Dynamics of Mechanical Systems 2009 Part 8 docx

332 Dynamics of Mechanical Systems If we multiply the terms of Eq. (10.6.12) by v G , the velocity of G in R, we have: (10.6.14) Similarly, if we multiply the terms of Eq. (10.6.13) by ωω ωω we have: (10.6.15) By adding the terms of Eqs. (10.6.14) and (10.6.15), we obtain: (10.6.16) From Eqs. (10.4.1), (10.4.2), and (10.4.3) we can recognize the left side of Eq. (10.6.16) as the derivative of the work W of the force system acting on B. Also, from Eq. (10.5.7) we recognize the right side of Eq. (10.6.16) as the derivative of the kinetic energy K of B. Hence, Eq. (10.6.16) takes the form: (10.6.17) Then, by integrating, we have: (10.6.18) where, as in Eqs. (10.6.5) and (10.6.10), K 2 and K 1 represent the kinetic energy of B at the beginning and end of the time interval that forces are acting on B. Equations (10.6.5), (10.6.10), and (10.6.18) are expressions of the principle of work and kinetic energy for a particle, a set of particles, and a rigid body, respectively. Simply stated, the work done is equal to the change in kinetic energy. In the remaining sections of this chapter we will consider several examples illustrating application of this principle. We will also consider combined application of this principle with the impulse–momentum principles of Chapter 9. 10.7 Elementary Example: A Falling Object Consider first the simple case of a particle P with mass m released from rest at distance h above a horizontal surface S as in Figure 10.7.1. The objective is to determine the speed v of P when it reaches S. Fv a v v⋅= ⋅=     GGG G M d dt M 1 2 2 TI I⋅=⋅ () ⋅+ ×⋅ () [] ⋅= ⋅⋅     ωωααωωωωωωωωωωωωI d dt 1 2 0 Fv T v I⋅+⋅=     +⋅⋅     GG d dt M d dt ωωωωωω 1 2 1 2 2 dW dt dK dt = WK K K=−= 21 ∆ 0593_C10_fm Page 332 Monday, May 6, 2002 2:57 PM Introduction to Energy Methods 333 Because gravity is the only force applied to P and because the path of movement of P is parallel to the weight force through a distance h, the work done W is simply: (10.7.1) Because P is released from rest, its initial kinetic energy is zero. When P reaches S, its kinetic energy may be expressed as: (10.7.2) Then, from the work–energy principle of Eq. (10.6.5), we have: (10.7.3) Solving for v, we obtain the familiar result: (10.7.4) 10.8 Elementary Example: The Simple Pendulum Consider next the simple pendulum depicted in Figure 10.8.1 where the mass m of the pendulum is concentrated in the bob P which is supported by a pinned, massless rod of length ᐉ as shown. Let θ measure the inclination of the pendulum to the vertical. Suppose the pendulum is held in a horizontal position and released from rest. The objective is to determine the speed v of P as it passes through the equilibrium position θ = 0. If we consider a free-body diagram of P as in Figure 10.8.2, we see that of the two forces applied to P the tension of the connecting rod does no work on P because its direction is perpen- dicular to the movement of P. From Eq. (10.2.18), we see that the work W done by the weight force is: (10.8.1) FIGURE 10.7.1 A particle released from rest in a gravitational field. h S O P(m) W mgh= Kmv= 1 2 2 W K mgh mv==∆ or 1 2 2 vgh= 2 Wmg= l 0593_C10_fm Page 333 Monday, May 6, 2002 2:57 PM 334 Dynamics of Mechanical Systems Because the pendulum is released from rest its initial kinetic energy is zero. Its kinetic energy K as it passes through the equilibrium position may be expressed as: (10.8.2) From the work–energy principle of Eq. (10.6.5), we then have: (10.8.3) or (10.8.4) Observe that this speed is the same as that of an object freely falling through a distance ᐉ (see Eq. (10.7.4)), even though the direction of the velocity is different. As a generalization of this example, consider a pendulum released from rest at an angle θ i with the objective of determining the speed of P when it falls to an angle θ f , as in Figure 10.8.3. The work i W f done by gravity as the pendulum falls from θ i to θ f is: (10.8.5) where ∆h is the change in elevation of P as the pendulum falls (see Eq. (10.2.20)). The kinetic energies K i and K f of P when θ is θ i and θ f are: (10.8.6) where i and f are the values of when θ is θ i and θ f , respectively. The work–energy principle then leads to: (10.8.7) FIGURE 10.8.1 A simple pendulum. FIGURE 10.8.2 Free body diagram of the pendulum bob. O θ P(m) ᐉ T mg P Kmv m== () 1 2 1 2 2 2 l ˙ θ W K mg mv m===∆ or ll 1 2 1 2 22 ˙ θ vmg= l i ff i Wmghmg== − () ∆ l cos cosθθ Km Km ii ff = () == () 12 0 1 2 2 2 ll ˙˙ θθand ˙ θ ˙ θ ˙ θ i ff i WKKK==−∆ 0593_C10_fm Page 334 Monday, May 6, 2002 2:57 PM Introduction to Energy Methods 335 or (10.8.8) Solving for f we have: (10.8.9) The result of Eq. (10.8.9) could also have been obtained by integrating the governing differential equations of motion obtained in Chapter 8. Recall from Eq. (8.4.4) that for a simple pendulum the governing equation is: (10.8.10) Then, by multiplying both sides of this equation by , we have: (10.8.11) Because may be recognized as being (1/2)d 2 /dt, we can integrate the equation and obtain: (10.8.12) Because is zero when θ is θ i , the constant is –(g/ᐉ)cosθ i . Therefore, we have: (10.8.13) When θ is θ f , Eqs. (10.8.9) and (10.8.13) are seen to be equivalent. The work–energy principle may also be used to determine the pendulum rise angle when the speed at the equilibrium position (θ = 0) is known. Specifically, suppose the angular speed of the pendulum when θ is zero is o . Then, the work W done on the pendulum as it rises to an angle θ f is: (10.8.14) FIGURE 10.8.3 A simple pendulum released from rest and falling through angle θ i – θ f . O P P θ i f θ mg m f i f llcos cos ˙ θθ θ− () = () 1 2 2 ˙ θ ˙ cos cos / θθθ ff i g=− () 2 12 l ˙˙ sinθθ+ () =g l 0 ˙ θ ˙˙˙ sin ˙ θθ θθ+ () =g l 0 ˙ θ ˙˙ θ ˙ θ 12 2 () − () = ˙ cosθθg l constant ˙ θ ˙ cos cosθθθ 2 2= () − () g i l ˙ θ Wmgh mg f ==−− () ∆ l 1 cosθ 0593_C10_fm Page 335 Monday, May 6, 2002 2:57 PM 336 Dynamics of Mechanical Systems where the negative sign occurs because the upward movement of the pendulum is opposite to the direction of gravity, producing negative work. The work–energy principle, then, is: (10.8.15) or (10.8.16) or (10.8.17) If the pendulum is to rise all the way to the vertical equilibrium position (θ = π), we have: (10.8.18) If is exactly 4gᐉ, the pendulum will rise to the vertical equilibrium position and come to rest at that position. If exceeds 4gᐉ, the pendulum will rise to the vertical position and rotate through it with an angular speed given by Eq. (10.8.18) (sometimes called the rotating pendulum). 10.9 Elementary Example — A Mass–Spring System For a third fundamental example, consider the mass–spring system depicted in Figure 10.9.1. It consists of a block B with mass m and a linear spring, with modulus k, moving without friction or damping in a horizontal direction. Let x measure the displacement of B away from equilibrium. Suppose B is displaced to the right (positive x displacement with the spring in tension) a distance δ away from equilibrium. Let B be released from rest in this position. Questions arising then are what is the speed v of B as it returns to the equilibrium position (x = 0), and how far to the left of the equilibrium position does B go? FIGURE 10.9.1 Ideal mass–spring system. 00 WKKK ff ==−∆ −− () = () () − () () mg m m ff lll112 12 2 0 2 cos ˙˙ θθ θ ˙˙ cosθθ θ ff g 2 0 2 21=− − () l ˙˙ θθ f g 2 0 2 4=−l ˙ θ 0 2 ˙ θ 0 2 m x B k 0593_C10_fm Page 336 Monday, May 6, 2002 2:57 PM Introduction to Energy Methods 337 To answer these questions using the work–energy principle, recall from Eq. (10.2.22) that when a linear spring is stretched (or compressed) a distance δ, the corresponding work W done by the stretching (or compressing) force is (1/2)kδ 2 . Because the force exerted on the spring is equal, but oppositely directed, to the force exerted on B, the work done on B as the spring is relaxed is also (1/2)kδ 2 . (That is, the work on B is positive because the force of the spring on B is in the same direction as the movement of B.) Because B is released from rest, its initial kinetic energy is zero. The kinetic energy at the equilibrium position is: (10.9.1) Then, from the work–energy principle, we have: (10.9.2) or (10.9.3) where the minus sign is selected because B is moving to the left. Next, as B continues to move to the left past the equilibrium position, the spring force will be directed opposite to the movement of B. Therefore, the work W done on B as B moves to the left a distance d from the equilibrium position is: (10.9.4) When B moves to its leftmost position, its kinetic energy is zero. From Eqs. (10.9.1) and (10.9.2), the kinetic energy of B at the equilibrium position is: (10.9.5) The work–energy principle then produces: (10.9.6) or (10.9.7) That is, the block moves to the left by precisely the same amount as it was originally displaced to the right. The usual explanation of this phenomenon is that when the spring is stretched (or compressed) the work done by the stretching (or compressing) force stores energy (potential energy) in the spring. This stored energy in the spring is derived from the kinetic energy of the block. Then, as the spring is relaxing, its potential energy is transferred back to kinetic energy of the block. There is thus a periodic transfer of energy between the spring Kmv mx= () = () 12 12 22 ˙ WK k mv= () = () −∆ or 12 12 0 22 δ vx km== ˙ δ Wkd=− () 12 2 Kmv k= () = () 12 12 22 δ W K kd k=− () =− () ∆ or 1 2 0 1 2 22 δ d =δ 0593_C10_fm Page 337 Monday, May 6, 2002 2:57 PM 338 Dynamics of Mechanical Systems and the block, with the sum of the potential energy of the spring and the kinetic energy of the block being constant. (We will discuss potential energy in the next chapter.) As another example of work–energy transfer, consider the mass–spring system arranged vertically as in Figure 10.9.2. Suppose B is held in a position where the spring is unstretched. If B is then released from rest from this position, it will fall and stretch the spring and eventually come to rest at an extreme downward position. Questions arising then include how far does B fall and what is the spring force when B reaches this maximum downward displacement? To answer these questions, consider that as B falls, the weight (or gravity) force on B is in the direction of the movement of B, whereas the spring force on B is opposite to the movement of B. Because B is at rest at both the beginning and the end of the movement, there is no change in the kinetic energy of B. The net work on B is then zero. That is, (10.9.8) where d is the distance B moves downward. By solving for d we obtain: (10.9.9) The spring force in this extended position is, then, (10.9.10) The result of Eq. (10.9.10) shows that a suddenly applied weight load on a spring creates a force twice that of the weight. This means that if a weight is suddenly placed on a machine or structure the force generated is twice that required to support the weight in a static equilibrium configuration. 10.10 Skidding Vehicle Speeds: Accident Reconstruction Analysis The work–energy principle is especially useful in determining speeds of accident vehi- cles by using measurements of skid-mark data. Indeed, the work–energy principle together with the conservation of momentum principles are the primary methods used by accident reconstructionists when attempting to determine vehicle speeds at various stages of an accident. FIGURE 10.9.2 A vertical mass–spring system. k Y m B W mgd kd== − () 012 2 dmgk= 2 Fkd mg==2 0593_C10_fm Page 338 Monday, May 6, 2002 2:57 PM Introduction to Energy Methods 339 To illustrate the procedure, suppose an automobile leaves skid marks from all four wheels in coming to an emergency stop. Given the length d of the skid marks, the objective is to determine the vehicle speed when the marks first began. Skid marks are created by abrading and degrading tires sliding on a roadway surface. The tire degradation is due to friction forces and heat abrading the rubber. The friction forces are proportional to the normal (perpendicular to the roadway surface) forces on the tires and to the coefficient of friction µ. The friction coefficient, ranging in value from 0 to 1.0, is a measure of the relative slipperiness between the tires and the roadway pavement. If F and N are equivalent friction and normal forces (see Section 6.5), they are related by the expression: (10.10.1) If an automobile is sliding on a flat, level (horizontal) roadway, a free-body diagram of the vehicle shows that the normal force N is equal to the vehicle weight w. Then, as the vehicle slides a distance d, the work W done by the friction force (acting opposite to the direction of the sliding) is: (10.10.2) where m is the mass of the automobile. Let v be the desired speed of the automobile when the skid marks first appear. Then, the kinetic energy K i of the vehicle at that point is: (10.10.3) Because the kinetic energy K f at the end of the skid marks is zero (the vehicle is then stopped), the work energy principle produces: (10.10.4) or (10.10.5) Observe that the calculated speed is independent of the automobile mass. To illustrate how the work–energy principle may be used in conjunction with the momentum conservation principles, suppose an automobile with mass m 1 slides a distance d 1 before colliding with a stopped automobile having mass m 2 . Suppose further that the two vehicles then slide together (a plastic collision) for a distance d 2 before coming to rest. The questions arising then are what were the speeds of the vehicles just before and just after impact and what was the speed of the first vehicle when it first began to slide? To answer these questions, consider first from Eq. (10.10.5) that the speed v a of the vehicles just after impact is: (10.10.6) FN=µ W Fd Nd wd mgd=− =− =− =−µµµ Kmv i = () 12 2 W K K K mgd mv f i ==− − =− () ∆ or µ 012 2 vgd= 2µ vgd a = 2 2 µ 0593_C10_fm Page 339 Monday, May 6, 2002 2:57 PM 340 Dynamics of Mechanical Systems Next, during impact, the momentum is conserved. That is, (10.10.7) where v b is the speed of the first vehicle just before impact (see Eq. (9.7.1)). From Eq. (10.10.2), the work W done by friction forces on the first vehicle as it slides a distance d 1 before the collision is: (10.10.8) If v 0 is the speed of the first vehicle when skidding begins, the change in kinetic energy of the vehicle from the beginning of skidding until the collision is: (10.10.9) The work–energy principle then gives: (10.10.10) Finally, using Eqs. (10.10.6) and (10.10.7), we can solve Eq. (10.10.10) for : (10.10.11) Observe from Eq. (10.10.7) that the speed v a of the first vehicle just after the collision with the second vehicle is reduced by the factor [m 1 /(m 1 + m 2 )]. That is, the change in speed ∆v is: (10.10.12) Observe further that because the velocity changes during the impact the kinetic energy also changes. That is, even though the momentum is conserved, the kinetic energy is not conserved. Indeed, the change in kinetic energy ∆K just before and just after the impact is: (10.10.13) In actual accidents, the vehicles do not usually leave uniform skid marks from all four wheels. Also, collisions are not usually perfectly plastic nor do the vehicles always move in a straight line on a level surface. However, with minor modifications, the work–energy and momentum conservation principles may still be used to obtain reasonable estimates of vehicle speeds at various stages of an accident. The details of these modifications and the corresponding application of the principles are beyond the scope of this text; the reader is referred to the references for further information. mv m m v b a112 =+ () Wmgd=−µ 11 ∆Kmv mv b = () − () 12 12 1 2 10 2 WK mgd mv mv b =−= () − () ∆ or µ 11 1 2 10 2 12 12 v 0 2 v v gd m m v gd m m gd gd b a0 22 121 2 2 1 21 2 21 21 2 122 =+ =+ () [] + =+ () [] + µµ µµ ∆vv v m m mv a bb =−=− + () [] 212 ∆Kmv mv m mmmm mm v a bb = () − () =− () ++ + ()         12 12 12 22 1 2 1 2 1 1 2 12 3 2 12 2 2 0593_C10_fm Page 340 Monday, May 6, 2002 2:57 PM Introduction to Energy Methods 341 10.11 A Wheel Rolling Over a Step For a second example illustrating the tandem use of the work–energy principle and the conservation of momentum principle, consider again the case of the wheel rolling over a step as in Figure 10.11.1 (recall that we considered this problem in Section 9.9). Let the wheel W have a radius r, mass m, and axial moment of inertia I. Suppose we are interested in knowing the speed v of the wheel center required for the wheel to roll over the step. Recall in Section 9.9 that when the wheel encounters the step its angular momentum about the corner (or nose) O of the step is conserved. By using the conservation of angular momentum principle we found that the angular speed of W just after impact is (see Eq. (9.9.8)): (10.11.1) where ω is the angular speed of W before impact and h is the height of the step. After impact, W rotates about the nose O of the step. For W to roll over the step it must have enough kinetic energy after impact to overcome the negative work of gravity as it rises up over the step. The work W g of gravity as W rolls completely up the step is: (10.11.2) If W just rolls over the step (that is, if W expends all its kinetic energy after impact in rolling over the step), it will come to rest at the top of the step and its kinetic energy K f at that point will be zero. The kinetic energy K i just after impact is: (10.11.3) The work–energy principle then leads to: (10.11.4) Solving we have: (10.11.5) Hence, from Eq. (10.11.1), the speed v of W just before impact for step rollover is: FIGURE 10.11.1 Wheel rolling over a step. W ω h n r O ˆ ω ˆ /ωω=+ − () [] + () {} Imrrh Imr 2 W mgh g =− K I mr I mr i = () + () = () + () 12 12 12 222 22 ˆˆ ˆ ωω ω W K K mgh I mr g f i =− − =− () + () or 0 1 2 22 ˆ ω ˆ ω ˆ / / ω= + () [] 2 2 12 mgh I mr 0593_C10_fm Page 341 Monday, May 6, 2002 2:57 PM [...]... ( Ω= 8 ) [ ] 1/2 ˆ 2 Ω = 96 2 g a (10.12.6) 0593_C10_fm Page 344 Monday, May 6, 2002 2:57 PM 344 Dynamics of Mechanical Systems 10.13 Closure The work–energy principle is probably the most widely used of all the principles of dynamics The primary advantage of the work–energy principle is that it only requires knowledge of velocities and not accelerations Also, calculation of the work done is often... called partial angular velocity vectors of B in R As with the ˙ partial velocity vectors of Eq (11.4.3), the partial angular velocity vectors may be thought of as being base vectors for the movement of B in the n-dimensional space of the qr Partial velocity and partial angular velocity vectors are remarkably easy to evaluate: from Eqs (11.4.5) and (11.4.15) we see that the partial velocity and partial... A particle P moving in a plane FIGURE 11.2.6 A particle moving in a Cartesian reference frame The number of degrees of freedom of a mechanical system is the number of coordinates of the system if it were unrestricted minus the number of constraint equations For example, if a particle P moves relative to a Cartesian reference frame R as in Figure 11.2.6, then it has, if unrestricted, three degrees of. .. use of kinetic energy and Lagrange’s equations) are precluded with nonholonomic systems With Kane’s equations, however, the analysis is essentially the same for both holonomic and nonholonomic systems Nevertheless, our focus will be on holonomic systems because, as noted earlier, the vast majority of mechanical systems of interest are holonomic systems 11.4 Vector Functions, Partial Velocity, and Partial... (SAE), Warrendale, PA, 1 989 10.3 Platt, F N., The Traffic Accident Handbook, Hanrow Press, Columbia, MD, 1 983 10.4 Moffatt, E A., and Moffatt, C A., Eds., Highway Collision Reconstruction, American Society of Mechanical Engineers, New York, 1 980 10.5 Gardner, J D., and Moffatt, E A., Eds., Highway Truck Collision Analysis, American Society of Mechanical Engineers, New York, 1 982 10.6 Adler, U., Ed.,... Then, x is said to be a coordinate of P Next, consider the simple pendulum of Figure 11.2.2 In this case, the configuration of the system and, as a consequence, the location of the bob P are determined by the angle θ Thus, θ is a coordinate of the system 353 0593_C11_fm Page 354 Monday, May 6, 2002 2:59 PM 354 Dynamics of Mechanical Systems θ ᐉ O L P x P(m) Figure 11.2.1 A particle moving on a straight line... O B Q 8 in P10.2.7: Repeat Problem P10.2.6 if the natural length of the spring is 4 in P10.2 .8: A motorist, in making a turn with a 15-in.-diameter steering wheel, exerts a force of 8 lb with each hand tangent to the wheel as in Figure P10.2 .8 If the wheel is turned through an angle of 150°, determine the work done by the motorist 8 lb 15 in diameter FIGURE P10.2 .8 Forces on a steering wheel 8 lb Section... the partial velocity of P relative to qr in R ˙ (11.5.1) 0593_C11_fm Page 364 Monday, May 6, 2002 2:59 PM 364 Dynamics of Mechanical Systems F F2 F1 B P P1 P2 Pi PN R Fi R FN FIGURE 11.5.1 A force F applied at a point P of a mechanical system S FIGURE 11.5.2 A rigid body B with applied forces Fi at points Pi Observe in Eq (11.5.1) that the number of generalized forces is the same as the degrees of. .. expressions of the form of Eq (11.3.2) are generally nonintegrable and as a consequence cannot be solved in terms of elementary functions Fortunately, the vast majority of mechanical systems of interest and importance in machine dynamics can be modeled as holonomic systems To illustrate these concepts, consider again the rolling circular disk (or “rolling coin”) discussed in Sections 4.12 and 8. 13 and as shown... the coefficients of the qr Specifically, we see from Eq ˙ ˙ (11.4.15) that analogous to Eq (11.4.4): ˙ ω qr = ∂ω ∂qr ˙ (11.4.16) To illustrate the ease of evaluation of these vectors, consider first the motion of a particle P in a three-dimensional inertia frame R as in Figure 11.4.1 Let p locate P relative to a 0593_C11_fm Page 362 Monday, May 6, 2002 2:59 PM 362 Dynamics of Mechanical Systems Y Z P(x,y,z) . 337 Monday, May 6, 2002 2:57 PM 3 38 Dynamics of Mechanical Systems and the block, with the sum of the potential energy of the spring and the kinetic energy of the block being constant. (We will. 335 or (10 .8. 8) Solving for f we have: (10 .8. 9) The result of Eq. (10 .8. 9) could also have been obtained by integrating the governing differential equations of motion obtained in Chapter 8. Recall. 332 Dynamics of Mechanical Systems If we multiply the terms of Eq. (10.6.12) by v G , the velocity of G in R, we have: (10.6.14) Similarly, if we multiply the terms of Eq. (10.6.13)