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Moments of inertia about pivot O: I b_o M b L 2 12 M b L 2 2 + := I SG_o I SG ML 2 + := I b_o I SG_o 0.014 = Total Mass moment of inertia about pivot O: I o 2I b_o I SG_o + := Equation of motion: From summation of moments about pivot: assume: no friction I o 2 t θ d d 2 M g L sin θ () 2M b g L 2 sin θ () = I o 2 t θ d d 2 gL sin θ () M M b + ( ) + 0 = [2] 2 t θ d d 2 g L M M b + ( ) I o sin θ () = sin θ () θ For small amplitude motions , Eqn(2) reduces to: I o 2 t θ d d 2 gL M M b + ( ) 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 r k .. 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 M b , I b =(1/12) M b L 2 , attach the SG CG to a shaft connected to a drive motor. The shaft is supported on frictionless bearings. The motor applies torque T m turning the bars and lifting the gondola to a release angle θ o . 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 θ o , derive an analytical expression (symbolic) for the SG angular velocity θ in terms of the system parameters, and angles θ and θ o . [10] c) Given, M=4000 kg, r k =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 θ o Data: θ o_ 125 := degree release angle M 4000 kg := L 10 m := r k 4m := M b 0.05 M := From Tables: (mass moment of inertia) I b_cg M b L 2 12 = I SG Mr k 2 := bar Gondola inertia about its cg

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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