Spring-Pendulum Dynamic System

Mechatronics K. Craig Spring-Pendulum Dynamic System 1 Measurements, Which Parameters to Identify? Calculations, Model Manufacturer's Specifications What Tests to Perform? Parameter ID

Physical Physical Math System Model Model

Physical Laws Assumptions Equation Solution: Experimental and Analytical Analysis Engineering Judgement Model Inadequate: and Numerical Modify Solution

Actual Predicted Dynamic Compare Dynamic Behavior Behavior

Make Design Design Modify Model Adequate, Model Adequate, Complete or Decisions Performance Inadequate Performance Adequate Augment Dynamic System Investigation

Mechatronics K. Craig Spring-Pendulum Dynamic System 2 Physical Model

l + r k

q m = pendulum mass = 1.815 kg m mspring = spring mass = 0.1445 kg l = unstretched spring length = 0.333 m k = spring constant = 172.8 N/m g = acceleration due to gravity = 9.81 m/s2

Ft = 5.71 N = pre-tension of spring rs = static spring stretch, i.e., rs = (mg-Ft)/k = 0.070 m rd = dynamic spring stretch r = total spring stretch = rs + rd

Mechatronics K. Craig Spring-Pendulum Dynamic System 3 Spring Calibration

Fspring (N)

mg = 17.805 N K = 172.8 N/m

Spring Pre-Tension

Ft = 5.71 N

Spring Displacement 0.070 m (meters)

Mechatronics K. Craig Spring-Pendulum Dynamic System 4 Physical Model Simplifying Assumptions

• pure spring, i.e., negligible inertia and damping • linear spring • frictionless pivot • neglect all material damping and air damping • point mass, i.e., neglect rotational inertia of mass • two degrees of freedom, i.e., r and q are the generalized coordinates (this assumes no out-of- plane motion and no bending of the spring) • support structure is rigid

Mechatronics K. Craig Spring-Pendulum Dynamic System 5 Polar Coordinates: Position, Velocity, Acceleration

r r = re$ r e$q de$ r r = e$q r dr dq e v = = rer + rq&eq = vr er + vqeq $ r dt &$ $ $ de $q dvr = -e$ r r 2 dq a = = cr - rq& her +br&q& + 2rq&geq dt && $ & $ r = a re$ r + aqe$ q PATH &r& magnitude change vr q &rq& direction change O r&q& + &rq& magnitude change vq rq& 2 direction change Mechatronics K. Craig Spring-Pendulum Dynamic System 6 Rigid Body Kinematics

Y XY: R reference frame (ground) y x xy: R1 reference frame (pendulum)

x cosq sinq 0 X q ML PO ML POML PO O X MyP = M- sinq cosq 0PMYP NMzQP NM 0 0 1QPNMZQP k L$i O L cosq sinq 0OL I$ O q M P M PM P l + r m P M$jP = M- sinq cosq 0PM J$ P NMk$QP NM 0 0 1QPNMK$ QP

R ar P =R ar O + R wr R1 ´ c R wr R1 ´ rOP h + R ar R1 ´ r OP + R1 ar P + 2 R wr R1 ´R1 vr P

Mechatronics K. Craig Spring-Pendulum Dynamic System 7 Rigid Body Kinematics

R ar O = 0 R wr R1 = q&k$ = q&K$ rOP r = -al + rf$j = -al + rf -sinqI$ + cosqJ$ R ar R1 = &q&k$ = &q&K$

R1 r P v = -&r$j = -&r -sinqI$ + cosqJ$ R1 r P a = -&r&$j = -&r& -sinqI$ + cosqJ$

After substitution and evaluation:

R r P 2 a = $i al + rf&q& + 2&rq& + $j -&r& + al + rfq&

Mechatronics K. Craig Spring-Pendulum Dynamic System 8 Mathematical Model

kr+Ft + r Free Body Diagram l

2 å Fr = ma r = m &r& - al + rfq& +q +q

å Fq = maq = m al + rf&q& + 2&rq& mg

2 +r -kr - Ft + mgcosq = m &r& - al + rfq& -mgsinq = m al + rf&q& + 2&rq&

2 m&r& - mal + rfq& + kr + Ft - mgcosq = 0 Nonlinear Equations of Motion al + rf&q& + 2&rq& + gsinq = 0

Mechatronics K. Craig Spring-Pendulum Dynamic System 9 d F ¶T I ¶T ¶V Mathematical Model: G J - + = Qi dt H ¶qi K ¶qi ¶qi Lagrange’s Equations & Lagrange’s Equations

q1 = r Q = -F Generalized Generalized Coordinates r t q = q 2 Qq = 0 Forces

1 2 T = m r 2 + a + rf q& 2 Kinetic Energy 2 & l 1 V = kr2 - mg a + rfcosq - Potential Energy 2 l l

2 m&r& - mal + rfq& + kr + Ft - mgcosq = 0 Nonlinear Equations of Motion al + rf&q& + 2&rq& + gsinq = 0

Mechatronics K. Craig Spring-Pendulum Dynamic System 10 Linearized Equations of Motion

kk &&rd+rd2=0 w==9.76 rad/sec=1.55 Hz mm

gg &&q+q=0 w1 ==4.93 rad/sec=0.78 Hz + ll+rrss

These equations do not predict the motion of the system except for particular sets of initial conditions !!

Mechatronics K. Craig Spring-Pendulum Dynamic System 11 Predicted Dynamic Behavior

sin(u) sin Product 1/s 1/s 9.81 Sum Product Integrate Integrate gravity (m/s^2) theta acc theta vel cos(u) Spring Pendulum Product 2 cos Dynamic System Gain Product

u^2 theta square theta position

Product r r position 1/s 1/s Integrate Integrate t r acc r vel Clock time Sum

95.21 0.333 u^(-1) spring length Sum2 inverse 5.710/1.815 k/m unstretched k=172.8 N/m Ft=5.71 N (meters) m=1.815 kg m=1.815 kg Matlab Simulink Block Diagram Mechatronics K. Craig Spring-Pendulum Dynamic System 12 Simulation Results

Simulation Results with Initial Conditions: theta = -0.274 rad, r = 0.046 m 0.3

0.2

0.1

0

-0.1

-0.2 radial and angular position (rad or m)

-0.3

-0.4 0 10 20 30 40 50 60 time (sec)

Mechatronics K. Craig Spring-Pendulum Dynamic System 13 Simulation Results

Simulation Results with Initial Conditions: theta = 0.021 rad, r = 0.115 m 0.25

0.2

0.15

0.1

0.05

0

-0.05

-0.1 radial and angular position (rad or m)

-0.15

-0.2

-0.25 0 10 20 30 40 50 60 time (sec)

Mechatronics K. Craig Spring-Pendulum Dynamic System 14 Sensor Calibration

Potentiometer Calibration Curve (Theta = 75.388 V - 196.51) 25

20

15

10

5

0

Theta (degrees) -5

-10

-15

-20

-25 2.3 2.4 2.5 2.6 2.7 2.8 2.9 volts

Mechatronics K. Craig Spring-Pendulum Dynamic System 15 Sensor Calibration

Ultrasonic Sensor Calibration Curve (X= -28.547 V + 206.41)

160

140

120

100

80 x (mm)

60

40

20

0 1 2 3 4 5 6 7 8 volts

Mechatronics K. Craig Spring-Pendulum Dynamic System 16 Actual Measured Dynamic Behavior

Experimental Results with Initial Conditions: theta = -0.274 rad, r = 0.046 m 0.3

0.2

0.1

0

-0.1

-0.2 radial and angular position (rad or m)

-0.3

-0.4 0 10 20 30 40 50 60 time (sec)

Mechatronics K. Craig Spring-Pendulum Dynamic System 17 Actual Measured Dynamic Behavior

Experimental Results with Initial Conditions: theta = 0.021 rad, r = 0.115 m 0.2

0.15

0.1

0.05

0

-0.05

radial and angular position (rad or m) -0.1

-0.15

-0.2 0 10 20 30 40 50 60 time (sec)

Mechatronics K. Craig Spring-Pendulum Dynamic System 18 Nonlinear Resonance

2 m&r& - mal + rfq& + kr + Ft - mgcosq = 0 Nonlinear Equations al + rf&q& + 2&rq& + gsinq = 0 of Motion

2 3 2 F q I q mr + kr = ma + rfq& - Ft + mg 1- sinq » q - + && l HG 2!KJ 3! L 2 F q3I q a + rf&q& + gGq - J = -2rq& cosq » 1- +L l H 3!K & 2! k F q2 r + r = a + rfq& 2 - t + g - g && m l m 2! q3 a + rf&q& + gq = -2rq& + g l & 3!

Mechatronics K. Craig Spring-Pendulum Dynamic System 19 2 k 2 Ft q$ r = r + r$ &$r&+ ar + $rf = al + r + r$fq$& - + g - g m m 2! q = q + q$ q$ 3 a + r + rf&q$& + gq$ = -2&rq&$ + g q = 0 l $ $ 3! mg - Ft 2 r = k q$ &r&+ r = a + r + rfq$& 2 - g k $ m $ l $ 2! g -2&rq$& g q$ 3 rq&$& q&$& + q$ = $ + - $ l + r l + r l + r 3! l + r q2 2 & 2 $ 2 k &$r&+ w r r$ = al + r + $rfq$ - g w r = 2! m 3 g 2 -2&rq&$ 2 q$ rq&$& 2 q&& +w q = $ + w - $ wq = $ q $ q + r l + r 3! l + r l

Mechatronics K. Craig Spring-Pendulum Dynamic System 20 Define: al + r + $rf = al + rfa1+ xf ® r$ = al + rfx q$ 2 a + rfx + w2 a + rfx = a + rfa1+ xfq$& 2 - g l && r l l 2! q$ 3 q&$& +w 2q$ = -2xq&$ + w2 - x&q$& q & q 3!

2 2 2 2 q$ x + w x = a1+ xfq$& - wq && r 2! 3 2 2 q$ a1+ xf&q$& + wqq$ = -2xq&$ + wq & 3! q2 2 & 2 2 $ & 2 &x& + w r x = q$ - wq + xq$ 1 2! » 1- x + x2 - x3 1+ x L w2q$ -2xq&$ w2 q$ 3 q&$& + q = & + q 1+ x 1+ x 1+ x 3!

Mechatronics K. Craig Spring-Pendulum Dynamic System 21 q$ 2 x + w 2x = q$& 2 - w2 + xq$& 2 && r q 2! 3 2 2 q$ q&$& + wqq$a1- xf = -2xq&$a1- xf + wq a1- xf & 3!

q$ 2 x + w 2x = q$& 2 - w2 + xq$& 2 && r q 2! 3 2 2 2 q$ q&$& +wqq$ = -2xq&$a1- xf + wqxq$ + wq a1- xf & 3!

q2 2 & 2 2 $ x + w r x = q$ - wq + Neglecting nonlinear terms && 2 L third order and higher && 2 & 2 q$ + wqq$ = -2x&q$ + wqxq$+L

Mechatronics K. Craig Spring-Pendulum Dynamic System 22