MCE 366 System Dynamics, Spring Problem Set 2. Solutions to Set 2

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1 MCE 366 System Dynamics, Spring 2012 Problem Set 2 Reading: Chapter 2, Sections 2.3 and 2.4, Chapter 3, Sections 3.1 and 3.2 Problems: 2.22, 2.24, 2.26, 2.31, 3.4(a, b, d), 3.5 Solutions to Set Using the equivalent mass approach, the equivalent mass referenced to the coordinate x is m e = m 1 + m 2 + I R 2 where I is the inertia of the cylinder about its center. The force acting on m e due to the weight of the cylinder is m 1 g sin β. The force acting on m e due to the weight of m 2 is m 2 g sin φ. See the following figure. The equation of motion of the equivalent system is m e ẍ = m 1 g sin β m 2 g sin φ Figure : for Problem 22 1

2 2.24 Summing moments about the pivot gives I O θ = T mgl sin θ where the effect of the motor torque is T = [2(1.5)]T m =3T m and the inertia is I O = ml 2 +I m [2(1.5)] 2 +I G1 [2(1.5)] 2 +I G2 (1.5) 2 +I G3 (1.5) 2 +I G4 =1.936 kg m 2 Thus θ =3T m sin θ 2.26 a) The equation of motion for rotation is I ω = Rf t where I = mr 2 /2=1.631 kg m 2 is the inertia of the cylinder about its center, and f t is the tangential force between the cylinder and the ground. The equation of motion for translation is mẍ = f cos φ f t = f cos φ I ω R Substituting ẍ = R ω we obtain ( mr 2 + I ) ω = Rf cos φ Noting that m = 800/9.81 = kg, and substituting the given values, we obtain ω = f cos φ b) Since we already have the answer from part (a), the simplest way is to just substitute ω = ẍ/r to obtain ẍ 0.2 = f cos φ or ẍ = f cos φ If we did not have the answer from part (a), we could use the equivalent mass approach. Using the coordinate x as the reference we express the kinetic energy as KE = 1 2 mẋ Iω2 = 1 2 mẋ2 + 1 ( ) 2 ẋ 2 I R or Thus the equivalent mass is KE = 1 2 (m + IR 2 ) ẋ 2 m e = m + I R 2 = m m =1.5m 2

3 where I = mr 2 /2 is the inertia of the cylinder about its center. The equation of motion is m e ẍ = f cos φ Substituting the given values, we obtain m e = kg and ẍ = f cos φ 3

4 2.31 See the following figure. Summing forces in the x and y directions gives: ma Gx = µ s N B mg sin θ (1) N A + N B = mg cos θ (2) Summing moments about G: Solve (2) and (3): N A L A N B L B + µ s N B H = 0 (3) N A = mg cos θ (µ sh L B ) = 16, 677 cos θ (µ s 2.1) µ s H L A L B µ s 4.6 N B = mgl A cos θ = 40, 025µ s cos θ µ s H L A L B 4.6 µ s The maximum acceleration is, from (1), a Gx = µ s cos θ 4.6 µ s 9.81 sin θ Figure : for Problem 31 4

5 3.4 NOTE: For this problem you may use the table on page 92. Note that the form of the free response is the solution of the differential equation in the table when its right hand side (b or c) iszero. The free response form is then evaluated for the given nonzero initial conditions. The form of the forced response is the solution of the differential equation given in the table if the right hand side is a nonzero constant. The forced response form is then evaluated using zero initial conditions. The total response is the sum of the free and the forced responses. The steady state response re- The transient response disappears eventually. mains. a) The roots are 3 and 5. Free Response: The form of the free response is x(t) =A 1 e 3t + A 2 e 5t Evaluate this with the given initial conditions. ẋ = 3A 1 e 3t 5A 2 e 5t x(0) = A 1 + A 2 =10 ẋ(0) = 3A 1 5A 2 =4 Solve these to obtain A 1 = 27 and A 2 = 17. This gives the free response x(t) =27e 3t 17e 5t Forced Response: The form of the forced response is x(t) =B 1 e 3t + B 2 e 5t + B 3 The steady-state solution is x ss =30/15 = 2, so B 3 = 2. Thus the form of the forced response becomes x(t) =2+B 1 e 3t + B 2 e 5t Evaluate this with zero initial conditions to obtain the forced response ẋ(t) = 3B 1 e 3t 5B 2 e 5t x(0) = 2 + B 1 + B 2 =0 ẋ(0) = 3B 1 5B 2 =0 These give B 1 = 5 and B 2 = 3. The forced response is x(t) =2 5e 3t +3e 5t Total Response: The total response is the sum of the free and the forced response. It is x(t) =27e 3t 17e 5t +2 5e 3t +3e 5t =2+22e 3t 14e 5t The transient response consists of the two exponential terms. The steady state response is 2. 5

6 3.4b) The roots are 5 and 5. Free Response: The form of the free response is x(t) =A 1 e 5t + A 2 te 5t Evaluate this with the given initial conditions ẋ(t) = 5A 1 e 5t 5A 2 te 5t + A 2 e 5t x(0) = A 1 =10 ẋ(0) = 5A 1 + A 2 =4 This give A 1 = 10 and A 2 = 54. The free response is x(t) =10e 5t +54te 5t Forced Response: The form of the forced response is x(t) =B 1 e 5t + B 2 te 5t + B 3 The steady-state solution is x ss =75/25 = 3, so B 3 = 3. Thus the form of the forced response becomes x(t) =3+B 1 e 5t + B 2 te 5t Evaluate this with zero initial conditions. ẋ(t) = 5B 1 e 5t 5B 2 te 5t + B 2 e 5t x(0) = 3 + B 1 =0 ẋ(0) = 5B 1 + B 2 =0 This give B 1 = 3 and B 2 = 15. The forced response is x(t) =3 3e 5t 15te 5t Total Response: The total response is the sum of the free and the forced response. It is x(t) =10e 5t +54te 5t +3 3e 5t 15te 5t =3+7e 5t +39te 5t The transient response consists of the two exponential terms. The steady state response is 3. 6

7 3.4d The roots are 4±7j. Free Response: The form of the free response is x(t) =A 1 e 4t sin 7t + A 2 e 4t cos 5t Evaluate this with the given initial conditions. ẋ(t) = 4A 1 e 4t sin 7t +7A 1 e 4t cos 7t 4A 2 e 4t cos 5t 5A 2 e 4t sin 5t x(0) = A 2 =10 ẋ(0) = 7A 1 4A 2 =4 These give A 1 =44/7 and A 2 = 10. The free response is x(t) = 44 7 e 4t sin 7t +10e 4t cos 7t Forced Response: The form of the forced response is x(t) =B 1 + B 2 e 4t sin 7t + B 3 e 4t cos 7t The steady-state solution is x ss = 130/65 = 2. Thus B 1 = 2. Evaluate this function with zero initial conditions. ẋ(t) = 4B 2 e 4t sin 7t +7B 2 e 4t cos 7t 4B 3 e 4t cos 7t 7B 3 e 4t sin 7t x(0) = 2 + B 3 =0 ẋ(0) = 7B 2 4B 3 =0 This gives B 2 = 8/7 and B 3 = 2. Thus the forced response is x(t) =2 8 7 e 4t sin 7t 2e 4t cos 7t Total Response: The total response is the sum of the free and the forced response. It is x(t) = 44 7 e 4t sin 7t+10e 4t cos 7t e 4t sin 7t 2e 4t cos 7t = e 4t sin 7t+8e 4t cos 7t The transient response consists of the last two terms. The steady state response is a) The root is s =5/3, which is positive. So the model is unstable. b) The roots are s = 5 and 2, one of which is positive. So the model is unstable. c) The roots are s =3± 5j, whose real part is positive. So the model is unstable. d) The root is s = 0, so the model is neutrally stable. e) The roots are s = ±2j, whose real part is zero. So the model is neutrally stable. f) The roots are s = 0 and 5, one of which is zero and the other is negative. So the model is neutrally stable. 7