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10 - 1 KINETICS OF PARTICLES: ENERGY AND MOMENTUM METHODS s2s2 A1A1 A2A2 A s1s1 s drdr F  ds Consider a force F acting on a particle A. The work of F.

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Presentation on theme: "10 - 1 KINETICS OF PARTICLES: ENERGY AND MOMENTUM METHODS s2s2 A1A1 A2A2 A s1s1 s drdr F  ds Consider a force F acting on a particle A. The work of F."— Presentation transcript:

1 10 - 1 KINETICS OF PARTICLES: ENERGY AND MOMENTUM METHODS s2s2 A1A1 A2A2 A s1s1 s drdr F  ds Consider a force F acting on a particle A. The work of F corresponding to the small displacement dr is defined as dU = F dr Recalling the definition of scalar product of two vectors, dU = F ds cos  where  is the angle between F and dr.

2 10 - 2 s2s2 A1A1 A2A2 A s1s1 s drdr F  ds dU = F dr = F ds cos  The work of F during a finite displacement from A 1 to A 2, denoted by U 1 2, is obtained U 1 2 = F dr ٍ A1A1 A2A2 by integrating along the path described by the particle. For a force defined by its rectangular components, we write U 1 2 = (F x dx + F y dy + F z dz) ٍ A1A1 A2A2

3 10 - 3 U 1 2 = - Wdy = Wy 1 - Wy 2 y1y1 y2y2 y dy W A1A1 A2A2 The work of the weight W of a body as its center of gravity moves from an elevation y 1 to y 2 is obtained by setting F x = F z = 0 and F y = - W. y1y1 ٍ y2y2 The work is negative when the elevation increases, and positive when the elevation decreases. A

4 10 - 4 U 1 2 = - k x dx x 1 ٍ x 2 =kx 1 - kx 2 B B x 1 A 1 A 2 A A O F B x 2 x The work of the force F exerted by a spring on a body A during a finite displacement of the body from A 1 ( x = x 1 ) to A 2 ( x = x 2 ) is obtained by writing 2 2 1 2 1 2 The work is positive when the spring is returning to its undeformed position.. spring undeformed dU = -Fdx = -kx dx

5 10 - 5 r1r1 r2r2 M m F A -F O A’A’ A1A1 A2A2 dr r  dd The work of the gravitational force F exerted by a particle of mass M located at O on a particle of mass m as the latter moves from A 1 to A 2 is obtained from ٍ r2r2 r1r1 U 1 2 = GMm r 2 dr GMm r 2 GMm r 1 = -

6 10 - 6 The kinetic energy of a particle of mass m moving with a velocity v is defined as the scalar quantity T = mv 2 1212 From Newton’s second law the principle of work and energy is derived. This principle states that the kinetic energy of a particle at A 2 can be obtained by adding to its kinetic energy at A 1 the work done during the displacement from A 1 to A 2 by the force F exerted on the particle: T 1 + U 1 2 = T 2

7 10 - 7 The power developed by a machine is defined as the time rate at which work is done: Power = = F v dU dt where F is the force exerted on the particle and v is the velocity of the particle. The mechanical efficiency, denoted by , is expressed as  = power output power input

8 10 - 8 When the work of a force F is independent of the path followed, the force F is said to be a conservative force, and its work is equal to minus the change in the potential energy V associated with F : U 1 2 = V 1 - V 2 The potential energy associated with each force considered earlier is Force of gravity (weight): Gravitational force: Elastic force exerted by a spring: V g = Wy V g = - GMm r V e = kx 2 1212

9 10 - 9 U 1 2 = V 1 - V 2 T 1 + V 1 = T 2 + V 1 This relationship between work and potential energy, when combined with the relationship between work and kinetic energy ( T 1 + U 1 2 = T 2 ) results in This is the principle of conservation of energy, which states that when a particle moves under the action of conservative forces, the sum of its kinetic and potential energies remains constant. The application of this principle facilitates the solution of problems involving only conservative forces.

10 10 - 10 O r v r0r0 P0P0 v0v0 P 00  When a particle moves under a central force F, its angular momentum about the center of force O remains constant. If the central force F is also conservative, the principles of conservation of angular momentum and conservation of energy can be used jointly to analyze the motion of the particle. For the case of oblique launching, we have (H O ) 0 = H O : r 0 mv 0 sin  0 = rmv sin  T 0 + V 0 = T + V : mv 2 - = mv 2 - 0 GMm r 0 GMm r 1212 1212 where m is the mass of the vehicle and M is the mass of the earth.

11 10 - 11 The linear momentum of a particle is defined as the product mv of the mass m of the particle and its velocity v. From Newton’s second law, F = ma, we derive the relation ٍ mv 1 + F dt = mv 2 t1t1 t2t2 where mv 1 and mv 2 represent the momentum of the particle at a time t 1 and a time t 2, respectively, and where the integral defines the linear impulse of the force F during the corresponding time interval. Therefore, mv 1 + Imp 1 2 = mv 2 which expresses the principle of impulse and momentum for a particle.

12 10 - 12 When the particle considered is subjected to several forces, the sum of the impulses of these forces should be used; Since vector quantities are involved, it is necessary to consider their x and y components separately. mv 1 +  Imp 1 2 = mv 2 The method of impulse and momentum is effective in the study of impulsive motion of a particle, when very large forces, called impulsive forces, are applied for a very short interval of time  t, since this method involves impulses F  t of the forces, rather than the forces themselves. Neglecting the impulse of any nonimpulsive force, we write mv 1 +  F  t = mv 2

13 10 - 13  mv 1 +  F  t =  mv 2 In the case of the impulsive motion of several particles, we write where the second term involves only impulsive, external forces. In the particular case when the sum of the impulses of the external forces is zero, the equation above reduces to  mv 1 =  mv 2 that is, the total momentum of the particles is conserved.

14 10 - 14 In the case of direct central impact, two colliding bodies A and B move along the line of impact with velocities v A and v B, respectively. Two equations can be used to determine their velocities v ’ A and v ’ B after the impact. The first represents the conservation of the total momentum of the two bodies, m A v A + m B v B = m A v’ A + m B v’ B A B vAvA vBvB Line of Impact A B v’Av’A v’Bv’B Before Impact After Impact

15 10 - 15 m A v A + m B v B = m A v’ A + m B v’ B The second equation relates the relative velocities of the two bodies before and after impact, v’ B - v’ A = e (v A - v B ) The constant e is known as the coefficient of restitution; its value lies between 0 and 1 and depends on the material involved. When e = 0, the impact is termed perfectly plastic; when e = 1, the impact is termed perfectly elastic. A B vAvA vBvB Line of Impact A B v’Av’A v’Bv’B Before Impact After Impact

16 10 - 16 A B vAvA vBvB In the case of oblique central impact, the velocities of the two colliding bodies before and after impact are resolved into n components along the line of impact and t components along the common tangent to the surfaces in contact. In the t direction, m A (v A ) n + m B (v B ) n = m A (v’ A ) n + m B (v’ B ) n Line of Impact A B v’Av’A v’Bv’B Before Impact After Impact n t vAvA vBvB (v A ) t = (v’ A ) t (v B ) t = (v’ B ) t (v’ B ) n - (v’ A ) n = e [(v A ) n - (v B ) n ] while in the n direction n t

17 10 - 17 A B vAvA vBvB m A (v A ) n + m B (v B ) n = m A (v’ A ) n + m B (v’ B ) n Line of Impact A B v’Av’A v’Bv’B Before Impact After Impact n t vAvA vBvB (v A ) t = (v’ A ) t (v B ) t = (v’ B ) t (v’ B ) n - (v’ A ) n = e [(v A ) n - (v B ) n ] n t Although this method was developed for bodies moving freely before and after impact, it could be extended to the case when one or both of the colliding bodies is constrained in its motion.


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