EX 1DOF Rot Gondola Swinging

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Transcript of EX 1DOF Rot Gondola Swinging
Moments of inertia about pivot O:
Ib_o MbL2
12⋅ Mb
L2
⎛⎜⎝
⎞⎟⎠
2⋅+:= ISG_o ISG M L2⋅+:= Ib_o
ISG_o0.014=
Total Mass moment of inertia about pivot O: Io 2 Ib_o⋅ ISG_o+:=
Equation of motion: From summation of moments about pivot:
assume:no frictionIo 2t
θd
d
2⋅ M− g⋅ L⋅ sin θ( )⋅ 2 Mb⋅ g⋅
L2⋅ sin θ( )⋅−=
Io 2tθd
d
2⋅ g L⋅ sin θ( )⋅ M Mb+( )⋅+ 0= [2]
2tθd
d
2 g− L⋅ M Mb+( )⋅
Iosin θ( )⋅=
sin θ( ) θ≡For small amplitude motions, Eqn(2) reduces to:
Io 2tθd
d
2⋅ g L⋅ M Mb+( )⋅ θ⋅+ 0= [3]
SWINGING GONDOLA  FUN RIDE! MEEN 363  SP09 (LSA)
The figure shows a SWINGING GONDOLA (SG) at an amusement park. The gondola’s mass and centroidal radius of gyration equal M and rk.. The total mass of the passengers does not affect significantly the magnitudes given above or CG location of SG. Two bars of length L and mass Mb, Ib=(1/12) MbL2 , attach the SG CG to a shaft connected to a drive motor. The shaft is supported on frictionless bearings. The motor applies torque Tm turning the bars and lifting the gondola to a release angle θo. At this time, the motor is disengaged and the gondola starts to swing, giving a good thrill to its riders. Determine: a) System EOM after the motor is disengaged. Express the angular acceleration θ in terms of the system
parameters & θ. [10] b) Given the (large) initial angle θo , derive an analytical expression (symbolic) for the SG angular velocity θ in
terms of the system parameters, and angles θ and θo. [10] c) Given, M=4000 kg, rk =4 m, L=10 m, Mb =0.05 M, and θo = 125°, calculate, when gondola crosses θ=0°, its
angular speed [Hz], and SG CG tangential speed and radial acceleration What is the type of “thrill” or sensation the riders feel when the SG passes through θ=0°? [2.5 x 4]
θ
O g
L
θo
Data:
θo_ 125:= degree release angle
M 4000 kg⋅:= L 10 m⋅:= rk 4 m⋅:=
Mb 0.05 M⋅:=
From Tables: (mass moment of inertia)
Ib_cg MbL2
12⋅= ISG M rk
2⋅:=
bar Gondola inertia about its cg
ωLin θ( ) ωn θo2
θ2
−⎛⎝
⎞⎠
0.5⋅:=
Thus, the linearized solution is
ω2
ωn2
θo2
θ2
−=and add both equations to give
a ωn t⋅=θ
2
θo2
cos a( )2=from these 2 equations, square both sides ω
2
θo2ωn
2⋅
sin a( )2=
ω t( ) θo− ωn⋅ sin ωn t⋅( )⋅:=andθ t( ) θo cos ωn t⋅( )⋅=
starting from rest. Thenωo 0=since
natural frequency, for small amplitude motions: Disregard bar masses
ωnagL
1
1rkL
⎛⎜⎝
⎞⎟⎠
2
+
⋅⎡⎢⎢⎢⎣
⎤⎥⎥⎥⎦
.5:=
ωng L⋅ M Mb+( )⋅
Io
⎡⎢⎣
⎤⎥⎦
0.5
:=
Natural periodTn
2 π⋅
ωn:=
Tna 2 π⋅Lg
1rkL
⎛⎜⎝
⎞⎟⎠
2
+⎡⎢⎢⎣
⎤⎥⎥⎦
⋅⎡⎢⎢⎣
⎤⎥⎥⎦
0.5
⋅:=
Tn 6.764s= Tna 6.834s=
Not much difference if bars are omitted from analysis
SOLUTION for SMALL ANGLES θ  LINEARIZED EOM:
θ t( ) θo cos ωn t⋅( )⋅ ωo sin ωn t⋅( )⋅+=The solution fo the linear EOM (3) is
where θo θo_π
180⋅:= ωo t
θdd
= are initial conditions
an 0( )
g2.769−= rider feels something in
his/her tummy!(a weighless sensation)
an 0( ) 27.156−m
s2=
an θ( ) L− ω θ( )2⋅:=
ω 0( )ωn
1.774=Vt 0( ) 16.479ms
=Vt θ( ) L ω θ( )⋅:=
f 0( ) 0.262Hz=Tangential speed and normal acceleration at θ=0 fn 0.148Hz=
in Hzf θ( ) ω θ( )2 π⋅
:=ω θ( ) ωn 2 cos θ( ) cos θo( )−( )⋅⎡⎣ ⎤⎦0.5
⋅:= fnωn2 π⋅
:=tθd
d=
Then, the gondola angular speed dθ/dt=ω:ωn
g L⋅ M Mb+( )⋅
Io
⎡⎢⎣
⎤⎥⎦
0.5
:=
where
12 t
θdd⎛⎜⎝
⎞⎟⎠
2⋅
M Mb+( ) g⋅ L⋅ cos θ( ) cos θo( )−( )⋅
Io= ωn
2 cos θ( ) cos θo( )−( )⋅=obtain:
Vo M Mb+( ) g⋅ L⋅ 1 cos θo( )−( )⋅=
V M Mb+( ) g⋅ L⋅ 1 cos θ( )−( )⋅=Potential EnergyT12
Io⋅ tθd
d⎛⎜⎝
⎞⎟⎠
2⋅=Kinetic energy
From Conservation of Energy, T+V = Vo; since gondola is released from initial angle θo with null speed
Find angular speed of gondola for large release angles
Graphs: variation of gondola angular speed and radial acceleration for entire ride:
125 75 25 25 75 1250
0.1
0.2
0.3
0.4
Non linearLinear
angle (deg)
Ang
ular
spee
d (H
z)θo_ 125= degrees
Angular speed of gondola derived from NONLINEAR solution is smaller than using LINEAR solution.
The differences increase as the release angle increases
Greatest diference when θο= 180 deg
125 75 25 25 75 1250
1
2
3
angle (deg)
Nor
mal
acc
eler
atio
n/g