ME 375 EXAM #1 Friday, March 13, 2015 SOLUTION

ME 375 EXAM #1 Friday, March 13, 2015 SOLUTION

PROBLEM 1 A system is made up of a homogeneous disk (of mass m and outer radius R), particle A (of mass m) and particle B (of mass m). The disk is pinned to ground at O and is able to roll without slipping on particle A. A spring (of stiffness k) connects A to ground. A second spring (also of stiffness k) connects A and B. A force f(t) acts on particle B. The system has two degrees of freedom, and is to be described in terms of the absolute coordinates x(t) and y(t). Derive the two differential equations of motion for the system in terms of the x and y coordinates. Your equations of motion should be written in terms of x and y, and their time derivatives, in addition to the parameters of m, k and f(t). Appropriate free body diagrams must be included with your solution in order to receive full credit for your work. SOLUTION From FBDs: Block A: Disk: Block B: Kinematics: F x = kx + k( y x) F = m!!x (1) M = FR = I!! θ = 1 2 mr2!! θ (2) F x = k( y x) + f = m!!y (3)

!!x = R!! θ!! θ =!!x / R (4) Combining equations (1), (2) and (4): F x = kx + k( y x) 1 1!!x R 2 mr2 R = m!!x 3 m!!x + 2kx ky = 0 2 From equation (3): m!!y kx + ky = f

PROBLEM 2 The differential equations of motion for a two-degree-of-freedom system in terms of coordinates x 1 and x 2 and input u(t) are known to be: m!!x 1 + c!x 1 + 2kx 1 kx 2 = f 0 u(t) (1) m!!x 2 kx 1 + kx 2 = 2 f 0 u(t) (2) Derive the single input/output differential equation of motion for the system with x 1 as the output and u(t) as the input. SOLUTION Taking the LT of equations (1) and (2) with zero initial conditions: ( ms 2 + cs + 2k) X 1 kx 2 = f 0 U (s) (3) ( ) X 2 = 2 f 0 U (s) X 2 = 2 f 0 U (s) + kx 1 kx 1 + ms 2 + k ms 2 + k Combining (3) and (4): ms 2 ( + cs + 2k) X 1 k 2 f 0 U (s) + kx 1 ms 2 + k = f 0 U (s) ( ms 2 + cs + 2k) ( ms 2 + k) X 1 k ( 2 f 0 U (s) + kx 1 ) = f 0 ( ms 2 + k)u (s) ( m 2 s 4 + cms 3 + 3mks 2 + cks + k 2 ) X 1 = f 0 ( ms 2 + 3k )U (s) Taking the inverse LT of the above gives: m 2!!!! x 1 + cm!!! x 1 + 3mk!!x 1 + ck!x 1 + k 2 x 1 = mf 0 u(t)!! + 3kf 0 u(t) (4)

PROBLEM 3 The poles for the second-order system shown below are shown in the above plot of the complex plane: m!!y + c!y + ky = f 0 u(t) where m = 10 and f 0 = 20. This system has a unit step input of u(t) = h(t) and is given zero initial conditions ( y(0) =!y(0) = 0 ). PART A i) Write down the response y(t) due to this input. ii) Determine the values for the percent overshoot (%OS) and 2% settling time ( t s ). iii) What are the values of c and k for this system? PART B i) Suppose that the original system is changed by reducing the %OS by 50% while keeping the settling time fixed. Write down the resulting response. ii) Suppose that the original system is changed by reducing the 2% settling time to a minimum while keeping the undamped natural frequency ω n unchanged. What are the numerical values for the two poles of the system as a result of this change?

SOLUTION Divide EOM by m :!!y + c m!y + k m y = f 0 m u(t)!!y + 2ζω n!y + ω 2 n y = Kω 2 n u(t) The transfer function is written down as: 2 Kω G(s) = n s 2 2 + 2ζω n s + ω = N(s) D(s) n From figure, we know that the characteristic equation for the system is: 0 = D(s) = ( s p 1 )( s p 1 ) = s 2 ( p 1 + p 1 )s + p 1 p 1 ( ) + ( 8 6 j) = s 2 8 + 6 j ( )( 8 6 j) s + 8 + 6 j = s 2 +16s +100 where: p 1 = σ + ω d j = 8 + 6 j. Comparing coefficients in the two expressions for D(s): ω n 2 = 100 ω n = 10 2ζω n = 16 ζω n = 8 Kω 2 n = f 0 m K = f 0 2 mω = 20 n 10 ( )( 100) = 0.02 PART A From lecture book (page V.10), the response of an underdamped second-order system to a step input is given by: y(t) = K 1 e σt cosω d t + σ sinω ω d t = 0.02 1 e 8t cos6t + 4 d 3 sin6t Also: and: PART B %OS = 100e πζ / 1 ζ 2 = 100e t s = 4 = 4 ζω n 8 = 0.5 k m = ω 2 n π( 0.8)/ 1 0.8 k = mω 2 n = ( 10) ( 100) = 1000 2ζω n = c m c = 2ζ mω n = 2 ( )2 = 1.51% ( )( 0.8) ( 10) ( 10) = 160 %OS = 100e πζ / 1 ζ 2 ln 2 %OS 100 = ζ 2 π 2 1 ζ 2 ζ = After the change in i), %OS = 1.51/ 2 = 0.76% ζ = ( ) ( ) = 0.841 ln 0.0076 π 2 + ln 2 0.0076 ( ) ( ) ln %OS / 100 π 2 + ln 2 %OS / 100

If t s is held constant ζω n = constant = 8 = σ ω n = 8 / ζ = 8 / 0.841= 9.51 ω d = ω n 1 ζ 2 = 9.51 1 0.841 2 = 5.14. Therefore: y(t) = K 1 e σt cosω d t + σ sinω ω d t = 0.02 1 e 8t ( cos5.14t +1.56sin5.14t) d For the change in ii): minimizing settling time while keeping ω n a constant moves the poles around a circle of radius ω n = 10 to where the circle intersects the real axis (which gives repeated roots corresponding to ζ = 1). Therefore, p 1 = p 2 = ζω n = 1 ( )( 10) = 10.

PROBLEM 4 Consider the following transfer function: G( s) = Y ( s) 0.6s = ( 200s + 2000 ) U ( s) 1.2s 2 +1440s +12000 where u t ( ) is the input and y( t) is the output. This system is given an input of u(t) = sinωt. On the next page, construct a straight-line approximation (asymptotes) for the db amplitude of the steady-state response y ss (t) as a function of ω. You are asked to provide details below (such as break frequencies, zero-intercepts, slopes in db/decade) related to the construction of your db amplitude plot. Without these details, you cannot receive full credit for your plot. Also, please provide numerical LABELS for your axes. SOLUTION G( s) = 0.6( 2000)s( s /10 +1) ( 12000) s 2 /100 2 + 3s / 25 +1 ( ) = 0.1 For constructing Bode plots: N 1 = 0.1 ( N 1 ) db = 20log( 0.1) = 20 = constant N 2 = s + 20dB / dec with intercept at ω = 1 ( )s( s /10 +1) D 1 s 2 /100 2 + 3s / 25 +1 = N 1 N 2 N 3 N 3 = s / 10 +1 0dB at low freqs, + 20dB / dec at high freqs and ω b = 10 ( ) 0dBat low freqs, 40dB / dec at high freqs 1/ D 1 = 1/ s 2 / 100 2 + 3s / 25+1 and ω b = 100