2.003SC Engineering Dynamics
Quiz 3 Solutions
Problem 1 Solution:
Modal Analysis Solution:
a) The natural frequencies of the system may be computed directly from the elements of the modal mass
and stifness matrices.
ω12 =
K1
⇒ω1 =
M1
190.4875
= 1.972 r/s
48.981
ω2 =
210.5125
= 4.269 r/s
11.5579
b) Again, directly from modal values:
ζ1 =
c)
q10
q20
9.5244
C1
=
= 0.0493
2(1.972)(48.981)
2ω1 M1
q̇10
q̇20
= 0,
˙ (0) =
= u−1 x
�
0.525
0.475
−0.0499
0.0499
1
0
d)
x
= uq =
�
1.0
−9.5125
(1)
q̇10 −ζ1 ω1d t
e
sin(ω1d t)
ω1d
where q̇10 = 0.525, ω1 = 1.972 r/s ≈ ω1d
q̇10
= 0.266
ω1d
1
=
0.525
0.475
Problem 2 Solution:
Vibration isolation:
fM = 3 Hz
ζ = 0.13
a) The total equivalent spring constant for springs in parallel is the sum of the individual ks:
Keq = 4k
b) For a linear system, steady state response:
frequency in=frequency out
fin = fout = 10Hz
c)
x
=
y
(1 + (2ζ ωωn )2 )1/2
(1 −
ω2 2
2 )
ωn
+ (2ζ ωωn )2
1/2
(1 + .8672 )1/2
2 2
2 1/2
[(1 − ( 10
3 ) ) + (0.867) ]
1.32
=
= 0.130
10.148
=
f
ω
10
=
=
fn
ωn
3
x
= 0.130
y
2
Problem 3 Solution:
r = 45 RPM
Izz,G = 100kg-m2
τ (t) = τ0 cos ωt, ω = 6π r/s
kt = 1600π 2 N-m/r
ct = 8π (N-m-s)/r
a).
τ/o = τo cos ωtk̂ − cT θ̇k̂ − kt θk̂ = Izz,G θ̈k̂
⇒ Izz,G θ¨ + cT θ̇ + kt θ = τo cos(ωt)
b).
ωn =
kt
Izz,G
=
1600π 2 N-m
= 4π r/s
100 kg-m2
ωd = ωn 1 − ζ 2
= .99995ωn
� ωn
ζ=
ct
8π
= 0.01
=
2 × 4π × 100
2Izz ωn
c). ωosc = ω = 6π r/s
d).
τ0
kT
= 0.2 radians
1/KT
|θ| = |τ0 | · |Hθ/τ | = τ0
(1 −
θ=
=
ω2 2
2 )
ωn
τ0
ω2
ω
(1 − 2 )2 + (2ζ )2
ωn
KT
ωn
0.2 rad
+ (2ζ ωωn )2
−1/2
[(1 − 1.52 )2 + (2(.01)(1.5))2 ]
.2
=
[1.56 + 0.0009]1/2
1/2
, which when evaluated at ω/ωn =
1/2
= 0.16 = θ
3
6π
= 1.5 yields:
4π
Problem 4 Solution:
a) The system has two degrees of freedom for the no-slip condition.
If has four degrees of freedom if slip is allowed.
b) Two generalized conditions are required.
I choose coordinates x1 and x2 , where
x1 = rθ1 , x2 = rθ2 .
c)
1
1
1
1
mẋ12 + mẋ22 + Ic1 θ̇12 + Ic2 θ̇22
2
2
2
2
for uniform disk, Ic = mr2 /2.
T =
1
1
1 m r2 ẋ21
1 m r2 ẋ22
mẋ21 + mẋ22 +
+
2 2 r2
2
2
2 2 r2
3
3
= mẋ21 + mẋ22
4
4
1 2 1 2 1
V = kx1 + kx2 + km (x2 − x1 )2
2
2
2
T =
d) Find EOMs:
From Lagrange
d
dt
�
�
∂T
∂q̇i
−
∂T
∂qi
+
∂V
∂qi
= Qi
�
∂T
3
∂T
=0
= mẍ1 ,
2
∂x1
∂ẋ1
∂V
= kx1 − km (x2 − x1 )
∂x1
⇒
d
dt
d
dt
3
mẍ1 + (k + km )x1 − km x2 = 0
2
�
∂T
3
∂T
=0
= mẍ2 ,
2
∂x2
∂ẋ2
∂V
= kx2 + km (x2 − x1 )
∂x2
⇒
3
mẍ2 − km x1 + (k + km )x2 = 0
2
4
MIT OpenCourseWare
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2.003SC / 1.053J Engineering Dynamics
Fall 2011
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