Dynamics of Mechanical Systems 2009 Part 6 pdf

232 Dynamics of Mechanical Systems and P7.2.8: See Problem P7.2.5. Find the x, y, z coordinates of the mass center G of S. Find P7.2.9: See Problem P7.2.8. Let G have an associated mass of 9 kg (equal to the sum of the masses of P 1 , P 2 , and P 3 ). Find P7.2.10: See Problems P7.2.5, P7.2.6, and P7.2.9. Show that: and P7.2.11: See Problem P7.2.5. Find a unit vector n perpendicular to the plane of P 1 , P 2 , and P 3 . Find also Show that is parallel to n. Section 7.3 Moments and Products of Inertia P7.3.1: See Problem P7.2.1. A particle P with mass of 3 slug has coordinates (2, –1, 3), measured in feet, in a Cartesian coordinate system as represented in Figure P7.3.1. Deter- mine the following moments and products of inertia: P7.3.2: See Problems P7.2.2 and P7.3.1. Let Q have coordinates (–1, 2, 4). Repeat Problem P7.3.1 with Q, instead of O, being the reference point. That is, determine P7.3.3: See Problem P7.2.5. Let a set S of three particles P 1 , P 2 , and P 3 be located at the vertices of a triangle as shown in Figure P7.3.3. Let the particles have masses 2, 3, and FIGURE P7.3.1 A particle P and a point Q. III b SO y SO z SO =+0 655 0 756 I III I x SG y SG z SG a SG b SG , , , , and . IIII I x GO y GO z GO a GO b GO , , , , and . III III III III x SO x SG x GO y SO y SG y GO z SO z SG z GO a SO a SG a GO =+ =+ =+ =+ III b SO b SG b GO =+ I n SO . I n S O I III III xx PO xy PO xz PO yy PO yz PO zz PO aa PO , ,,, ,,, II I bb PO ab PO ba PO , , and . III xx PQ xy PQ xz PQ , , , I IIIII I yy PQ yz PQ zz PQ aa PQ bb PQ ab PQ ba PQ , , , , , , and . 0593_C07_fm Page 232 Monday, May 6, 2002 2:42 PM Inertia, Second Moment Vectors, Moments and Products of Inertia, Inertia Dyadics 233 4 kg, respectively. Find the following moments and products of inertia of S relative to the origin O of the X-, Y-, Z-axis system of Figure P7.3.3: P7.3.4: See Problems P7.2.5, P7.2.8, P7.2.9, and P7.3.3. For the system S shown in Figure P7.3.3, find the following moments and products of inertia: where G is the mass center of S, as determined in Problem P7.2.8. (Compare the magnitudes of these results with those of Problem P7.3.3.) P7.3.5: See Problems P7.2.9 and P7.3.3. For the system S shown in Figure P7.3.3, find the following moments and products of inertia: P7.3.6: See Problems P7.2.10, P7.3.3, P7.3.4, and P7.3.5. Show that: P7.3.7: See Problems P7.2.5, P7.2.6, P7.2.7, and P7.3.3. As in Problem P7.2.6 let n a and n b be the unit vectors: and Find Section 7.4 Inertia Dyadics P7.4.1: Let vectors a, b, and c be expressed as: FIGURE P7.3.3 Particles at the vertices of a triangle. IIIII I xx SO xy SO xz SO yy SO yz SO zz SO , , , , , and . I III I xx SG xy SG xz SG yy SG yz SG , ,,, , and I zz SG I IIII I xx GO xy GO xz GO yy GO yz GO zz GO , , , , , and . III ij SO ij SG ij GO ij xyz=+ = () ,,, nnn n axy z =−+0 75 0 5 0 433 nnn b yz =+0 655 0 756 II I aa SO ab SO bb SO ,, .and an n n bnnn cnn n =−+ =− + − =−+ 634 547 39 123 123 12 3 0593_C07_fm Page 233 Monday, May 6, 2002 2:42 PM 234 Dynamics of Mechanical Systems where n 1 , n 2 , and n 3 are mutually perpendicular unit vectors. Compute the following dyadic products: (a) ab, (b) ba, (c) ca + cb, (d) c(a + b), (e) (a + b)c, and (f) ac + bc. P7.4.2: See Problem 7.2.1. A particle P with mass 3 slug has coordinates (2, –1, 3), measured in feet, in a Cartesian coordinate system as represented in Figure P7.4.2. Determine the inertia dyadic of P relative to the origin O, I P/O . Express the results in terms of the unit vectors n x , n y , and n z . P7.4.3: See Problem P7.2.2. Let Q have coordinates (–1, 2, 4). Repeat Problem P7.4.2 with Q instead of O being the reference point. That is, find I P/Q . P7.4.4: See Problems P7.2.5 and P7.3.3. Let S be the set of three particles P 1 , P 2 , and P 3 located at the vertices of a triangle as shown in Figure P7.4.4. Let the particles have masses: 2, 3, and 4 kg, respectively. Find the inertia dyadic of S relative to O, I S/O . Express the results in terms of the unit vectors n x , n y , and n z . P7.4.5: See Problems P7.2.8, P7.2.9, P7.3.4, and P7.4.4. Let G be the mass center of S. Find the inertia dyadic of S relative to G, I S/G . Express the results in terms of the unit vectors n x , n y , and n z . P7.4.6: See Problems P7.4.4 and P7.4.5. Let G have an associated mass of 9 kg. Find the inertia dyadic of G relative to the origin O, I G/O . Express the result in terms of the unit vectors n x , n y , and n z . P7.4.7: See Problems P7.4.5 and P7.4.6. Show that: FIGURE P7.4.2 A particle P in a Cartesian reference frame. FIGURE P7.4.4 Particles at the vertices of a triangle. O Z Y X Q(-1,2,4) P(2,-1,3) n n n z y x O Z Y X n n n z y x P (0,5,2) P (1,1,1) (units in meters) 2 3 1 P (2,2,4) III SO SG GO =+ 0593_C07_fm Page 234 Monday, May 6, 2002 2:42 PM Inertia, Second Moment Vectors, Moments and Products of Inertia, Inertia Dyadics 235 P7.4.7: See Problems P7.2.5 and P7.4.4. Find the second moments of S relative to O for the directions of n x , n y , and n z . P7.4.8: See Problems P7.3.3 and P7.4.4. Find the following moments and products of inertia of S for O: P7.4.9: See Problems P7.2.6 and P7.4.4. Let n a and n b be unit vectors with coordinates relative to n x , n y , and n z as: Find the second moment vectors P7.4.10: See Problems P7.2.5, P7.2.6, P7.3.7, P7.4.4, and P7.4.9. Let n a and n b be the unit vectors of Problem P7.4.9. Find the following moments and products of inertia of S relative to O: Section 7.5 Transformation Rules P7.5.1: Let S be a set of eight particles P i (i = 1,…, 8) located at the vertices of a cube as in Figure P7.5.1. Let the masses m i of the P i be as listed in the figure. Determine the second- moment vectors for the directions of the unit vectors n 1 , n 2 , and n 3 shown in Figure P7.5.1. P7.5.2: See Problem P7.5.1. Let n a , n b , and n c be unit vectors with components relative to n 1 , n 2 , and n 3 as: FIGURE P7.5.1 Particles at the vertices of a cube. IIII I xx SO xz SO yy SO yz SO zz SO , , , , and . nnn n nnn axy z b yz =−+ =+ 0 75 0 5 0 433 0 655 0 756 II a SO b SO and . II I aa SO ab SO bb SO ,, .and II I 12 3 SO SO S O , , and O Z X Y 2m 2m 2m P n n n P P P P P P P 3 4 8 5 6 2 7 1 1 2 3 m = 2 kg, 1 m = 3 kg, 2 m = 1 kg, 3 m = 5 kg, 4 m = 4 kg 5 m = 6 kg 6 m = 3 kg 7 m = 2 kg 8 nn n nnnn nn nn a b c =+ =− + + =− + 05 0866 0 433 0 25 0 866 075 0433 05 12 12 3 123 . 0593_C07_fm Page 235 Monday, May 6, 2002 2:42 PM 236 Dynamics of Mechanical Systems Determine the second moment vectors P7.5.3: See Problem P7.5.1. Determine the moments and products of inertia (i, j = 1, 2, 3). P7.5.4: See Problem P7.5.2. Let the transformation matrix between n a , n b , n c and n 1 , n 2 , n 3 have elements S jα (j = 1, 2, 3; α = a, b, c) defined as: Find the S jα . P7.5.5: See Problems P7.5.1 to P7.5.4. Find the moments and products of inertia (α, β = a, b, c). Also verify that: and P7.5.6: See Problem P7.5.3. Find the inertia dyadic I S/O . Express the results in terms of the unit vectors n 1 , n 2 , and n 3 of Figure P7.5.1. P7.5.7: See Problems P7.5.5 and P7.5.6. Verify that (α, β = a, b, c) is given by: P7.5.8: See Problems P7.5.3 and P7.5.5. Verify that: P7.5.9: A 3-ft bar B weighs 18 pounds. Let the bar be homogeneous and uniform so that its mass center G is at the geometric center. Let the bar be placed on an X–Y plane so that it is inclined at 30° to the X-axis as shown in Figure P7.5.9. It is known that the moment of inertia of a homogeneous, uniform bar relative to its center is zero for directions parallel to the bar and mᐉ 2 /12 for directions perpendicular to the bar where m is the bar mass and ᐉ is its length (see Appendix II). It is also known that the products of inertia for a bar for directions parallel and perpendicular to the bar are zero. Determine the moments and products of inertia: FIGURE P7.5.9 A homogeneous bar in the X–Y plane with center at the origin. II I a SO b SO c SO ,, .and I ij S O S jjαα =⋅nn I αβ SO II αβ α β SO i j ij SO SS= II ij SO i j SO SS= α βαβ I αβ SO InIn αβ α β SO SO =⋅ ⋅ IIIIII 11 22 33 SO SO SO aa SO bb SO cc SO ++=++ IIIII I xx BG yy BG zz BG xy BG xz BG yz BG ,,,,, .and Y X B 30° O G ᐉ = 3 ft weight = 18 lb 0593_C07_fm Page 236 Monday, May 6, 2002 2:42 PM Inertia, Second Moment Vectors, Moments and Products of Inertia, Inertia Dyadics 237 P7.5.10: A thin uniform circular disk D with mass m and radius r is mounted on a shaft S with a small misalignment, measured by the angle θ as represented in Figure P7.5.10. Knowing that the moments of inertia of D for its center for directions parallel to and perpendicular to its axis are mr 2 /2 and mr 2 /4, respectively, and that the corresponding products of inertia of D for its axis and diameter directions are zero (see Appendix II), find the moment of inertia of D for its center G for the shaft axis direction x: Section 7.6 Parallel Axis Theorems P7.6.1: Consider the homogeneous rectangular parallepiped (block) B shown in Figure P7.6.1. From Appendix II, we see that the moments of inertia of B for the mass center G for the X, Y, and Z directions are: where m is the mass of B and a, b and c are the dimensions as shown in Figure P7.6.1. Let B have the following properties: Determine the moments of inertia of B relative to G for the directions of X, Y, and Z. P7.6.2: Repeat Problem P7.6.1 with B having the following properties: P7.6.3: See Problems P7.6.1 and P7.6.2. For the properties of Problems P7.6.1 and P7.6.2, find the moments of inertia of B for Q for the direction X, Y, and Z where Q is a vertex of B with coordinates (a, b, c) as shown in Figure P7.6.1. FIGURE P7.5.10 A misaligned circular disk on a shaft S. FIGURE P7.6.1 A homogeneous rectangular block. I xx DG . X D G S ImbcImacImab xx BG yy BG zz BG =+ () =+ () =+ () 1 12 1 12 1 12 22 22 22 , , mabc====12 2 4 3kg, m m m, , O Z Y X b a c G Q ma bc== ==15 2 5 5 3lb, ft ft ft., , 0593_C07_fm Page 237 Monday, May 6, 2002 2:42 PM 238 Dynamics of Mechanical Systems P7.6.4: A body B has mass center G with coordinates (1, 3, 2), in meters, in a Cartesian reference frame as represented in Figure P7.6.4. Let the mass of B be 0.5 kg. Let the inertia dyadic of B for the origin O have the matrix given by: where n 1 , n 2 , and n 3 are parallel to the X-, Y-, and Z-axes. Determine the components of the inertia dyadic of B for point Q, where the coordinates of Q are (2, 6, 3), in meters. P7.6.5: A thin, rectangular plate P weighs 15 lb. The dimensions of the plate are 20 in. by 10 in. See Figure P7.6.5, and determine the moments of inertia of P relative to corner A for the X, Y, and Z directions (see Appendix II). P7.6.6: Repeat Problem P7.6.5 for a plate with a 5-in diameter circular hole centered in the left half of the plate as represented in Figure P7.6.6. FIGURE P7.6.4 A body B in a Cartesian reference frame. FIGURE P7.6.5 A rectangular plate in a Cartesian reference frame. FIGURE 7.6.6 A rectangular plate with an offset circular hole. I ij BO Inn BO ij BO ij ij BO II== −− −− −−−           , . . 9 2 634 1 2 634 4 3 137 kgm 2 I i j BQ G(1,3,2) Q(2,6,3) B Z Y X O n n n 3 2 1 Y A X O P G 10 in. 20 in. Y A X O G 10 in. 20 in. 5 in 0593_C07_fm Page 238 Monday, May 6, 2002 2:42 PM Inertia, Second Moment Vectors, Moments and Products of Inertia, Inertia Dyadics 239 Sections 7.7, 7.8, 7.9 Principal Moments of Inertia P7.7.1: Review again the example of Section 7.8. Repeat the example for an inertia matrix given by: P7.7.2: A 2 × 4-ft rectangular plate OABC is bonded to a 2-ft-square plate CDEF, forming a composite body S as in Figure P7.7.2. Let the rectangular plate weigh 40 lb and the square plate 20 lb. a. Determine the x, y, z components of the mass center G of S. b. Find the inertia dyadic of S for G. Express the results in terms of the unit vectors n x , n y , and n z shown in Figure P7.7.2. c. Find the principal moments of inertia of S for G. d. Find the principal unit vectors of S for G. Express the results in terms of n x , n y , and n z . P7.7.3: Repeat Problem P7.7.2 if the square plate CDEF weighs 30 lb. P7.7.4: Repeat Problem P7.7.3 if the square plate CDEF weighs 10 lb. FIGURE P7.7.2 A composite plate body. I ij =           32 2 6 2329 6916 slugft 2 X Y E D F 2 ft 2 ft C B A O S 2 ft Z 2 ft 4 ft 0593_C07_fm Page 239 Monday, May 6, 2002 2:42 PM 0593_C07_fm Page 240 Monday, May 6, 2002 2:42 PM 241 8 Principles of Dynamics: Newton’s Laws and d’Alembert’s Principle 8.1 Introduction Dynamics is a combined study of motion (kinematics), forces (kinetics), and inertia (mass distributing). By using the principles of dynamics we can obtain mathematical models of the behavior of mechanical systems. In this chapter, and in subsequent chapters, we will explore the principles of dynamics and their applications. The development of dynamics principles dates back to at least the 14th century, long before the development of calculus and other widely used analytical procedures. One of the earliest statements of a dynamics principle in the Western world is attributed to John Buridan in (1358) [8.1]: From this theory also appears the cause of why the natural motion of a heavy body downward is continually accelerated. For from the beginning only the gravity was moving it. Therefore, it moved more slowly, but in moving it impressed in the heavy body an impetus. This impetus now together with its gravity moves it. Therefore, the motion becomes faster, and by the amount it is faster so the impetus becomes more intense. Therefore, the movement evidently becomes continually faster. While this statement seems to be intuitively reasonable, it is not strictly correct, as we now understand the physics of falling bodies. Moreover, the statement does not readily lead to a quantitative analysis. The earliest principles that adequately describe the physics and lead to quantitative analysis are generally attributed to Isaac Newton. His principles, first published in 1687, are generally stated in three laws [8.2]: First law (law of inertia): In the absence of forces applied to a particle, the particle will remain at rest or it will move along a straight line at constant velocity. Second law (law of kinetics): If a force is applied to a particle, the particle accelerates in the direction of the force. The magnitude of the acceleration is proportional to the magnitude of the force and inversely proportional to the mass of the particle. Third law (law of action–reaction): If two particles exert forces on each other, the respective forces are equal in magnitude and oppositely directed along the line joining the particles. 0593_C08_fm Page 241 Monday, May 6, 2002 2:45 PM [...]... on B consist of the system of forces made up of the inertia forces on the particles of B This system of forces (usually a very large number of forces) may be represented by an equivalent force system (see Section 6. 5) consisting of a single force F* FIGURE 8 .6. 1 A rigid body moving in an inertial reference frame 0593_C08_fm Page 250 Monday, May 6, 2002 2:45 PM 250 Dynamics of Mechanical Systems passing... T3 = 0 FIGURE 8.12.3 A free-body diagram of rod B (8.12.12) 0593_C08_fm Page 266 Monday, May 6, 2002 2:45 PM 266 Dynamics of Mechanical Systems where a2 is the n2 component of a and T3 is the n3 component of T*; a2 may be obtained directly from Eq (8.12.8), and, because n1, n2, and n3 are parallel to principal inertia directions of B for G, T3 is given by Eq (8 .6. 12) as: T3 = −α 3 I 33 + ω 1ω 2 ( I11... parallel to the axis of S as shown Let the radius of S be small Let B be attached to S at one of its ends O by a frictionless pin whose axis is fixed on a radial 0593_C08_fm Page 264 Monday, May 6, 2002 2:45 PM 264 Dynamics of Mechanical Systems FIGURE 8.12.1 Rotating, pinned rod FIGURE 8.12.2 Unit vectors of the shaft and rod line of S Let B have length ᐉ and mass m, and let the orientation of B be defined... stated, the sum of the applied and inertia forces on a particle is zero When d’Alembert’s principle is extended to a set of particles, or to rigid bodies, or to a system of particles and rigid bodies, the principle may be stated simply: the combined system of applied and inertia forces acting on a mechanical system is a zero system (see Section 6. 4) When sets of particles, rigid bodies, or systems are... diagram of B Using d’Alembert’s principle, the governing equations of motion of B are, then, W + F* = 0 (8.7.2) T* = 0 (8.7.3) a = − gN 3 (8.7.4) T1 = T2 = T3 = 0 (8.7.5) and or R G and Suppose the acceleration of G is expressed in the form: ˙˙ ˙˙ ˙˙ a = xN1 + yN 2 + zN 3 R G (8.7 .6) 0593_C08_fm Page 252 Monday, May 6, 2002 2:45 PM 252 Dynamics of Mechanical Systems where (x, y, z) are the coordinates of. .. the mass center of B, ri is the position vector of Pi relative to G, α is the angular acceleration of B in R, and ω is the angular velocity of B in R Let B be considered to be composed of particles such as the crystals of a sandstone Let Pi be a point of a typical particle having mass mi Then, from Eq (8.2.5), the inertia force on the particle is: Fi* = − mi R a Pi (no sum on i) (8 .6. 2) The inertia... fundamental principles of mechanics Newton’s laws directly provide a means for studying dynamical systems They also provide a means for developing other principles of dynamics These other principles are often in forms that are more convenient than Newton’s laws for the analysis of some classes of systems Some of these other principles have been formulated independently of Newton’s laws, but all of the principles... parallel to the axis of rotation and perpendicular to the unit vectors nx, ny , nr , and nθ as shown in Figure 8.9.1 FIGURE 8.9.1 The rod pendulum 0593_C08_fm Page 2 56 Monday, May 6, 2002 2:45 PM 2 56 Dynamics of Mechanical Systems Oy Ox ¨ (1/2)mᐉ 2 θ ¨ mᐉ θ /2 mg FIGURE 8.9.2 Free-body diagram of the rod pendulum 2 mᐉ θ /2 The mass center G moves in a circle with radius ᐉ/2 The acceleration of G is: ˙˙ ˙... R a Pi (8 .6. 3) i =1 and N T* = ∑ N ri × Fi* = − i =1 ∑m r × a R i i Pi (8 .6. 4) i =1 where N is the number of particles of B Recall that we already examined the summation in Eqs (8 .6. 3) and (8 .6. 4) in Section 7.12 Specifically, by using the definitions of mass center and inertia dyadic we found that F* and T* could be expressed as (see Eqs (6. 9.9), (7.12.1), and (7.12.8)): F * = −M R aG (8 .6. 5) and (... Because the system of forces in a free-body diagram is a zero system (see Section 6. 4), the forces must produce a zero resultant in all directions Hence, by adding force components in the radial and tangential directions, we obtain: ˙ T − mg cos θ − mlθ2 = 0 (8.4.1) 0593_C08_fm Page 2 46 Monday, May 6, 2002 2:45 PM 2 46 Dynamics of Mechanical Systems FIGURE 8.4.4 A free-body diagram of the pendulum mass . nnn b yz =+0 65 5 0 7 56 II I aa SO ab SO bb SO ,, .and an n n bnnn cnn n =−+ =− + − =−+ 63 4 547 39 123 123 12 3 0593_C07_fm Page 233 Monday, May 6, 2002 2:42 PM 234 Dynamics of Mechanical Systems where. the moments of inertia of B relative to G for the directions of X, Y, and Z. P7 .6. 2: Repeat Problem P7 .6. 1 with B having the following properties: P7 .6. 3: See Problems P7 .6. 1 and P7 .6. 2. For the. the principles of dynamics we can obtain mathematical models of the behavior of mechanical systems. In this chapter, and in subsequent chapters, we will explore the principles of dynamics and their