A Differential Look at the Watt s Governor
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1 Differential Equations Spring /25 A Differential Look at the Watt s Governor by Tim Honn & Seth Stone College of the Redwoods Eureka,CA Math dept. timhonn@cox.net lamentofseth@hotmail.com
2 Introduction Invented by James Watt in the late 1700s, a governor is an automated speed control that ushered in the industrial revolution. 2/25 Mathematical model. Bifurcation. Damping.
3 3/25 The Watt s governor controlling a steam engine.
4 A Simplified Version 4/25 Ball-bearing in a rotating hoop.
5 10 5 ω 0 5 5/ Phase plane for Ω = 1 rad/sec vs. t for Ω = 1 rad/sec t
6 6/25 For Ω > 12 rad/sec the ball moves towards a new equilibrium point.
7 /25 ω vs. t for Ω = 13 rad/sec.
8 Identifying the Forces Identify the forces that always balance. Identify the forces that do not always balance. Sum the forces to derive the equations. 8/25
9 Ω (0, R) 9/25 T R cos (0, 0) F cent cos R sin mg sin Forces opposing the normal force. mg mg cos F cent sin F cent
10 Ω (0, R) 10/25 T (0, 0) F cent cos R cos R sin mg sin mg Tangential forces in the vertical plane. mg cos F cent sin F cent
11 Ω (0, R) 11/25 T (0, 0) F cent cos R cos C R sin mg sin The horizontal path of the ball. mg mg cos F cent sin F cent
12 Finding F cent Recall the kinematic identities, and our values. 12/25 v lin = rv ang a r = v2 lin r F cent = ma r v lin = (R sin )Ω [(R sin )Ω]2 a r = R sin = (R sin )Ω 2 F cent = m(r sin )Ω 2 In our case, Ω is the angular velocity v ang, about the center of C and the radius is R sin. The centrifugal force acting on the ball is the mass times a r. F cent = mω 2 R sin.
13 Ω (0, R) 13/25 T (0, 0) F cent cos R cos R sin F cent mg sin mg cos mg F cent sin ma T = F cent cos mg sin mr = mω 2 R sin cos mg sin mr = mω 2 R sin cos mg sin = Ω 2 sin cos g sin (1) R
14 = Ω 2 sin cos g R sin In order to use this equation we must first transpose it into two first order equations. 14/25 { = ω = = ω ω = = Ω 2 sin cos g R sin An equilibrium angle means that the forces are balanced and the acceleration is zero.
15 Set the right side equal to zero. = 0 Ω 2 sin cos g R sin = 0 15/25 sin (Ω 2 cos g R ) = 0 Therefore, sin = 0 or Ω 2 cos g R = 0. When sin = 0, = 0 or π. To find other equilibrium angles we set the other factor equal to zero.
16 Ω 2 cos g R = 0 cos = g/r Ω 2 (2) 16/25 Cosine is never greater than 1 so we seek Ωs that make the right side less than or equal to 1. g RΩ g R Ω2 0 g R Ω 0 (3)
17 In our case the Ω where bifurcation occurs is, Ω Ω 0. (4) 17/25 Now we find the Ω that produces = π/4. cos = g/r Ω 2 cos π 4 = 9.8/.06 Ω cos π = Ω = Ω. (5)
18 6 4 2 ω 0 18/ For Ω = 15.2 rad/sec. 1 vs. t for Ω = 15.2 rad/sec t
19 19/25
20 Now we have a governor that will maintain the desired angle but oscillates perpetually. How can we improve this performance? = Ω 2 sin cos g mg sin k R m (6) 20/25 The damping term is proportional to the angular velocity (in the vertical plane) and is divided by the mass.
21 6 4 ω / Ω = 15.2 rad/sec with damping term t vs. t for Ω = 15.2 and damping term.
22 / t vs. t for Ω = 15.2, damping term, and m = 50g.
23 / t vs. t for Ω = 15.2, damping term, and m = 5g.
24 / t 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 the parameters and would undoubtedly be somewhere between 5g and 1/4g.
25 Putting It All Together We have a governor design that will maintain the desired Changing R only effects the where the critical Ωs occur but not the oscillatory behavior. Increasing the mass reduces the effects of damping, reducing mass increases the effects of damping. Changing the damping term has an inverse effect as changing the mass. 25/25
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