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CR MATH 55 - Differential Equations

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IntroductionA Simplified VersionIdentifying the ForcesFinding FcentPutting It All Together1/25 Differential EquationsSpring 2003A Differential Look at theWatt’s Governorby Tim Honn & Seth StoneCollege of the RedwoodsEureka,CAMath dept.email: [email protected]: [email protected]/25 IntroductionInvented by James Watt in the late 1700s, a governor is an automatedspeed control that ushered in the industrial revolution.• Mathematical model.• Bifurcation.• Damping.3/25 The Watt’s governor controlling a steam engine.4/25 A Simplified VersionBall-bearing in a rotating hoop.5/25 −1 −0.5 0 0.5 1−10−50510θωPhase plane for Ω = 1 rad/sec.−1 −0.5 0 0.5 1−0.500.5tθθ vs. t for Ω = 1 rad/sec.6/25 For Ω > 12 rad/sec the ball moves towards a new equilibrium point.7/25 −0.1 −0.05 0 0.05 0.1−0.2−0.15−0.1−0.0500.050.10.150.2θωθ vs. t for Ω = 13 rad/sec.8/25 Identifying the Forces• Identify the forces that always balance.• Identify the forces that do not always balance.• Sum the forces to derive the equations.9/25 θ−mg−mg cos θFcentsin θFcent−mg sin θTΩR sin θR cos θ(0, 0)(0, R)Fcentcos θForces opposing the normal force.10/25 θ−mg−mg cos θFcentsin θFcent−mg sin θTΩR sin θR cos θ(0, 0)(0, R)Fcentcos θTangential forces in the vertical plane.11/25 θ−mg−mg cos θFcentsin θFcent−mg sin θTΩR sin θR cos θ(0, 0)(0, R)Fcentcos θCThe horizontal path of the ball.12/25 Finding FcentRecall the kinematic identities, and our values.vlin= rvangar=v2linrFcent= marvlin= (R sin θ)Ωar=[(R sin θ)Ω]2R sin θ= (R sin θ)Ω2Fcent= m(R sin θ)Ω2In our case, Ω is the angular velocity vang, about the center of C andthe radius is R sin θ. The centrifugal force acting o n the ball is the masstimes ar.Fcent= mΩ2R sin θ.13/25 θ−mg−mg cos θFcentsin θFcent−mg sin θTΩR sin θR cos θ(0, 0)(0, R)Fcentcos θmaT= Fcentcos θ − mg sin θmR¨θ = mΩ2R sin θ cos θ − mg sin θmR¨θ = mΩ2R sin θ cos θ − mg sin θ¨θ = Ω2sin θ cos θ −gRsin θ (1)14/25 ¨θ = Ω2sin θ cos θ −gRsin θIn order to use this equation we must first transpose it into two firstorder equations.(θ = θω =˙θ˙θ = ω˙ω =¨θ = Ω2sin θ cos θ −gRsin θAn equilibrium angle means that the forces are balanced and the ac-celeration is zero.15/25 Set the right side equal to zero.¨θ = 0Ω2sin θ cos θ −gRsin θ = 0sin θ(Ω2cos θ −gR) = 0Therefore,sin θ = 0 or Ω2cos θ −gR= 0.When sin θ = 0, θ = 0 or π. To find other equilibrium angles we setthe other factor equal to zero.16/25 Ω2cos θ −gR= 0cos θ =g/RΩ2(2)Cosine is never greater than 1 so we seek Ωs that make the right sideless than or equal to 1.gRΩ20≤ 1.gR≤ Ω20rgR≤ Ω0(3)17/25 In our case the Ω where bifurcation occurs is,r9.8.06≤ Ω012.78 ≤ Ω0. (4)Now we find the Ω that produces θ = π/4.cos θ =g/RΩ2cosπ4=9.8/.06Ω2s9.8.06 cosπ4= Ω15.2 = Ω. (5)18/25 0.4 0.5 0.6 0.7 0.8 0.9 1 1.1 1.2−4−20246θωFor Ω = 15.2 rad/sec.−1.5 −1 −0.5 0 0.5 1 1.50.50.60.70.80.91tθθ vs. t for Ω = 15.2 rad/sec.19/25 20/25 Now we have a governor that will maintain the desired angle butoscillates perpetually. How can we improve this performance?¨θ = Ω2sin θ cos θ −gRmg sin θ − k˙θm(6)The damping term is proportional to the angular velocity (in the ver-tical plane) and is divided by the mass.21/25 0.4 0.5 0.6 0.7 0.8 0.9 1 1.1 1.2−4−20246θωΩ = 15.2 rad/sec with damping term.0 0.5 1 1.5 2 2.5 30.50.550.60.650.70.750.80.850.9tθθ vs. t for Ω = 15.2 and damping term.22/25 0 2 4 60.50.60.70.80.91tθθ vs. t for Ω = 15.2, damping term, and m = 50g.23/25 0 0.5 1 1.50.50.550.60.650.70.750.80.85tθθ vs. t for Ω = 15.2, damping term, and m = 5g.24/25 0 1 2 3 4 5 60.50.550.60.650.70.75tθθ vs. t for Ω = 15.2, damping term, and m = .25g.As you can see this also would not be a governor of optimum design.When designing a governor one would have to experiment with theparameters and would undoubtedly be s omewhere between 5g and 1/4g.25/25 Putting It All TogetherWe have a governor design that will maintain the desired• Changing R only effects the where the critical Ωs occur but not theoscillatory behavior.• Increasing the mass reduces the effects of damping, reducing massincreases the effects of damping.• Changing the damping term has an inverse effect as changing


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CR MATH 55 - Differential Equations

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