Name:
______________SOLUTION_________________
Section Time (please circle):
9:30
3:30
ME 375
EXAM I
Wednesday, 25 September 2013
6:30 pm-7:30 pm
CL 50
Problem 1 ________________/35
Problem 2_________________/35
Problem 3_________________/30
Total Score:_____________/100
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Don’t forget your name and your section time.
Exam is 7 pages including this cover page: make sure you’re not missing any pages.
If you use extra pages, indicate on the problem page that you’re continuing onto an
extra page.
Pay attention to units if they are given.
Show your work. Correct answer with wrong explanation gets no credit.
1
Problem 1.
a) Model the pendulum connected to a motor. Consider the mass at the end a “point mass” (i.e. ignore the
mass’ dimensions). The motor applies a torque. There is also rotational damping in the system. The rod
itself is massless and has length L. Gravity is acting downwards. Assume that system moves by very small
amounts about its equilibrium point θ=0. Put your answer in the form:
A1θ + A2θ + A3θ =
τ
where the A’s are constants that are in terms of the system parameters.
Summing the moments about the pivot point gives:
−mgL sin θ − bθ + τ =
mL2θ
For small angles, we have:
mL2θ + bθ + mgLθ =
τ
2
Problem 1 continued.
b) Now, the system on the previous page is modified. There are now a linear damper and a linear spring
acting on the point mass, and the rod itself now has an inertia “I” about the pivot point (center of the
motor). Small angle approximation is still valid. Also assume that spring is uncompressed/unstretched when
θ=0. Modify the original differential equation with the additional terms (fill in the blanks):
τ
( A1 + ____)θ + ( A2 + ___)θ + ( A3 + ___)θ =
Hint: Be careful with units. For example: the damper constants ‘b’ and ‘c’ don’t have the same units.
If you sum the moments about the pivot point, you’ll get
−mgL sin θ −
3g
I sin θ − kL sin θ L cos θ − cL sin θL cos θ − bθ +=
τ (mL2 + I )θ
2L
The second term is due to the moment caused by the weight of the rod pulling it down at L/2. Mass of the
rod in terms of I is 3I/L2.
2
2
For small angles, we have: (mL + I )θ + (b + cL )θ + (mgL +
3g
I + kL2 )θ =
τ
2L
It is also possible to answer this question by noticing how the new elements affect the system. The I will
add to the existing mL2 term of A1, since it’s already defined about the pivot point. The linear damper
applies a force, not directly a torque (you can’t just add c to A2). So you need to calculate the torque of the
damper force which is cLdθ/dt*L, so the additional term to A2 will be cL2. Similarly, to account for the linear
spring, you’ll need to add kL2 to the A3 term. The mass of the rod applies a moment at the center of the rod
which is equal to: -mrodgθL/2 which is (3/2L)gIθ, so you’ll also need to add (3/2L)gI to the A3 term.
3g
( A1 + ___ I __)θ + ( A2 + __ cL2 __)θ + ( A3 +
I + kL2 )θ =
τ
2L
3
Problem 2.
You are given the following 2nd order system:
4 
y (t ) + 2 y (t ) + y (t ) =
4u (t )
a) First, obtain an expression for Y(s) in terms of the initial conditions, and U(s). Organize your final answer
so that it is clear what part is free response and what part is forced. Indicate the free and forced parts.
Taking the Laplace transform of the ODE gives:
4(Y ( s ) s 2 − sy (0) − y (0)) + 2(Y ( s ) s − y (0)) + Y ( s ) =
4U ( s )
Rearranging gives:
=
Y (s)
4
(4 s + 2) y (0) + 4 y (0)
U (s) +
4s + 2s + 1
4s 2 + 2s + 1
2
Forced
Free
4
b) Now, for zero initial conditions, and a unit ramp input (u=t), determine the total response y(t) using the
Laplace/Inverse Laplace method.








4
1
1
1
1
1


=
Y ( s ) =



 2 =
2
2
2
1
1
 4s + 2s + 1  s
 (s + 1 )2 + 3  s
 s2 + s +  s
2
4
4
16 


3
16
1
D( s + )
A B
4
= + 2+
+
s s (s + 1 )2 + 3 (s + 1 )2 + 3
4
16
4
16
C
Equalizing denominators and comparing the numerators gives:
( A + D) s 3 + (
A
3 D 2 A B
B
+ B+C
+ )s + ( + )s + =
1
2
16 4
4 2
4
Which gives: A=-8, B=4, C=-8/31/2 and D=8, which yields:
y (t ) =−8 + 4t −
t
−
8 − 4t
3
3
e sin
t + 8e 4 cos
t
4
4
3
5
Problem 3.
Comment on the stability of the systems in parts a, b and c.
a) 
y (t ) + y (t ) + y (t ) =
u (t )
poles are −0.5 ± 0.87 j so the system is stable.
Alternatively, you could say: this is a 2nd order system and all coefficients have the same sign, so it’s stable.
b) y (t ) − y (t ) =
u (t )
1st order system with varying signs: unstable.
Pole of the system is: +1
t + t (where t + t 4 is the input to the system)
c) y (t ) + y (t ) =
4
Pole of this system is -1 so it’s stable. The right hand side of the equation constitutes the input, so it doesn’t
determine stability.
You can also say: First order system with same signs: stable.
d) If a system is not excited by any input (i.e. u(t)=0), that automatically means that the system is stable.
True or False?
False. Stability doesn’t have anything to do with the input.
e) If a system is not excited by any input (i.e. u(t)=0), that necessarily means that the total response will be
equal to zero.
True or False?
False….. Not necessarily. If there are non-zero initial conditions, there will be a free response. The free
response may eventually decay to zero (if system is stable) but that doesn’t mean, y(t)=0.
f) Stability of a system depends only on the system’s poles.
True or False?
Very true.
6
Laplace Transform Table
f(t)
F(s)
1.
δ(t)
2.
1
3.
Comment
1
Unit impulse
1
s
Unit step
t for t>0
1
s2
Unit ramp
4.
e-at
1
s+a
Exponential
5.
sin ωt
ω
s +ω2
Sine
6.
cos ωt
s
s +ω2
Cosine
7.
f(t)
8.
df ( t )
dt
for t>0
2
F(s)
d n f (t )
9.
dt n
10.
2
sF(s) − f (0 − )
s F(s) − s
n
− at
n −1
−
f (0 ) − s
First Derivative
n −2
df (0 − )

dt
nth Derivative
n!
tn
11. e
Function
s
f (t )
n +1
F(s + a)
1
(s + a )2
12. te-at
13. tne-at
n!
(s + a ) n +1
14. e-at sin ωt
ω
(s + a )2 + ω 2
15. e-at cos ωt
16. Initial Value Theorem:
17. Final Value Theorem:
(s + a )
(s + a )2 + ω 2
f (0 + ) = lim sF(s)
s→∞
lim f ( t ) = lim sF(s)
t →∞
s→ 0
7