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Workshop G-W-0: The slide - Self-study

Problem analysis - Negelect the air drag

Modeling

restart :

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(3.4)(3.4)

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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

3

4

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