Lesson 5: Kinematics and Dynamics of Particles

Transcription

1 Lesson 5: Knematcs and Dynamcs of Partcles hs set of notes descrbes the basc methodology for formulatng the knematc and knetc equatons for multbody dynamcs. In order to concentrate on the methodology and not on the detals and the complexty of the equatons, partcles are used nstead of bodes. Snce partcles do not have rotatonal degrees-of-freedom, unlke bodes, the correspondng mult-partcle dynamc equatons are much smpler. For multbody systems, a smlar process s descrbed n the textbook. One Partcle Knematcs Poston (coordnate), velocty, and acceleraton vectors for partcle : r (x ) x r (x ) x r (x ) x r = r (y ) = y, v r = r (y ) = y, a r = r (y ) = y r (z) z r (z) z r (z) z Knetcs f (x ) Mass: m Force: f = f (y ) f (z ) Equaton of moton (Newton's second law): m r = f or m 0 0 x f (x ) m 0 0 M r = f ; or 0 m 0 y = f (y ) where M = 0 m m z f (z ) 0 0 m A System of n p Partcles Knematcs f r f n p x z r y r j r n p j f j f n p Page

2 Coordnate, velocty, and acceleraton arrays: each s a 3n p array r r r r =, r =, r = r n p r n p r n p Knetcs Mass of each partcle: m ; =,..., n p Mass matrx ( 3 3) for each partcle: M ; =,..., n p f Array of appled forces: f = f n p Equatons of Moton: M r = f M 0 0 where M = M n p = dag M M n p [ ] wo Partcles Connected by A Rgd (massless) Rod A d z r A B y r B x x B x A Appled forces are not shown; d = r B r A = y B y A ; magntude of d s. z j z A Knematcs Poston constrant: Expanded form: 2 (xb x A ) 2 + (y B y A ) 2 + (z B z A ) 2 ( ) = 2 ()2 Compact form: 2 d d = 2 ()2 Note: he factor 2 s ncluded n order to elmnate the 2 factor from the velocty and acceleraton equatons, and also from the elements of the Jacoban matrx. Velocty constrant: Expanded form: (x B x A )( x B x A ) + (y B y A )( y B y A ) + (z B z A )( z B z A ) = 0 Compact form: d d =0 Acceleraton constrant: Expanded form: (x B x A )( x B x A ) + (y B y A )( y B y A ) + (z B z A )( z B z A ) = ( x B x A ) 2 ( y B y A ) 2 ( z B z A ) 2 Expanded matrx form: Page 2

3 x A y A z [ x B + x A y B + y A z B + z A x B x A y B y A z B z A ] A x B = y B z B ( x B x A ) 2 ( y B y A ) 2 ( z B z A ) 2 [ ] Compact matrx form: d d Jacoban matrx: Expanded form: r A r B = d d (x B x A ) (y B y A ) (z B z A ) (x B x A ) (y B y A ) (z B z A ) Compact form: [ d d ] Knetcs Equatons of moton: m A r A = f A + f B,A m B r B = f B + f A,B or, M A 0 r A 0 M B r B = f A f B + f B,A f A,B Vectors f B,A and f A,B are the unknown reacton forces the two partcles apply on each other. We expect f B,A = f A,B. How do we determne the reacton forces? We frst descrbe the reacton forces n terms of the transpose of the Jacoban matrx tmes a multpler (an unknown coeffcent) as: f B,A f A,B = d d hs yelds f A,B = f B,A = d. If we determne the value of the multpler we can then compute the components of the reacton forces. hs technque s known as the method of Lagrange multplers. he number of multplers must be equal to the number of constrants! We append the acceleraton constrant(s) to the equatons of moton to get M A 0 d r A f A 0 M B d r B d d 0 = f B d d Now we have as many equatons as the number of unknowns. hree Partcles Connected by wo Rgd Rods z A d C y B x Page 3

4 Knematcs All equatons n compact form! Poston constrants: 2 d d = 2 ( )2 2d 2 = ( 2 2 )2 Velocty constrants: r A d r B = 0 0 r C 0 Acceleraton constrants: r A d r B 0 r C = d d d 2 Jacoban matrx: d 0 Knetcs Equatons of moton: M A 0 0 r A f A d 0 M B 0 r B = f B M C r C f C d 0 2 If we append the acceleraton constrants to the equatons of moton, we get M A 0 r A f A 0 M B 0 0 r B f B 0 0 M C d 0 r C = f C d 0 0 d d d 2 Note that the angle between the two rods can change when the partcles move. hree Partcles, wo Rgd Rods, and One Angle Constrant Knematcs All equatons are presented only n compact form! Poston constrants: Page 4

5 2d d = 2( ) 2 d 2 2 = ( 2 2 )2 d = 2 cos Velocty constrants: d r A 0 0 r B = 0 d d r C 0 Acceleraton constrants: d r A d d 0 r B = d 2 d d r C 2 d d 2 Jacoban matrx: d 0 d d Knetcs Equatons of moton: M A 0 0 r A f A d d 0 M B 0 r B = f B M C r C f C d 0 d 3 If we append the acceleraton constrants to the equatons of moton, we get M A 0 d r A f A 0 M B 0 0 r B f B 0 0 M C d 0 d r C f C = d d d d 2 d d d General Problem; A System of n p Partcles and m constrants Knematcs All equatons n compact form! Poston constrants: (r) = 0 m constrants (mostly nonlnear n coordnates) Velocty constrant: Dr = 0 m constrants (lnear n veloctes) Acceleraton constrant: D r + D r = 0 m constrants (lnear n acceleratons) Jacoban matrx: D An m 3n p matrx Page 5

6 Knetcs Equatons of moton: M r = f + D 3n p equatons If we append the acceleraton constrants to the equatons of moton, we get M D r D 0 = f D 3n r p + m equatons n 3n p + m unknowns Page 6