ME36500 Homework #5 Due: 10/02/2014 Problem#1( Homework #5 Due: 10/02/2014 2/3...
Transcript of ME36500 Homework #5 Due: 10/02/2014 Problem#1( Homework #5 Due: 10/02/2014 2/3...
ME36500 Homework #5 Due: 10/02/2014
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Problem #1 (30%) A disk type flywheel with moment of inertia J of mass 10 kg and radius of 0.5 m is driven by a gasoline engine which generates a constant torque of T=10 Nm. The bearings in the shaft may be modeled as a viscous rotary dampers with damping coefficient of B=0.1 N-‐m-‐s/rad. If the flywheel is at rest at t=0 sec and the power is suddenly applied to the engine, the 1st order ODE can be expressed as
𝐽𝐵𝑑𝜔𝑑𝑡 + 𝜔 =
1𝐵 𝑇(𝑡)
[Hint: moment of inertia J is= mR2/2] (A) Calculate the time constant
τ=12.5 sec (B) Calculate the total solution (homogeneous and particular solution)
homogeneous solution: 𝜔! = 𝐶𝑒!!
!".!!; particular solution: 𝜔! = 100 rad/s
total solution: 𝜔 = 𝜔! + 𝜔! = 𝐶𝑒!!
!".!! + 100
plug in t=0, 𝜔 = 0; C=-‐100. Thus 𝜔 = −100𝑒!!
!".!! + 100 (C) What is the steady state angular velocity?
𝜔! = 100 rad/s (D) Plot the step response for m=10 kg (given) and also for m=20 kg. What is the
difference?
Change in m values cause change in time constants, which leads to different response rates.
(E) Plot the step response for m=10 kg, radius =0.5 m (given) and compare the response when your radius increases to 0.8 m. What is the difference?
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Time [S]
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ME36500 Homework #5 Due: 10/02/2014
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Change in radius values cause change in time constants, which leads to different response rates.
Problem #2 (40%) (A) Reduce the following complex numbers to the form 𝐴 + 𝑗𝐵:
a. 1+ 𝑗𝜔 !=1− 3𝜔! + (3𝜔 − 𝜔!)j
b. (!!!")(!!!")!! ! =− !
!"− !!
!"
c. !!!!!!!!
=− !!"+ !!
!"𝑗
(B) Convert the following exponentials to the form 𝐴 cos 𝜔𝑡 + 𝑗 sin 𝜔𝑡 :
a. 7𝑒!!.!!=7[cos(0.4t)+jsin(0.4t)]
b. 2 𝑗 𝑒!!!! + 𝑒!!! =2 2 cos(6𝑡)𝑗
(C) Find the magnitude and phase (in degrees) of the following:
a. −4+ 𝑗2 mag: 2 5; phase: 153.43 deg
b. 6.21𝑒 !!! !! ! mag:339; phase: !! 𝑡
c. !!!(!")!!! !" !!" !" ! mag:
!"!!"!!
(!!!"!!)!!!"!!; phase: −atan !!
!−atan( !!
!!!"!!)
(D) Let 𝑧! = 5+ 𝑗8 and 𝑧! = −2− 𝑗7.
a. Solve 𝑧!×𝑧! by expanding terms and reducing your answer to the form A+jB.
𝑧!×𝑧!=(5+8j) ×(-‐2-‐7j)=46-‐51j
b. Compute 𝑧!×𝑧! by converting 𝑧! and 𝑧! to the form 𝐴𝑒!" , then solving for an answer in the same form.
z1= 25+ 64𝑒!.!"!; z2= 4+ 49𝑒 !.!"±! !; 𝑧!×𝑧!=68.68𝑒(!.!±!)!=46-‐51j
c. Convert your answer from part (ii) into the form 𝐴 cos𝜙 + 𝑗 sin𝜙 .
68.68[cos(2.3±𝜋)+jsin(2.3±𝜋)] d. Draw a diagram on the complex plane that shows why your answers for parts
(a) and (c) are different representations of the same value.
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Problem #3 (30%) An accelerometer’s behavior can be modeled by a second order differential equation with a natural frequency (fn) of 120 Hz, damping ratio (ζ) of 0.1, and a static sensitivity of 5 mV/g. The accelerometer is used in a feedback system on a shaker and is measuring the shaker’s acceleration. The shaker table is initially at rest and then a sinusoidal excitation is turned on with an acceleration level of 4 m/sec2 peak-‐to-‐peak and frequency 12 Hz. You may assume there are no bias voltages. (A) Write down the differential equation that relates output voltage, v(t), to input
accelerations a(t). The coefficient of the differential equation should have numerical values. Please make sure the units are consistent.
1𝜔!!
𝑦 +2𝜍𝜔!
𝑦 + 𝑦 = 𝐾𝑥
plug in parameters: 𝜔! = 2𝜋𝑓 = 754 𝑟𝑎𝑑/𝑠; K=5.1×10-‐4 V/m*s2 1.76×10!!𝑦 + 2.65×10!!𝑦 + 𝑦 = 5.1×10!!𝑥
(B) Calculation the steady-‐state response of the accelerometer to the 12 Hz input excitation described above using both of the following approaches: Approach 1: Assume the steady state response is vSS (t) = A ⋅sin(ωt)+ B ⋅ cos(ωt) where ω = 2π f and f is the frequency of the excitation. Substitute the above representation into the differential equation and solve for A and B. 1 − !!
!!! A − !"#
!!B = K (1)
!"#!!
A + 1 − !!
!!! B = 0 (2)
A=0.00101 B=-‐0.00002 thus, Vss = 0.00101sin(𝜔𝑡) -‐0.00002cos(𝜔𝑡) = 0.00103sin(ωt – 0.02) [V] Approach 2: Calculate the frequency response function (T) by assuming the acceleration is a(t)= A ⋅ e jωt and the output voltage is v(t)=V ⋅ e jωt . Substitute these representations into the differential equation and then determine the steady-‐state response. Your answer should be the same as in Approach 1. Make sure the steps to get this result are clearly provided in your solution.
plug in a(t) and v(t), we could get − !!
!!!∙ 𝑣𝑒!"# + !!
!!v jω 𝑒!"# + v𝑒!"# = KAe!!!
by rearrangement: !"!"#
!!!"#= !
!!!!
!!!! !"!!
!!=G(j𝜔)
𝐺(𝑗𝜔) = !
(!!!!
!!!)!!( !"!!!)
!
=5.15*10-‐4;
∠ 𝐺(𝑗𝜔) = −atan ( !"!!ω/(1 − !!
!!!))=-‐0.02 rad
Vss = 0.2*5.15*10-‐4*sin(ωt – 0.02) = 0.000103* sin(ωt – 0.02) [V] = 0.00101sin(𝜔𝑡) –0.00002cos(𝜔𝑡)