Dynamics of Mechanical Systems 2009 Part 9 potx

382 Dynamics of Mechanical Systems (11.10.14) Let the inertia forces on B 1 and B 2 be represented by forces and passing through G 1 and G 2 together with couples having torques and . Then , , , and are: or (11.10.15) or (11.10.16) (11.10.17) and (11.10.18) From Eq. (11.10.12) the partial velocities of G 1 and G 2 and the partial angular velocities of B 1 and B 2 are (see also Eqs. (11.4.23)): FIGURE 11.10.4 Unit vector geometry for the double- rod pendulum. n n n θ 1 1 z 2 2 n n n n 1 1r 2 2r θ θ θ nnnnn nnnnn 22112112122 2211211 2122 rr r =− () +− () =+ =− () +− () =− + cos sin cos sin sin cos sin cos θθ θθ θ θ θθ θθ θ θ θ θθ F 1 * F 2 * T 1 * T 2 * F 1 * F 2 * T 1 * T 2 * Fm m m G r r 11 2 111 22 * ˙˙˙ =− = () − () annllθθ θ Fnn 11 2 11 11 1 2 11 12 22 * ˙ cos ˙˙ sin ˙ sin ˙˙ cos= () + () + () − () mmllθθθθ θθθθ Fa n n n n 21 2 111 2 2 222 2 22 * ˙˙˙ ˙ ˙˙ =− = − + () − () mm m m m G rr ll l lθθ θ θ θθ Fn n 2111 2 12 22 2 21 111 2 12 22 2 22 22 22 * ˙˙ sin cos ˙˙ sin ˙ cos ˙˙ cos ˙ sin ˙˙ cos ˙ sin =++ () + () [] +− + − () + () [] m m l l θθθθθ θθ θ θθθθθ θθ θ Tn 1 2 13 12 * ˙˙ =− () ml θ Tn 2 2 23 12 * ˙˙ =− () ml θ 0593_C11_fm Page 382 Monday, May 6, 2002 2:59 PM Generalized Dynamics: Kinematics and Kinetics 383 (11.10.19) and (11.10.20) Finally, from Eq. (11.9.6) the generalized inertia forces are: (11.10.21) and (11.10.22) Observe how routine the computation is; the principal difficulty is the detail. We will discuss this later. Example 11.10.4: Spring-Supported Particles in a Rotating Tube Consider again the system of Section 11.7 consisting of a cylindrical tube T containing three spring-supported particles P 1 , P 2 , and P 3 (or small spheres) as in Figure 11.10.5. As before, T has mass M and length L and it rotates in a vertical plane with the angle of rotation being θ as shown. The particles each have mass m, and their positions within T are defined by the coordinates x 1 , x 2 , and x 3 as in Figure 11.10.6, where ᐉ is the natural length of each of the springs. FIGURE 11.10.5 A rotating tube containing spring-supported particles. FIGURE 11.10.6 Coordinates of particles within the tube. vn n n vn n n v vn n n ˙ ˙ ˙ ˙ sin cos sin cos sin cos θ θ θ θ θ θ θ θθ θθ θθ 1 1 1 2 2 1 2 2 22 2 0 22 2 11112 11112 22122 G G G G = () =− () + () ==− + = = () =− () + () ll l ll l ll l ωωωωωωωω ˙˙˙˙ ,,, θθθθ 1 1 1 2 2 1 2 2 33 00 BBBB ====nn F θ θθθθ θθθ 1 43 12 12 2 1 2 221 2 2 2 21 * ˙˙ ˙˙ cos ˙ sin=− () − () − () + () − () mm mll l F θ θθθθ θθθ 2 13 12 12 2 2 2 121 2 1 2 21 * ˙˙ ˙˙ cos ˙ sin=− () − () − () − () − () mm mll l P P P n n n T 2 1 3 3 1 2 R j i 3ᐉ 2ᐉ ᐉ P P P n n n T x x x 2 1 3 3 3 1 1 2 2 0593_C11_fm Page 383 Monday, May 6, 2002 2:59 PM 384 Dynamics of Mechanical Systems The velocities and accelerations of P 1 , P 2 , and P 3 in the fixed or inertial frame R are (see Eqs. (11.7.2), (11.7.3), and (11.7.4)): (11.10.23) (11.10.24) (11.10.25) where n 1 , n 2 , and n 3 are the unit vectors shown in Figures 11.10.5 and 11.10.6. Let G be the mass center of T. Then, from Eq. (11.7.1), the velocity and acceleration of G in R are: (11.10.26) Finally, the angular velocity and the angular acceleration of T in R are: (11.10.27) As noted in Section 11.7, the system has four degrees of freedom represented by the coordinates x 1 , x 2 , x 3 , and θ. The corresponding partial velocities and partial angular velocities are recorded in Eqs. (11.7.5), (11.7.6), and (11.7.7) as: (11.10.28) From Eqs. (11.10.23) to (11.10.27), the inertia forces on the particles and the tube may be represented by: (11.10.29) vn na n n PP xx xx xx 11 11 1 2 1 1 2 11 12 2=++ () =−+ () [] ++ () + [] ˙ ˙ , ˙˙ ˙˙˙˙ ˙ lllθθθθ vn na n n PP xx xx xx 22 21 2 2 2 2 2 1222 2222=++ () =−+ () [] ++ () + [] ˙ ˙ , ˙˙ ˙˙˙˙ ˙ lllθθθθ vn na n n PP xx xx xx 33 31 3 2 3 3 2 1332 3332=++ () =−+ () [] ++ () + [] ˙ ˙ , ˙˙ ˙˙˙˙ ˙ lllθθθθ vna nn GG LLL= () =− () + () 222 2 2 12 ˙ , ˙˙˙ θθθ ωωαα== ˙˙˙ θθnn 33 and vn v v v n vvnvv n vvvnv ˙˙˙˙ ˙˙ ˙˙ ˙˙˙ ˙ ,, , ,,, ,, , x P x P x PP x P x P x PP x P x P x PP x x 1 1 2 1 3 11 1 2 2 2 3 22 1 3 2 3 3 3 112 122 1 00 002 00 ====+ () ====+ () === θ θ θ l l 33 123 123 3 000 2 000 32 2 3 =+ () ==== () ==== l x L x G x G x GG xxx n vvv vn n ˙˙˙ ˙ ˙˙˙ ˙ ,,, ,,, θ θ ωωωωωωωω Fa n n Fa n n Fa P P P P P P mmxx mx x mmx x mx x m 1 1 2 2 3 11 2 1112 22 2 1222 2 222 * * * ˙˙ ˙˙˙˙ ˙ ˙˙ ˙˙˙˙ ˙ =− =− − + () [] −+ () + [] =− =− − + () [] −+ () + [] =− ll ll θθθ θθθ 33 33 2 1332 2 12 2 3 332 22 12 =− − + () [] −+ () + [] =− = () − () =− ⋅ − × ⋅ () =− () mx x m x x MML ML ML T G TG G ˙˙ ˙˙˙˙ ˙ ˙˙˙ ˙˙ * * llθθθ θθ θ nn Fa n n TI nααωωωωI 0593_C11_fm Page 384 Monday, May 6, 2002 2:59 PM Generalized Dynamics: Kinematics and Kinetics 385 Finally, from Eq. (11.9.6), the generalized inertia forces are: (11.10.30) (11.10.31) (11.10.32) or (11.10.33) Example 11.10.5: Rolling Circular Disk As a final example consider again the rolling circular disk D with mass m and radius r of Figure 11.10.7. (We first considered this system in Section 4.12 and later in Sections 8.13 and 11.3.) As we observed in Section 11.3, this is a nonholonomic system having three FIGURE 11.10.7 A circular disk rolling on a horizontal surface. FvFvFvFvF T x x P P x P P x P P x T x T mx x 1 1 1 1 1 2 2 1 3 3 11 11 2 * ˙ * ˙ * ˙ * ˙ ** ˙ * ˙˙ ˙ =⋅+⋅+⋅+⋅+⋅ =− − + () [] ω θl F vF F vF vF T x x P P x P P x P P x G T x T v mx x 2 2 1 1 2 2 2 2 3 3 22 22 2 2 * ˙ * ˙ * ˙ * ˙ * ˙ * ˙˙ ˙ =⋅+⋅+⋅+⋅+⋅ =− − + () [] ω θl F vF F vF vF T x x P P x P P x P P x G T x T v mx x 3 3 1 1 3 2 2 3 3 3 33 33 2 3 * ˙ * ˙ * ˙ * ˙ * ˙ * ˙˙ ˙ =⋅+⋅+⋅+⋅+⋅ =− − + () [] ω θl F mx x xm x x x mx P P P P P P G TTθ θθ θ θθ ω θθ θθ * ˙ * ˙ * ˙ * ˙ * ˙ * ˙ ˙˙ ˙ ˙ ˙˙ ˙ ˙ =⋅+⋅+⋅+⋅+⋅ =− + () + () + [] −+ () + () + [] −+ ( vF vF vF vF T 1 1 2 2 3 3 11 1 2 2 2 3 222 2 3 ll l l l )) + () + [] − () − () 32 2 12 33 2 2 l x x ML ML ˙˙ ˙ ˙ ˙˙ ˙˙ θθ θ θ Fmx x x ML mxx xx xx θ θθ θ * ˙˙ ˙˙ ˙˙˙ ˙ =− + () ++ () ++ () [] − () −+ () ++ () ++ () [] lll ll l 1 2 2 2 3 2 2 11 22 33 23 3 223 D Y N Z N n G c n L N X θ φ n 3 2 1 1 2 3 ψ 0593_C11_fm Page 385 Monday, May 6, 2002 2:59 PM 386 Dynamics of Mechanical Systems degrees of freedom. Recall that if we introduce six parameters (say, x, y, z, θ, φ, and ψ) to define the position and orientation of D we find that the conditions of rolling lead to the constraint equations (see Eq. (11.3.8), (11.3.9), and (11.3.10)): (11.10.34) (11.10.35) (11.10.36) where x, y, and z are the Cartesian coordinates of G and θ, φ, and ψ are the orientation angles of D as in Figure 11.10.7. The last of these equations is integrable leading to the expression: (11.10.37) This expression simply means that D must remain in contact with the surface S. It is therefore a geometric (or holonomic) constraint. Equations (11.10.34) and (11.10.35), how- ever, are not integrable in terms of elementary functions. These equations are kinematic (or nonholonomic) constraints. They ensure that the instantaneous velocity of the contact point C, relative to S, is zero. To obtain the generalized inertia forces for this nonholonomic system we may simply select three of the six parameters as our independent variables. Then the partial velocities and partial angular velocities may be determined from the coefficients of the derivatives of these three variables in the expressions for the velocities and angular velocity. That is, observe that the coordinate derivatives are linearly related in Eqs. (11.10.34), (11.10.35), and (11.10.36). This means that we can readily solve for the nonselected coordinate deriv- atives in terms of the selected coordinate derivatives. To illustrate this, suppose we want to describe the movement of D in terms of the orientation angles θ, φ, and ψ. We may express the velocity of the mass center G as: (11.10.38) where N 1 , N 2 , and N 3 are unit vectors parallel to the X-, Y-, and Z-axes as in Figure 11.10.7. Then, from Eqs. (11.10.34), (11.10.35), and (11.10.36), we may express v G as: (11.10.39) ˙ ˙ sin cos ˙ cos sinxr=+ () + [] ψφ θ φθ θ θ ˙ ˙ sin sin ˙ cos cosyr=+ () − [] ψφ θ φθ θ φ ˙ ˙ sinzr=− θ θ zr= cosθ vNNN G xy z=++ ˙ ˙ ˙ 123 vN N N G r r r =+ () + [] ++ () − [] − ˙ ˙ sin cos ˙ cos sin ˙ ˙ sin sin ˙ cos cos ˙ sin ψφ θ φθ θ φ ψφ θ φθ θ φ θθ 1 2 3 0593_C11_fm Page 386 Monday, May 6, 2002 2:59 PM Generalized Dynamics: Kinematics and Kinetics 387 From Figure 11.10.7, we see that N 1 , N 2 , and N 3 may be expressed in terms of the unit vectors n 1 , n 2 , and n 3 as: (11.10.40) Hence, by substituting into Eq. (11.10.39) v G becomes: (11.10.41) Also, the angular velocity ωω ωω of D relative to S may be expressed as (see Eq. (4.12.2)): (11.10.42) Observe that the result of Eq. (11.10.41) could have been obtained directly as ωω ωω × rn 3 , the expression for velocities of points of rolling bodies (see Section 4.11, Eq. (4.11.5)). The partial velocities of G and the partial angular velocities of D with respect to θ, φ, and ψ are then: (11.10.43) and (11.10.44) The inertia force system on D may be represented by a single force F * passing through G together with a couple with torque T * where F * and T * are (see Section 8.13, Eqs. (8.13.7) to (8.13.12)): (11.10.45) and (11.10.46) where (11.10.47) Nn n n Nn n n Nnn 11 1 3 21 2 3 323 =− + =+ − =+ cos sin cos sin sin sin cos cos cos sin sin cos φφθθφ φφθφθ θθ vnn G rr=+ () − ˙ ˙ sin ˙ ψφ θ θ 12 ωω= + + () + ˙ ˙ ˙ sin ˙ cosθψφθφθnnn 123 vnv nvn ˙˙ ˙ , sin , θ φ ψ θ GG G rr r=− = = 211 ωωωωωω ˙˙ ˙ , sin cos , θ φ ψ θθ==+ =nnnn 1232 Fa nnn * =− =− + + () mmaaa G 11 22 33 Tnnn * =++TTT 11 22 33 T T T 1 1 11 2 3 22 33 2 2 22 3 1 33 11 3 3 33 1 2 11 22 =− + − () =− + − () =− + − () αωω αωω αωω ΙΙΙ ΙΙΙ ΙΙΙ 0593_C11_fm Page 387 Monday, May 6, 2002 2:59 PM 388 Dynamics of Mechanical Systems where a G is the acceleration of G relative to the fixed surface S (the inertia frame); where ω i and α i (i = 1, 2, 3) are the n i components of ωω ωω and the angular acceleration αα αα of D relative to S; and, finally, where: (11.10.48) From Eqs. (4.12.7) and (4.12.8), a G and αα αα are: (11.10.49) and (11.10.50) Hence, F * and T * may be written as: (11.10.51) and (11.10.52) Finally, by using Eq. (11.9.6), the generalized inertia forces become: (11.10.53) ΙΙ Ι 11 33 2 22 2 42== =mr mr, an n G rr r =+ + () +−+ + () +− − − () ˙˙ ˙˙ sin ˙ ˙ cos ˙˙ ˙ ˙ cos ˙ sin cos ˙ ˙ sin ˙ sin ˙ ψφ θ φθ θ θψφ θφ θ θ ψφ θ φ θ θ 2 1 2 22 2 3 αα= − () ++ + () +− + () ˙˙ ˙ ˙ cos ˙˙ ˙˙ sin ˙ ˙ cos ˙˙ cos ˙ ˙ sin ˙ ˙ θψφθ ψφθθφθ φθφθθψθ nn n 12 3 Fn nn * ˙˙ ˙˙ sin ˙ ˙ cos ˙˙ ˙ ˙ cos ˙ sin cos ˙ ˙ sin ˙ sin ˙ =− + + () +− + ( [ + ) +− − − () ] mr ψφ θ φθ θ θψφ θ φθθ ψφθφ θθ 2 1 2 2 22 2 3 Tn n n * ˙˙ ˙ ˙ cos ˙ sin cos ˙˙ ˙˙ sin ˙ ˙ cos ˙˙ cos ˙ ˙ =− () −− () [ ++ + () ++ () ] mr 22 1 2 3 42 22 2 2 θψφθφ θθ ψφθθφθ φθψθ F mr mr mr G θ θθ θψθ θφ θ θ θψφθφ θθ θψφθφθθ * ˙ * ˙ * ˙˙ ˙ ˙ cos ˙ sin cos ˙˙ ˙ ˙ cos ˙ sin cos ˙˙ ˙ ˙ cos ˙ sin cos =⋅+⋅ =−+ + () − () −− () =− () + () + () [] vF Tωω 22 22 22 42 54 32 54 0593_C11_fm Page 388 Monday, May 6, 2002 2:59 PM Generalized Dynamics: Kinematics and Kinetics 389 (11.10.54) and (11.10.55) As an aside, we can readily develop the generalized applied (or active) forces for this system. Indeed, the only applied forces are gravity and contact forces, and of these only the gravity (or weight) forces contribute to the generalized forces. (The contact forces do not contribute because they are applied at a point of zero velocity.) The weight forces may be represented by a single vertical force W passing through G given by: (11.10.56) Then, from Eq. (11.10.43), the generalized forces are: (11.10.57) 11.11 Potential Energy In elementary mechanics, potential energy is often defined as the “ability to do work.” While this is an intuitively satisfying concept it requires further development to be com- putationally useful. To this end, we will define potential energy to be a scalar function of the generalized coordinates which when differentiated with respect to one of the coordi- nates produces the negative of the generalized force for that coordinate. Specifically, we define potential energy P(q r ) as the function such that: (11.11.1) F mr mr mr mr G φ φφ θψ φ θ φθ θ θψ φ θ θφ θ θφ θ ψθ ψθ φ * ˙ * ˙ * sin ˙˙ ˙˙ sin ˙ ˙ cos sin ˙˙ ˙˙ sin ˙ ˙ cos cos ˙˙ cos ˙ ˙ ˙˙ sin ˙˙ sin =⋅+⋅ =− + + () − () ++ () − () + () =− () − () vF Tωω 2 2 2 22 2 422 2 42 32 32 θθφθθθ φθ ψθθ − () [ − () + () ] 52 14 12 2 ˙ ˙ sin cos ˙˙ cos ˙ ˙ cos Fmr mr mr ψ ψφ θ φθ θ ψφθθφθ ψφθθφθ * ˙˙ ˙˙ sin ˙ ˙ cos ˙˙ ˙˙ sin ˙ ˙ cos ˙˙ ˙˙ sin ˙ ˙ cos =− + + () − () ++ () =− () + () + () [] 2 2 2 2 42 2 2 32 32 52 WN n n=− =− + () mg mg 323 sin cosθθ Fmg F F xθφ θ===sin , ,00 Fqrn rr = −∂ ∂ = … () D P 1, , 0593_C11_fm Page 389 Monday, May 6, 2002 2:59 PM 390 Dynamics of Mechanical Systems where as before, n is the number of degrees of freedom of the system. (The minus sign is chosen so that P is positive in the usual physical applications.) To illustrate the consistency of this definition with the intuitive concept, consider a particle Q having a mass m in a gravitational field. Let Q be at an elevation h above a fixed level surface S as in Figure 11.11.1. Then, if Q is released from rest in this position, the work w done by gravity as Q falls to S is: (11.11.2) From a different perspective, if h is viewed as a generalized coordinate, the velocity v of Q and, consequently, the partial velocity v h of Q (relative to h) are: (11.11.3) where k is the vertical unit vector as in Figure 11.11.1. The generalized force due to gravity is then: (11.11.4) Let P be a potential energy defined as: (11.11.5) Then, from Eq. (11.11.1), F h is: (11.11.6) which is consistent with the results of Eq. (11.11.4). As a second illustration, consider a linear spring as in Figure 11.11.2. Let the spring have modulus k and natural length ᐉ. Let the spring be supported at one end, O, and let its other end, Q, be subjected to a force with magnitude F producing a displacement x of Q, as depicted in Figure 11.11.2. The movement of Q has one degree of freedom represented by the parameter x. The velocity v and partial velocity of Q are then: (11.11.7) FIGURE 11.11.1 A particle Q above a level surface S. FIGURE 11.11.2 A force applied to a linear spring. w mgh= vk k== ˙ hv h and Fmg mg hh =− ⋅ =−kv P ==w mgh Fhmg h =−∂ ∂ =−P v ˙ x vn vn== ˙ ˙ x x and k h Q(m) S n k Q F x ᐉ O 0593_C11_fm Page 390 Monday, May 6, 2002 2:59 PM Generalized Dynamics: Kinematics and Kinetics 391 The force S exerted on Q by the spring is: (11.11.8) The generalized force F x relative to x is then: (11.11.9) Let a potential energy function P be defined as: (11.11.10) Then, from Eq. (11.11.1), we have: (11.11.11) which is consistent with Eq. (11.11.9). It happens that Eqs. (11.11.5) and (11.11.10) are potential energy functions for gravity and spring forces in general. Consider first Eq. (11.11.5) for gravity forces. Suppose Q is a particle with mass m. Let Q be a part of a mechanical system S having n degrees of freedom represented by the coordinates q r (r = 1,…, n). Recall from Eq. (11.6.5) that the contribution of the weight force on Q to the generalized force F r for the coordinate q r is: (11.11.12) where h is the elevation of Q above a reference level as in Figure 11.11.3. From Eq. (11.11.2), if a potential energy function P is given by mgh, Eq. (11.11.1) gives the contribution to the generalized force of q r for the weight force as: (11.11.13) which is consistent with Eq. (11.11.12). Consider next Eq. (11.11.10) for spring forces. Suppose Q is a point at the end of a spring which is part of a mechanical system S as depicted in Figure 11.11.4. Let S have n degrees FIGURE 11.11.3 Elevation of a particle Q of a mechanical system S. FIGURE 11.11.4 A spring within a mechanical system S. Sn n=− =−Fkx Fkx x x =⋅ =−Sv ˙ P = () 12 2 kx −∂ ∂ =− =P xkxF x ˆ F r ˆ Fmghq rr =− ∂ ∂ ˆ F r ˆ Fqmghq rr r =−∂ ∂ =− ∂ ∂P Q(m) r h(q ,t) R S k R O Q n S 0593_C11_fm Page 391 Monday, May 6, 2002 2:59 PM [...]... York, 198 5, p 100 0 593 _C11_fm Page 402 Monday, May 6, 2002 2: 59 PM 402 Dynamics of Mechanical Systems 11.3 Kane, T R., Analytical Elements of Mechanics, Vol 1, Academic Press, New York, 195 9, p 128 11.4 Huston, R L., and Passerello, C E., Another look at nonholonomic systems, J Appl Mech., 40, 101–104, 197 3 11.5 Huston, R L., Multibody Dynamics, Butterworth-Heinemann, Stoneham, MA, 199 0 11.6 Huston, R... (11.10.22) 0 593 _C11_fm Page 398 Monday, May 6, 2002 2: 59 PM 398 Dynamics of Mechanical Systems j O θ n3 i P1 n2 P2 P3 3ᐉ 2ᐉ n3 n ᐉ 1 T x3 n2 x2 n1 x1 R P1 FIGURE 11.12.5 Rotating tube containing spring-supported particles P2 P3 T FIGURE 11.12.6 Coordinates of particles in the tube Example 11.12.4: Spring-Supported Particles in a Rotating Tube For another example illustrating the use of kinetic energy... Consider next a set of particles Pi (i = 1,…, N) as parts of a mechanical system S having n degrees of freedom Then, by superposing (or adding together) equations as Eq (11.12.5) for each of the particles, we obtain an expression identical in form to Eq (11.12.5) and valid for the set of particles Finally, if the set of particles is a rigid body, Eq (11.12.5) also holds To illustrate the use of Eq (11.12.5),... ) sin θ (11.11.23) FIGURE 11.11.6 Coordinates of the particles within the tube 0 593 _C11_fm Page 394 Monday, May 6, 2002 2: 59 PM 394 Dynamics of Mechanical Systems In the development of these generalized forces we discovered that the only forces contributing to these forces were gravity and spring forces The contact forces between the smooth tube and the particles did not contribute to the generalized... identical with that of Eq (11.10.5) R O n θ ᐉ nr FIGURE 11.12.1 The simple pendulum P θ 0 593 _C11_fm Page 396 Monday, May 6, 2002 2: 59 PM 396 Dynamics of Mechanical Systems FIGURE 11.12.2 The rod pendulum Example 11.12.2: Rod Pendulum Consider next the rod pendulum of Figure 11.12.2, using the same notation as before Recall that the mass center velocity vG and the angular velocity of the rod are (see...0 593 _C11_fm Page 392 Monday, May 6, 2002 2: 59 PM 392 Dynamics of Mechanical Systems of freedom represented by the coordinates qr (r = 1,…, n) Then, from Eq (11.6.7), we ˆ recall that the contribution Fr of the spring force to the generalized force on S, for the coordinate qr , is: ∂x ˆ Fr = − f ( x) ∂qr (11.11.14) where f(x) is the magnitude of the spring force due to a spring... (11.11. 19) Unfortunately, it is not always possible to perform the integration indicated in Eq (11.11. 19) Indeed, the integral represents an anti-partial differentiation with respect to 0 593 _C11_fm Page 393 Monday, May 6, 2002 2: 59 PM Generalized Dynamics: Kinematics and Kinetics 393 FIGURE 11.11.5 A rotating tube containing springconnected particles ˙ each of the qr Also, if the Fr involve derivatives of. .. envisioned in 196 1 Indeed, Kane’s equations are currently the equations of choice for automated (numerical) formulation of the governing equations of motion for large mechanical systems [12.5, 12.6] Intuitively, Kane’s equations may be interpreted as follows: if the partial velocity vectors define the directions of motion of a mechanical system, then Kane’s equations represent a projection of the applied... pendulum 0 593 _C11_fm Page 410 Monday, May 6, 2002 2: 59 PM 410 Dynamics of Mechanical Systems P11 .9. 2: A rod B with length ᐉ and mass m is pinned to a vertical rotating shaft S as depicted in Figure P11 .9. 2 Let the pin axis be along a radial line of S Let θ define the angle between B and S Let the rotation speed of S be specified as Ω This system has one degree of freedom represented by the angle θ Determine... equations provide an elegant formulation of the dynamical equations of motion Kane’s equations simply state that the sum of the generalized forces (both applied and inertia forces), for each generalized coordinate, is zero That is, for a mechanical system 415 0 593 _C12_fm Page 416 Monday, May 6, 2002 3:11 PM 416 Dynamics of Mechanical Systems S having n degrees of freedom, represented by generalized coordinates . ,00 Fqrn rr = −∂ ∂ = … () D P 1, , 0 593 _C11_fm Page 3 89 Monday, May 6, 2002 2: 59 PM 390 Dynamics of Mechanical Systems where as before, n is the number of degrees of freedom of the system. (The minus sign. l 0 593 _C11_fm Page 397 Monday, May 6, 2002 2: 59 PM 398 Dynamics of Mechanical Systems Example 11.12.4: Spring-Supported Particles in a Rotating Tube For another example illustrating the use of. φθ θ φ ˙ ˙ sinzr=− θ θ 0 593 _C11_fm Page 399 Monday, May 6, 2002 2: 59 PM 400 Dynamics of Mechanical Systems The first two of these equations are nonintegrable in terms of elementary functions; therefore,