mechanics 6.6 Impact • General concepts • Direct central impact • Oblique central impact • Sample examples Chapter 6. Dynamics of Particle system Chapter 6. Dynamics of particle systems

  4. Line of impact 1. General concepts Impact(碰撞): A collision between two bodies whichoccurs in a veryshort time period, and during which the two bodies exert relatively large impulsive forces on each other. Examples: a hammer striking a nail, a bat striking a ball, … Line of impact(碰撞线): The common normal to the surfaces in contact during the impact. Central impact(中心碰撞) If the mass centers on the two colliding bodies are located on the line of impact, then the impact is a central impact, otherwise, the impact is said to be eccentric(偏心的) we’ll discuss central impact only Chapter 6. Dynamics of particle systems

  5. Direct central impact(正碰)occurs when the directions of motion of the two colliding particles are along the line of impact. Oblique impact(斜碰)occurs when the direction of motion of one or both of the particles is at an angle to the line of impact. 1. General concepts In general, there are twotypes of central impact: Chapter 6. Dynamics of particle systems

  6. Mechanics 6.6 Impact • General concepts • Direct central impact • Oblique central impact • Sample examples Chapter 6. Dynamics of Particle system Chapter 6. Dynamics of particle systems

  7. Before impact:both particles are moving in the same straight line and to the right with known velocity v10 and v20, |v10| > |v20| v10 v20 After impact: both particles are moving in the same straight line as before with velocities v1 and v2 respectively. v1 v2 2. Direct central impact Direct Central impacthappens when the velocities of the two objects are along the line of impact. Consider two particles of masses m1 and m2 Chapter 6. Dynamics of particle systems

  8. u 2. Direct central impact How to determine v1 and v2?  Need to study the impact process which can be divided in to two stages: Deformation stage(压缩阶段): Once the particles contact, they will be compressed by the impact force and will begin to deform. At the end of the period of deformation, the two particles will have the same velocity u Restitution stage(恢复阶段): During this stage, the particles return totally or partially to its undeformed shape. At the end, the velocities of the particles reach to v1 and v2 respectively. Chapter 6. Dynamics of particle systems

  9. 2. Direct central impact 1.1) basic formulas • First, consider the two particles as a single system Since the impact forces between the two particles are impulsive, the resultant impulse of the external forces(non-impulsive) can be neglected the total momentum of the two particles is conserved Since the motions of the particles are rectilinear chose ox axis with positive directed to the right Chapter 6. Dynamics of particle systems

  10. m1u m1v1 m1v10 m1u Pdt Rdt + + = = • Find a second relation between the scalar quantities v1 and v2: 2. Direct central impact Consider the motion of the particle m1: Deformation stage: the impulsive force acting on m1 is the force P exerted by m2 Restitution stage: R is the force exerted by m2 on m1 Chapter 6. Dynamics of particle systems

  11. Define: 2. Direct central impact Coefficient of restitution(恢复系数): the ratio of the magnitudes of the impulse corresponding, respectively, to the restitution stage and to the deformation stage In general, |Rdt| |Pdt|, so that0 e  1  A similar analysis of particle m2 leads to the relation Chapter 6. Dynamics of particle systems

  12. 2. Direct central impact vsep = v2 –v1: the velocity of separationthe rate at which the distance between the particles increase after the impact vapp = v10 –v20: the velocity of approachthe rate at which the distance between the particles decrease before the impact • For an impact to occur, vapp > 0 • If the particles are to separate after the impact(omitting the possibility that m1 could pass through m2), vsep > 0  e will generally be a nonnegative number Chapter 6. Dynamics of particle systems

  13. 2. Direct central impact Solving equation (1) and (2) for v1 and v2 Chapter 6. Dynamics of particle systems

  14. 2. Direct central impact The change in the total kinetic energy of the two particles after the impact Suppose there is no external forces acting on the two particles, then according to the work-energy principle of a particle system Aint = E Where Aint is the work done by internal forces between the particles Chapter 6. Dynamics of particle systems

  15. 2. Direct central impact According to the value of the coefficient of restitution e, the impact between two particles can be divided into three cases: • Perfectly elastic impact： e = 1 • Perfectly inelastic impact：e = 0 • Imperfectly elastic impact：0 < e < 1 Chapter 6. Dynamics of particle systems

  16. 2. Direct central impact 1.2) Perfectly elastic impact:e = 1 According to the definition of the coefficient of restitution, when e =1 From momentum conservation: Multiply these two equations member by member, rearranging the terms and multiplying by ½  The kinetic energy of the particles is conserved. Chapter 6. Dynamics of particle systems

  17. 2. Direct central impact Dynamics analysis: The forces between the particles during the impact are conservative: Elastic force: Deformation stage: kinetic energy partially converted to elastic potential energy Restitution stage: elastic potential energy completely converted into kinetic energy Chapter 6. Dynamics of particle systems

  18. 2. Direct central impact The velocities of the particles after the impact: Special cases: (1) m1 = m2 The two particles exchange their velocities after the impact Chapter 6. Dynamics of particle systems

  19. 2. Direct central impact The velocities of the particles after the impact: Special cases: (2) m1 >> m2，v20 = 0 Collision between a bowling ball and a ping-pang ball Chapter 6. Dynamics of particle systems

  20. 2. Direct central impact The velocities of the particles after the impact: Special cases: (3) m1 << m2，v20 = 0 Collision between a ping-pang ball and bowling ball Chapter 6. Dynamics of particle systems

  21. 2. Direct central impact 1.3) Perfectly inelastic impact:e = 0 Also called perfectly plastic impact According to the definition of coefficient, when e = 0 v1 = v2 = v In a plastic impact, the relative separation velocity is zero. The particles stick together and move with a common velocity after the impact. From the momentum conservation: = the velocity of the mass center Chapter 6. Dynamics of particle systems

  22. 2. Direct central impact Dynamic analysis: During the impact, the forces between the two particles are non-conservative • Deformation stage: the internal non conservative forces do negative work and the total kinetic energy decreasesconverted to other forms of energy, such as heat, sound,.. • Restitution stage: none, the particles stick together and stay permanently deformed Kinetic energy loss: Chapter 6. Dynamics of particle systems

  23. 2. Direct central impact 1.4) Imperfectly elastic impact：0 < e < 1 The two particles will separate and have different velocities after the impact • During the impact, the forces acting on the two particles include conservative forces(elastic forces) and non-conservative forces • The particles partially regain their original shape after the impact Kinetic energy loss: Chapter 6. Dynamics of particle systems

  24. Mechanics 6.6 Impact • General concepts • Direct central impact • Oblique central impact • Sample examples Chapter 6. Dynamics of Particle system Chapter 6. Dynamics of particle systems

  25. v2 n t Line of impact m2 m1 v1 v20 v10 3. Oblique central impact Oblique central impactoccurs when the direction of motion of one or both of the particles is at an angle to the line of impact. Consider two particles of mass m1 and m2 Before impact: v10and v20 After impact: v1 and v2 (unknown) Choose a n-t coordinate system: • n-axis: along the line of impact • t-axis: along the common tangent of the two particles  perpendicular to n-axis Four unknowns: v1n,v1t, v2n, v2t Chapter 6. Dynamics of particle systems

  26. m2v2 t n t n t n m2 m2 m2 + = m1 m1 -Ft m1 m2v20 m1v1 Ft m1v10 3. Oblique central impact During impact: Assume: 1) the particles are perfectly smooth and frictionless 2) There are no external forces The only impulses exerted on the particles are due to the internal forces directed along the line of impact Chapter 6. Dynamics of particle systems

  27. 3. Oblique central impact • Momentum of each particle is conserved in the direction along t-axis • The total momentum of the particles in the direction along n-axis is conserved • Define the coefficient of restitution e as Chapter 6. Dynamics of particle systems

  28. 3. Oblique central impact Solving these four independent equations for the components of the velocities of the two particles: Chapter 6. Dynamics of particle systems

  29. v2 t n Line of impact m2 m1 v1  v10 3. Oblique central impact Special case: Before impact, m2 is at rest, v20 = 0 : the angle between v10 and t direction |v10|=v0 Chapter 6. Dynamics of particle systems

  30. 3. Oblique central impact If e = 1 and m1 = m2 (perfectly elastic impact) the velocities of the two particles after the impact are always mutually perpendicular Chapter 6. Dynamics of particle systems

  31. Mechanics 6.6 Impact • General concepts • Direct central impact • Oblique central impact • Sample examples Chapter 6. Dynamics of Particle system Chapter 6. Dynamics of particle systems

  32. PROCEDURE FOR ANALYSIS 4. Sample examples • In many impact problems, the initial velocities of the particles and the coefficient of restitution, e, are known, with the final velocities to be determined. • Define the n-t axes. Typically, then-axisis definedalongthe line of impact and thet-axisis in the plane of contactperpendicular to the n-axis. • For bothdirect and obliqueimpact problems, the following equations applyalongthe line of impact (n-dir.):  m(vi0)n =  m(vi)n and e = [(v2)n – (v1)n]/[(v10)n – (v20)n] • Forobliqueimpact problems, the following equations are also required, appliedperpendicularto the line of impact (t-dir.): m1(v10)t = m1(v1)t and m2(v20)t = m2(v2)t Chapter 6. Dynamics of particle systems

  33. EXAMPLE 6.6-1 4. Sample examples Given:A 0.5 kg ball is ejected from the tube at A with a horizontal velocity vA = 2 m/s. The coefficient of restitution at B is e = 0.6. Find:The horizontal distance R where the ball strikes the smooth inclined plane and the speed at which it bounces from the plane. Plan: 1) Usekinematicsto find the distance R (projectile motion). 2) The collision at B is anoblique impact, with the line of impact perpendicular to the plane. 3) Thus, the coefficient of restitution applies perpendicular to the incline and the momentum of the ball is conserved along the incline. Chapter 6. Dynamics of particle systems

  34. vx = vxo = 2 m/s ( ) vy = vyo – gt = 0 – 9.81(1.028) = -10.0886 m/s ( ) => v = 10.285 m/s 78.8° Solution: 4. Sample examples 1)Apply the equations of projectile motionto determine R. Place the origin at A (xo = yo = 0) with the initial velocity of vyo = 0, vxo = vA = 2 m/s: x = xo + vxot => R = 0 + 2t y = yo + vyot – ½ gt2 => -(4 + R tan30°) = 0 + 0 – 0.5(9.81)t2 Solving these equations simultaneously yields t = 1.028 s and R = 2.06 m It is also necessary to calculate the velocity of the ball just before impact: Chapter 6. Dynamics of particle systems

  35. vA1 y x 48.8° vA2 30° The speed is themagnitudeof the velocity vector: vA2 = ((vAx)2)2 + ((vAy)2)2 = 8.21 m/s 2)Solve the impact problem by usingx-y axes defined along and perpendicular to the line of impact, respectively: 4. Sample examples Denoting the ball as A and plane as B, the momentum of the ball is conserved in the y-dir: mA(-vAy)1 = mA(-vAy)2 (vAy)2 = (vAy)1 = vA cos48.8° = 6.77 m/s The coefficient of restitution applies in thex-dirand(vBx)1 = (vBx)2 = 0: e = [(vBx)2 – (vAx)2]/[(vAx)1 – (vBx)1] => 0.6 = [0 - (vAx)2]/[-10.285 sin48.8 – 0] => (vAx)2 = 4.64 m/s Chapter 6. Dynamics of particle systems

  36. Example 6.6-2 4. Sample examples Given:A 2 kg block A is released from rest, falls a distance h = 0.5 m, and strikes plate B (3 kg mass). The coefficient of restitution between A and B is e = 0.6, and the spring stiffness is k = 30 N/m. Find:The velocity of block A just after the collision. Which directions do they move? Plan:1) Determine the speed of the block just before the collision using projectile motion or an energy method. 2) Analyze the collision as a central impact problem. Chapter 6. Dynamics of particle systems

  37. v2 = 3.132 m/s 4. Sample examples Solution: 1)Determine the speed of block Ajust before impact by using conservation of energy. Defining the gravitational datum at the initial position of the block (h1 = 0) and noting the block is released from rest (v1 = 0): ½ m(v1)2 + mgh1 = ½ m(v2)2 + mgh2 0 + 0 = 0.5(2)(v2)2 + (2)(9.81)(-0.5) This is the speed of the block just before the collision. Plate (B) is at rest, velocity of zero, before the collision. Chapter 6. Dynamics of particle systems

  38. (vA)2 (vA)1 = 3.132 m/s A B + mA(vA)1 + mB(vB)1 = mA(vA)2 + mB(vB)2 (2)(-3.132) + 0 = (2)(vA)2 + (3)(vB)2 (vB)2 (vB)1 = 0 Using the coefficient of restitution: + e = [(vB)2 – (vA)2]/[(vA)1 – (vB)1] => 0.6 = [(vB)2 – (vA)2]/[-3.132 – 0] => -1.879 = (vB)2 – (vA)2 Solving the two equations simultaneously yields (vA)2 = -0.125 m/s , (vB)2 = -2.00 m/s Both the block and plate will travel down after the collision. 4. Sample examples 2) Analyze the collision as a central impact problem. Apply conservation of momentum to the system in the vertical direction: Chapter 6. Dynamics of particle systems

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