User:Egm6321.f10.team03.sigillo/Hwk1

=Homework One Summary= Problems assigned in the first two weeks of class.

= Problem 1 - Maglev First and Second Derivative Derivations =

Reading convetion
A function $$\,f^1$$ or $$\,f^1(.)$$ has to be read as: "$$f$$ superscritp 1". A function $$\,(f^1)^2$$ or $$\,(f^1(.))^2$$has to be read as: "$$f$$ superscritp 1, power 2".

Given
Based on the definitions and equations in Maglev Train EOM the axial deformation of the guidance for a High-Speed Train is represented by a function $$\,u^1(s,t)$$. The argument $$\,s$$ itself is also a function of the timevariable $$\,t$$, so that for a general function $$\,f$$ the following valid is:
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$$\,{{\left. f(s,t) \right|}_{s={{y}^{1}}(t)}}=f({{y}^{1}}(t),t)$$
 * $$\,(Eq.1.1)$$
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Find
The first and second derivation of $$\,f({{y}^{1}}(t),t)$$ has to be calculated with respect to $$\,t$$.

Solution
First derivation $$\frac{df}{dt}$$: Let's set
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$$\,z:=f({{y}^{1}}(t),t)$$ $$ and
 * $$\,(Eq. 1.2)
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$$\,s:={{y}^{1}}(t)$$ $$ thus we get for $$z$$ :
 * $$\,(Eq. 1.3)
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$$\,z=f(s,t)$$ $$ The absolute derivation of $$z$$ is:
 * $$\,(Eq. 1.4)
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$$\,dz={{f}_{s}}\ ds+{{f}_{t}}\ dt$$ $$ Following the suggestion in Papula, we devide $$(Eq.1.5)$$ by $$\,dt$$ and get:
 * $$\,(Eq. 1.5)
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$$\frac{dz}{dt}={{f}_{s}}\ \frac{ds}{dt}+{{f}_{t}}$$ $$ substituting $$z$$ gives:
 * $$\,(Eq. 1.6)
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$$\frac{df(s,t)}{dt}=\frac{\partial f(s,t)}{\partial s}\frac{\partial s}{\partial t}+\frac{\partial f(s,t)}{\partial t}$$ $$ substituting $$s$$ this time, gives:
 * $$\,(Eq. 1.7)
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$$\frac{df({{y}^{1}}(t),t)}{dt}=\frac{\partial f({{y}^{1}}(t),t)}{\partial s}\frac{\partial {{y}^{1}}(t)}{\partial t}+\frac{\partial f({{y}^{1}}(t),t)}{\partial t}$$ $$ So that we can finally say:
 * $$\,(Eq. 1.8)
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$$\frac{df}{dt}={{f}_{s}}\cdot \dot{y}^{1}+{{f}_{t}}$$ $$
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 * $$\,(Eq. 1.9)
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'''Second derivation $$\frac{{{d}^{2}}f}{d{{t}^{2}}}$$''': The procedure to calculate the second derivative of $$f$$ is the same used to calculate the first derivative. For a better overview let's assigne $$(Eq. 1.9)$$ to a new variable and also parts of the equation to new variables like follows:
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$$\,q:= {{f}_{s}}\cdot \dot{y}^{1}+{{f}_{t}}$$ $$\,r:= {{f}_{s}}$$ $$\,m:= \dot{y}$$ $$\,h:= {{f}_{t}}$$ Equation $$(Eq. 1.9)$$ can then be written as:
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$$\,q=r\cdot m+h$$ $$ Similarly to the calculation of the first derivative, we calculate the total derivative of $$q$$, paying attention to the fact that now we have to derive the prduct $$(r\cdot m)$$ with $$r$$ being a function of $$s$$ and $$t$$ :
 * $$\,(Eq. 1.10)
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$$dq=({{r}_{s}}\cdot ds+{{r}_{s}}\cdot dt)\cdot m+r\cdot {{m}_{t}}\cdot dt+{{h}_{t}}\cdot dt+{{h}_{s}}\cdot ds$$ $$ multiplying $$m$$ into the braket expression gives:
 * $$\,(Eq. 1.11)
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$$dq={{r}_{s}}\cdot ds\cdot m+{{r}_{s}}\cdot dt\cdot m+r\cdot {{m}_{t}}\cdot dt+{{h}_{t}}\cdot dt+{{h}_{s}}\cdot ds$$ $$ Dividing also here $$dq$$ by $$dt$$ and substituiting $$r, m, h$$ with their original terms we get:
 * $$\,(Eq. 1.12)
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$$\frac{dq}{dt}=\frac{\partial {{f}_{s}}}{\partial s}\ \frac{ds}{dt}\ {{{\dot{y}}}^{1}}+\frac{\partial {{f}_{s}}}{\partial t}\ {{{\dot{y}}}^{1}}+{{f}_{s}}\frac{\partial {{{\dot{y}}}^{1}}}{\partial t}+\frac{\partial {{f}_{t}}}{\partial t}+\frac{\partial {{f}_{t}}}{\partial s}\ \frac{ds}{dt}$$ $$ That in fact after deriving becomes:
 * $$\,(Eq. 1.13)
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$$\frac{dq}{dt}={{f}_{ss}}\cdot {{({{{\dot{y}}}^{1}})}^{2}}+{{f}_{st}}\cdot {{{\dot{y}}}^{1}}+{{f}_{s}}\cdot {{{\ddot{y}}}^{1}}+{{f}_{tt}}+{{f}_{ts}}\cdot {{{\dot{y}}}^{1}}$$ $$ Since the derivation sequence doesn't matter for partial derivations, we finally after ordring the terms in $$(Eq. 1.14) $$ get for the second derivative:
 * $$\,(Eq. 1.14)
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$$\frac{{{d}^{2}}f}{d{{t}^{2}}}={{f}_{ss}}\cdot {{({{{\dot{y}}}^{1}})}^{2}}+2{{f}_{st}}\cdot {{{\dot{y}}}^{1}}+{{f}_{s}}\cdot {{{\ddot{y}}}^{1}}+{{f}_{tt}}$$ $$
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 * $$\,(Eq. 1.15)
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= Problem 2 - Dimensional Analysis=

Given
From the Maglev Train EOM we are given the horizontal force on the wheel/magnet $$\,{{c}_{0}}({{y}^{1}},t)$$ :
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$$\,{{c}_{0}}({{y}^{1}},t)=-{{F}^{1}}[1-\bar{R}\cdot u_{ss}^{2}({{y}^{1}},t)]-{{F}^{2}}\cdot u_{s}^{2}-\frac{T}{R}+M[[1-\bar{R}\cdot u_{ss}^{2}][u_{tt}^{1}-\bar{R}\cdot u_{stt}^{2}]+u_{s}^{2}\cdot u_{tt}^{2}]$$ $$ with :
 * $$\,(Eq. 2.1)
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$$\,[{{F}^{1}}]=F(force)$$ $$\,[T]=T(torque)$$ $$\,[M]=M(mass)$$ $$\,[\bar{R}]=L(lenght)$$ $$\,[t]=t(time)$$ $$\,[{{u}^{2}}]=L(length)$$ $$\,[u_{ss}^{2}]=\frac{L}={{L}^{-1}}(1/length)$$ $$\,[1]=1$$
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Find
Dimensional Analysis of the terms in $$(Eq. 2.1)$$ and physical meaning.

Solution
Analysis of the first term $$\,(-{{F}^{1}}[1-\bar{R}\cdot u_{ss}^{2}({{y}^{1}},t)])$$ :
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$$\,F[1-L\cdot {{L}^{-1}}]=F[1-1]\to$$ Dimension results to be a Force Analysis of the second term $$\,(-{{F}^{2}}\cdot u_{s}^{2}-\frac{T}{R})$$ :
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$$F\cdot \frac{L}{L}-\frac{T}{L}=F-F\to $$ Dimension results to be a Force Analysis of third term $$\,(M[[1-\bar{R}\cdot u_{ss}^{2}][u_{tt}^{1}-\bar{R}\cdot u_{stt}^{2}]+u_{s}^{2}\cdot u_{tt}^{2}])$$ :
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$$\,M\left[ \left[ 1-L\cdot {{L}^{-1}} \right]\left[ \frac{L}-L\cdot \frac{L}{L\cdot {{t}^{2}}} \right]+\frac{L}{L}\cdot \frac{L} \right]=M\cdot \frac{L}+M\cdot \frac{L}=F+F\to $$ Dimension results to be a Force
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