Example: Inverted pendulum on cart

Transcription

1 Chapter 11 Eample: Inverted pendulum on cart The figure to the right shows a rigid body attached by an frictionless pin (revolute) joint to a cart (modeled as a particle). Thecart slides on a horizontal frictionless track. The track is fied in a ewtonian frame. Right-handed orthogonal unit vectors n, n y, n z and b, b y, b z are fied in and respectively, with: n horizontally-right and n y vertically-upward n z b z parallel to s ais of rotation in b y directed from to the distal end of Complete the figure to the right by adding the identifiers,,, cm,, F c,,, n, n y, n z, b, b y, b z. Quantity Symbol Value Mass of m 1. kg Mass of m 1. kg Distance between and cm ( s center of mass).5 m s moment of inertia about cm for b z I zz.8333 kg m 2 Earth s gravitational constant g 9.8 m/s 2 bn measure of feedback-control force applied to F c Specified bn measure of s position vector from o (a point fied in ) Variable ngle from bn y to b y with bn z sense Variable 11.1 Kinematics (space and time) Kinematics is the study of the relationship between space and time, independent of the influence of mass or forces. The kinematic quantities normally needed for motion analysis are listed below. In many circumstances, it is efficient to form rotation matrices, angular velocities, and angular accelerations before position vectors, velocities, and accelerations. Kinematic Quantity Rotation matri ngular velocity ngular acceleration Quantities needed for analyzing the inverted pendulum on a cart b R n, the rotation matri relating b, b y, b z and n, n y, n z ω, s angular velocity in α, s angular acceleration in F m a Position vectors r /o and r cm/, the position vector of from o and of cm from Velocity v and v cm, s velocity in and cm s velocity in cceleration a and a cm, s acceleration in and cm s acceleration in R r ω v α a Copyright c Paul Mitiguy. ll rights reserved. 69 Chapter 11: Eample: Inverted pendulum on cart

2 11.2 Rotation matri The relative orientation of two sets of right-handed orthogonal unit vectors b, b y, b z and n, n y, n z is frequently stored in a 3 3 rotation matri denoted b R n whose elements are b i n j (i, j, y, z). ll rotation matrices are orthogonal, which means its inverse is equal to its transpose, and which means the matri can be written as a table read horizontally or vertically. To relate b, b y, b z and n, n y, n z, redraw these unit vectors in the geometrically-suggestive way shown below. To determine the 1 st row of the b R n rotation matri, b is epressed in terms of n, n y, n z as shown below. Similarly, the 2 nd and 3 rd rows of b R n are found by epressing b y and b z in terms of n, n y, n z. b y b R n n n y n z b cos() n sin() n y b cos() sin() b b y sin() n + cos() n y b z n z b y sin() cos() b z ngular velocity (special 2D case) Since 3D angular velocity is complicated, many tetbooks only define simple angular velocity, which is useful for two-dimensional analysis. 1 When a unit vector λ λ λ λ λ λ λ λ λ λ λ λ λ is fied in both reference frames and, has a simple angular velocity in that can be calculated via equation (1). ω (simple) ± λ λ λ λ λ λ λ λ λ λ λ λ λ (1) The sign of λ λ λ λ λ λ λ λ λ λ λ λ λ is determined by the right-hand rule. If increasing causes a right-hand rotation of in about + λ λ λ λ λ λ λ λ λ λ λ λ λ, the sign is positive, otherwise it is negative Simple angular velocity eample: Step-by-step process to calculate ω Due to the pin joint, b z is fied a in both and, so has a simple angular velocity in. b z is a unit vector fied in both and (parallel to the pin joint) n y is fied in and perpendicular to b z b y is fied in and perpendicular to b z is the angle between n y and b y,and is its time-derivative fter pointing the four fingers of your right hand in the direction of n y and curling them in the direction of b y, your thumb points in the bz direction. Since the right-hand rule produces a sign of b z that is negative: ω bz a vectorissaiobefied in reference frame if its magnitude is constant and its direction does not change in ngular velocity and vector differentiation For any vector v, thegolden rule for vector differentiation relates d v of v in a reference frame ) to: ω, s angular velocity in d v, the ordinary time-derivative of v in 1 One of the major obstacles in three-dimensional kinematics is properly calculating angular velocity. ote: See 3D dynamics tetbooks for undergraduates/professionals at d v (the ordinary time-derivative d v + ω v (2) Copyright c Paul Mitiguy. ll rights reserved. 7 Chapter 11: Eample: Inverted pendulum on cart

3 Equation (2) is one of the most important formulas in kinematics because v can be any vector, e.g., a unit vector, a position vector, a velocity or acceleration vector, a linear/angular momentum vector, or a force or torque vector ngular acceleration Equation (3) defines the angular acceleration of a reference frame in a reference frame. α also happens to be equal to the time-derivative in of ω. ote: Calculate with d ω if it is easier to compute than d ω. s angular acceleration in is most easily calculated with its alternate definition, i.e., α α d ω d ω d ( bz ) d ω bz (3) 11.5 Position vectors (inspection and vector addition) Position vectors are usually formed by inspection and vector addition. Inspection of the figure: r /o n ( s position vector from ). Inspection of the figure: r cm/ b y ( cm s position vector from ). Vector addition: r cm/o r cm/ + r /o b y + n ( cm s position vector from o) 11.6 Velocity and acceleration v cm (the velocity of a point cm in a reference frame ) is defined as the time-derivative in of r cm/o ( cm s position vector from o). v cm d r cm/ o d ( n ) ẋ n + + d ( by ) d ( n + b y ) d ( by ) + ω b y ẋ n + + bz b y v cm d r cm/o (4) Point o is any point fied in ẋ n + b a cm (the acceleration of point cm in reference frame ) is defined as the time-derivative in of v cm ( cm s velocity in ). a cm d v cm (5) a cm d v cm ẍ n + d (ẋ n + b ) d (ẋ n ) d ( b ) + d ( b ) + ω ( b ) ẍ n + b + 2 b y Copyright c Paul Mitiguy. ll rights reserved. 71 Chapter 11: Eample: Inverted pendulum on cart

4 11.7 Forces, moments, and free-body diagrams (2D) To draw a free-body diagram (FD), isolate a single body (or system S of and ) and draw all the eternal contact and distance forces that act on it. Shown right are FDs with all the eternal forces on the cart and pendulum. a Quantity Description Type F c n measure of control force applied to Contact n y measure of the resultant normal force on from Contact R n measure of the force on from across the revolute joint Contact R y n y measure of the force on from across the revolute joint Contact m g ny measure of Earth s gravitational force on Distance m g ny measure of Earth s gravitational force on Distance F m a m g R Resultant force on : F ( R ) n + ( m g R y ) n y Ry FD of Resultant force on : F R n + (R y m g) n y Resultant force on S: F S n + [ (m + m ) g] n y a lternately, to use the efficient MG road-map/d lembert method ) to eliminate constraint forces R and R y, draw a FD of the system S consisting of and (no need to draw alone). Since the revolute joint between and is ideal, action/reaction is used to minimize the number of unknowns. The b z component of the moment of all forces on about cm is a Ry m g R FD of / M cm z r /cm (R n + R y n y ) + r cm/cm ( m ) g n y b y (R n + R y n y ) [ cos()r sin()r y ] b z a ote: The rotation table to useful for calculating the cross-products ( b b y bn ) and ( b b y bn y) Mass, center of mass, inertia (required by dynamics) Mass of each particle and body, e.g., m (mass of particle ) and m (mass of body ). ocation of each particle and body center of mass, e.g., r /o and r cm/. Inertia dyadic of each rigid body about a point fied on the body. Since s angular velocity in is simple, I zz ( s moment of inertia about cm for b z) suffices for this analyses. F m a 11.9 ewton/euler laws of motion for and separately (inefficient) n inefficient way to form this system s equations of motion is with separate analyses of and. Using F m a for particle and body [in conjunction with the previous free-body diagrams (FDs)] yields, F m a (F c R ) n + ( m g R y ) n y m ẍ n Dot-multiply with bn : F c R m ẍ Dot-multiply with bn y: m g R y F m a cm R n + (R y m g) n y m (ẍ n + b 2 b y ) Dot-multiplication with n and n y (use the rotation table to calculate dot-products) gives R m [ ẍ + ( b n ) 2 ( b y n ) ] (R y m g) m [ ( b n y ) 2 ( b y n y ) ] R m [ ẍ + cos() 2 sin() ] (R y m g) m [ sin() 2 cos() ] ote: Separate analyses of and is less efficient than the MG road-map/d lembert method and Homework Copyright c Paul Mitiguy. ll rights reserved. 72 Chapter 11: Eample: Inverted pendulum on cart

5 Dynamics for a rigid body with a simple angular velocity (special 2D case) Euler s equation for a rigid body with a simple angular velocity in a ewtonian reference frame is: M /cm z (8.3) I zz α m g M /cm z is the b z n z component of the moment of all forces on about cm. I zz is s moment of inertia about the line passing through cm and parallel to b z. α is s angular acceleration in. R ssembling these terms and subsequent dot-multiplication with b z produces Ry cos() R sin() R y I zz Summary of ewton/euler equations of motion (inefficient) F c R m ẍ m g R y R m [ẍ + cos() 2 sin()] (R y m g) m [ sin() 2 cos()] cos() R sin() R y I zz There are 5 unknown variables in the previous set of equations, namely R, R y,,,. ote: Once (t) isknown, (t) and (t) are known. Similarly, once (t) isknown,ẋ(t) andẍ(t) areknown Equations of motion via MG road-maps/d lembert (efficient) For various purposes (e.g., control system design), it is useful to eliminate the unknown constraint forces R, R y,. lthough tedious linear-algebra can reduce the previous set of 5 equations in 5 unknowns to 2 equations in 2 unknowns (ẍ, ), itismore efficient to use MG road-maps or the methods of agrange or Kane as they automatically eliminate R, R y,. MG road-map for F c (m + m )ẍ + m cos() m sin() 2 MG road-map for m g sin() m cos()ẍ +(I zz + m 2 ) Matri form of nonlinear and linearized equations of motion For numerical solution and various control-systems techniques, it can be useful to write this system s nonlinear equations of motion in matri form as [ ] [ 1 [ ] m + m m ][ẍ ] [ cos() m m cos() I zz + m 2 + sin() 2 ] m g sin() Certain stability analyses and control-system techniques depend on linear ODEs. Using the small angle approimations cos() 1, sin(), and the small rate approimation 2, this system s equations of motion can be linearized and put into various matri forms, e.g., [ ] [ inearized 1 [ ] m matri form + m m ][ẍ ] [ ][ ] m I zz + m 2 + m g 1 State-space form d I zz (m + m ) + m m 2 ẋ 1 ( g 2 m 2 ) /d + (I zz + m 2 ) /d F c [gm (m + m )] /d ( m ) /d Copyright c Paul Mitiguy. ll rights reserved. 73 Chapter 11: Eample: Inverted pendulum on cart