Robot Dynamics
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Robot Dynamics Hesheng Wang Dept. of Automation Shanghai Jiao Tong University What is Robot Dynamics? • Robot dynamics studies the relation between robot motion and forces and moments acting on the robot. 2 l2 2 l 1 1 1 x Rotation about a Fixed Axis The velocity v can be determined from the cross product of and rp . Here rp is a vector from any point on the axis of rotation to P. v = x rp = x r The direction of v is determined by the right-hand rule. Rotation about a Fixed Axis (continued) The acceleration of P can also be defined by differentiating the velocity. a = dv/dt = d/dt x rP + x drP/dt = x rP + x ( x rP) Tangent accel Normal/centripetal acceleration It can be shown that this equation reduces to 2 a = a x r –wr = at + an 2 2 The magnitude of the acceleration vector is a = (at) + (an) Rotation of a Vector • Consider rotation of a vector about a axis. P r Point P is rotating about axis u. r is the position vector of point P. u : the speed of rotation The angular velcotiy : ω u The rate of change of position vector r : d r ωr dt Rotation of a Frame Frame A y B x • Consider a frame B rotating B about an unit vector u. u i B , jB ,k B : unit directional vectors of the axes z of frame B w.r.t. the reference frame A B d d d i ωi , j ω j , k ωk dt B B dt B B dt B B The derivative of the rotation matrix of frame B : d R i j k ω i ω j ωk dt B B B B B B R ω R General Motion • A general motion can be considered a combination of a translation with a z1 point and motion about the point. B A O xo yo zo : A fixed reference frame z y1 A x y z : A frame that translate with A 0 1 1 1 x1 O The velocity relation: y0 x0 VB VA ωr The acceleration relation: aB a A α r ω(ωr) Introduction to Dynamics Newton’s Laws of Motion First Law: A particle originally at rest, or moving in a straight line at constant velocity, will remain in this state if the resultant force acting on the particle is zero. F Second Law: If the resultant force on the particle is not zero, the a particle experiences an acceleration in the same direction as the resultant F ma force. This acceleration has a m : the mass magnitude proportional to the F : the net force a : the acceleration resultant force. Third Law: Mutual forces of action and reaction between two particles F F’ are equal, opposite, and collinear. Example m Find the accelerations of the ball 1 Frictionless and the wedge. surface m2 Solution: (1) Draw the free-body diagram of the particles (2) Apply Newton’s 2nd Law N For the ball R m1xb N sin (1) m1yb N cos m1g (2) For the wedge m2g m1g y N m2 xw N sin (3) x Example (continued) • Consider acceleration relationship between the ball and the wedge ab ab aw ab/ w ab / w : relative velocity of the ball a w.r.t. the wedge w Note that the relative velocity is along the surface of the wedge ab/w From the diagram, we have xb xw ab/W cos (4) yb ab/ w sin (5) From eqs. (1)-(5), we can solve the five unknowns, i.e. the acceleration Linear Momentum • Linear momentum: product of mass and velocity: L mV mV – It is a vector, in the same direction as velocity - SI Unit : Kg m/s V From Newton's 2nd Law : d F ma m V L dt • Principle of Linear Momentum: The rate of change of the linear momentum of a particle is equal to the result force acting on the particle Angular Momentum • r: the position vector of a particle w.r.t. a V reference point O. • V: the velocity of the particle r • m: the mass of the particle • The angular momentum of the particle O about reference O: Ho r mV Ho is a vector perpendicular to both V and r. Its direction is determined by the right- hand rule Its unit is Kgm2 / s Principle of Angular Momentum Consider a force F acting on the particle V Differentiating the angular momentum F r H r mV r mV r F o O H o r F Mo Principle of angular momentum: The rate of change of the angular momentum of particle about a fixed point O is equal to the resultant moment of forces acting on the particle about point O Dynamics of a System of Particles Consider a system of n particles. f1 fi fi: the external force exerted on particle i m1 eij: the internal force exerted on particle i mi by particle j m2 eij mi: mass of particle i. r : position vector of particle i eji i f d 2 m Linear momentum: L m V m r j i i i i fj dt Total mass mr The center of mass : r i L m r MV c i c C mi Velocity of The linear momentum of a system of particles is center of mass equal to the product of the total mass and the velocity of the center of mass Dynamics of a System of Particles Differentiating the linear momentum 2 d f1 fi L m r (f e ) m1 i dt 2 i i ij i ij mi eij 0 L fi m2 ei ij i The rate of change the linear momentum of a ej f2 system of particles is equal to the resultant of all mj EXTERNAL forces acting on the particles fj ac: acceleration of center of mass L MVC Mac fi i Equation of motion of Center of mass The center of mass of the system moves as if all the forces and masses are concentrated at the center of mass Angular Momentum of a System of Particles Similarly, by differentiating the f f 1 m i angular momentum 1 mi d m2 ei H o ri mVi MO i dt ej f 2 m The resultant moment of all j fj EXTERNAL forces acting on the system about O O The reference point must be a fixed point. However, the center of mass of the system can be the reference point even when it is moving Example: m2 Ry Example: Calculate the angular acceleration l2 of the massless link Rx l O Solution 1 Consider the two particles and the link as a system m2g m (1) Analyze the external forces 1 (2) Consider the angular momentum about O m1g 0 0 d H o Mo 0 0 dt 2 2 m gl cos m gl cos (m1l1 m2l2 ) 2 2 1 1 (m1l1 m2l2 )cos 2 2 m1l1 m2l2 Linear and Angular Momentums for Rigid Body • Since a rigid body can be considered as a system with infinite number of VC particles, the linear momentum L mV C Center of mass •The angular momentum about the center of mass ω HC Iω I : Inertia tensor matrix about C. ω : angular velocity of the body Newton’s Equation and Euler’s Equation • A general motion can be considered a f1 combination of a translation with the n z center of mass and motion about the i 1 center of mass C O xo yo zo : A fixed reference frame y z0 ri 1 C x1 y1z1 : A frame that translates with C x1 f O i The equation of translation: y0 x ma F f 0 c i Newton’s equation The equation of motion about C: Euler’s equation d A A A HC MC Iω ω( Iω) ri fi ni dt Newton’s Equation and Euler’s Equation (Cont’) • The angular velocity is with respect to the translating frame. • The inertia tensor matrix is with respect to the translating frame, so it will change its value with rotation of the body. • The force is the resultant of the EXTERNAL force • The moment is the resultant moment of the EXTERNAL forces and moments. Example • Derive the dynamic equation of the 2 DOF manipulator. Here, the masses of links 1 and 2 are m1 and m2 respectively. Assume that the mass is uniformly distributed over the link. 2 Solution: l2 (1) Analyzing forces acting on the links 2 l 1 1 2 Rx O 1 1 x0 Ry Nx Rx Ry m g m2g Ny 1 Example (continued) R (2) Dynamics of link 1. As x 1 link 1 is rotating about O, Nx 1 Ry N m1g IMOo1 y Moment of Resultant moment about O inertia about O l I m g 1 cos R l cos R l sin (1) O 1 1 1 2 1 y 1 1 x 1 1 Example (3) Dynamics of link 2. 2 As link 2 is in a general plane motion C2 m a F Ry 2 C2 C C I ω ω I ω M Rx 2 2 2 2 2 C2 B Acceleration of C2: a a a a m g C2 B C / Bt C / Bn 2 2 l11 c1 l11s1 aBt aB aBn aBt 2 ... l11 s1 l11c1 l 2 c l a 2 ( ) 12 2 s12 C / Bn 1 2 s aC / Bt (1 2 ) 1 2 12 2 c12 l l a l ( s 2c ) 2 ( )s 2 ( )2 c Bn 1 1 1 1 1 2 1 2 12 2 1 2 12 aC 2 2 l2 l2 2 l1(1c1 1 s1) (1 2 )c12 (1 2 ) s12 a 2 2 C / Bt 1 2 Newton’s m a R (2) 2 c2 x x equation: m a R m g (3) 2 c2 y y 2 aC / Bn Example (continued) Consider Euler’s equation.