Dynamics of Mechanical Systems 2009 Part 2 potx

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32 Dynamics of Mechanical Systems Returning to the product A × D, let A and D be depicted as in Figure 2.7.5a, where θ is the angle between the vectors. As before, let n A be a unit vector parallel to A. Then, n A × D is a vector perpendicular to n A and with magnitude Dsinθ. From Figure 2.7.5b, we see that: (2.7.18) By similar reasoning we have: (2.7.19) Therefore, by comparing Eqs. (2.7.17) and (2.7.19), we have: (2.7.20) This establishes the distributive law. Finally, suppose that n 1 , n 2 , and n 3 are mutually perpendicular unit vectors, and suppose that vectors A and B are expressed in the forms: (2.7.21) Then, by repeated use of Eqs. (2.3.6), (2.7.5), and (2.7.20) we see that A × B may be expressed as: (2.7.22) By recalling the elementary rules for expanding determinants, we see that Eq. (2.7.22) may be written as: (2.7.23) This is a useful algorithm for computing the vector product. FIGURE 2.7.5 Vectors A, D, n A , D  , and D ⊥ . (a) (b) θ θ A D n D ll D D A n A ⊥ DDsinθ= ⊥ ABAB ACAC×=× ×=× ⊥⊥ and ADA BC ABAC×=×+ () =×+× An n n n n n=++ =++AAA BBBB 11 22 33 11 22 33 and AB n n n n ×= − () +− () +− () = === ∑∑∑ AB AB AB AB AB AB eAB ijk ii k kji 23 32 1 31 31 2 12 21 3 1 3 1 3 1 3 AB×= nnn AAA BBB 123 123 123 0593_C02_fm Page 32 Monday, May 6, 2002 1:46 PM Review of Vector Algebra 33 Example 2.7.1: Vector Product Computation and Geometric Properties of the Vector Product Let vectors A and B be expressed in terms of mutually perpendicular dextral unit vectors n 1 , n 2 , and n 3 as: (2.7.24) Let C be the vector product A × B. a. Find C. b. Show that C is perpendicular to both A and B. c. Show that B × A = –C. Solution: a. From Eq. (2.7.24), C is: (2.7.25) b. If C is perpendicular to A, with the angle θ between C and A being 90˚, C • A is zero because cosθ is zero. Conversely, if C • A is zero, and neither C nor A is zero, then cosθ is zero, making C perpendicular to A. From Eq. (2.6.21), C • A is: (2.7.26) Similarly, C · B is (2.7.27) c. From Eq. (2.7.23), B × A is: (2.7.28) which is seen to be from Eq. (2.7.25). 2.8 Vector Multiplication: Triple Products On many occasions it is necessary to consider the product of three vectors. Such products are called “triple” products. Two triple products that will be helpful to use are the scalar triple product and the vector triple product. An n n Bn n n=−+ =+−724 38 123 123 and C nnn nnn=− − =+ + 123 123 724 13 8 46023 CA⋅= ()() + () − () + ()() =4 7 60 2 23 4 0 CB⋅= ()() + ()() + () − () =4 1 60 3 23 8 0 BA nnn nnn×= − − =− − − 123 123 13 8 724 46023 0593_C02_fm Page 33 Monday, May 6, 2002 1:46 PM 34 Dynamics of Mechanical Systems Given three vectors A, B, and C, the scalar triple product has the form A · (B × C). The result is a scalar. The scalar triple product is seen to be a combination of a vector product and a scalar product. Recall from Eqs. (2.6.21) and (2.7.23) that if n 1 , n 2 , and n 3 are mutually perpendicular unit vectors, the algorithms for evaluating the scalar and vector products are: (2.8.1) and (2.8.2) where the A i and B i (i = 1, 2, 3) are the n i components of A and B. By comparing Eqs. (2.8.1) and (2.8.2), we see that the scalar triple product A · (B × C) may be obtained by replacing n 1 , n 2 , and n 3 in: (2.8.3) Recall from the elementary rules of evaluating determinants that the rows and columns may be interchanged without changing the value of the determinant. Also, the rows and columns may be cyclically permuted without changing the determinant value. If the rows or columns are anticyclically permuted, the value of the determinant changes sign. Hence, we can rewrite Eq. (2.8.3) in the forms: (2.8.4) By examining Eq. (2.8.4), we see that the dot and the cross may be interchanged in the product. Also, the parentheses are unnecessary. Finally, the vectors may be cyclically permuted without changing the value of the product. An anticyclic permutation changes the sign of the result. Speciﬁcally, (2.8.5) AB⋅= + +AB AB AB 11 22 33 AB A A A BBB ×= nnn 123 123 123 ABC⋅× () = AAA BBB CCC 123 123 123 ABC⋅× () === =− =− =− AAA BBB CCC BBB CCC AAA CCC AAA BBB mp BBB AAA CCC AAA CCC BBB CCC BB 123 123 123 123 123 123 123 123 123 123 123 123 123 123 123 123 12 BB AAA 3 123 ABC ABCABCCAB BC A BA C CB A AC B ⋅× () =⋅×=×⋅=⋅× =⋅×=−⋅×=−⋅×=−⋅× 0593_C02_fm Page 34 Monday, May 6, 2002 1:46 PM Review of Vector Algebra 35 If the vectors A, B, and C coincide with the edges of a parallelepiped as in Figure 2.8.1, the scalar triple product of A, B, and C is seen to be equal to the volume of the parallel- epiped. That is, the volume V is: (2.8.6) Example 2.8.1: Veriﬁcation of Interchangeability of Terms of Triple Scalar Products Let vectors A, B, and C be expressed in terms of mutually perpendicular unit vectors n 1 , n 2 , and n 3 as: (2.8.7) Verify the equalities of Eq. (2.8.5). Solution: From Eq. (2.8.2), the vector products A × B and B × C are: (2.8.8) (2.8.9) Then, from Eq. (2.6.22), the triple scalar products A × B · C and B × C · A are: (2.8.10) and (2.8.11) The other equalities are veriﬁed similarly (see Problem P2.8.1). Next, the vector triple product has one of the two forms: A × (B × C) or (A × B) × C. The result is a vector. The position of the parentheses is important, as the two forms generally produce different results. To explore this, let the vectors A, B, and C be expressed in terms of mutually perpen- dicular unit vectors n i with scalar components A i , B i , and C i (i = 1, 2, 3). Then, by using FIGURE 2.8.1 A parallelepiped with vectors A, B, and C along the edges. B A C V =×⋅ABC AnnnBnnnCnnn=− − + = + − =− + −324735 123 1 2 3 1 2 3 , , AB nnn nnn×= − − =+ + 123 123 311 24 7 32314 BC nnn nnn×= − −− =+ + 123 123 24 7 13 5 17 10 ABC×⋅= () − () + ()() + () − () =−3 1 23 3 14 5 4 BCA×⋅= ()() + () − () + ()() =−1 3 17 1 10 1 4 0593_C02_fm Page 35 Monday, May 6, 2002 1:46 PM 36 Dynamics of Mechanical Systems the algorithms of Eqs. (2.8.1) and (2.8.2), we see that the vector triple products may be expressed as: (2.8.12) and (2.8.13) Observe that the last terms in these expressions are different. Example 2.8.2: Validity of Eqs. (2.8.6) and (2.8.7) and the Necessity of Parentheses Verify Eqs. (2.8.6) and (2.8.7) using the vectors of Eq. (2.8.7). Solution: From Eqs. (2.8.2) and (2.8.9), A × (B × C) is: (2.8.14) From Eq. (2.6.22), (A · C)B – (A · B)C is: (2.8.15) Similarly, (A × B) × C and (A · C)B – (B · C)A are: (2.8.16) and (2.8.17) ABC ACBABC×⋅ () =⋅ () −⋅ () AB C ACB BCA× () ×= ⋅ () −⋅ () ABC nnn nnn×× () =− =−−+ 123 123 311 11710 27 29 52 ACB ACC n n n nnn nnn n ⋅ () −⋅ () = () − () +− ()() + () − () [] +− () − ()() +− ()() + () − () [] −+ − () =− () +− () −− () −+ 31 13 152 4 7 32 14 1 7 3 5 11 2 4 7 5 123 123 123 1 335 27 29 52 23 123 nn nnn − () =− − + AB C nnn nn n× () ×= −− =− + + 123 12 3 32314 13 5 157 32 ACB BCA n n n nnn nnn n ⋅ () −⋅ () = () − () +− ()() + () − () [] +− () − () − () + ()() +− () − () [] −+ () =− () +− () − () − 31 13 152 4 7 21 43 753 11 2 4 7 45 3 123 123 123 1 nnn nn n 23 12 3 157 32 + () =− + + 0593_C02_fm Page 36 Monday, May 6, 2002 1:46 PM Review of Vector Algebra 37 Observe that the results in Eqs. (2.8.14) and (2.8.15) are identical and thus consistent with Eq. (2.8.12). Similarly, the results of Eqs. (2.8.16) and (2.8.17) are the same, thus verifying Eq. (2.8.13). Finally, observe that the results of Eqs. (2.8.14) and (2.8.16) are different, thus demonstrating the necessity for parentheses on the left sides of Eqs. (2.8.12) and (2.8.13). Recall from Eq. (2.6.10) that the projection A  of a vector A along a line L is: (2.8.18) where n is a unit vector parallel to L, as in Figure 2.8.2. The vector triple product may be used to obtain A ⊥ , the component of perpendicular to L. That is, (2.8.19) To see this, use Eqs. (2.8.6) and (2.8.8) to expand the product. That is, (2.8.20) 2.9 Use of the Index Summation Convention Observe in the previous sections that expressing a vector in terms of mutually perpendic- ular unit vectors results in a sum of products of the scalar components and the unit vectors. Speciﬁcally, if v is any vector and if n 1 , n 2 , and n 3 are mutually perpendicular unit vectors, we can express v in the form: (2.9.1) Because these sums occur so frequently, and because the pattern of the indices is similar for sums of products, it is convenient to introduce the “summation convention”: if an index is repeated, there is a sum over the range of the index (usually 1 to 3). This means, for example, in Eq. (2.9.1), that the summation sign may be deleted. That is, v may be expressed as: (2.9.2) FIGURE 2.8.2 Projection of a vector A parallel and perpendicular to a line. L n A A A ⊥ ࿣ AAnn || =⋅ () AnAn ⊥ =× × () nAn An nAA A×× () =−⋅ () =− = ⊥ A || v nnn n=− + + = = ∑ vvv v ii i 11 21 33 1 3 Vn= v ii 0593_C02_fm Page 37 Monday, May 6, 2002 1:46 PM 38 Dynamics of Mechanical Systems In using this convention, several rules are useful. First, in an equation or expression, a repeated index may not be repeated more than one time. Second, any letter may be used for a repeated index. For example, in Eq. (2.9.2) we may write: (2.9.3) Finally, if an index is not repeated it is a “free” index. In an equation, there must be a correspondence of free indices on both sides of the equation and in each term of the equation. In using the summation convention, we can express the scalar and vector products as follows: If n 1 , n 2 , and n 3 are mutually perpendicular unit vectors and if vectors a and b are expressed as: (2.9.4) then the products a · b and a × b are: (2.9.5) where, as before, e ijk is the permutation symbol (see Eq. (2.7.7)). With a little practice, the summation convention becomes natural in analysis procedures. We will employ it when it is convenient. Example 2.9.1: Kronecker Delta Function Interpreted as a Substitution Symbol Consider the expression δ ij V j where the δ ij (i, j = 1, 2, 3) are components of the Kronecker delta function deﬁned by Eq. (2.6.7) and where V j are the components of a vector V relative to a mutually perpendicular unit vector set n i (i = 1, 2, 3). From the summation convention, we have: (2.9.6) In this equation, i has one of the values 1, 2, or 3. If i is 1, the right side of the equation reduces to V 1 ; if i is 2, the right side becomes V 2 ; and, if i is 3, the right side is V 3 . Therefore, the right side is simply V i . That is, (2.9.7) The Kronecker delta function may then be interpreted as an index operator, substituting an i for the j, thus the name substitution symbol. 2.10 Review of Matrix Procedures In continuing our review of vector algebra, it is helpful to recall the elementary procedures in matrix algebra. For illustration purposes, we will focus our attention primarily on square vn n n===vvv ii jj kk an bn==ab ii ii and ab a b n⋅= ×=a b and e a b ii ijk ij k δδδ δ ij j i i i VVVVi=++ = () 11 22 33 123 , , δ ij j i VV= 0593_C02_fm Page 38 Monday, May 6, 2002 1:46 PM Review of Vector Algebra 39 matrices and on row and column arrays. Recall that a matrix A is simply an array of numbers a ij (i = 1,…, m i ; j = 1,…, n) arranged in m rows and n columns as: (2.10.1) The entries a ij are usually called the elements of the matrix. The ﬁrst subscript indicates the row position, and the second subscript designates the column position. Two matrices A and B are said to be equal if they have equal elements. That is, (2.10.2) If all the elements of matrix are zero, it is called a zero matrix. If a matrix has only one row, it is called a row matrix or row array. If a matrix has only one column, it is called a column matrix or column array. A matrix with an equal number of rows and columns is a square matrix. If all the elements of a square matrix are zero except for the diagonal elements, the matrix is called a diagonal matrix. If all the diagonal elements of a diagonal matrix have the value 1, the matrix is called an identity matrix. If a square matrix has a zero determinant, it is said to be a singular matrix. The transpose of a matrix A (written A T ) is the matrix obtained by interchanging the rows and columns of A. If a matrix and its transpose are equal, the matrix is said to be symmetric. If a matrix is equal to the negative of its transpose, it is said to be antisymmetric. Recall that the algebra of matrices is based upon a few simple rules: First, the multipli- cation of a matrix A by a scalar s produces a matrix B whose elements are equal to the elements of A multiplied by s. That is, (2.10.3) Next, the sum of two matrices A and B is a matrix C whose elements are equal to the sum of the respective elements of A and B. That is, (2.10.4) Matrix subtraction is deﬁned similarly. The product of matrices is deﬁned through the “row–column” product algorithm. The product of two matrices A and B (written AB) is possible only if the number of columns in the ﬁrst matrix A is equal to the number of rows of the second matrix B. When this occurs, the matrices are said to be conformable. If C is the product AB, then the element c ij is the sum of products of the elements of the ith row of A with the corresponding elements of the jth column of B. Speciﬁcally, (2.10.5) where n is the number of columns of A and the number of rows of B. A aa an aa an aa a mn mn =             11 12 1 21 22 2 12 AB a b i nj m ij ij ==== , , ; , , if and only if 1 1 BsA b sa ij ij == if and only if CAB c a b ij ij ij =+ =+ if andonlyif cab ij ik kj k n = = ∑ 1 0593_C02_fm Page 39 Monday, May 6, 2002 1:46 PM 40 Dynamics of Mechanical Systems Matrix products are distributive over addition and subtraction. That is, (2.10.6) and (2.10.7) Matrix products are also associative. That is, for conformable matrices A, B, and C, we have: (2.10.8) Hence, the parentheses are unnecessary. Next, it is readily seen that the transpose of a product is the product of the transposes in reverse order. That is: (2.10.9) Finally, if A is a nonsingular square matrix, the inverse of A, written as A –1 , is the matrix such that: (2.10.10) where I is the identity matrix. A –1 may be determined as follows: Let M ij be the minor of the element a ij deﬁned as the determinant of the matrix occurring when the ith row of A and the jth column of a are deleted. Let Â ij be the adjoint of a ij deﬁned as: (2.10.11) Then A –1 is the matrix with elements: (2.10.12) where detA designates the determinant of A and [Â ij ] T is the transpose of the matrix of adjoints. If it happens that: (2.10.13) then A is said to be orthogonal. In this case, the rows and columns of A may be considered as components of mutually perpendicular unit vectors. AB C AB AC+ () =+ BCA BACA+ () =+ AB C A BC ABC () = () = AB B A T TT () = AA A A I −− == 11 ˆ AM ij ij ij =− () + 1 AA A ij T − = [] 1 ˆ det AA T− = 1 0593_C02_fm Page 40 Monday, May 6, 2002 1:46 PM Review of Vector Algebra 41 2.11 Reference Frames and Unit Vector Sets In the analysis of dynamical systems, it is frequently useful to express kinematical and dynamical quantities in several different reference frames. Orthogonal transformation matrices (as discussed in this section) are useful in obtaining relationships between the representations of the quantities in the different reference frames. To explore these ideas consider two unit vector sets n i and i and an arbitrary V as in Figure 2.11.1. Let the sets be inclined relative to each other as shown. Recall from Eq. (2.6.19) that V may be expressed in terms of the n i as: (2.11.1) where the V i are the scalar components of V. Similarly, V may be expressed in terms of the i as: (2.11.2) Given the relative inclination of the unit vector sets, our objective is to obtain relations between the V i and the i . To that end, let S be a matrix with elements S ij deﬁned as: (2.11.3) Consider the n i : from Eq. (2.11.2), we can express n i as: (2.11.4) Similarly, n 2 and n 3 may be expressed as: (2.11.5) Thus, in general, we have: (2.11.6) FIGURE 2.11.1 Two unit vector sets. ˆ n V Vn n Vn n Vn n Vn n Vn=⋅ () +⋅ () +⋅ () =⋅ () = 11 22 33 ii ii ˆ n VVnnVn=⋅ () = ˆˆ ˆ ˆ ii ii ˆ V S ij i j =⋅nn ˆ nnnn n 11 1 =⋅ () = ˆˆ ˆ ii ii S nn nn 22 33 ==SS ii ii ˆˆ and nn iijj S= ˆ V n n n n n n 3 1 2 3 2 1 ˆ ˆ ˆ 0593_C02_fm Page 41 Monday, May 6, 2002 1:46 PM [...]... between n2 and n2 (and also between n3 and n3) is α, as shown ˆ in Figure 2. 11 .2 Then, by inspection of the ﬁgure, ni and ni are related by the expressions: ˆ n1 = n1 ˆ n1 = n1 ˆ n 2 = cα n 2 + sα n 3 ˆ ˆ n 3 = sα n 2 + cα n 3 and ˆ n 2 = cα n 2 + sα n 3 ˆ n 3 = − sα n 2 + cα n 3 (2. 11.13) 0593_C 02_ fm Page 43 Monday, May 6, 20 02 1:46 PM Review of Vector Algebra 43 ˆ n3 n3 ˆ n2 α α n2 FIGURE 2. 11 .2 ˆ Unit... the scalar product A · B |A| = 8 B n |B| = 5 2 120 ° A FIGURE P2.6.1 Vectors A and B n 1 P2.6 .2: See Problem P2.6.1 and Figure P2.6.1 Express A and B in terms of the unit vectors n1 and n2 as shown in the ﬁgure Use Eq (2. 6 .2) to evaluate A · B P2.6.3: See Problems P2.6.1 and P2.6 .2 Let C be the resultant of A and B Find the magnitude of C P2.6.4: Let n1, n2, and n3 be mutually perpendicular unit vectors...0593_C 02_ fm Page 42 Monday, May 6, 20 02 1:46 PM 42 Dynamics of Mechanical Systems ˆ ˆ ˆ ˆ Similarly, if we express n1, n2, and n3 in terms of the ni, we have: ˆ n i = Sij n j (2. 11.7) Observe the difference between Eqs (2. 11.6) and (2. 11.7): in Eq (2. 11.6), the sum is taken over the second index of Sij, whereas in Eq (2. 11.7) it is taken over the ﬁrst index This is consistent with Eq (2. 11.3), where... (see Eq (2. 10.11)) Compute A–1 Compute the products A–1A and AA–1 P2.10 .2: Given the matrix S (an orthogonal matrix):  12  S = − 3 2  0  a b c d Find detS Find S–1 Find ST Compare S–1 and ST 6 4 2 4 − 2 2 6 4  2 4 2 2  0593_C 02_ fm Page 55 Monday, May 6, 20 02 1:46 PM Review of Vector Algebra 55 Section 2. 11 Reference Frames and Unit Vector Sets P2.11.1: Determine the numerical values of the... Figure P2.7.7 Q 1(-1, -2, 7) Z P2 (0,6,4) LP nz Q 2( 4,5 ,2) P1 (3,-1,3) LQ Y FIGURE P2.7.7 Lines LP and LQ in an X, Y, Z reference frame ny X nx P2.7.8: See Figure P2.7.8 Let points P and Q have the coordinates (in feet) as shown Let L be a line passing through P and Q, and let F be a force acting along L Let the magnitude of F be 7 lb 0593_C 02_ fm Page 52 Monday, May 6, 20 02 1:46 PM 52 Dynamics of Mechanical. .. the resultant R of F1 Express R in terms of i, j, and k e Find the magnitude of R k 4 in H C F6 D F 1 3 in O FIGURE P2.4.7 Forces acting along the edges and diagonals of a parallelepiped F4 A i F2 G B 12 in E j F3 F5 P2.4.8: Let forces be given as: F1 = 3n1 − 2n 2 + n 3 N F2 = 4n1 + 6n 2 − 2n 3 N F3 = − n1 + 3n 2 + 4n 3 N F4 = 7 n1 − 8n 2 + 3n 3 N F5 = −5n1 + 4n 2 − 9n 3 N where n1, n2, and n3 are mutually... given by: d = P1P2 ⋅ n 0593_C 02_ fm Page 53 Monday, May 6, 20 02 1:46 PM Review of Vector Algebra 53 Using these concepts, ﬁnd the distance d between the lines L1 and L2 shown in Figure P2.8.3, where the coordinates are expressed in feet (Observe that n may be obtained from the vector product (P1Q1 × P2Q2)/P1Q1 × P2Q2.) Z nz P1 (2, -2, 4) n Q 2( 0,6,3) P (6,-1,4) 2 ny d O Y Q 1(1,5,0) FIGURE P2.8.3 Distance... Components P2.4.1: Given the unit vectors i and j and the vectors A and B as shown in Figure P2.4.1, let the magnitudes of A and B be 10 N and 15 N, respectively Express A and B in terms of i and j P2.4 .2: See Problem P2.4.1 Let C be the resultant (or sum) of A and B a Express C in terms of the unit vectors i and j b Find the magnitude of C 0593_C 02_ fm Page 46 Monday, May 6, 20 02 1:46 PM 46 Dynamics of Mechanical. .. transformation matrix elements of Eq (2. 11.19) if α, β, and γ have the values: α = 30°, β = 45°, γ = 60° P2.11 .2: See Problem P2.11.1 Show that ST = S–1 and that detS = 1 0593_C 02_ fm Page 56 Monday, May 6, 20 02 1:46 PM 0593_C03_fm Page 57 Monday, May 6, 20 02 2:03 PM 3 Kinematics of a Particle 3.1 Introduction Kinematics is a study of motion without regard to the cause of the motion Often this motion occurs... F Y ny nx X P2.7.9: Let eijk and δjk be the permutation symbol and the Kronecker delta symbol as in Eqs (2. 7.7) and (2. 6.7) Evaluate the sums: 3 3 ∑∑e ijk δ jk i = 1, 2, 3 j =1 k =1 Section 2. 8 Vector Multiplication: Triple Products P2.8.1: See Example 2. 8.1 Verify the remaining equalities of Eq (2. 8.5) for the vectors A, B, and C of Eq (2. 8.7) P2.8 .2: Use Eq (2. 8.6) to ﬁnd the volume of the parallelepiped . =− AAA BBB CCC BBB CCC AAA CCC AAA BBB mp BBB AAA CCC AAA CCC BBB CCC BB 123 123 123 123 123 123 123 123 123 123 123 123 123 123 123 123 12 BB AAA 3 123 ABC ABCABCCAB BC A BA C CB A AC B ⋅× () =⋅×=×⋅=⋅× =⋅×=−⋅×=−⋅×=−⋅× 0593_C 02_ fm Page 34 Monday, May 6, 20 02 1:46. 4 2 F 6 i j Fnnn Fnnn Fnnn Fnnn Fnnn 1 123 21 23 3 123 4 123 5 123 32 4 62 34 783 549 =−+ =+− =− + + =−+ =− + − N N N N N 0593_C 02_ fm Page 47 Monday, May 6, 20 02 1:46 PM 48 Dynamics of Mechanical. AB AB eAB ijk ii k kji 23 32 1 31 31 2 12 21 3 1 3 1 3 1 3 AB×= nnn AAA BBB 123 123 123 0593_C 02_ fm Page 32 Monday, May 6, 20 02 1:46 PM Review of Vector Algebra 33 Example 2. 7.1: Vector Product
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