MSMS 214-215 Basilio Bona DAUIN PoliTo Problem 2 The planar system illustrated in Figure 1 consists of a bar B and a wheel W moving (no friction, no sliding) along the bar; the bar can rotate around an axis perpendicular to the plane at the pivot point A. The wheel can rotate around its center and maintains always a contact point with the bar. No elastic elements are present. Questions Answer to the following questions: 1. Defineapossible set ofgeneralized coordinatesq i (t). Trytolinkq 1 tothebarrotation and q 2 to the wheel motion. 2. Write the homogeneous transformation T 2 between R and R 2. 3. Compute the constraint between the angular motion of W and the linear motion of its center. 4. Find the total angular velocity of the wheel W. 5. Compute the kinetic co-energy of the system. 6. Compute the potential energy of the system if g = [ G ] T. 7. Compute the total virtual work due to force F and torque N. 8. Write the two second order differential equations obtained using the Lagrange approach. 9. Using the Lagrange equations and the data below, compute the force F required to keep the bar in equilibrium, assuming that the bar is horizontal, that τ =, and the that the wheel is at the left extremity of the bar. 1
MSMS Problem 214-215 2 Data Description Symbol Value Unit Bar mass m b 2 kg Bar length L 2 m Bar height ε negligible.7 Bar inertia matrix Γ b ε kg m 2.7 Wheel mass m w 1 kg Wheel radius R.1 m.25 Wheel inertia matrix Γ w.25 kg m 2.5 Applied force Applied torque F N [ f(t) ] T N [ τ ] T N m Solution 1A: Generalized coordinates are angles The most immediate choice for q 1 is the angle between i and i 1 ; q 2 can be chosen as the rotation angle of the wheel, see Figure 1. The first component of the position vector of the wheel center along the axis i 1 is constrained by the relation R(q 2 +θ) = x 1 ; the negative sign is necessary since for positive q 2 the wheel center moves to the left (negative values of component along i 1 ); θ is a constant angle that depends on the orientation of the wheel when x 1 =. We assume for simplicity θ = ; this means that x 1 = when q 2 =. Solution 2: Homogeneous transformations c 1 s 1 L T 1 = +c 2 s 1 c 1 1 T1 2 = 1 c 2 s 2 Rq 2 s 2 c 2 R 1 1 = where c is a constant that depends on the position of the origin of R with respect to the left side of the bar. For simplicity we assume c =.
MSMS Problem 214-215 3 Therefore c 12 s 12 Rq 2 c 1 Rs 1 + L 2 T 2 = s 12 c 12 Rq 2 s 1 +Rc 1 1 1 We can compute the position vectors of the points A,B,W with respect to the reference frame R ; they will be used to answer the various questions: L L (1+c 2 2 1) Rq 2 c 1 Rs 1 + L r A = r B = L s 2 2 1 r W = Rq 2 s 1 +Rc 1 from T 2 We have used the symbols r instead of p for convenience (see also Solution 7 paragraph) Solution 3 See discussion in Solution 1A paragraph. Solution 4: Angular velocity A simple geometrical consideration gives the total angular velocity of the wheel as ω W = q 1 + q 2 A formal approach consists in taking the rotation matrix R 2 from T 2 c 12 s 12 R 2 = s 12 c 12 1 derive it with respect to time s 12 c 12 Ṙ 2 = c 12 s 12 ( q 1 + q 2 ) and use the relation Ṙ 2 = S(ω W)R 2, or Ṙ 2 (R 2 )T = S(ω W ), that produces
MSMS Problem 214-215 4 s 12 c 12 c 12 s 12 Ṙ 2 (R 2 )T = c 12 s 12 s 12 c 12 ( q 1 + q 2 ) 1 s 12 c 12 +s 12 c 12 (s 2 12 +c2 12 ) = s 2 12 +c 2 12 s 12 c 12 s 12 c 12 ( q 1 + q 2 ) 1 = 1 ( q 1 + q 2 ) Recalling the meaning of S(ω) as ω 3 ω 2 S(ω) = ω 3 ω 1 ω 2 ω 1 one finds again the ω W given above. Solution 5: Kinetic co-energy In order to compute the kinetic co-energy C it is necessary to compute the norms of the linear velocities of the centers-of-mass and the angular velocities; in particular we have ṙ A = ω A = q 1 R(q 2 s 1 q 1 c 1 q 1 c 1 q 2 ) ṙ W = R(q 2 c 1 q 1 +s 1 q 1 +s 1 q 2 ) ω W = q 1 + q 2 Ls 2 1 q 1 ṙ B = L 2 1 q 1 The norm ṙ W 2 is ṗ W 2 = R 2( (q 2 s 1 q 1 c 1 q 1 c 1 q 2 ) 2 +(q 2 c 1 q 1 +s 1 q 1 +s 1 q 2 ) 2) = R 2( q ) 1 2 + q 2 2 +2 q 1 q 2 +q2 2 q 1 2 = R 2( ) ( q 1 + q 2 ) 2 +q2 2 q2 1 (1)
MSMS Problem 214-215 5 Hence C A = 1 2 q2 1Γ b,z C W = 1 2 m wr 2 (( q 1 + q 2 ) 2 +q 2 2 q2 1 )+ 1 2 ( q 1 + q 2 ) 2 Γ w,z C total = C A +C W where Γ b,z =.7 and Γ w,z =.5. Solution 6: Potential energy Since no elastic elements are present in the system, the potential energy is only due to the gravitational field. The formula for potential energy gives P A = m b g T r A = P W = m w g T r [ ] Rq 2 c 1 Rs 1 + L 2 W = m w G Rq 2 s 1 +Rc 1 = m w GR(c 1 q 2 s 1 ) P total = P A +P W = P W Solution 7: Virtual work There are two external generalized forces applied to the system, F and N; therefore the total virtual work is given by δw total = N T δα A +F T δr B (2) where with α A one indicates the virtual angular motion associated to the bar rotation and with δr B the virtual displacement of the right side of the bar, where F is applied. If one introduces the pose vector p = [ r α ] T and the generalized force vector Φ = [ F N ] T, one can write δw total = δw A +δw B = δp T Φ = Φ T δp (3) It appears that eqn. (2) and eqn. (3) are the same, apart from different symbols used. We have now the following vectors ṗ A = δp A = q 1 δq 1
MSMS Problem 214-215 6 and and the virtual work is L s 2 1 q 1 L L c 2 1 q 1 s 2 1δq 1 L ṗ B = c 2 1δq 1 δp B = q 1 δq 1 Φ A = τ f(t) Φ B = δw total = τδq 1 f(t) L 2 c 1δq 1 = F 1 δq 1 +F 2 δq 2 where, equating the corresponding symbols, we obtain F 1 = τ f(t) L 2 c 1 F 2 = Solution 8: Lagrange equations Equation 1: angular coordinate, torques therefore C q 1 = Γ b,z q 1 +(m w R 2 +Γ w,z )( q 1 + q 2 )+m w R 2 q 2 2 q 1 d C = Γ b,z q 1 +(m w R 2 +Γ w,z )( q 1 + q 2 )+m w R 2 q dt q 2 q 2 1 +2m w R 2 q 2 q 1 q 2 1 C q 1 = P q 1 = m w GR( s 1 q 2 c 1 ) (m w R 2 +m w R 2 q2 2 +Γ b,z +Γ w,z ) q 1 +(m w R 2 +Γ w,z) q 2 +2m w R 2 q 2 q 1 q 2 = τ f(t) L }{{}}{{} c }{{} 2 1 Γ 1 Γ 2 Coriolis acc.
MSMS Problem 214-215 7 Equation 2: angular coordinate, torques therefore C q 2 = (m w R 2 +Γ w,z )( q 1 + q 2 ) d C = (m w R 2 +Γ w,z )( q 1 + q 2 ) dt q 2 C q 2 = m w R 2 q 2 q 2 1 P q 2 = m w GRs 1 (m w R 2 +Γ w,z ) q }{{} 1 +(m w R 2 +Γ w,z) q }{{} 2 m w R 2 q 2 q 1 2 }{{} Γ 2 Γ 2 centrifugal acc. m w GRs }{{} 1 = weight force Summarizing, we have can write the second order differential equations as Γ 1 q 1 +Γ 2 q 2 +2m w R 2 q 2 q 1 q 2 = τ f(t) Lc 2 1 Γ 2 q 1 +Γ 2 q 2 m w R 2 q 2 q 2 1 = m wgrs 1 Notice that the system is linear with respect to q 1 and q 2, but contains also the nonlinear terms (2m w R 2 q 2 q 1 q 2 ), (m w R 2 q 2 q 2 1), and the two right-hand torques, with trigonometric functions. Solution 9: Equilibrium Considering the given conditions 1. the bar is in equilibrium (static). The static equilibrium is intended as mechanical engineers define it, i.e., q 1 = q 2 =, but generic nonzero accelerations. 2. the bar is horizontal, i.e., q 1 =, 3. τ =, 4. the wheel is at the left extremity of the bar, i.e., q 2 = L 2R The two Lagrange equations become Γ 1 q 1 +Γ 2 q 2 = f(t) L 2 Γ 2 q 1 +Γ 2 q 2 =
MSMS Problem 214-215 8 When condition 1) is relaxed and the equilibrium is dynamic, i.e., velocities and accelerations are generic ones, the two equations become where Γ 1 q 1 +Γ 2 q 2 m w RL q 1 q 2 = f(t) L 2 Γ 2 q 1 +Γ 2 q 2 1 2 m wrl q 2 1 = Γ 1 = (m w R 2 +m w R 2L2 4 +Γ b,z +Γ w,z ) Γ 1 The two equations are still nonlinear due to the terms q 1 q 2 and q 1, 2 but if we assume a further condition, i.e., that the velocities are small, so that q 1 q 2 ε and q 1 2 ε, we can write Γ 1 q 1 +Γ 2 q 2 ε = f(t) L 2 Γ 2 q 1 +Γ 2 q 2 ε = Setting ε =, we can write the resulting equations as Γ 1 q 1 +Γ 2 q 2 = f(t) L 2 Γ 2 q 1 +Γ 2 q 2 = or [ ] Γ 1 Γ 2 ][ q1 Γ 2 Γ 2 = q 2 [ ].5 f(t)l (4) Additional note: linearization and state variable equations The problem does not require to linearize the system or write the state variable equations, but we will do it, linearizing around the following equilibrium point q 1,eq =, q 2,eq = l R = C, q 1,eq, q 2,eq where l,c, q 1,eq and q 2,eq are given constant values. The main nonlinear parts are the terms 2m w R 2 q 2 q 1 q 2 and m w R 2 q 2 q 1 2 ; recalling that a generic nonlinear function f(x 1,x 2,x 3 ) = x 1 x 2 x 3 can be linearized around the values x 1, x 2, x 3 as f + f + f = x 2 x 3 + x 1 x 3 + x 1 x 2 x 1 x1, x 2, x 3 x 2 x1, x 2, x 3 x 3 x1, x 2, x 3 we can conclude that the linearized Lagrange equations are Γ 1 q 1 +Γ 2 q 2 +2m w R 2 ( q 1,eq q 2,eq +C q 2,eq +C q 1,eq ) = τ f(t) Lc 2 1 Γ 2 q 1 +Γ 2 q 2 +m w R 2 ( q 1,eq 2 +2C q 1,eq) = m w GRs 1
MSMS Problem 214-215 9 Since c 1 = 1, s 1 = an τ = we have the final linearized form Γ 1 q 1 +Γ 2 q 2 +2m w R 2 ( q 1,eq q 2,eq +C q 2,eq +C q 1,eq ) = f(t) L 2 Γ 2 q 1 +Γ 2 q 2 +m w R 2 ( q 2 1,eq +2C q 1,eq) = (5) When q 1,eq =, q 2,eq = l R = C, q 1,eq =, q 2,eq = the equations (5) reduce to [ ] Γ1 Γ 2 ][ q1 = Γ 2 Γ 2 q 2 [ ].5 f(t)l that are similar to (4), apart from the inertia values. In matrix term, we have Since Γ is invertible Γ 1 = 1 Γ 2 (Γ 1 Γ 2 ) Γ q(t) = bf(t) [ ] Γ2 Γ 2 = Γ 2 Γ 1 1 1 Γ 1 Γ 2 Γ 1 Γ 2 1 Γ 1 Γ 2 Γ 1 Γ 2 (Γ 1 Γ 2 ) we can write where B = Γ 1 b is q(t) = Γ 1 bf(t) = Bf(t).5 Γ B = 1 Γ 2.5 Γ 1 Γ 2 Setting the states as x 1 = q 1, x 2 = q 2, x 3 = q 1, x 4 = q 2 the resulting linearized differential state equation is ẋ(t) = A 1 x(t)+bf(t) where 1 1 A 1 = 1
MSMS Problem 214-215 1 A different choice of states, as x 1 = q 1, x 2 = q 1, x 3 = q 2, x 4 = q 2 will produce a different A 2 matrix 1 A 2 = 1 In any case the A matrix has four null eigenvalues, and the linearized system is unstable. Figure 1: Problem 2: choice of q i.