(6.1) CHAPTER 6 IMPULSE, MOMENTUM, AND ENERGY 6.1 INTRODUCTION In Chapters 4 and 5, efforts were directed toward analyzing the transient response and parametric relationships of a dynamic system under impact and/or excitation conditions. The basis for modeling such a dynamic system is Newton’s Second Law. In this chapter, the principle of impulse and momentum and the principle of energy derived from Newton’s Second Law are utilized to solve impulsive loading problems. The solutions to such dynamic problems do not directly involve the time variable. On the subject of impulse and momentum, the basic principles are reviewed first. This is followed by the application of the CG (center of gravity) motion theorem in an analysis of multiple vehicle collisions and the circle of constant acceleration (COCA) for the g-loading of vehicles. Impulsive loading can be concentrated or distributed. The distributed loading can be analyzed because the superposition method is applicable to linear systems. Specific design analysis, such as the selection of a location for an air bag crash sensor, is presented. In the analysis of vehicle/occupant collisions without using time as a reference variable, it is imperative that both principles be used. This is because collisions between multiple objects involve energy loss, and the principle of impulse and momentum provides information about the kinetic and the absorbed structural energies being transferred. In modeling component tests, the parametric relationships among the effective weight of the energy absorber, the mass ratio, and the coefficient of the restitution are presented. Methods of determining vehicle inertia properties, such as the CG height and moment of inertia of a vehicle, are covered. The formulations of critical sliding velocity (CSV), rollover dynamics, and detection of an incipient rollover using a simple vehicle model are introduced. Extending the local coordinate based COCA method, a vector operation in a global coordinate system is used to analyze the acceleration response at any location in a vehicle subjected to eccentric loading. The steps used to construct a vector diagram to analyze the loading distribution in the vehicle are described. The outcome of the eccentric loading analysis helps identifying where the largest loading occurs so that counter-measures can be taken. The vector method is also presented to determine the roll over kinematics including the geometric conditions for a vehicle to stop rolling after one side of vehicle hitting on the ground. 6.2 BACKGROUND Before presenting applications of the principles of impulse and momentum, definitions of these important dynamic variables will be made: Momentum is a vector quantity pertaining to the motion of an object; its magnitude is the mass of the object times its speed, and its direction is the direction of motion. The units of momentum are mass times speed [kg @ m/s]. The symbol usually used for momentum is “p”. Impulse is a vector quantity pertaining to a force; it is the integral of the force over a specified period of time. Usually, the time period is very short, as in the impact occurring during a crash. The units are the same as for momentum. The symbol for impulse is usually an “ I ”. The effect of an impulse on a body is a change in the momentum (a vector quantity) of the body: )p = I. Then, for the momentum before and after an impulse, one can write: © 2002 by CRC Press LLC Fig. 6.1 Resultant Force and Acceleration (6.2) (6.3) Fig. 6.2 Vector Addition of Momentum and Impulse 6.2.1 Impulse and Momentum for a Single Particle For a body acted on by several forces, as shown in Fig. 6.1, the forces can be summed vectorially, giving one resultant force. This produces an acceleration in the direction of the resultant force. Then, using Newton’s Second Law: For a single particle, with an impulse acting on it, we have : p 1 + I = p 2 . This also can be stated as : Graphically, this equation can be depicted as shown below: Case Study: Batting A Baseball. A baseball player sees a ball thrown at him horizontally. He intends to hit the ball in such a way that the ball will take off at an angle of 40° and at a speed of 120 ft/sec, as shown in Fig. 6.3. The given information is as follows. weight of the ball (W) = 4 oz incoming speed (v 1 ) = 80 ft/sec outgoing speed (v 2 ) = 120 ft/sec ball-bat contact duration ( )t) = .015 sec © 2002 by CRC Press LLC Fig. 6.3 Batting a Baseball (6.4) Fig. 6.4 Vector Operation of Batting a Baseball (6.5) Determine the upward angle " that he has to swing the bat and what the impulse force F should be to achieve the desired outcome. Using the principle of impulse and momentum, one gets a vector relationship. The triangle which defines the vector relationships is shown in Fig. 6.4. Since the initial and final momentum vectors are given, the two unknowns are the magnitude of the force and angle ". Assuming the force is constant, the magnitude of the impulse can then be measured from the plot or computed by using Eq. (6.4): Note on statistics: 1. An official major league baseball weighs 5.25 ounces and is sewn with 116 stitches. The coefficient of restitution of the ball, striking a board made of white ash at a speed of 85 ft/s, is 0.546 ± 3.2%. The percent of kinetic energy loss due to impact is (1 - e 2 ) x 100% = 70.2 %. © 2002 by CRC Press LLC Fig. 6.5 External and Internal Forces Acting on a Rigid Body Fig. 6.6 External and Internal Forces Acting on Particle i (6.6) (6.7) For average major league players: 2. Ball speed = 90 mph = 132 ft/s 3. Duration of flight = distance/time = 54 ft / 132 ft/s = .41 sec 4. Time taken swinging bat (until contact with ball) = .28 sec 5. Time available for decision to swing or not = .13 sec Since it takes two-thirds of the flight time of the ball to swing a bat, there is very little time left to make a decision. 6.2.2 Impulse and Momentum for a System of Particles For a system of particles, a similar expression can be stated, except that the velocities are replaced by the velocity of the center of mass of the particles. The center of mass is a point in the system which will be defined later. There are both internal and external forces acting on each particle in the system as shown in Fig. 6.5. Internal forces are those exerted by other particles in the system. External forces are those produced by elements not part of the system (e.g., for a vehicle, the external forces are the force of gravity, frictional forces arising from tires contacting the road, impact forces due to a collision with an obstacle, etc.). The equation of motion for particle i in the system shown in Fig. 6.6 is: The sum over j is for the internal forces associated with other particles. Summing up over all particles in the system gives: Note that as a result of Newton’s Third Law, the internal forces all cancel out, since for every internal force given by (i, j) there is an internal force with subscript (j, i) that is of equal magnitude but in the opposite direction. 6.3 CENTER OF GRAVITY AND MOTION THEOREM By recognizing the relationship between the formula for the coordinates of the center of mass (CM or center of gravity, CG) and the CG motion of a system of particles, the solution to some difficult accident reconstruction problems can be undertaken with ease. This is due to the fact that if the external impulse to a system of particles is negligible; the CG motion of the system with and without collision is the same. © 2002 by CRC Press LLC (6.8) (6.9) (6.10) (6.11) Fig. 6.7 Vector Operation of Linear Impulse and Momentum Changes 6.3.1 Location and Motion of Center of Mass In the following, the terms “center of mass” (CM) and “center of gravity” (CG) will be used interchangeably. The vector position of the center of mass is given by: Substituting these expressions of the position vectors into the CG (center of gravity) formula, one gets for the components of the position vector of the CG: Differentiating the CG formula shown in Eq.(6.8) twice yields its velocity and acceleration: The acceleration equation, (2) of Eq.(6.10), can be written as: In words, this states that the final momentum is equal to the initial momentum plus the sum of all the impulses. Note that each term is a vector quantity. Graphically, this equation can be presented by the vector operation shown in Fig. 6.7. © 2002 by CRC Press LLC (6.12) Fig. 6.8 CG Motion of Two Particles (6.13) (6.14) When the sum of all impulses is zero, we get the classical conservation of momentum law: (1) of Eq. (6.12) is in terms of the center of mass, and (2) is in terms of a system of n particles: 6.3.2 Conservation of Momentum and CG Formula Assuming a collision of a system of particles is totally inelastic and the external frictional impulse is negligible, then the movement of the center of mass of the system of particles after collision is the same as that without collision. If there is an inelastic impact between m 1 and m 2 , then the distance traveled by both particles in a given time t is equal to v c t, as shown in Fig. 6.8. If there were no collision, the two particles would simply pass by each other and travel in time t a distance of v 1 t and v 2 t, respectively. It will be shown that the center of mass of the two particles in free travel at any given time t is at the same location as when the two particles travel together. Since the initial momentum of the particles is the same as that at the time of common velocity, we have the momentum relationship shown in Eq. (6.13). I. One-Dimensional Analysis (a) Conservation of Momentum (before and after collision) The common velocity of the two particles after impact can be derived as shown in Eq. (6.14). (b) Without Collision (CG location at time t) The x coordinate of the CG position of a system of particles at time t is shown in Eq. (6.15). In the derivation, the position of each particle is equal to its speed multiplied by the time elapsed. © 2002 by CRC Press LLC (6.15) (6.16) (6.17) (6.18) Eq. (6.15) shows that the displacement of the CG of a system of particles with collision is the same as that without a collision. II. Multi-Dimensional (Vector) Analysis The position vector relationship between the particles and the CG of the system of particles is shown in (1) of Eq. (6.16). After differentiating the position vector formula, one gets the relationship shown in (2) of Eq. (6.16) which is the conservation of momentum for a system of two particles. 6.3.3 CG Motion Theorem Theorem: In the absence of any external forces, the center of mass of a system of particles will move in a straight line at a constant speed. Any collisions between the particles, elastic or inelastic, will have no effect on this motion. Proof : The position vector of the center of mass is defined as follows: The total momentum of the system is equal to the product of the total mass and the velocity of the center of mass. This can be called the momentum of the center of mass. Keep in mind that the center of mass is not a mass but a point. It is possible that a real mass doesn’t even exist at the point! Taking the next derivative: © 2002 by CRC Press LLC (6.19) (6.20) (6.21) (6.22) (6.23) Fig. 6.9 Two Vehicles with and without Collision Now applying Newton’s Second Law to each term on the right: F i is the net external force acting on particle i and f i is the net internal force acting on it. Summing over all particles gives: The f ij ’s are internal forces and, based on Newton’s Third Law, for each of these there is an equal and opposite reaction force acting on another particle. This includes any forces developed in a collision between any two or more particles in the system. In summing over i and j, all of these forces cancel out, leaving no net internal force. Thus, dropping the sum involving the internal forces, one gets: In the case wherein there are no external forces, one gets: Solving this differential equation leads to: This is the equation of a straight line whose coefficient is a constant speed. Thus: In the absence of external forces, the center of mass of a system of particles will move in a straight line at constant speed, independently of any collisions of the particles with each other. Case Study 1: CG of Two Vehicles With and Without Collision . Two vehicles having masses m 1 and m 2 are traveling on a horizontal icy surface with velocities of v 1 and v 2 , respectively. Assume that they collide and after the collision, their separation velocities are v 1 ' and v 2 ' , respectively. Show that the velocities of the center of the mass (CM or CG) before and after collision are equal. © 2002 by CRC Press LLC (6.24) Fig. 6.10 Car & Truck Head-On Fig. 6.11 Displacement (Travel) versus Time Case Study 2: CG Motion of Two Cars With And Without Collision . The following problem is an exercise for students: The reference values of the mass and velocity of vehicle #2 are m 2 = 1 and v 2 = 1, respectively. The values of those for the subject vehicle #1 are relative to #1 and are to be filled in. m 1 = v 1 = m 2 = 1. v 2 = 1. (1) Write down below the values of m 1 and v 1 from the vehicle-to-vehicle crash experiment. (2) On the x vs. t (displacement vs. time) plot, draw a line segment from the lower left corner (0,0) to the right border where the slope is velocity v 2 . Similarly, draw another line segment from the upper left corner (0,20) to the right border for the velocity v 1 . (3) From the intersection of the two line segments, draw another line segment where the slope is the common velocity of the two vehicles, v c , which is defined as follows: © 2002 by CRC Press LLC (6.25) (6.26) Fig. 6.12 Intersection Collision (4) Extending the common velocity line segment to the left and right borders, one gets a line of C.M. (or CG) motion of the two vehicles before and after the inelastic impact where no vehicle separation occurs. (5) To verify the line of CG motion, compute the CG location of the two vehicles at any point in time using the following CG position formula. where x 1 and x 2 are the x positions of the two vehicles measured along x from (0,0) at any point in time. Check whether x CG falls right on the common velocity line segment, and what does Eq. (6.26) become after being differentiated w.r.t. time? Why ? Case Study 3: Collision at Intersection. Car A and Cruiser B of equal mass collide at a right angle at the intersection of two icy roads. The cars become entangled and move off together with a common velocity v c in the direction shown. If the Cruiser B was traveling 30 mph at the instant of impact, compute the corresponding velocity of car A just prior to the collision. [Ans.: v A =52 mph] 6.3.4 Use of CG Motion Theorem in a Three !Car Collision Analysis Car A was a police car traveling east at 60 mph; it was hit by car B, which was going south at a high speed. The collision occurred at the middle of an intersection (the origin in the accompanying diagram). The pavement was wet and the two cars slid together and collided with another police car (car C). Car C was traveling north at 45 mph. The three cars stuck together and skidded into a wall and came to a stop at point D in the figure. Point D was 42 feet east and 46.5 feet south of the intersection. Cars A and C weighed 3000 lbs and car B weighed 3600 lbs. Car C was 63 feet south of the intersection when the first collision occurred. Car C was determined to be 20 feet to the right of the intersection before and during its collision with car A and B. Determine the speed of car B, v B , and the time elapsed from the time of the first collision to the stop at point D. Neglect any forces exerted on the cars by the wet pavement. © 2002 by CRC Press LLC [...]... of a vehicle subjected to side, frontal, and offset impulsive loadings can be examined by the use of COCA Under the same impulsive loading to the vehicle, different locations in the vehicle would bear different acceleration magnitudes and different acceleration directions By plotting the acceleration magnitudes and directions, it becomes easy to compare the impact severity at any point on the vehicle. .. speed of the CG of the three vehicles can also be computed from each of the two phases of collisions: the first phase from F to E has a constant speed of |FE| / ªt1 = 17.2 ft / 5 s = 34.4 ft/s, and the second phase from E to D has a constant speed of |ED| / ªt2 = 27.5 ft / 8 s = 34.4 ft/s The constant speed of the CG of the three vehicles is the common speed of all the three vehicles involved in the collision... superposition technique Therefore, offset impact severity can be analyzed by the COCA method In developing an air bag crash sensor system, judging the impact severity at any point on the vehicle can be expedited by examining the COCA for the magnitude and direction Fig 6.29 shows a vehicle subjected to a concentrated loading at the right front corner similar to the tree impact accident described in... front and right near-center of the vehicle For each impulsive loading, two COCA’s, passing through points Q1 and Q2 at the left and right sides of the vehicle, are shown The resultant of the right front and right near-center loadings is shown in Fig 6.34 Two COCA’s due to the resultant loading are shown passing through points Q1 and Q2, at the left and right sides of the vehicle Fig 6.35 shows a distributed... 0 to 1, the velocity of car B, vB0, before impact can then be computed (6.63) Case Study 3: Vehicle- to-Barrier (VTB) Impact Using the principle of work and energy, derive the relationship between the dynamic crush, barrier impact speed, and the circular natural frequency of the vehicle system Just before the vehicle contacts the rigid barrier, its velocity is v0 , as shown in position 1 of Fig 6.39... ratios (c) of 0.7, 1.0, and 1.3 The outer circle, having a c of 1.3, passes through point N on the right side of the vehicle, and the inner circle, having a c of 0.7, passes through a point opposite to N on the left side The middle circle, having a c of 1.0, simply passes through the CG of the vehicle Note that from the similar triangles, the ratio of acceleration at point N to that at CG is equal to the... on the flight recorder data) is low, the crash sensor located in the right hand side (on the same side as the impulsive loading) near point N in Fig 6.29 was activated This is evident from the outer COCA plot which shows that the acceleration at point N is relatively high, 30 % higher than that at the CG and 2 times higher than that at the opposite side If the crash sensor were located on the opposite... than that at the opposite side If the crash sensor were located on the opposite side of the impulsive loading, the acceleration at the sensor location would be low relative to that of the CG, and the crash sensor might not be activated This type of accident scenario does happen and has been observed in a real world accident An L-type left front side impact and COCA’s with acceleration ratios of 0.5,... second equation can be solved for vB , yielding vB = 110 feet/second Then, the speed of car B is 110 feet/second or 75 MPH According to the CG motion theorem presented in Section 6.3.3, the CG of the three vehicles moves in a straight line (FED) as shown in Fig 6.14 at constant speed The constant speed will be computed using the following method for the detailed analysis Method II: In-Depth Analysis Using... forces Therefore, applying the energy formula (1) of Eq (6.60) yields the relationship between the kinetic energy and energy lost due to frictional force The computations are shown in Eq (6.62) Fig 6.37 Vehicle Skidding to Stop (6.62) Case Study 2: Shown in Fig 6.38 are two cars, A and B, each having a weight of 4000 lb, that collide on an icy pavement at an intersection The direction of motion of each . of determining vehicle inertia properties, such as the CG height and moment of inertia of a vehicle, are covered. The formulations of critical sliding velocity (CSV), rollover dynamics, and detection. v 1 = m 2 = 1. v 2 = 1. (1) Write down below the values of m 1 and v 1 from the vehicle- to -vehicle crash experiment. (2) On the x vs. t (displacement vs. time) plot, draw a line segment. motion of the two vehicles before and after the inelastic impact where no vehicle separation occurs. (5) To verify the line of CG motion, compute the CG location of the two vehicles at any point