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Mechanics

Mechanics. Chapter 6 Dynamics of particle system 6.1 Kinetics: Force-Mass-Acceleration Method 动力学：力 - 质量 - 加速度方法. Section 15.4. 6.1 Kinetics: Force-Mass-Acceleration Method. Using Newton’s second law to study the motion of a particle system:.

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Mechanics

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  1. Mechanics Chapter 6 Dynamics of particle system 6.1 Kinetics: Force-Mass-Acceleration Method 动力学：力-质量-加速度方法 Section 15.4 Chapter 6. Dynamics Of particle systems

  2. 6.1 Kinetics: Force-Mass-Acceleration Method Using Newton’s second law to study the motion of a particle system: • Use theFBDandMADfor each particle in the system to obtain the equations of motion of the individual particles  n vector equations of motion • Kinematic constraints: due to massless connections,such as ropes joining the particles • Solve the n vector equations and the constraint equations to get the unknownsnot easy Chapter 6. Dynamics Of particle systems

  3. 6.1 Kinetics: Force-Mass-Acceleration Method If we are only interested in its overall motion, we can treat the particle system as a point particle with all of its mass located at the center of mass point. The motion of the center of mass is exactly the same as that of a true point particle of equal mass subjected to a force equal to the vector sum of the external forces. Chapter 6. Dynamics Of particle systems

  4. Mechanics 6.1 Kinetics: Force-Mass-Acceleration Method • Center of mass(质心) • External and internal forces • Equation of motion of mass center Chapter 6 Dynamics of particle system Chapter 6. Dynamics Of particle systems

  5. 1. Center of mass mi C ri rC o Consider a system of n particles. O-xyz is an inertial reference frame System of n particles For the i th particle in the system Mass: mi Position vector:ri = xii+yij+zik Velocity:vi Acceleration:ai The center of mass C of the system is defined to be the point whose position vector is Total mass of the particle system Chapter 6. Dynamics Of particle systems

  6. 1. Center of mass The rectangular components ofrc: Differentiatingrcwith respect to time • Velocity of center of mass: • Acceleration of center of mass: Chapter 6. Dynamics Of particle systems

  7. 1. Center of mass m1 m2 rC r1 r2 y x If the system is made up of only two particles: If m1 = m2 , m = m1 + m2 = 2m So: CoM is halfway between the masses. Chapter 6. Dynamics Of particle systems

  8. 1. Center of mass Sample example 6.1-1 GIVEN:Three particles of masses m1 = 1.2 kg, m2 = 2.5 kg, and m3 = 3.4 kg form an equilateral triangle of edge length a = 140 cm. FIND: Where is the center of mass of this three-particle system? Chapter 6. Dynamics Of particle systems

  9. 1. Center of mass Particle   x (cm)   y (cm)   Mass (kg)   1   1.2   0   0   2   2.5   140   0   3   3.4   70   121   SOLUTION Chapter 6. Dynamics Of particle systems

  10. 1. Center of mass Chapter 6. Dynamics Of particle systems

  11. Mechanics 6.1 Kinetics: Force-Mass-Acceleration Method • Center of mass • External and internal forces • Equation of motion of mass center Chapter 6 Dynamics of particle system Chapter 6. Dynamics Of particle systems

  12. 2. External and internal forces F2 Fn f f´ F1 Fi F3 The forces acting on each of the particles in the system may be classified into external and internal forces: External forces:any force on the particles inside the system from bodies outside the system  the sources of the forces are external to the system Internal forces:any force caused by the interactions between particles inside the system Example:two particles connected by a spring,.. Chapter 6. Dynamics Of particle systems

  13. 2. External and internal forces i fij z fji j y o x Characteristic of internal force: According to Newton’s third law, internal forces always occur as pairs of forces that are equal in magnitude, opposite in direction, and have collinear line of action. (Strong law of action-reaction) Consider two particles i and j in the system: fij: internal force acting on the ith particle caused by jth particle fji: internal force acting on the jth particle caused by ith particle fijandfji are along the line joining the two particles fij= -fji ( i  j ) Chapter 6. Dynamics Of particle systems

  14. 2. External and internal forces z Fi mi ri o y x The free body diagram for the ith particle Fi: The resultant external force fi : The resultant internal force sum of the forces applied to the ith particle by all the other particles in the system i = jis excluded becausefii, the force exerted on the ith by the ith particle, is meaningless. The resultant internal force acting on the system Chapter 6. Dynamics Of particle systems

  15. 2. External and internal forces The resultant internal force acting on the system is zero:Fint.= 0 Proof: • Simple consideration: • The internal forces occur in equal and opposite pairs • The resultant of each pair is zero • Therefore, their sum vanishesFint.= 0 Chapter 6. Dynamics Of particle systems

  16. 2. External and internal forces • Vector operation If we setfii=0, i = 1,2,…,n, then the sum runs over all particles in the system Since i and j are being summed over they are dummy indices and hence we can i  j and write Since summation order doesn’t matter Chapter 6. Dynamics Of particle systems

  17. 2. External and internal forces Sincefij= -fji by Newton’s 3rd law Chapter 6. Dynamics Of particle systems

  18. 2. External and internal forces An example with three particles F13 F31 1 3 F12 F32 F23 F21 2 Fint,i = Fij = (F12+F21) + (F23+F32) + (F13+F31) = 0 + 0 + 0 = 0 Chapter 6. Dynamics Of particle systems

  19. Mechanics 6.1 Kinetics: Force-Mass-Acceleration Method • Center of mass • External and internal forces • Equation of motion of mass center Chapter 6 Dynamics of particle system Chapter 6. Dynamics Of particle systems

  20. 3. Equation of motion of mass center z Fi mi z ri mi o y ri x o y x For the ith particle in the system ==  n vector equations Chapter 6. Dynamics Of particle systems

  21. 3. Equation of motion of mass center Resultant external force Resultant internal force = zero Summing all n equations the equation of motion of the mass center Chapter 6. Dynamics Of particle systems

  22. 3. Equation of motion of mass center Thecenter of massof a system of particles moves as if it were aparticleof mass equal to the total mass of the system, acted on by the resultant of the external forces acting on the system Chapter 6. Dynamics Of particle systems

  23. 3. Equation of motion of mass center • A fireworks rocket explodes in flight. In the absence of air drag, the center of mass of the fragments would continue to follow the original parabolic path, until fragments began to hit the ground. Chapter 6. Dynamics Of particle systems

  24. 3. Equation of motion of mass center Note: • Fiis the net force of allexternal forces that act on the system. Internal forces are not included. • m is the total mass of the system. We assume that no mass enters or leaves the system as it moves, so that m remains constant. The system is said to be closed. • aC is the acceleration of thecenter of massof the system. The equation gives no information about the acceleration of any other point of the system Chapter 6. Dynamics Of particle systems

  25. 3. Equation of motion of mass center Things need to be remembered: • Center of mass of the particle system: • Velocity and acceleration of center of mass: • The equation of motion of the mass center 6.2 Work Energy principle Next: Chapter 6. Dynamics Of particle systems

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