82 Dynamics of Mechanical Systems In matrix form these equations may be expressed as: (4.3.8) where N and n represent the columns of the N i and n i , and S is the matrix triple product: (4.3.9) where, as before, A, B, and C are: (4.3.10) Observe that by carrying out the product of Eq. (4.3.9) with A, B, and C given by Eq. (4.3.10) leads to Eq. (4.2.3) (see Problem 4.3.1). Finally, observe that a body B may be brought into a general orientation in a reference frame R by successively rotating B an arbitrary sequence of vectors as illustrated in the following example. Example 4.3.1: A 1–3–1 (Euler Angle) Rotation Sequence Consider rotating B about n 1 , then n 3 , and then n 1 again through angles θ 1 , θ 2 , and θ 3 . In this case, the conﬁguration graph takes the form as shown in Figure 4.3.9. With the rotation angles being θ 1 , θ 2 , and θ 3 , the transformation matrices are: (4.3.11) and the general transformation matrix becomes: (4.3.12) FIGURE 4.3.7 A rigid body B with a general orientation in a reference frame R. B n n n N R N N 3 3 2 2 1 1 Nn n N==SS T and S ABC= Acs sc B cs sc C cs sc=− = − = − 10 0 0 0 0 010 0 0 0 001 αα αα ββ ββ γγ γγ , , Acs sc B cs sc C c s sc =− =− =− 10 0 0 0 0 0 001 10 0 0 0 11 11 22 22 3 3 33 ,, S ABC csc ss cs ccc ss ccc sc ss scc cs scs cc == − −− () −− () −+ () −+ () 223 23 12 123 13 123 13 12 123 13 123 13 0593_C04*_fm Page 82 Monday, May 6, 2002 2:06 PM Kinematics of a Rigid Body 83 Hence, the unit vectors are related by the expressions: (4.3.13) and (4.3.14) The angles θ 1 , θ 2 , and θ 3 are Euler orientation angles and the rotation sequence is referred to as a 1–3–1 sequence. (The angles α, β, and γ are called dextral orientation angles or Bryan orientation angles and the rotation sequence is a 1–2–3 sequence.) 4.4 Simple Angular Velocity and Simple Angular Acceleration Of all kinematic quantities, angular velocity is the most fundamental and the most useful in studying the motion of rigid bodies. In this and the following three sections we will study angular velocity and its applications. We begin with a study of simple angular velocity, where a body rotates about a ﬁxed axis. Speciﬁcally, let B be a rigid body rotating about an axis Z–Z ﬁxed in both B and a reference frame R as in Figure 4.4.1. Let n be a unit vector parallel to Z–Z as shown. Simple angular velocity is then deﬁned to be a vector parallel to n measuring the rotation rate of B in R. To quantify this further, consider an end view of B and of axis Z–Z as in Figure 4.4.2. Let X and Y be axes ﬁxed in R and let L be a line ﬁxed in B and parallel to the X–Y plane. Let θ be an angle measuring the inclination of L relative to the X-axis as shown. Then, the angular velocity ω (simple angular velocity) of B in R is deﬁned to be: (4.4.1) where is sometimes called the angular speed of B in R. FIGURE 4.3.8 Conﬁguration graph deﬁning the orientation of B in R (see Figure 4.3.7). FIGURE 4.3.9 Conﬁguration graph for a 1–3–1 rotation. N i 1 2 3 N N α ˆ i i n n i i ˆ β γ N i 1 2 3 N N ˆ i i n n i i ˆ θ θ θ 1 2 3 Nn n n Nn n n Nn n n 121232233 2 12 1 123 13 2 12
3 13 3 3 12 1 123 13 2 123 13 3 =+ − =− + − () +− − () =− + + () +− + () cscss cs ccc ss ccc sc ss scc cs scs cc nN N N nN N N nN N N 121122123 2 23 1 123 13 2 123 13 3 3 23 1 123 13 2 123 13 3 =+ − =+− () ++ () =− + − − () +− + () ccsss sc ccc ss scc cs ss ccs sc scs cc ωω = D ˙ θn ˙ θ 0593_C04*_fm Page 83 Monday, May 6, 2002 2:06 PM 84 Dynamics of Mechanical Systems Observe that θ measures the rotation of B in R. It is also a measure of the orientation of B in R. In this context, the angular velocity of B in R is a measure of the rate of change of orientation of B in R. The simple angular acceleration α of B in R is then deﬁned as the time rate of change of the angular velocity. That is, (4.4.2) If we express α and β in the forms: (4.4.3) then α, ω, and θ are related by the expressions: (4.4.4) By integrating we have the relations: (4.4.5) where θ o and ω o are values of θ and ω at some initial time t o . From the chain rule for differentiation we have: (4.4.6) Then by integrating we obtain: (4.4.7) where, as before, ω o is an initial value of ω when θ is θ o . If α is constant, we have the familiar relation: (4.4.8) FIGURE 4.4.1 A body B rotating in a reference frame R. FIGURE 4.4.2 End view of body B. B Z Z n R Y L X Z θ B n ααωω==ddt ˙˙ θn ααωω==αωnn and ωθ αωθ=== ˙ ˙ ˙˙ and θωθ ωαω=+ =+ ∫∫ dt dt oo and αω ω ωθ θ ωωθ ω== = ()() == () ˙ d dt d d d dt d d d dt 2 2 ωαθω 22 2=+ ∫ d o ωω αθ 22 2=+ o 0593_C04*_fm Page 84 Monday, May 6, 2002 2:06 PM Kinematics of a Rigid Body 85 Example 4.4.1: Revolutions Turned Through During Braking Suppose a rotor is rotating at an angular speed of 100 rpm. Suppose further that the rotor is braked to rest with a constant angular deceleration of 5 rad/sec 2 . Find the number N of revolutions turned through during braking. Solution: From Eq. (4.4.8), when the rotor is braked to rest, its angular speed ω is zero. The angle θ turned through during braking is, then, Hence, the number of revolutions turned through is: 4.5 General Angular Velocity Angular velocity may be deﬁned intuitively as the time rate of change of orientation. Generally, however, no single quantity deﬁnes the orientation for a rigid body. Hence, unlike velocity, angular velocity cannot be considered as the derivative of a single quantity. Nevertheless, it is possible to deﬁne the angular velocity in terms of derivatives of a set of unit vectors ﬁxed in the body. These unit vector derivatives thus provide a measure of the rate of change of orientation of the body. Speciﬁcally, let B be a body whose orientation is changing in a reference frame R, as depicted in Figure 4.5.1. Let n 1 , n 2 , and n 3 be mutually perpendicular unit vectors as shown. Then, the angular velocity of B in R, written as R ω B , is deﬁned as: (4.5.1) The angular velocity vector has several properties that are useful in dynamical analyses. Perhaps the most important is the property of producing derivatives by vector multipli- cation: speciﬁcally, let c be any vector ﬁxed in B. Then, the derivative of c in R is given by the single expression: (4.5.2) FIGURE 4.5.1 A rigid body changing orientation in a reference frame. θωα π=− =− ()( ) [] () − () [] == o
2 2 2 2 100 60 2 5 10 966 628 3 radians degrees N = 1 754. RB d dt d dt d dt ωω = ⋅ +⋅ +⋅ D n nn n nn n nn 2 31 3 12 1 23 ddt RB cc=×ωω B R n n n 3 2 1 0593_C04*_fm Page 85 Monday, May 6, 2002 2:06 PM 86 Dynamics of Mechanical Systems To see this, let c be expressed in terms of the unit vectors n 1 , n 2 , and n 3 . That is, let (4.5.3) Because c is ﬁxed in B, the scalar components c i (i = 1, 2, 3) are constants; hence, the derivative of c in R is: (4.5.4) Next, consider the product R ω B × n 1 : (4.5.5) Recall that: (4.5.6) By differentiating these expressions we have: (4.5.7) Hence, Eq. (4.5.5) may be written as: (4.5.8) Because any vector V may be expressed as (V · n 1 )n 1 + (V · n 2 )n 2 + (V · n 3 )n 3 , we see that the right side of Eq. (4.5.8) is an identity for dn 1 /dt. Thus, we have the result: (4.5.9) Similarly, we have the companion results: (4.5.10) Finally, by using these results in Eq. (4.5.4) we have: (4.5.11) cn n n=++ccc 11 22 33 d dt c d dt c dt c dtcnnn=++ 1 1 22 33 RB d dt d dt d dt d dt d dt ωω× = ⋅ +⋅ +⋅ × =− ⋅ +⋅ n n nn n nn n nn n n nn n nn 1 2 31 3 12 1 2
3 1 3 13 1 22 nn nn 11 13 10⋅= ⋅= and d = 0 and 1 n n n nn n 11 31 3 dt d dt d dt ⋅⋅=−⋅ RB d dt d dt d dt ωω× = ⋅ +⋅ +⋅ n n nn n nn n nn
1 1 11 1 22 1 33 RB ddtωω× =nn 11 RB RB ddt ddtωωωω×= ×=nn nn 22 33 and ddtc c c ccc RB RB RB RB RB cnnn nnn c =×+×+× =× + + () =× 112233 11 22 33 ωωωωωω ωω ωω 0593_C04*_fm Page 86 Monday, May 6, 2002 2:06 PM Kinematics of a Rigid Body 87 Observe the pattern of the terms in Eq. (4.5.1). They all have the same form, and they may be developed from one another by simply permuting the indices. Example 4.5.1: Simple Angular Velocity We may also observe that Eq. (4.5.1) is consistent with our earlier results on simple angular velocity. To see this, let B rotate in R about an axis parallel to, say, n 1 , as shown in Figure 4.5.2. Let X–X be ﬁxed in both B and R. Then, from Eq. (4.4.1), the angular velocity of B in R is: (4.5.12) where, as before, θ measures the rotation angle. From Eq. (4.5.1), we see that R ω B may be expressed as: (4.5.13) Because n 1 is ﬁxed, parallel to axis X–X, its derivative is zero; hence, the third term in Eq. (4.5.13) is zero. The ﬁrst two terms may be evaluated using Eq. (3.5.7). Speciﬁcally, (4.5.14) By substituting into Eq. (4.5.13), we have: (4.5.15) which is identical to Eq. (4.5.11). 4.6 Differentiation in Different Reference Frames Consider next the differentiation of a vector with respect to different reference frames. Speciﬁcally, let V be the vector and let R and be two distinct reference frames. Let i be mutually perpendicular unit vectors ﬁxed in , as represented in Figure 4.6.1. Let V be expressed in terms of the i as: (4.6.1) FIGURE 4.5.2 Rotation of a body about a ﬁxed axis X–X . X X R n n n B 2 1 3 RB ωω= ˙ θn 1 RB d dt d dt d dt ωω= ⋅ +⋅ +⋅ n nn n nn n nn 2 31 3 12 1 23 ddt ddtnnnn nnnn 2123 3132 =×= =×=− ˙˙ ˙˙ θθ θ θ and RB ωω= ˙ θn 1 ˆ R ˆ n ˆ R ˆ n Vn n n=++ ˆ ˆ ˆ ˆ ˆ ˆ VVV 11 22 33 0593_C04*_fm Page 87 Monday, May 6, 2002 2:06 PM 88 Dynamics of Mechanical Systems Because the i are ﬁxed in , the derivative of V in is obtained by simply differen- tiating the scalar components in Eq. (4.6.1). That is, (4.6.2) Next, relative to reference frame R, the derivative of V is: (4.6.3) where the second equality is determined from Eq. (4.6.2). Because the i are ﬁxed in , the derivatives in R are: (4.6.4) Hence, R dV/dt becomes: or (4.6.5) Because there were no restrictions on V, Eq. (4.6.5) may be written as: (4.6.6) where any vector quantity may be inserted in the parentheses. FIGURE 4.6.1 Vector V and reference frames R and ˆ R. R R V n n n ˆ ˆ ˆ ˆ 3 2 1 ˆ n ˆ R ˆ R ˆ ˆ ˆ ˆ ˆ ˆ ˆ R ddt dVdt dVdt dVdtVnnn= () + () + () 11 22 33 RR RR RR R R ddt dVdt Vd dtdVdt V d dt dV dt V d dt d dtVd dtVd dtVd dt Vnn n nnn Vn n n = () ++ () ++ () + =+ + + ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆˆ ˆ ˆ 1111 22 23 3 333 11 22 33 ˆ n ˆ R RRR i ddt i ˆˆ ,, ˆ nn 1 123=× = () ωω R R RR RR RR RRR ddt ddtV V V ddt V V V VV n n n Vnnn =+×+× +× =+×++ () ˆˆˆ ˆ ˆˆ ˆˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ 1122 33 11 22 33 ωωωω ωω ωω RRRR ddt ddtVV V=+× ˆˆ ωω RRRR ddtddt () = () +× () ˆˆ ωω 0593_C04*_fm Page 88 Monday, May 6, 2002 2:06 PM Kinematics of a Rigid Body 89 Example 4.6.1: Effect of Earth Rotation Equation (4.6.6) is useful for developing expressions for velocity and acceleration of particles moving relative to rotating bodies. To illustrate this, consider the case of a rocket R launched vertically from the Earth’s surface. Let the vertical speed of R relative to the Earth (designated as E) be V 0 . Suppose we want to determine the velocity of R in an astronomical reference frame A in which E rotates for the cases when R is launched from (a) the North Pole, and (b) the equator. Solution: Consider a representation of R, E, and A as in Figure 4.6.2a and b, where i, j, and k are mutually perpendicular unit vectors ﬁxed in E, with k being along the north/south axis, the axis of rotation of E. In both cases the velocity of R in A may be expressed as: (4.6.7) where O is the center of the Earth which is also ﬁxed in A. a. For launch at the North Pole, the position vector OR is simply (r + h)k, where r is the radius of E (approximately 3960 miles) and h is the height of R above the surface of E. Thus, from Eq. (4.6.6), A V R is: (4.6.8) Observe that angular velocity of E in A is along k with a speed of one revolution per day. That is, (4.6.9) Hence, by substituting into Eq. (4.6.8), we have: (4.6.10) FIGURE 4.6.2 Vertical launch of a rocket from the surface of the Earth. (a) Launch at the North Pole; (b) launch at the Equator. O N S k j i A R V 0 (a) O N S k j i A R V 0 (b) AR A ddtVOR= AR E AE dr h dt r hVk k=+ () [] +×+ () ωω AE ωω= = ()( ) =× − ω π kkk 2 24 3600 727 10 5 . rad sec AR o hVV== ˙ kk 0593_C04*_fm Page 89 Monday, May 6, 2002 2:06 PM 90 Dynamics of Mechanical Systems b. For the launch at the Equator, the position vector OR is (r + h)i. In this case, Eq. (4.6.7) becomes: or (4.6.11) Observe that h is small (at least, initially) compared with r. Thus, a reasonable approximation to A V R is: (4.6.12) Observe also the differences in the results of Eqs. (4.6.10) and (4.6.11). 4.7 Addition Theorem for Angular Velocity Equation (4.6.6) is useful for establishing the addition theorem for angular velocity — one of the most important equations of rigid body kinematics. Consider a body B moving in a reference frame , which in turn is moving in a reference frame R as depicted in Figure 4.7.1. Let V be an arbitrary vector ﬁxed in B. Using Eq. (4.6.6), the derivative of V in R is: (4.7.1) Because V is ﬁxed in B its derivatives in R and may be expressed in the forms (see Eq. (4.5.2)): (4.7.2) FIGURE 4.7.1 Body B moving in reference frame R and ˆ R. AR E AE o dr h dt r h hrh Vh Vi i ij ij =+ () +×+ () =++ () =+ ()() + [] × () − ω ω ˙ .3960 5280 7 27 10 5 AR o VhVi j=+ + () 1520 ω ft sec AR O VVi j=+1520 ft sec ˆ R RRRR ddt ddtVV V=+× ˆˆ ωω ˆ R RRR RRB ddt ddtVVVV=× =×ωωωω ˆˆ and B R R ˆ V 0593_C04*_fm Page 90 Monday, May 6, 2002 2:06 PM Kinematics of a Rigid Body 91 Hence, by substituting into Eq. (4.7.1) we have: or (4.7.3) Because V is arbitrary, we thus have: or (4.7.4) Equation (4.7.4) is an expression of the addition theorem for angular velocity. Because body B may itself be considered as a reference frame, Eq. (4.7.4) may be rewritten in the form: (4.7.5) Equation (4.7.5) may be generalized to include reference frames. That is, suppose a reference frame R n is moving in a reference frame R 0 and suppose that there are (n –1) intermediate reference frames, as depicted in Figure 4.7.2. Then, by repeated use of Eq. (4.7.5), we have: (4.7.6) The addition theorem together with the conﬁguration graphs of Section 4.3 are useful for obtaining more insight into the nature of angular velocity. Consider again a body B moving in a reference frame R as in Figure 4.7.3. Let the orientation of B in R be described by dextral orientation angles α, β, and γ. Let n i and N i (i = 1, 2, 3) be unit vector sets ﬁxed in B and R, respectively. Then, from Figure 4.3.8, the conﬁguration graph relating n i and FIGURE 4.7.2 A set of n + 1 reference frames. FIGURE 4.7.3 A body B moving in a reference frame R. RB RB RR ωωωωωω×= ×+ ×VVV ˆˆ RBRBRR ωωωωωω−− () ×= ˆˆ V 0 RBRBRR ωωωωωω−− = ˆˆ 0 RB RBRR ωωωωωω=+ ˆˆ RR RRRR 02 0112 ωωωωωω=+ RR RRRR R R nnn00112 1 ωωωωωωωω=+++ − R R R R R n-1 n 2 1 0 R N N N B n n n 3 2 1 3 2 1 0593_C04*_fm Page 91 Monday, May 6, 2002 2:06 PM [...]... × 0. 433 n y + 0.25n z ) 05 93_ C04*_fm Page 100 Monday, May 6, 2002 2:06 PM 100 Dynamics of Mechanical Systems 1 rev/sec nz nx L R 30 ° ny 30 ° O 6 in FIGURE 4.9 .3 Sports car steering wheel W and left and right hands L and R W or S V L = 22 .38 1n x − 1.571n y + 2.721n z ft sec (4.9.17) and S ( ) − 0.88n ) × (0. 433 n a L = −19 .36 n y + ( −5.529)n y × 0. 433 n y + 0.25n z [ +(6.283n x − 0.88n z ) × (6.283n x z... n2 and d3 parallel to LR ˆ ˆ ˆ Let S (N1, N2, N3), R1 ( n1, n2, n3), R2 (n1, n2, n3), and D (d1, d2, d3) be reference frames containing the unit vector sets as indicated Then, the conﬁguration graph deﬁning the orientation of D in S is shown in Figure 4.12.2 05 93_ C04*_fm Page 108 Monday, May 6, 2002 2:06 PM 108 Dynamics of Mechanical Systems ˆ n3 LD n3 Z N3 θ LR N1 D i n2 ψ r φ N2 n1 R2 ψ 3 ˆ n2 Y... x + 3 y − 12α z + 0 aB = 0. 632 3 x + 0α y − 4α z + 0 aB = 7.1 12α x + 4α y + 0α z − aB = 1.776 (4.9 .35 ) −4α x + 12α y + 3 z + 0 aB = 0 Solving for αx, αy , αz, and aB, we obtain: α x = −2.0 83 rad sec 2, α y = −0.641 rad sec 2, α z = −0.2 13 rad sec 2 (4.9 .36 ) aB = −29 .33 m s (4.9 .37 ) and 05 93_ C04*_fm Page 1 03 Monday, May 6, 2002 2:06 PM Kinematics of a Rigid Body 1 03 4.10 Points Moving on a Rigid... and (4.9 .32 ) into (4.9 .30 ), we have: ( ) aBn z = −3n x + 3 y − 12α z n x + ( 3 x − 4α z )n y ( ) + 12α x + 4α y n z − 8ω y n x + (8ω x − 6ω z )n y + 6ω y n z (4.9 .33 ) Also, from Eq (4.9.20) we have: −4α x + 12α y + 3 z = 0 (4.9 .34 ) Using Eq (4.9.29) for values of ωx, ωy , and ωz, Eqs (4.9 .33 ) and (4.9 .34 ) are equivalent to the following four scalar equations for αx, αy , αz, and aB: 0α x + 3 y −... ω1, ω2, and 3 are to be determined To this end, observe that because B rolls on R at both C2 and C3, the velocity of its center G may be expressed both as: R V G = R ω B × rn 2 and as R V G = R ω B × ( − rn1 ) (4. 13. 3) By substituting from Eq (4. 13. 2) into (4. 13. 3) and by comparing the results, we obtain the relation: rω 1n 3 − rω 3n1 = rω 2n 3 − rω 3n 2 (4. 13. 4) Hence, we have: D ω 3 = 0 and ω 1... being p
art of an imaginary cone with axis along C2C3 and rolling on the conical end of S This is depicted in Figure 4. 13. 3, where O is at the apex of the rolling cones Let h measure the elevation of C1 above O, as shown Then, from the ﬁgure, we see that b and h are: b = a − r cos θ = h tan θ (4. 13. 19) 05 93_ C04*_fm Page 1 13 Monday, May 6, 2002 2:06 PM Kinematics of a Rigid Body 1 13 θ b G r r C 3 h C2... have: B ωS ⋅ n⊥ = 0 (4. 13. 13) where BωS is the angular velocity of the shaft S relative to the ball B From Eqs (4.7.5), (4. 13. 1), and (4. 13. 6), BωS is: B ω S = R ω S − R ω B = Ωn 2 − ωn1 − ωn 3 = −ωn1 + (Ω − ω )n 2 (4. 13. 14) From Figure 4. 13. 2, we see that n⊥ may be expressed in terms of n1 and n2 as: n ⊥ = − cos θn1 + sin θn 2 (4. 13. 15) Hence, the pure rolling criterion of Eq (4. 13. 13) becomes: ω cos θ... R C2 n C3 a FIGURE 4. 13. 1 A conical thrust ball bearing C3 a FIGURE 4. 13. 2 Ball and shaft geometry * This problem is discussed by Kane (4.1, 4.2) and Cabannes (4.5); see also Ramsey (4.6) R 1 3 05 93_ C04*_fm Page 111 Monday, May 6, 2002 2:06 PM Kinematics of a Rigid Body 111 Let the angular velocity of ball B in R be expressed in terms of n1, n2, and n3 as: R ω B = ω 1n1 + ω 2n 2 + ω 3n 3 (4. 13. 2) where... Cabannes, H., General Mechanics, Blaisdell, Boston, 1968, pp 38 4 38 7 4.6 Ramsey, A S., Dynamics, Part II, Cambridge University Press, London, 1 937 , p 160 05 93_ C04*_fm Page 114 Monday, May 6, 2002 2:06 PM 114 Dynamics of Mechanical Systems Problems Section 4.2 Orientation of Rigid Bodies P4.2.1: In Eq (4.2 .3) , suppose that the orientation angles are α = 30 °, β = –45°, and γ = 60° Determine the transformation... Section 4 .3 Conﬁguration Graphs P4 .3. 1: Consider the conﬁguration graph of Figure P4 .3. 1 Find the relations between the Ni and the ni Express the results in the matrix forms: N = Sn and n = TN Show that: S = T T = S−1 i ˆ n ˆ Ni Ni n i i 1 2 3 FIGURE P4 .3. 1 Conﬁguration graph θ1 θ2 3 P4 .3. 2: Repeat Problem P4 .3. 1 for the conﬁguration graph of Figure P4 .3. 2 i ˆ Ni Ni ˆ n n i i 1 2 3 FIGURE P4 .3. 2 Conﬁguration . + ˙˙ ˙ ˙˙˙ ωω ωωω ωωω ωωω 11 1 1 22 22 33 33 11 22 33 11 22 33 nn n nnn nnn nnn R i RB i ddt inn=× = () ωω ,,1 23 ωωω ωωω ωωω ωωω 11 22 33 112 233 11 22 33 11 22 33 0 RRR RB RB RB RB RB RB RB RBRB d. N nN N N 1211221 23 2 23 1 1 23 13 2 1 23 13 3 3 23 1 1 23 13 2 1 23 13 3 =+ − =+− () ++ () =− + − − () +− + () ccsss sc ccc ss scc cs ss ccs sc scs cc ωω = D ˙ θn ˙ θ 05 93_ C04*_fm Page 83 Monday, May. i n n i i ˆ β γ N i 1 2 3 N N ˆ i i n n i i ˆ θ θ θ 1 2 3 Nn n n Nn n n Nn n n 121 232 233 2 12 1 1 23 13 2 1 23 13 3 3 12 1 1 23 13 2 1 23 13 3 =+ − =− + − () +− − () =− + + () +−