User:Egm6321.f11.team3/Hwk1

Given
From lecture note 2-4, first total time derivative is given by,
 * $$\displaystyle

\frac{\mathrm{d} }{\mathrm{d} t}f(Y^{1}(t),t)=\frac{\partial f(Y^{1}(t),t)}{\partial S}\dot{Y^{1}}+\frac{\partial f(Y^{1}(t),t)}{\partial t} $$ with
 * $$\displaystyle

\dot{Y^{1}}:=\frac{\mathrm{d} Y^{1}(t)}{\mathrm{d} x} $$

Find
Show that
 * $$\displaystyle

\frac{\mathrm{d^2}f }{\mathrm{d} t^2}=f_{,S}(Y^{1},t)\ddot{Y}^1+f_{,SS}(Y^{1},t)(\dot{Y}^{1})^{2}+2f_{,St}(Y^{1},t)\dot{Y}^{1}+f_{,tt}(Y^{1},t) $$ with
 * $$\displaystyle

f_{,S}(Y^{1},t):=\frac{\partial f(Y^{1},t)}{\partial S} $$
 * $$\displaystyle

f_{,St}(Y^{1},t):=\frac{\partial^{2} f(Y^{1},t)}{\partial S \partial t} $$ Similarly for the other partial derivatives

Solution
We solved this problem on our own Let
 * $$\displaystyle

\frac{\mathrm d^{2}f}{\mathrm dt^{2}}=\frac{\mathrm d}{\mathrm dt}(\frac{\mathrm d}{\mathrm dt}f(Y^{1}(t),t))=\frac{\mathrm d}{\mathrm dt}(\frac{\partial f(Y^{1}(t),t)}{\partial s}\dot{Y}^{1}+\frac{\partial f(Y^{1}(t),t)}{\partial t}) $$
 * $$\displaystyle

=\frac{\partial f(Y^{1}(t),t)}{\partial S}\frac{\mathrm d^{2}Y^{1}(t)}{\mathrm dt^{2}}+\frac{\partial }{\partial s}(\frac{\partial f(Y^{1}(t),t)}{\partial s}\frac{\mathrm dY^{1}(t)}{\mathrm dt}+\frac{\partial f(Y^{1}(t),t)}{\partial t})\frac {\mathrm dY^{1}(t)}{\mathrm dt} $$
 * $$\displaystyle

+\frac{\partial}{\partial t}(\frac{\partial f(Y^{1}(t),t)}{\partial S}\frac{\mathrm dY^{1}(t)}{\mathrm dt}+\frac{\partial f(Y^{1}(t),t)}{\partial t}) $$
 * $$\displaystyle

=\frac{\partial^{2}f(Y^1(t),t)}{\partial S^{2}}(\frac{\mathrm dY^{1}(t)}{\mathrm dt})^{2}+2\frac{\partial^{2}f(Y^{1}(t),t)}{\partial S \partial t}\frac{\mathrm dY^{1}(t)}{\mathrm dt}+\frac{\partial^{2}f(Y^{1}(t),t)}{\partial t^{2}} $$
 * $$\displaystyle

+\frac{\mathrm d^{2}Y^{1}(t)}{\mathrm dt^{2}}\frac{\partial f(Y^{1}(t),t)}{\partial S} $$
 * $$\displaystyle

=f_{,S}(Y^{1},t) \ddot{Y}^{1}+f_{,SS}(Y^{1},t)(\dot{Y}^{1})^{2}+2f_{,St}(Y^{1},t)\dot{Y}^{1}+f_{,tt}(Y^{1},t) $$

Given
The equation of motion (EOM) for the maglev train is modeled by:


 * $$\displaystyle

\displaystyle f(S,t)| _{S=Y^1(t)}=f(Y^1(t),t) $$ From lecture notes on page 2-4, the 1st total time derivative is given by:

and the 2nd total time derivative, on page 2-5, is given by:

Find

 * Derive the 1st total time derivative.
 * Derive the 2nd total time derivative.
 * Show the similarity with the derivation of the Coriolis force.

Solution, 1st Total Time Derivative
Applying the chain rule to
 * $$\displaystyle

f(S,t) $$

Yields:
 * $$\displaystyle

\frac{df(S,t)}{dt}=\frac{\partial f}{\partial S}\frac{\partial S}{\partial t}+\frac{\partial f}{\partial t}\frac{\partial t}{\partial t} = \frac{\partial f(S,t)}{\partial S}\frac{\partial S}{\partial t} + \frac{\partial f(S,t)}{\partial t}\frac{\partial t}{\partial t} $$ Given that $$ \displaystyle S $$ is evaluated at $$ \displaystyle Y^1(t) $$ :
 * $$\displaystyle

f(S,t)| _{S=Y^1(t)}=f(Y^1(t),t) $$ Then the equation can be rewritten as:

And simplified to result in equation 2.1:
 * $$\displaystyle

\frac{d}{dt}f(Y^1(t),t)= \frac{\partial f(Y^1(t),t)}{\partial S} \dot Y^1+\frac{\partial f(Y^1(t),t)}{\partial t} $$

Solution, 2nd Total Time Derivative
The general expression for the 2nd total time derivative is:


 * $$\displaystyle

\frac{d^2}{dt^2}f=\frac{d}{dt}\left(\frac{d}{dt}f\right) $$ Substituting from equation 2.3 results in:
 * $$\displaystyle

\frac{d^2}{dt^2}f(Y^1(t),t)=\frac{d}{dt}\left(\frac{\partial f}{\partial S}\frac{\partial S}{\partial t} + \frac{\partial f}{\partial t}\underbrace{\frac{\partial t}{\partial t}}_{1}\right) $$ Applying the chain rule yields:
 * $$\displaystyle

\frac{d^2}{dt^2}f(Y^1(t),t)=\frac{\partial^2f}{\partial S^2}\left(\frac{\partial S}{\partial t}\right)^2

+\left(\frac{\partial^2 f}{\partial S \partial t} \cdot \frac{\partial S}{\partial t} \cdot \underbrace{\frac{\partial t}{\partial t}}_{1}\right)

+\left( \frac{\partial f}{\partial S} \cdot \underbrace{\frac{\partial^2 S}{\partial S \partial t}}_{0} \cdot \frac{\partial S}{\partial t}\right)

+\left(\frac{\partial f}{\partial S} \cdot \frac{\partial^2 S}{\partial t^2} \cdot \underbrace{\frac{\partial t}{\partial t}}_{1}\right)

+\left(\frac{\partial^2 f}{\partial S \partial t} \cdot \frac{\partial S}{\partial t}\right)

+\left(\frac{\partial^2 f}{\partial t^2} \cdot \underbrace{\frac{\partial t}{\partial t}}_{1}\right)

$$ Collecting terms and simplifying creates the equation:
 * $$\displaystyle

\frac{d^2}{dt^2}f(Y^1(t),t)=\underbrace{\frac{\partial^2f}{\partial S^2}}_{f_{,SS}}\left(\underbrace{\frac{\partial S}{\partial t}}_{\dot Y^1}\right)^2

+\left(2 \cdot \underbrace{\frac{\partial^2 f}{\partial S \partial t}}_{f_{,St}} \cdot \underbrace{\frac{\partial S}{\partial t}}_{\dot Y^1}\right)

+ \left( \underbrace{\frac{\partial f}{\partial S}}_{f_{,S}} \cdot \underbrace{\frac{\partial^2 S}{\partial t^2}}_{\ddot Y^1} \right)

+ \underbrace{\frac{\partial^2 f}{\partial t^2}}_{f_{,tt}}

$$ Then substituting using the denoted subscripts, and rearranging, results in equation 2.2:
 * $$\displaystyle

\displaystyle \frac{d^2f}{dt^2}=f_{,S}\left({Y^1,t}\right)\ddot Y^1+f_{,SS}(Y^1,t)(\dot Y^1)^2+2f_{,St}(Y^1,t)\dot Y^1+f_{,tt}(Y^1,t) $$

Similarity to Coriolis Force Derivation
Suppose that a given object has position vector r in the rotating reference frame with instantaneous velocity v. Then it is obviously that $$\frac{\mathrm{d} r}{\mathrm{d} t}=\frac{\partial r}{\partial t}+\dot{\theta}\times r=v+\omega\times r $$ $$\frac{\mathrm{d}^{2} r}{\mathrm{d} t^{2}}=\frac{\mathrm{d}}{\mathrm{d}t}(\frac{\partial r}{\partial t}+\dot{\theta}\times r)=\frac{\partial}{\partial t}(\frac{\partial r}{\partial t}+\dot{\theta}\times r)+\frac{\mathrm d}{\mathrm d t}(\dot{\theta}\times r)$$ $$ =\frac{\partial^{2}r}{\partial t^{2}}+2\dot{\theta}\times\frac{\partial r}{\partial t}+\ddot{\theta}\times r+\dot{\theta^{2}}\times r $$ Compare 1st and 2nd Total Time Derivatives with Coriolis equation we can find that

Given
The horizontal force $$\displaystyle c_0 $$ for the wheel/magnet on the maglev train is:

Find
Analyze the dimensions of all terms in equation 3.1 and provide the physical meaning.

Solution
We solved this problem on our own The horizontal force $$\displaystyle c_0 $$ for the wheel/magnet on the maglev train is:

Dimensions of all the quantities:

[F1] = Force (F) $$[\bar R] = Length (L)$$ [u2,ss] = L-1 [F2] = F [u2,s] = 1 [T] = FL [R] = L [M] = Mass (M) [u1,tt] = Lt-2 [u2,stt] = t-2 [c0] = F A := $$\displaystyle F^1[1-\bar R u^2_{,SS} (Y^1,t)] $$ Dimension of A = [A] [A] = -F [1-LL-1] [A] = F B := F2u2,s Dimension of B = [B] [B] = F x 1 = F [B] = F C := $$\displaystyle \frac{T}{R} $$ Dimension of C = [C] [C] = FL x L-1 [C] = F

D := $$\displaystyle M \left [(1-\bar R u^2_{,SS})(u^1_{,tt}-\bar R u^2_{,Stt})+u^2_{,S} u^2_{,tt} \right]$$ Dimension of D = [D] [D] = M{( 1-LL-1)(Lt-2-Lt-2) + 1 x Lt-2 } [D] = M {1 x Lt-2 + Lt-2 } [D] = MLt-2 [D] = F  [A]=[B]=[C]=[D]=F [c0(Y1,t)] = F Physical Meaning A is the transverse force due to deformation in the transverse direction B is the transverse force due to deformation of guideway slope C is the force due to rotation of wheel D is the transverse force due to acceleration of wheel c0(Y1,t) is the sum of all forces acting on the wheel in the transverse direction. Prashant Gopichandran 00:47 September 14th,2011 (UTC)

Given
ξ=(ξ1,ξ2,ξ3)

X(ξ)=X1(ξ1)X2(ξ2)X3(ξ3)

Find
Draw the polar coordinate lines, in a 2-D plane emanating from a point, not at the origin.

Solution
We solved this problem on our own In a 2-D plane, choose a point p and call it the pole, as shown in figure below. Then we select an x axis through that pole and let x axis parallel to x1 axis. It is polar coordinate lines emanating from a point, not at the origin.

Reference
Polar Coordinates

Find
Show that (5.1) becomes:
 * $$\displaystyle

y''+\underbrace{\frac{g'(x)}{g(x)}}_{a_1(x)}y'+a_0(x)\,y=0 $$ Change symbols to simplify (lecture notes 4-5): $$\displaystyle \xi_i \rightarrow x$$ $$X_i(\xi_i)\rightarrow y(x)$$ $$g_i(\xi_i)\rightarrow g(x)$$ $$f_i(\xi_i)\rightarrow a_0(x)$$

Solution
Substitute into equation 5.1 with these new variables: $$\displaystyle \underbrace{\frac{1}{g_i(\xi_i)}}_{\color{blue}\frac{1}{g(x)}} \underbrace{\frac{d}{d \xi_i}}_{\color{blue}\frac{d}{dx}} \left[\underbrace{g_i(\xi_i)}_{\color{blue}g(x)} \underbrace{\frac{d X_i(\xi_i)}{d \xi_i}}_{\color{blue}\frac{d y(x)}{dx}} \right]+ \underbrace{f_i(\xi_i)}_{\color{blue} a_0 (x)} \underbrace{X_i(\xi_i)}_{\color{blue} y(x)}=0 $$ $$\displaystyle \frac{1}{g(x)} \cdot \frac{d}{dx} \cdot \left[g(x) \cdot \frac{dy(x)}{dx}\right]+a_0 (x) \cdot y(x)=0 $$ $$\displaystyle \frac{1}{g(x)} \cdot \left[g'(x) \cdot \frac{dy(x)}{dx} + g(x)\frac {d^2y(x)}{dx^2}\right]+a_0 (x) \cdot y(x)=0 $$ $$\displaystyle \frac{g'(x)}{g(x)} \cdot \frac{dy(x)}{dx} + \frac {d^2y(x)}{dx^2}+a_0 (x) \cdot y(x)=0 $$ $$\displaystyle \underbrace{\frac{g'(x)}{g(x)}}_{a_1(x)} \cdot \underbrace{\frac{dy(x)}{dx}}_{y'} + \underbrace{\frac {d^2y(x)}{dx^2}}_{y''}+a_0 (x) \cdot y(x)=0 $$ $$\displaystyle y''+a_1(x)y'+a_0(x)\,y=0 $$

Given:
The equation of motion of the wheel/magnet provide in Lecture 3, p3-3 is given as:

where $$ \displaystyle Y^1 $$ is the nominal position of the wheel.

The mass component of the equation of motion is defined as

Find:
Show that $$\displaystyle c_3(Y^1,t)\ddot Y^1$$ is nonlinear with respect to $$\displaystyle Y^1 $$.

Solution
We solved this problem on our own and checked it against previous class solutions A linear seoond order equation is one that can be written in the form:

or with any function y with a continuous second derivative on the interval I, then

To demonstrate $$\displaystyle c_3(Y^1,t)\ddot Y^1$$ to be non-linear with respect to $$\displaystyle Y^1 $$, it suffices to show that the equation 6.6 is not always satisfied.

Now we insert Equation 6.2

and change the notation of the second derivative

Now we pull out the constant on the left side of the equation.

It is now clear that the two sides of the equation are not equal due to the constant on the second internal term. Therefore, $$\displaystyle c_3(Y^1,t)\ddot Y^1$$ is nonlinear with respect to $$\displaystyle Y^1 $$.

Given

 * $$\displaystyle

L_{2}(\cdot )=\frac{d^{2}(\cdot )}{dx^{2}}+a_{1}(x)\frac{d(\cdot )}{dx}+a_{0}(x)(\cdot ) $$ You can see from [PEA1.F11.mtg6 p6-3 (1)]

Find
Show that :$$\displaystyle L_{2}(\cdot )$$ is linear

Solution
We solved this problem on our own Let u(x) and v(x) be any 2 functions of x Let :$$\displaystyle \alpha \, and \, \beta $$ be and 2 real numbers We write :$$\displaystyle \forall u,v:\mathbb{R}\rightarrow \mathbb{R} \, x \mapsto u(x)  \,\,  x \mapsto v(x)  \,\, \forall \alpha ,\beta \in \mathbb{R} $$

So it is linear.

Given

 * $$\displaystyle

y(x)=y^{_{H}}(x)+y^{_{P}}(x)=c_{1}y_{H}^{1}(x)+c_{2}y_{H}^{2}(x)+y_{P}(x) $$

Find
Find the integration constants in terms of the initial conditions :$$\displaystyle y(a)=\alpha \,\, \, y{}'(a)=\beta $$

Solution
We solved this problem on our own


 * $$\displaystyle You\,\, can\, get\,  c_{1}\, by\,\,  equation(8.1)\times \dot{y}_{H}^{2}(a)-equation(8.2)\times y_{H}^{2}(a)$$


 * $$\displaystyle You\,\, can\, get\,  c_{2}\, in\,\,   the\,  similar\,  way$$


 * $$\displaystyle \Rightarrow c_{2}= \frac{\alpha \dot{y}_{H}^{1}(a)-\beta y_{H}^{1}(a)-y_{P}(a)\dot{y}_{H}^{1}(a)+\dot{y}_{P}y_{H}^{1}(a)}{y_{H}^{2}(a)\dot{y}_{H}^{1}(a)-\dot{y}_{H}^{2}(a)y_{H}^{1}(a)}       $$

 (8.4)