Post on 27-Jun-2015
description
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Workshop G-W-0: The slide - Self-study
Problem analysis - Negelect the air drag
Modeling
restart :
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1. Inertia force diff m s t , t$2 ;
m d2
dt2 s t
2. Gravity force (weight)m$g;
m g
3. Boundary force: Reaction (normal force)
m$g$cos theta ;
m g cos θ Friction (reaction × friction coefficient)
f $m$g$cos theta ;
f m g cos θ
In the direction along the slope
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Sum F = 0;
> F = 0
inertia force + gravity component (in the direction of slope) + friction = 0
we define the displacement as a function s of time t
equ dKm$diff s t , t$2 Cm$g $sin theta K f$m$g$cos theta = 0
equ := Km d2
dt2 s t Cm g sin θ K f m g cos θ = 0
Question 1: Calculation of the speed Initial conditions, both speed and displacement are zero
icsd s 0 = 0, D s 0 = 0 :
Solve it analyticallysold dsolve equ, ics , s t ;
sol := s t =12
sin θ gKcos θ g f t2
Assign values
md 20.0 : gd 9.8 : theta d30$3.1415926
180.0: fd 0.2 :
Get the function out of solutionsd unapply rhs sol , t ;
s := t/1.601295062 t2
Plot the function out with the time range from 0 to 2plot s t , diff s t , t , diff s t , t$2 , t = 0 ..2, color = blue, red, green , style = point,
line, point
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t0 0.5 1 1.5 2
0
1
2
3
4
5
6
Find a solutionActualTime d solve s t = 5 , t
ActualTime := 1.767051960, K1.767051960
Actual timeActualTime 1
1.767051960
Actual Speed at the bottomevalf diff s t , t t= ActualTime 1
5.659143156
Question 2: Get the leaning angle of the slide (Trial & error method)Method - Trial and error
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-- Use the result in Question 1, given different leaning angle until the speed is lower than 3.5 meter/s
restart :
The process starts in the same wayequ dKm$diff s t , t$2 Cm$g $sin theta K f$m$g$cos theta = 0
equ := Km d2
dt2 s t Cm g sin θ K f m g cos θ = 0
icsd s 0 = 0, D s 0 = 0 :
sold dsolve equ, ics , s t ;
sol := s t =12
sin θ gKcos θ g f t2
Give the same parameters and propose a value for θmd 20.0 : gd 9.8 : fd 0.2 :
theta d18$3.1415926
180.0;
θ := 0.3141592600
sd unapply rhs sol , t
s := t/0.5821478600 t2
Analyse the resultsplot s t , diff s t , t , diff s t , t$2 , t = 0 ..3, color = blue, red, green , style = point,
line, point
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t0 1 2 3
0
1
2
3
4
5
ActualTime d solve s t = 5 , tActualTime := 2.930679656, K2.930679656
ActualTime 12.930679656
ActuallSpeedd evalf diff s t , t t= ActualTime 1
ActuallSpeed := 3.412177780
The proposed value for θ was right, because the speed is less than 3.5
Question 2: Get the leaning angle of the slide (find the exact solution)Method - use the equation of speed
-- Equate the derivative to the value that you need
Same as above, but here we introduce the lean angle as a parameter
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restart;
equ dKm$diff s t , t$2 Cm$g $sin theta K f$m$g$cos theta = 0
equ := Km d2
dt2 s t Cm g sin θ K f m g cos θ = 0
icsd s 0 = 0, D s 0 = 0 :
sold dsolve equ, ics , s t ;
sol := s t =12
sin θ gKcos θ g f t2
md 20.0 : gd 9.8 : fd 0.2 : Please note that the lean angle theta is a parameter in the above solution
Differentiate to get the equation of speedspeedd diff sol, t ;
speed :=ddt
s t = 9.8 sin θ K1.96 cos θ t
Use the given parameters (displacement and speed) for the equationseq1 and eq2 set a system of equations with 2 unknow parameters (t and theta). That is, we have 2 equations and 2 unkown parameters.
eq1d rhs sol = 5;eq2d rhs speed = 3.5;
eq1 :=12
9.8 sin θ K1.96 cos θ t2 = 5
eq2 := 9.8 sin θ K1.96 cos θ t = 3.5
Solve the equations as usual (there are two solutions in this case)solve eq1, eq2 , theta, t ;
θ = 0.3202771606, t = 2.857142857 , θ =K3.067078695, t = 2.857142857
Use the first solution (positive value of angle)resd op 1, solve eq1, eq2 , theta, t ;
res := θ = 0.3202771606, t = 2.857142857
op 1, res ;op 2, res ;
θ = 0.3202771606
t = 2.857142857
The result for θ is given in radians, thus we change it to degrees in the following way
ActualAngled rhs op 1, res $180
3.1415926;
ActualTimed rhs op 2, res ; ActualAngle := 18.35052989
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ActualTime := 2.857142857
Verify the speed, Use those values in the equation of speedeval diff sol, t , res ;
ddt
s t
t = 2.857142857
= 3.499999997
The result is correct, because the speed is less or equal to 3.5. Note that a slight decrease in the angle will lower speed as well. It is actually desirable to round it to 18° for manufacturing purposes.
Advanced: Consider the air dragConsider the air drag:
NAME SYMBOL VALUE
Drag force Fdrag Unknown
Density of air ρ 1.2 kg
m3
Drag coefficient fdrag 0.5
Front area A 0.09 m2
Speed ddt
s tUnknown
restart;
Fdrag = 12$'density of air '$ 'Speed' 2$'Area ';
Fdrag =12
rho$diff s t , t 2$A;
Fdrag =12
density of air Speed2 Area
Fdrag =12
ρ ddt
s t2
A
What is the speed for a leaning angle of 30°?
Solution considering the air drag
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equ dKm$diff s t , t$2 Cm$g $sin theta K fground$m$g$cos theta K12$rho$diff s t , t 2
$fdrag$A = 0
equ := Km d2
dt2 s t Cm g sin θ K fground m g cos θ K
12
ρ ddt
s t2
fdrag A = 0
Initial conditions, both speed and displacement are zeroicsd s 0 = 0, D s 0 = 0 :
Solve it analyticallysold dsolve equ, ics , s t ;
sol := s t =K1
m3/2 ρ fdrag A2 g sin θ Kfground cos θ ρ fdrag A t m2
K2 m5/2 ln12
e
2 g sin θ K fground
cos θ ρ fdrag
A t
m C12
Assign values
md 20.0 : gd 9.8 : theta d30$3.1415926
180.0: fgroundd 0.2 : rho d 1.2 : A d 0.09 : fdrag
d 0.5 :
Get the function out of solutionsd unapply rhs sol , t
s := t/K34.44044847 2 tC740.7407409 ln12
e0.09298921083 2 tC12
Plot the function out with the time range from 0 to 2
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> > plot s t , diff s t , t , diff s t , t$2 , t = 0 ..2, color = blue, red, green , style = point,line, point
t0 0.5 1 1.5 2
0
1
2
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5
6
Find the solutionActualTime d fsolve s t K5, t
ActualTime := 1.769040566
ActualSpeedd evalf diff s t , t t= ActualTime
ActualSpeed := 5.64009716