试卷答案 (2013至2014学年第2学期) 课程名称复杂系统动力学 1.Consider the mechanical system shown in the figure.Disk B2 makes rolling contact with link B without slip, which can be seen as a rack and pinion. The arc lengths of OD and O2D are equal.The distance between point Ci B and O1 is given by a.Point O of link B B is connected to the ground with a revolute joint.Point A of link B can 1 slide on the link B3 without friction. 2公 Point B of link B3 can slide on the horizontal ground without friction.The B X radius of B2 is r,and the length of Bi and B3 are both 1.The origins of the body-fixed frames of B1,B2 and B3 are all located at the centroids.The generalized coordinate vector is given byq=[x为4x2为4x3为4]' Write kinematic constraint equations and calculate degrees of freedom.(25) Solution: x1-c0S01 y-ising -sing (x-x2)+cos(y-y2)-r Φ= =0 cos0(x2-x)+sin0,(y2-)+a-r(02-9) -sing;(x3-x-cosp)+coso(y3-y-sino) 为-sinp+b D0F=9-6=3
1. Consider the mechanical system shown in the figure. Disk B2 makes rolling contact with link B1 without slip, which can be seen as a rack and pinion. The arc lengths of Q D1 and Q D2 are equal. The distance between point C1 and Q1 is given by a. Point O of link B1 is connected to the ground with a revolute joint. Point A of link B1 can slide on the link B3 without friction. Point B of link B3 can slide on the horizontal ground without friction. The radius of B2 is r, and the length of B1 and B3 are both l. The origins of the body-fixed frames of B1, B2 and B3 are all located at the centroids. The generalized coordinate vector is given by [ 1 11 2 2 2 3 33 ] T q = xy xy xy φ φ φ Write kinematic constraint equations and calculate degrees of freedom. (25 分) Solution: () ( ) () ( ) ( ) ( ) ( ) 1 1 2 1 1 2 11 2 11 2 12 1 12 1 2 1 33 1 1 33 1 1 2 2 3 3 2 cos sin sin cos cos sin sin cos cos sin sin l l l l l x y xx yy r x x y y ar xx yy y b ϕ ϕ ϕ ϕ ϕ ϕ ϕϕ ϕ ϕϕ ϕ ϕ ⎡ ⎤ − ⎢ ⎥ − − −+ −− = = − + − +− − − −− + −− ⎣ ⎦ − + Φ 0 DOF =−= 963 试 卷 答 案 （ 2013 至 2014 学年 第 2 学期 ） 课程名称 复杂系统动力学 x G y G O φ1 1 y G 1 x G 2 x G 2 y G C1 C2 φ2 B1 a α 2 Q2 Q1 D B2 φ3 3 x G 3 y G b A B B3
2.As shown in the figure,Point 4 of B2 can slide without friction on the ground along the vertical axis.Br is connected to the ground with a revolute joint at point O.B2 can slide along axis relative to B without friction. y The length of the link Bi is 21.The mass of B and B2 are both m.The inertia of B2 with 2 respect to the centroid C2 is J.A perpendicular force F is applied on Bi at point B.The origins of the body-fixed frames s() of Bi and B2 are located at the centroids.The generalized coordinate vector is given by =[x y2.The system is driven bys() (1)Write the kinematics and driving a constraint equations. (2)Write the Lagrange equations of the first kind for the purpose of inverse dynamic analysis (3)Give physics interpretation for each Lagrange multiplier.(25) Solution: x:-Icoso y-Ising -sing (x-x2)+cose(y-y2)-b (1)Φ= =0 P2-P x2 +csinpz-a cose (-x2)+sin (y-y2)-1+so+at2 (2)Mi+更'元=Q M=diag(m,m,ml-,m,m,J) 1 0 Isin 0 0 0 0 1 -Icos 0 0 0 西= -sino cos a s1n01 -cos 0 0 0 -1 0 0 1 0 0 0 1 0 ccos coso sinp a -cos -sin 0 a =-cos (x-x2)-sing (y-2)=-d a2 =-sing (x-x2)+coso(y-2)=b =[a2元] Q=[Fsino -mg-Fcoso -FI 0 -mg 0]
2. As shown in the figure, Point A of B2 can slide without friction on the ground along the vertical axis. B1 is connected to the ground with a revolute joint at point O. B2 can slide along 1 x G axis relative to B1 without friction. The length of the link B1 is 2l. The mass of B1 and B2 are both m. The inertia of B2 with respect to the centroid C2 is J. A perpendicular force F is applied on B1 at point B. The origins of the body-fixed frames of B1 and B2 are located at the centroids. The generalized coordinate vector is given by [ ] 1 11 2 2 2 T q = xy xy φ φ . The system is driven by 2 0 1 ( ) 2 s t s at = + . (1) Write the kinematics and driving constraint equations. (2) Write the Lagrange equations of the first kind for the purpose of inverse dynamic analysis. (3) Give physics interpretation for each Lagrange multiplier. (25 分) Solution: (1) () ( ) () ( ) 1 1 1 1 11 2 11 2 2 1 2 2 1 2 11 2 11 2 0 2 cos sin sin cos sin cos sin x l y l xx yy b xc a x x y y l s at ϕ ϕ ϕ ϕ ϕ ϕ ϕ ϕ ϕ ⎡ ⎤ − ⎢ ⎥ − − −+ −− = = − + − − + − −+ + ⎣ ⎦ Φ 0 (2) = T A Mq Q + q Φ λ 1 2 3 M = diag(,, ,,,) m m ml m m J 1 1 1 11 1 1 2 1 12 1 1 1 0 sin 0 0 0 0 1 cos 0 0 0 sin cos sin cos 0 00 1 0 0 1 0 0 0 1 0 cos cos sin cos sin 0 l l a c a ϕ ϕ ϕϕ ϕ ϕ ϕ ϕϕ ϕ ϕ ⎡ ⎤ ⎢ ⎥ − − − = − ⎣ ⎦ − − Φq () ( ) 1 11 2 11 2 a xx yy d =− − − − =− cos sin ϕ ϕ () ( ) 2 11 2 11 2 a xx yy b =− − + − = sin cos ϕ ϕ = [ 123456 ] T λ λ λλλλλ [ sin cos 0 0 1 1 ] A T Q = −− − − F mg F Fl mg ϕ ϕ s (t) 1 x G O A B C B2 2 B1 C1 b c F G a g G 1 y G 2 y G 2 x G x G y G
-元+sinp,入-cosp6 -元-cosp,-sinp,% (3)0=-'1= -Isin+Icos+d+b -sinp,乙3-入+co0sp% cosp乙+sinp,% -元4-cc0sp2 F -sing N-cos F F=- F,+cosN-sinF F,=- Isin F,-Icos F,-dN,+M-bF N1=-元 sinN-N2 +coso F M=元4 -cos N +sin F N2=5 -M-ccoso2N2 F= d C2 N, A
(3) 1 13 16 2 13 16 11 12 3 4 6 13 5 16 13 16 4 25 sin cos cos sin sin cos sin cos cos sin cos C T l l db c λ ϕλ ϕλ λ ϕλ ϕλ ϕ λ ϕλ λ λ λ ϕλ λ ϕλ ϕλ ϕλ λ ϕλ ⎡ ⎤ −+ − ⎢ ⎥ −− − − + + +− =− = − −+ + ⎣ ⎦ − − Q Φ λ q 11 1 11 1 1 1 11 11 2 1 11 1 1 22 sin cos cos sin sin cos sin cos cos sin cos x d y d C x y d d d FN F F NF l F l F d N M bF NN F N F Mc N ϕ ϕ ϕ ϕ ϕ ϕ ϕ ϕ ϕ ϕ ϕ ⎡ ⎤ − − ⎢ ⎥ + − − − +− = − + − + ⎣ ⎦ − − Q , 1 2 1 3 1 4 2 5 6 x y d F F N M N F λ λ λ λ λ λ = − = − = − = = = O B C2 B1 C1 b Fx G A B2 C2 C1 b c Fd G N2 G N1 G M1 N1 ′ G Fd ′ G M1 ′ F y G d
3.Consider the mechanism shown in the figure.Gear B2 makes rolling contact inside gear Bi without slip. Gear Bi is connected to the ground with a translational spring at point A. The coefficient and free length of the A spring are k and lo,respectively.The masses of Bi and B2 are both m,and the radius of Bi is r and the radius of C B2 is r2.The origins of the body-fixed M frames of Bi and B2 are located at the B centroids.A torque M is applied on B2. The generalized coordinate vector is given by 4=y4x2y24, D (1)Write the kinematics constraint equations,calculate degree of freedom. (2)Calculate M,2,r (3)Write the mixed differential-algebraic equations for the purpose of dynamic analysis. (4)Give the physics interpretation for each Lagrange multiplier.(25) Solution: (-xP+0-y-G-=0 cose(x2x)+sine(y2-y) r(0-)=5(02-0),日=9-52 1-2 DOF=6-2=4 (2)M=diag(m,m,mr2,m,m,mr,2) 2=[0mg00mgM]+2, e=-[a'【5o]Rd00o] d=cos B.=RA.I=Vd'd,f=k(1-b) [y-rsing 「-2(x2-x)-2(-y)02(2-x)2(y2-)0 -cos0 -sine a cos sine a, 4=-sm6(6-）+cs0s-]725 a=-[sn0s-+0-265 「-2(6-x)-2(-1 sine0(:-)-cos0(3-)」
3．Consider the mechanism shown in the figure. Gear B2 makes rolling contact inside gear B1 without slip. Gear B1 is connected to the ground with a translational spring at point A. The coefficient and free length of the spring are k and l0, respectively. The masses of B1 and B2 are both m, and the radius of B1 is r1 and the radius of B2 is r2. The origins of the body-fixed frames of B1 and B2 are located at the centroids. A torque M is applied on B2. The generalized coordinate vector is given by [ ]T x y x y = 1 1 φ1 2 2 φ 2 q . (1) Write the kinematics constraint equations, calculate degree of freedom. (2) Calculate M , A Q , Φq , γ (3) Write the mixed differential-algebraic equations for the purpose of dynamic analysis. (4) Give the physics interpretation for each Lagrange multiplier. (25 分) Solution: (1) ( )( )( ) () ( ) 2 22 2 1 2 1 12 21 21 cos sin x x y y rr θ θ xx yy ⎡ ⎤ − + − −− = = ⎢ ⎥ −+ − ⎣ ⎦ Φ 0 ( )( ) 11 2 2 r r ϕ −= − θ ϕ θ ， 11 2 2 1 2 r r r r ϕ ϕ θ − = − DOF =−= 624 (2) 2 2 1 1 2 2 M = diag(,, ,,, ) m m mr m m mr [0 00 ] A A T Q Q = + mg mg M s [ ] 1 1 0 000 T AT T s f r l =− − ⎡ ⎤ ⎣ ⎦ Q d Bd 11 1 11 1 cos sin x r y r ϕ ϕ ⎡ ⎤ − = ⎢ ⎥ ⎣ ⎦ − d ， B RA 1 1 = ， T l = d d ， ( ) 0 f = kl l − ( )( ) 21 21 21 21 ( ) ( ) 1 2 2 2 02 2 0 cos sin cos sin xx yy xx yy θ θ θθ a a ⎡ ⎤ − −− − − − = ⎢ ⎥ ⎣ ⎦ − − Φq () () 1 1 21 21 1 1 2 sin cos r a xx yy r r r =− − + − = ⎡ ⎤ θ θ ⎣ ⎦ − () ( ) 2 2 21 21 2 1 2 sin cos r a xx yy r r r =− − − + − =− ⎡ ⎤ θ θ ⎣ ⎦ − ( )( ) () ( ) 2 2 21 21 21 21 2 2 sin cos xx yy θθ θθ x x yy ⎡ ⎤ − −− − = ⎢ ⎥ −− − ⎣ ⎦ γ x G y G O g K φ1 1 y G 1 x G 2 x G 2 y G C1 C2 φ2 α 2 Q2 Q1 D α 1 B1 θ B2 M A
&- 2(x2-x)1+c0sM2 2(2-y)2+sin队2 (4)0=-0T= -5 -2(x2-x)2-c0s肌 -2(y2-y)2-sin队 52 -sinON +cosF cosON+sin@F -nF N=2(G-5)2 sinON-cosOF F,= -cosON-sinOF 5F, C M B2 F N
(3) T A ⎡ ⎤⎡ ⎤ ⎡ ⎤ ⎢ ⎥⎢ ⎥ = ⎢ ⎥ ⎣ ⎦⎣ ⎦ ⎣ ⎦ q q M Φ q Q Φ 0 λ γ (4) ( ) ( ) ( ) ( ) 2 11 2 2 11 2 1 2 2 11 2 2 11 2 2 2 2 cos 2 sin 2 cos 2 sin C T x x y y r x x y y r λ θλ λ θλ λ λ θλ λ θλ λ ⎡ ⎤ − + ⎢ ⎥ − + − =− = −− − −− − ⎣ ⎦ Q Φ λ q 1 2 sin cos cos sin sin cos cos sin f f C f f f f N F N F r F N F N F r F θ θ θ θ θ θ θ θ ⎡ ⎤ − + ⎢ ⎥ + − = − − − ⎣ ⎦ Q , ( 1 21 ) 2 2 f N rr F λ λ = − = x G y G g K φ1 1 x G 2 x G C1 C2 φ2 D B1 θ B2 M A Ff G N G N′ G Ff ′ G
X B F B, 4.An equilibrium system is shown in the figure.B can roll along the fixed inclined plane without slip.Link B2 is connected to B with a revolute joint.The masses of B and B2 are both m.The length of link B2 is 21,and the radius of Bi is r.The origins of the body-fixed frames of B and B2 are located at the centroids.Force F is applied on point 4 of B1.which is parallel to the inclined plane.A horizontal force F2 is applied on the tip point D of the link.The generalized coordinate vector is given by 4=出4x2y24了 (1)Write the kinematic constraint equations,calculate degree of freedom. (2)Write the equilibrium equations. (3)Find the equilibrium condition.(25) Solution: x+ro-c (1) Φ= 片-r =0 x2-Isin-x Ly2+Icoso2-片J DOF=6-4=2 (2) Equilibrium equations:= -F+mg sin a 「10r00 0 -mg cosa 7 01000 0 Frsin ，= Q= λ= -1001 0 -1cos2 F cosa+mg sina -1001-lsin92 F,sina-mg cosa Flcos(p2-a) (3)
4. An equilibrium system is shown in the figure. B1 can roll along the fixed inclined plane without slip. Link B2 is connected to B1 with a revolute joint. The masses of B1 and B2 are both m. The length of link B2 is 2l, and the radius of B1 is r. The origins of the body-fixed frames of B1 and B2 are located at the centroids. Force F1 is applied on point A of B1, which is parallel to the inclined plane. A horizontal force F2 is applied on the tip point D of the link. The generalized coordinate vector is given by [ ]T x y x y = 1 1 φ1 2 2 φ 2 q . (1) Write the kinematic constraint equations, calculate degree of freedom. (2) Write the equilibrium equations. (3) Find the equilibrium condition. (25 分) Solution: (1) 1 1 1 2 21 2 21 sin cos xr c y r xl x yl y ϕ ϕ ϕ ⎡ ⎤ + − ⎢ ⎥ − = = ⎢ ⎥ ⎢ ⎥ − − ⎢ ⎥ ⎣ ⎦ + − Φ 0 DOF =−= 64 2 (2) Equilibrium equations ： T A Φ λq = Q 2 2 1 0 00 0 0 1 000 0 1 0 0 1 0 cos 0 1 0 0 1 sin r l l ϕ ϕ ⎡ ⎤ ⎢ ⎥ = − − ⎣ ⎦ − − Φq ， ( ) 1 1 1 2 2 2 2 sin cos sin = cos sin sin cos cos A F mg mg F r F mg F mg F l α α ϕ α α α α ϕ α ⎡ − + ⎤ ⎢ ⎥ − ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ + ⎢ ⎥ ⎢ − ⎥ ⎢ ⎥ ⎢⎣ − ⎥⎦ Q ， 1 2 3 4 λ λ λ λ ⎡ ⎤ ⎢ ⎥ = ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎣ ⎦ λ (3) F2 G F1 G x K 1 x K 1 y G 2 x K 2 y G g K α α y G B1 B2 A D ϕ1 ϕ 2
=-F+mgsina (1) 2-元4=-mg cosa(2) r=Frsing (3) 元3=F cosa+mg sina(4) =F,sina-mg cosa (5) -Icos-Ising=Flcos(2-a)(6) =Fsin =F cosa+mg sin a 元=F,sina-mg cosa =F sina-2mg cosa Substitute into(1)and(6),the equilibrium condition: Fsin+F=F cosa+2mg sina mgsin(o-a)=2F cos(o2-a)
( ) 13 1 2 4 11 1 3 2 4 2 23 24 2 2 sin (1) cos (2) sin (3) cos sin (4) sin cos (5) cos sin cos (6) F mg mg r Fr F mg F mg l l Fl λ λ α λλ α λ ϕ λ αα λαα ϕλ ϕλ ϕ α − =− + − =− = = + = − −−= − 11 1 3 2 4 2 2 2 sin cos sin sin cos sin 2 cos F F mg F mg F mg λ ϕ λ α α λ α α λ α α = = + = − = − Substitute into (1) and (6), the equilibrium condition: 1 11 2 F F F mg sin cos 2 sin ϕ += + α α () () 2 22 mg F sin 2 cos ϕ −= − α ϕ α
