Dynamics of Mechanical Systems 2009 Part 7 pps

0593_C09_fm Page 282 Monday, May 6, 2002 2:50 PM 282 Dynamics of Mechanical Systems where the final expression is obtained by recognizing M as the total mass of the particles of B and by recalling that G is the mass center of B so that: N ∑m r = (9.3.6) i i i =1 (See Section 6.8.) Equation (9.3.5) shows that for a rigid body the computation of the linear momentum is in essence as simple as the computation of linear momentum for a single particle as in Eq (9.3.1) 9.4 Angular Momentum Consider again a particle P having mass m and velocity v in a reference frame R as in Figure 9.4.1 Let Q be an arbitrary reference point Then, the angular momentum AQ of P about Q in R is defined as: D AQ = p × mv (9.4.1) where p locates P relative to Q Observe that AQ has the units mass–length–velocity, or mass–(length)2 per unit time Angular momentum is sometimes called moment of momentum Indeed, if we recognize mv as the linear momentum L of P, we can write Eq (9.4.1) in the form: AQ = p × L (9.4.2) Consider next a set S of N particles Pi (i = 1,…, N) having masses mi and velocities v Pi in R as in Figure 9.4.2 Again, let Q be an arbitrary reference point Then, the angular momentum of S about Q in R is defined as: N AS Q = ∑p × m v i i Pi (9.4.3) i =1 where pi locates Pi relative to Q Finally, consider a rigid body B with mass m and mass center G moving in a reference frame R as in Figure 9.4.3 Consider B to be composed of N particles Pi with masses mi FIGURE 9.4.1 A particle P with mass m, velocity v, and reference point Q 0593_C09_fm Page 283 Monday, May 6, 2002 2:50 PM Principles of Impulse and Momentum 283 FIGURE 9.4.2 A set S of moving particles FIGURE 9.4.3 A rigid body B with mass m, mass center G, and reference point Q (i = 1,…, N) Again, let Q be an arbitrary reference point Then, as in Eq (9.4.3), the angular momentum of B about Q in R is defined as: N AB Q = ∑p × m v i i Pi (9.4.4) i =1 where pi locates Pi relative to Q Observe that because Pi and mass center G are both fixed in B their velocities in R are related by the expression (see Eq (4.9.4)): v Pi = v G + ω × ri (9.4.5) where ω is the angular velocity of B in R and where ri locates Pi relative to G (see Figure 9.4.3) Observe further from Figure 9.4.3 that pi and ri are related by the expression: pi = QG + ri (9.4.6) By substituting from Eqs (9.4.5) and (9.4.6) into (9.4.4) AB/Q becomes: N AB Q = ∑ m (QG + r ) × (v i + ω × ri G i i =1 N = ∑ N mi QG × v G + i =1 ∑ i  ∑ m  QG × v i i =1 N ∑ i =1 ∑ m r × (ω × r ) i i i i =1 N  +  i N mi ri × v G + i =1  =  ∑ m QG × (ω × r ) i =1 N + ) G   + QG × ω ×      mi ri  × v G +  (9.4.7)  N ∑ m r   i i i =1 N ∑ m r × (ω × r ) i i i i =1 N = MQG × v G + + + ∑ m r × (ω × r ) i i i =1 i 0593_C09_fm Page 284 Monday, May 6, 2002 2:50 PM 284 Dynamics of Mechanical Systems where the middle two terms in the last line are zero because G is the mass center of B (see Section 6.8) The first term may be recognized as the moment of the linear momentum of a particle with mass M at G about Q That is, D D MQG × v G = QG × Mv G = QG × LG = A G Q (9.4.8) The last term of Eq (9.4.7) may be expressed in terms of the central inertia dyadic (see Sections 7.4 and 7.6) as follows: Let nω be a unit vector with the same direction as ω Then, ω may be expressed as: ω = ωn ω (9.4.9) where ω is the magnitude of ω Hence, we have: N ∑ mi ri × (ω × ri ) = i =1 N ∑ mi ri × (ωnω × ri ) = ω i =1 N ∑ m r × (n i i ω × ri ) (9.4.10) i =1 From Eqs (7.2.2) and (7.4.11), this last expression in turn may be expressed in terms of the inertia dyadic as: N ω ∑ m r × (n i i ω B × ri ) = ωIω G = ωIB G ⋅ nω (9.4.11) i =1 = IB G ⋅ (ωnω ) = IB G ⋅ ω We can obtain yet a different form of the last term of Eq (9.4.7) by recognizing that the ω parenthetical term (ω × ri) may be identified as v Pi /G or as v Pi − v G (see Eq (4.9.4)) Hence, we have: N ∑ i =1 mi ri × (ω × ri ) = N ∑ m r × (v i i Pi − vG i =1 N = ∑ ) N mi ri × v Pi − i =1 ∑m r × v i i G (9.4.12) i =1 = AB G − where the last equality is obtained from Eq (9.4.4) and by recalling that G is the mass center of B Then, by comparing the results in Eqs (9.4.11) and (9.4.12), we have: N ∑ m r × (ω × r ) = I i i i =1 i BG ⋅ ω = AB G (9.4.13) 0593_C09_fm Page 285 Monday, May 6, 2002 2:50 PM Principles of Impulse and Momentum 285 Finally, by substituting from Eqs (9.4.8) and (9.4.13) into Eq (9.4.7) we obtain the addition theorem for angular momentum for a rigid body: A B Q = AG Q + A B G 9.5 (9.4.14) Principle of Linear Impulse and Momentum Consider again a particle P with mass m moving in an inertial reference frame R as in Figure 9.5.1 Let P be acted upon by a force F as shown Then, by Newton’s law (see Eq (8.3.1)), F is related to the acceleration a of P in R by the expression: F = ma (9.5.1) Recalling that the acceleration is the derivative of the velocity we can use the definition of linear impulse of Eqs (9.2.1) and (9.2.3) to integrate Eq (9.5.1) That is, t2 ∫ t2 t2 ∫ ∫ I = Fdt = madt = m(dv dt)dt t1 t1 t1 (9.5.2) = mv(t2 ) − mv(t1 ) = L(t2 ) − L(t1 ) or I = ∆L (9.5.3) where I represents the impulse applied between t1 and t2 Equation (9.5.3) states that the linear impulse is equal to the change in linear momentum This verbal statement is often called the principle of linear momentum This principle is readily extended to systems of particles and to rigid bodies Consider first the system S of N particles Pi with masses mi (i = 1,…, N) and moving in an inertial FIGURE 9.5.1 A particle P with mass m moving in an inertial reference frame 0593_C09_fm Page 286 Monday, May 6, 2002 2:50 PM 286 Dynamics of Mechanical Systems FIGURE 9.5.2 A set of particles moving in an inertial reference frame R frame R as in Figure 9.5.2 Let the particles be acted upon by forces Fi (i = 1,…, N) as shown Then, from Newton’s law, we have for each particle: (i = 1,K , N ) (no sum) Fi = mia i (9.5.4) where is the acceleration of Pi in R Let G be the mass center of S Then, a i = aG + d ri dt (9.5.5) where ri locates Pi relative to G as in Figure 9.5.2 Let the system of forces Fi be represented by an equivalent force system (see Section 6.5) consisting of a single force F passing through G together with a couple with torque T Then, F and T are: N N F= ∑ Fi and T = i =1 ∑r × F i (9.5.6) i i =1 Hence, from Eqs (9.5.4) and (9.5.5), we have: N F= N ∑ m a = ∑ m (a i i i =1  =  N ∑ i =1 i G + d ri dt i =1  d2  mi  aG +  dt   N ∑ i =1 )  mi ri  = MaG  (9.5.7) where M is the total mass of S and where the operations of summation and integration may be interchanged because N is finite (The sum N ∑m r i i i =1 is zero because G is the mass center.) 0593_C09_fm Page 287 Monday, May 6, 2002 2:50 PM Principles of Impulse and Momentum 287 Finally, by integrating in Eq (9.5.7) we have: t2 t2 ∫ I = F dt = t1 t2 ∫ Ma dt = ∫ M(dv G t1 G dt)dt t1 (9.5.8) = Mv G (t2 ) − Mv G (t1 ) = L(t2 ) − L(t1 ) or I = ∆L (9.5.9) where L represents the linear momentum of S as in Eq (9.3.2) We can develop a similar analysis for a rigid body B as in Figure 9.5.3 where, as before, we consider B to be composed of a set of N particles Pi having masses mi (i = 1,…, N) Let G be the mass center of B and let R be an inertial reference frame in which B moves Let ri locate Pi relative to G Then, because Pi and G are both fixed in B, their accelerations are related by the expression (see Eq (4.9.6)): a i = aG + α × ri + ω × (ω × ri ) (9.5.10) where as before represents the acceleration of Pi in R and where α and ω are the angular acceleration and angular velocity of B in R Let Pi be acted upon by a force Fi as shown in Figure 9.5.3 Let the set of forces Fi (i = 1,…, N) be represented by an equivalent force system consisting of a single force F passing through G together with a couple with torque T Then, F and T are: N F= ∑F i N and i =1 T= ∑r × F i i (9.5.11) i =1 Again, from Newton’s law we have: Fi = mia i FIGURE 9.5.3 A rigid body B moving in an inertia reference frame R (no sum) (9.5.12) 0593_C09_fm Page 288 Monday, May 6, 2002 2:50 PM 288 Dynamics of Mechanical Systems By substituting from Eqs (9.5.10) and (9.5.12) into Eq (9.5.11) we have: F = MaG (9.5.13) where M is the mass of B Finally, by integrating in Eq (9.5.13) we obtain (as in Eqs (9.5.8) and (9.5.9)): I = ∆L (9.5.14) where now L represents the linear momentum of B Observe the identical formats of Eqs (9.5.3), (9.5.9), and (9.5.14) for a single particle, a set of particles, and a rigid body 9.6 Principle of Angular Impulse and Momentum We can develop expressions analogous to Eqs (9.5.3), (9.5.9), and (9.5.14) for angular impulse and angular momentum The development here, however, has the added feature of involving a reference point (or object point) Because angular momentum is always computed relative to a point, the choice of that point may affect the form of the relation between angular impulse and angular momentum Consider again a particle P with mass m moving in an inertial reference frame R as in Figure 9.6.1 Let P be acted upon by a force F as shown Let Q be an arbitrarily chosen reference point Consider a free-body diagram of P as in Figure 9.6.2, where F* is the inertia force on P given by (see Eq (8.3.2)): F * = − ma P = − mdv P dt (9.6.1) where vP and aP are the velocity and acceleration of P in R From d’Alembert’s principle, we have: F + F* = or F = mdv P dt FIGURE 9.6.1 A particle P moving in an inertial reference frame R with applied force F and reference point Q FIGURE 9.6.2 Free-body diagram of P (9.6.2) 0593_C09_fm Page 289 Monday, May 6, 2002 2:50 PM Principles of Impulse and Momentum 289 By setting moments about Q equal to zero we have: QP × F + QP × F * = or QP × F = QP × mdv P dt (9.6.3) Consider the final term in Eq (9.6.3) By the product rule for differentiation we have: ( ) QP × mdv P dt = d QP × mdv P dt − (dQP dt) × mv P = dAQ dt − v P Q × mv P ( (9.6.4) ) = dAQ dt − v P − vQ × mv P = dAQ dt + vQ × mv P Observe that QP × F is the moment of F about Q Hence, by substituting from Eq (9.6.4) into (9.6.3) we obtain: MQ = dAQ dt + vQ × mv P (9.6.5) If the velocity of Q is zero (that is, if Q is fixed in R), then we have: MQ = dAQ dt (9.6.6) By integrating over the time interval in which F is applied, we have: t2 ∫ t1 t2 MQ dt = ∫ (dA dt) dt = A (t ) − A (t ) Q Q Q (9.6.7) t1 or JQ = ∆AQ (9.6.8) where JQ is the angular impulse of F about Q during the time interval (t1, t2) That is, the angular impulse about a point Q fixed in an inertial reference frame is equal to the change in angular momentum about Q This verbal statement of Eq (9.6.8) is called the principle of angular momentum Note, however, it is valid only if Q is fixed in the inertial reference frame Consider next a set S of N particles Pi with masses mi (i = 1,…, N) moving in an inertial reference frame R as in Figure 9.6.3 Let the particles be acted upon by forces Fi (i = 1,…, N) as shown Consider a free-body diagram of a typical particle Pi as in Figure 9.6.4 where Fi* is the inertia force on Pi given by: Fi* = − mia Pi = − mdv Pi dt (no sum on i) (9.6.9) 0593_C09_fm Page 290 Monday, May 6, 2002 2:50 PM 290 Dynamics of Mechanical Systems FIGURE 9.6.3 A set S of particles moving in an inertial reference frame R with reference point Q FIGURE 9.6.4 Free-body diagram of typical particle Pi where v Pi and a Pi are the velocity and acceleration of Pi in R Then, from d’Alembert’s principle, we have: Fi + Fi* = or Fi = mi dv Pi dt (9.6.10) By setting moments about Q equal to zero, we have: QPi × Fi + QPi × Fi* = or QPi × Fi = QPi × mi dv Pi dt (9.6.11) Consider the final term in Eq (9.6.11) By the product rule for differentiation, we have (see Eq (9.6.4)): QPi × mi dv Pi dt = d(QPi × mi v Pi ) dt − dQPi dt × mi v Pi P = dAQi dt + vQ × mi v Pi (9.6.12) By substituting into Eq (9.6.11) we have: M Fi Q = dA Pi Q dt + vQ × mi v Pi (9.6.13) By adding the effects from all the particles, we have: N MQ = dAS Q dt + vQ × ∑m v i Pi (9.6.14) i =1 If the velocity of Q is zero (that is, if Q is fixed in R), then we have: MQ = dAS Q dt (9.6.15) 0593_C09_fm Page 291 Monday, May 6, 2002 2:50 PM Principles of Impulse and Momentum 291 By integrating over the time interval in which the forces are applied, we have: t2 t2 ∫ M dt = ∫ (dA Q t1 SQ ) dt dt = AS Q (t2 ) − AS Q (t1 ) t1 or JQ = ∆AS Q (9.6.16) where here JQ represents the sum of the angular impulses of the applied forces about Q during the time interval (t1, t2) Hence, as with a single particle, the angular impulse about a point Q fixed in an inertial reference frame is equal to the change in angular momentum of the set of particles about Q Finally, consider a rigid body B moving in an inertial reference frame R as in Figure 9.6.5 Let G be the mass center of B, let Q be a reference point, and let O be the origin of R Then, from Eqs (9.4.8), (9.4.12), and (9.4.13), the angular momenta of B about O, Q, and G are: A B O = A B G + AG O = IB G ⋅ ω + PG × mv G (9.6.17) A B Q = A B G + AG Q = IB G ⋅ ω + QG × mv G (9.6.18) A B G = IB G ⋅ ω (9.6.19) where PG and QG locate G relative to O and Q as in Figure 9.6.5, and where as before ω is the angular velocity of B in R, IB/G is the central inertia dyadic of B, and m is the mass of B Consider the derivatives of these momenta For AB/O we have: R dA B O dt = R d(IB G ⋅ ω ) dt + R d(PG × mv G ) dt = B d(IB G ⋅ ω ) dt + ω × (IB G ⋅ ω ) + v G × mv G + PG × maG FIGURE 9.6.5 A rigid body B moving in an inertial reference frame R and a reference point Q (9.6.20) 0593_C09_fm Page 317 Monday, May 6, 2002 2:50 PM Principles of Impulse and Momentum 317 O ᐉ h m FIGURE P9.7.9 A bullet fired into a hanging bar V n2 C n1 B n3 G 10 m/sec 4m FIGURE P9.7.11 A rectangular plate striking a fixed ledge O A 3m D P9.7.12: See Problem P9.7.11 Determine the velocities of points A, B, and C and the mass center G of the plate just after impact P9.7.13: See Problem P9.7.10 In Problem 9.7.10 find the distance h so that the bullet strikes the bar at the center of percussion so that there is no horizontal reaction at the pin at O Repeat Problem 9.7.10 using this value of h instead of 0.8 m P9.7.14: Consider again the pin-connected double bar of Figure 9.8.7 and shown again in Figure P9.7.14 Let the bars each have length 1.0 m and mass 0.5 kg Let the impulse of the force P be 150 Nm Determine the velocity of end O and the angular velocities of the bars just after impact 1m FIGURE P9.7.14 Pin-connected bars struck at one end Q 1m O P P9.7.15: Repeat the analysis of Example 9.8.4 if the bars are no longer identical but instead have lengths ᐉ1 and ᐉ2 and masses m1 and m2 where the masses are proportional to the lengths P9.7.16: An automobile wheel and tire W weighing 50 lb with diameter 24 in and axial radius of gyration in is rolling at 15 mph toward a 6-in curb Find the angular velocity of W immediately after impact with the curb Section 9.10 Impact: Coefficient of Restitution P9.10.1: A car A traveling at 20 mph strikes the rear of a car B which is at rest as in Figure P9.10.1 Let the weights of A and B be 3220 lb and 2800 lb, respectively Determine the speeds of A and B immediately after collision under the following assumptions 0593_C09_fm Page 318 Monday, May 6, 2002 2:50 PM 318 Dynamics of Mechanical Systems a A perfectly plastic collision: e = b A semi-elastic collision: e = 0.5 c A perfectly elastic collision: e = 1.0 20 mph (stopped) A B FIGURE P9.10.1 A rear-end collision where e is the coefficient of restitution P9.10.2: Repeat Problem P9.10.1 if B is moving to the right at 10 mph just before impact P9.10.3: See Problems 9.10.1 and 9.10.2 Repeat Problem P9.10.1 if B is moving to the left at 10 mph (that is, B is backing) P9.10.4: A basketball B is dropped from a height h onto a fixed horizontal surface S It is known that the speed v of B just before it strikes S is gh (g is gravity acceleration) Also ˆ it is known that if B is projected upward from S, after bouncing, with a speed v it will ˆ ˆ reach a height h given by v 2 g Let e be the coefficient of restitution between B and ˆ ˆ S Let h be ft and e be 0.9 Find v, v and h ˆ P9.10.5: See Problem P9.10.4 If h is ft and h is 6.5 ft, what is e? Section 9.11 Oblique Impact P9.11.1: Two spheres, A and B, collide as in Figure P9.11.1 Let their masses be mA = 0.5 kg and mB = -.75 kg Let the collision be nearly elastic so that the coefficient of restitution e is 0.9 Determine the velocities (speeds and directions) of A and B immediately after impact A B (Impact) m/s 60° FIGURE P9.11.1 Colliding spheres m/sec 30° B A P9.11.2: Repeat Problem P9.11.1 if mA = 0.75 kg, mB = 0.5 kg, and e = 0.85 P9.11.3: A test car A collides with a fixed, angled barrier as in Figure P9.11.3 Let the barrier surface S be smooth and rigid so that it does not resist motion in its tangential direction Determine the velocity (speed and direction) of A immediately after impact if the coefficient of restitution between A and C is (a) (plastic), (b) 0.75 (semi-elastic), and (c) 1.0 (elastic) P9.11.4: Repeat Problem P9.11.3 if the barrier angle is at 45° P9.11.5: Two cars A and B have an intersection collision at 45° to their pre-impact direction of travel as in Figure P9.11.4 Determine the post-impact velocity of the cars for the following conditions: mA = 100 slug, mB = 80 slug, VA = 30 mph, VB = 25 mph, and the coefficient of restitution e = 0.75 Assume the colliding surfaces are smooth (Express the results in terms of the unit vectors nn and ne of Figure P9.11.4.) P9.11.6: Repeat Problem P9.11.4 if the collision is (a) perfectly plastic (e = 0), and (b) perfectly elastic (e = 1) 0593_C09_fm Page 319 Monday, May 6, 2002 2:50 PM Principles of Impulse and Momentum 319 P9.11.7: Repeat Problems P9.11.4 and P9.11.5 if mA = 150 kg, mB = 1200 kg, VA = 45 km/hr, and VB = 37 km/hr S 35 mph FIGURE P9.11.3 A test car colliding with a fixed rigid barrier 30° A nn A A VA ne B 45° VB FIGURE P9.11.4 Intersection collision of cars B 0593_C09_fm Page 320 Monday, May 6, 2002 2:50 PM 0593_C10_fm Page 321 Monday, May 6, 2002 2:57 PM 10 Introduction to Energy Methods 10.1 Introduction In this chapter, we consider energy methods with a focus on the work–energy principle Energy methods are very convenient for a broad class of systems — particularly those with relatively simple geometrics and those for which limited information is desired Energy methods, like impulse–momentum principles, are formulated in terms of velocities, thus avoiding the computation of accelerations as is required with Newton’s laws and d’Alembert’s principle But, unlike the impulse–momentum principles, energy methods are formulated in terms of scalars By thus avoiding vector operations, energy methods generally involve simpler analyses The information gained, however, may be somewhat limited because often only one equation is obtained with the work–energy method We begin our study with a brief discussion of “work” and its computation We then discuss power and kinetic energy and their relation to work The balance of the chapter is then devoted to examples illustrating applications and combined use with impulse–momentum principles In the next chapter we will discuss more advanced energy methods and the concepts of generalized mechanics 10.2 Work Intuitively, work is related to expended effort or expenditure of energy In elementary physics, work is defined as the product of a force (effort) and the displacement (movement) of an object to which the force is applied To develop these concepts, let P be a particle and let F be a force applied to P as represented in Figure 10.2.1 Let P move through a distance d in the direction of F as shown Then, the work W done by F is defined as: D W = Fd (10.2.1) Generally when a force is applied on a particle (or object) the force does not remain constant during the movement of the particle (or object) Both the magnitude and the direction of the force may change Also, the particle (or object) will in general not move on a straight line 321 0593_C10_fm Page 322 Monday, May 6, 2002 2:57 PM 322 Dynamics of Mechanical Systems P P d F P C F FIGURE 10.2.1 A force F moving a particle P FIGURE 10.2.2 A force F applied to a particle P moving on a curve C In view of these observations, it is necessary to generalize the definition of Eq (10.2.1): specifically, let a force F be applied to a particle P which moves along a curve C as in Figure 10.2.2 Then, the work W done by F as P moves along C is defined as: δ D W= ∫ F ⋅ ds (10.2.2) where ds is a differential arc length vector tangent to C at the position of P and where δ is the distance P moves along C under the action of F From this generalized definition of work we see that if F is always directed perpendicular to the movement of P, the work is zero Also, we see that if P moves in a direction opposed to F, the work is negative As an example, consider a particle P moving on a curve C defined by the parametric equations: x = r cos t , y = r sin t , z = t (10.2.3) where C may be recognized as a circular helix as depicted in Figure 10.2.3 Let F be a force defined as: F = tn x + t 2n y + t 3n z (10.2.4) Z nz P C p O FIGURE 10.2.3 A particle moving on a circular helix X nx Y ny 0593_C10_fm Page 323 Monday, May 6, 2002 2:57 PM Introduction to Energy Methods 323 where nx, ny , and nz are unit vectors parallel to the coordinate axes as shown We can determine the work done by F on P as follows: knowing F, we need an expression for ds, the differential arc length along C Because the direction of ds is tangent to C, ds may be expressed as: ds = τ ds (10.2.5) where τ is a unit vector tangent to C at the position of P and ds is the differential arc length We can obtain expressions for both τ and ds from Eq (10.2.3), the defining equations of C Recall that the velocity of a particle is tangent to its path (or curve) of travel Hence, by considering the parameter t in Eq (10.2.3) as time, the velocity of P is: ˙ ˙ ˙ v = xn x + yn y + zn z = − r sin t n x = r cos t n y + n z (10.2.6) Then, τ becomes: ( ) τ = v / v = − r sin t n x + r cos t n y + n z / (1 + r ) 1/2 (10.2.7) The differential arc length is: ˙ ˙ ˙ ds = ( x + y + z ) 1/2 dt = (1 + r ) 1/2 dt (10.2.8) By substituting from Eqs (10.2.4), (10.2.7), and (10.2.8) into (10.2.2), we find the work of F to be: t* ∫ W = F ⋅ ds = t* ∫ (−rt sin t + rt cos t + t ) 1/2 (10.2.9) dt where t* is the value of t locating the ending position of P For example, if t* is π, W becomes: { [ ( ) ] } W = − r [sin t − t cos t] + r 2t cos t + t − sin t + t 4/4 π (10.2.10) = −3πr By comparing Eqs (10.2.5) and (10.2.7), we see that ds may be written as: ds = vds/ v (10.2.11) By recognizing v as ds/dt, ds becomes: ds = vdt Hence, the definition of work of Eq (10.2.2) takes the form: (10.2.12) 0593_C10_fm Page 324 Monday, May 6, 2002 2:57 PM 324 Dynamics of Mechanical Systems t* ∫ W = F ⋅ v dt (10.2.13) where t* is the time of action of F The integrand of Eq (10.2.13), F • v, is often called the power of F (see Section 10.4) As a second example, consider the work done by gravity on a simple pendulum as it falls from a horizontal position to the vertical equilibrium position as in Figure 10.2.4 Let the pendulum mass be m and let its length be ᐉ as shown The gravity (or weight) force is, then, w = mgk (10.2.14) where k is a vertically downward directed unit vector as shown in Figure 10.2.4 To apply Eq (10.2.2), consider that the differential arc vector may be expressed as: ds = ldφn φ (10.2.15) where φ measures the angle of the pendulum to the horizontal and nφ is a unit vector tangent to the circular arc of the pendulum as shown in Figure 10.2.4 Hence, the work of the weight force is: φ = π/2 W= ∫ π/2 w ⋅ ds φ=0 ∫ mg k ⋅ ln dφ φ (10.2.16) π/2 = mgl π/2 ∫ cos φ d φ = mgl sin φ = mgl Observe from the next to last expression of Eq (10.2.16) that the work done by gravity as the pendulum falls through an arbitrary angle φ is: W = mgl sin θ (10.2.17) The distance ᐉ sinφ may be recognized as the vertical drop h of the pendulum; hence, the gravitational work is: W = mgh k O (10.2.18) ᐉ P(m) φ ᐉ nφ FIGURE 10.2.4 A falling pendulum P(m) 0593_C10_fm Page 325 Monday, May 6, 2002 2:57 PM Introduction to Energy Methods 325 P1 k P(m) θ C h dh ds τ ˆ C P2 FIGURE 10.2.5 A particle P moving from P1 to P2 under the action of gravity FIGURE 10.2.6 ˆ A differential arc of an arbitrary curve C It happens that Eq (10.2.18) is a valid expression for the work of the gravity force regardless of the path of descent of the particle P In the pendulum example, the particle ˆ path was a circular arc However, if P had moved on some other path (say, C , as in Figure 10.2.5), the work done by gravity would still be mgh, where h is the change in elevation ˆ ˆ of P as it descends from P1 to P2 along C To see this, consider a differential arc of C as ˆ τ is a unit vector tangent to C and k is a vertically downward in Figure 10.2.6 where directed unit vector Then, the integrand of Eq (10.2.2) becomes: w ⋅ ds = mg k ⋅ τ ds = mg cosθ ds = mg dh (10.2.19) ˆ where θ is the angle between C and the vertical as in Figure 10.2.6 and dh is the differential elevation change as shown Hence, integrating the work done by gravity is: ∫ W = mgdh = mgh (10.2.20) For a third example, consider the work done by a force in stretching or compressing a linear spring Specifically, consider a spring as depicted in Figure 10.2.7 Let F be the magnitude of the stretching force F Let the natural length of the spring be ᐉ, and let its stretched length, under the action of F, be ᐉ + δ Then, from Eq (10.2.2), the work is: δ W= ∫ δ Fdx = ∫ kx dx = (1 2)kδ (10.2.21) where k is the spring modulus, and x is the end displacement along the axis of the spring Similarly, if the spring is compressed, the work done by the compressing force is (1/2)kδ2, where now δ is a measure of the shortening of the spring F FIGURE 10.2.7 A linear spring stretched by a force F 0593_C10_fm Page 326 Monday, May 6, 2002 2:57 PM 326 Dynamics of Mechanical Systems d θ F d/2 F B F FIGURE 10.3.1 A couple applied to a flywheel FIGURE 10.3.2 Displacement at the point of application of the force 10.3 Work Done by a Couple Suppose a couple is applied to a body B, causing B to rotate Then, from our intuitive understanding of work, as an exertion creating a movement, we expect that work has been performed by the couple To develop and quantify this, consider first a flywheel B supported by a frictionless pin at its center Let a couple C consisting of two equal magnitude, but oppositely directed, forces be applied to B as in Figure 10.3.1 If B is initially at rest, the application of the couple will cause B to rotate about its pin Specifically, the points of application of the forces will experience displacements with the forces, thus doing work Let the forces of C change orientation as B rotates (that is, let C “rotate” with B) Consider the work of one of these forces: from Figure 10.3.2, we see that the displacement at the point of application of the force is (d/2)θ where d is the distance between the couple forces and θ is the rotation angle Then, from our definition of work of Eq (10.2.2) we see that the work done by one of the couple forces is: W = F ( d 2) θ (10.3.1) Hence, the work done by the couple is: W = F d θ = Tθ (10.3.2) where T is the torque of the couple (see Section 6.4) Consider now a generalization of this example where a couple C is applied to a body B with B instantaneously rotating through a differential angle dθ about an axis parallel to a unit vector n Then, from Eq (10.3.2), we see that the work done by C is: θ* ∫ W = T ⋅ n dθ (10.3.3) where θ* is the total angle of rotation while C is being applied, and T is the torque of C 0593_C10_fm Page 327 Monday, May 6, 2002 2:57 PM Introduction to Energy Methods 327 A question that arises, however, is how we determine n, the unit vector parallel to the axis of rotation of B? To answer this question, recall the example of the previous section where a particle P was pushed along a circular helix In that example, we obtained the direction of motion of P from the velocity vector In like manner, the instantaneous axis of rotation of body B is parallel to the angular velocity ω of B Then, in Eq (10.3.3), n is given by: n = ω/ ω (10.3.4) Because angular velocity is a measure of the rate of change of orientation, we see that ω ω may be identified with dθ/dt Hence, we may also express the work of Eq (10.3.3) in the form: t* ∫ W = T ⋅ ω dt (10.3.5) where t* is the time of action of T Observe the similarity of Eqs (10.2.13) and (10.3.5) 10.4 Power Power is defined as the rate at which work is done That is, P = dW dt (10.4.1) For a force F doing work on a particle or body, the power of the force may be obtained by differentiating in Eq (10.2.13): P = F⋅v (10.4.2) where v is the velocity of the point of application of F Similarly, for a couple doing work on a body, the power developed by the couple may be obtained by differentiating in Eq (10.3.5): P = T ⋅ω (10.4.3) where, as before, T is the torque of the couple and ω is the angular velocity of the body 10.5 Kinetic Energy Kinetic energy is probably the most familiar and most widely used of all the energy functions Kinetic energy is sometimes described as energy due to motion 0593_C10_fm Page 328 Monday, May 6, 2002 2:57 PM 328 Dynamics of Mechanical Systems If P is a particle with mass m, the kinetic energy K of P is defined as: m v2 D K= (10.5.1) where v is the velocity of P in an inertial reference frame R The factor 1/2 is introduced for convenience in relating kinetic energy to work and to other energy functions If S is a set of particles, the kinetic energy of S is defined as the sum of the kinetic energies of the individual particles Specifically, if S contains N particles Pi with masses mi and velocities vi (i = 1,…, N) in inertial frame R, the kinetic energy of S is defined as: N D K= ∑2mv i i (10.5.2) i =1 For a rigid body B we can define the kinetic energy of B as the sum of the kinetic energies of the particles making up B To see this, consider a depiction of B as in Figure 10.5.1 Let G be the mass center of B (see Section 6.8) Then, from Eq (4.9.4), the velocity vi of typical particle Pi of B in an inertial reference frame R is: v i = v G + ω × ri (10.5.3) where vG is the velocity of G in R, ω is the angular velocity of B in R, and ri locates Pi relative to G Then, from Eq (10.5.2), the kinetic energy of B is: N K= ∑ i =1 m v2 = i i N ∑ m (v i G i =1 + ω × ri ) (10.5.4) By expanding the terms in Eq (10.5.4), the kinetic energy becomes: K= = = N ∑ N mi v G + i =1 1  2 N ∑ i =1 ∑ mi v G ⋅ ω × ri + i =1   mi  v G + v G ⋅ ω ×    1 Mv G + + 2 N ∑ i =1 N ∑ m (ω × r ) i N ∑ m (ω × r ) i i i =1  mi ri  +  N ∑ m (ω × r ) i (10.5.5) i =1 i i =1 P2 (m 2) P1 (m 1) G ri PN (m N ) FIGURE 10.5.1 A rigid body composed i R Pi (m i ) B 0593_C10_fm Page 329 Monday, May 6, 2002 2:57 PM Introduction to Energy Methods 329 where M is the mass of B (the total mass of all the particles composing B), and the middle term is zero because G is the mass center of B Consider the last term of Eq (10.5.5) By expanding and rearranging the terms we have: N ∑ m (ω × r ) i i =1 i = = N ∑ i =1 ω⋅ mi (ω × ri ) ⋅ (ω × ri ) = N ∑ i =1 mi ri × (ω × ri ) = ∑ m ω ⋅[r × (ω × r )] ω⋅ = i i i i =1 N ∑ m [r × (ωn i i =1 i ω ] × ri ) (10.5.6) 1 B ω ⋅ Iω G ω = ω ⋅ I B G ⋅ n ω ω 2 = N ω ⋅ IB G ⋅ ω B where ω is the magnitude of ω, nω is a unit vector parallel to ω, Iω G is the second moment B/G is the central inertia dyadic of B (see vector of B for G for the direction of nω, and I Sections 7.4 and 7.6.) By substituting from Eq (10.5.6) into (10.5.5), the kinetic energy of B becomes: K= v + ω ⋅ IB G ⋅ ω G (10.5.7) Finally, suppose that n1, n2, and n3 are mutually perpendicular principal unit vectors of B for G (see Section 7.7), then K may be expressed as: K= 1 2 Mv G + (I11 ω + I22 ω + I33 ω ) 2 (10.5.8) where I11, I22, and I33 are the corresponding principal moments of inertia of B for B, and the ωi (i = 1, 2, 3) are the ni components of ω Observe that in all the expressions for kinetic energy no accelerations appear, only velocities 10.6 Work–Energy Principles We can obtain a relationship between work and kinetic energy by integrating equations obtained from Newton’s laws To this end, consider first a particle P with mass m subjected to a force F as in Figure 10.6.1 Then, from Newton’s laws (see Section 8.2), we have: F = ma (10.6.1) where a is the acceleration of P in an inertial frame R If we project the terms of Eq (10.6.1) along the velocity v of P in R we have: F ⋅ v = ma ⋅ v (10.6.2) 0593_C10_fm Page 330 Monday, May 6, 2002 2:57 PM 330 Dynamics of Mechanical Systems P2 (m ) P1 (m ) P(m) S P i (m i ) F2 FN F1 R F PN(m N) R FIGURE 10.6.1 A force applied to a particle FIGURE 10.6.2 A set of particles subjected to forces The left side of Eq (10.6.2) may be recognized as the power of F (see Eq (10.4.2)) The right side of Eq (10.6.2) may be expressed in terms of a derivative of the square of the velocity of P That is, ( ) ma ⋅ v = md v 2 dt = d 1  mv  dt  (10.6.3) From Eqs (10.4.1) and (10.4.2) we recognize the left side of Eq (10.6.2) as the derivative of the work W of F In like manner, from Eq (10.5.1), we recognize the right side of Eq (10.6.3) as the derivative of the kinetic energy K of P Hence, we have: dW dK = dt dt (10.6.4) W = K2 − K1 = ∆K (10.6.5) or where K2 and K1 represent the kinetic energy of P at the beginning and end of the time interval during which F is applied to P Next, consider a set S of particles Pi (i = 1,…, N) subjected to forces Fi as in Figure 10.6.2 Then, for a typical particle Pi, Newton’s laws become: Fi = mia i (no sum on i) (10.6.6) where is the acceleration of Pi in inertia frame R By multiplying the terms of Eq (10.6.6) by vi, the velocity of Pi in R, we obtain: Fi ⋅ v i = mia i ⋅ v i = d 1  dKi m v2 = dt  i i  dt (10.6.7) where Ki is the kinetic energy of Pi We can recognize the left side of Eq (10.6.7) as the derivative of the work Wi of Fi Hence, Eq (10.6.7) becomes: dWi dKi = dt dt (10.6.8) 0593_C10_fm Page 331 Monday, May 6, 2002 2:57 PM Introduction to Energy Methods B P2 (m ) F P1 (m ) 331 B i P i (m i ) F2 F1 G G T F PN(m N) R R FIGURE 10.6.3 A rigid body B subjected to forces FIGURE 10.6.4 Equivalent force system on rigid body B By adding together the terms of similar equations for each of the particles, we have: dW dK = dt dt (10.6.9) where W is the work on S from all the forces, and K is the kinetic energy of S By integrating Eq (10.6.9), we have: W = K2 − K1 = ∆K (10.6.10) where, as in Eq (10.6.5), K2 and K1 represent the kinetic energy of S at the beginning and end of the time interval that the forces are applied to S Finally, consider a rigid body B acted upon by a system of forces as in Figure (10.6.3) As before, consider B to be composed of fixed particles Pi with masses mi as shown Let the force system applied to B be replaced by an equivalent force system (see Section 6.5) consisting of a single force F passing through the mass center G of B together with a couple with torque T as in Figure 10.6.4 From d’Alembert’s principle (see Section 8.3) we can represent the inertia force system on B by a force F* passing through G together with a couple with torque T* where F* and T* are given by (see Eqs (8.6.5) and (8.6.6)): F * = − MaG and T * = −I ⋅ α − ω × (I ⋅ ω ) (10.6.11) where, as before, M is the mass of B, I is the central inertia dyadic of B, aG is the acceleration of G in inertia frame R, and ω and α are the angular velocity and angular acceleration of B in R Then, from Newton’s laws and d’Alembert’s principle, we have: F = MaG (10.6.12) T = I ⋅ α + ω × (I ⋅ ω ) (10.6.13) and ... 6, 2002 2: 57 PM 324 Dynamics of Mechanical Systems t* ∫ W = F ⋅ v dt (10.2.13) where t* is the time of action of F The integrand of Eq (10.2.13), F • v, is often called the power of F (see Section... FIGURE P9 .7. 5 Side impact to the rear of a stopped vehicle A 3500 lb stopped 0593_C09_fm Page 316 Monday, May 6, 2002 2:50 PM 316 Dynamics of Mechanical Systems of B be 64 in in front of the rear... 2002 2: 57 PM 322 Dynamics of Mechanical Systems P P d F P C F FIGURE 10.2.1 A force F moving a particle P FIGURE 10.2.2 A force F applied to a particle P moving on a curve C In view of these