User:Egm6321.f11.team1.zheng.zx/teampage/report1

= R1.1 Second Total Time Derivative =

Given
$$\displaystyle \dot Y :=\frac{dY^1(t)}{dt}$$

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

$$\displaystyle f_{,S t}(Y^1,t):=\frac{\partial^2{f(Y^1,t)}}{\partial{S}\partial{t}}$$

Find
Show that the following equality is true.

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

Basic Equations
First Total Derivative:


 * {| style="width:100%" border="0" align="left"

$$ $$
 * $$\displaystyle \frac{d}{dt}f(Y^1(t),t)=\frac{\partial{f(Y^1(t),t)}}{\partial{S}}\dot{Y}^1+\frac{\partial{f(Y^1(t),t)}}{\partial{t}}
 * $$\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}}
 * $$\displaystyle (Equation\;1.1.1)
 * }
 * }

Chain Rule: 

If $$y=f(u),u=g(x)$$, and the derivatives $$\frac{dy}{du}$$ and $$\frac{du}{dx}$$ both exist, then the composite function defined by $$y=f(g(x))$$ has a derivative given by

$$ \frac{dy}{dx}=\frac{dy}{du}\frac{du}{dx}=f'(u)g'(x)=f'(g(x))g'(x) $$

Solution
The following steps make use of algebra of derivatives and the chain rule.

$$ \displaystyle \frac{d^2f}{dt^2}=\frac{d}{dt}\left [ \frac{df}{dt} \right ] $$

$$ \displaystyle \frac{d^2f}{dt^2}=\frac{d}{dt}\left [ \frac{\partial{f(Y^1(t),t)}}{\partial{S}}\dot{Y}^1+\frac{\partial{f(Y^1(t),t)}}{\partial{t}} \right ] $$

$$ \displaystyle \frac{d^2f}{dt^2}=\frac{d}{dt}\left[\frac{\partial}{\partial S}f(Y^1(t),t)\dot Y^1 \right ]+\frac{d}{dt}\left[\frac{\partial}{\partial t}f(Y^1(t),t) \right ] $$

$$ \displaystyle \frac{d^2f}{dt^2}=\left [\frac{\partial^2}{\partial S\partial t}f(Y^1(t),t)\dot Y^1+\frac{\partial}{\partial S}\frac{\partial f(Y^1(t),t)}{\partial t}\dot Y^1+\frac{\partial}{\partial S}f(Y^1(t),t)\ddot Y^1 \right]+\left [\frac{\partial^2}{\partial t^2}f(Y^1(t),t)+\frac{\partial}{\partial t}\frac{\partial f(Y^1(t),t)}{\partial t}\right] $$

NOTE: $$ \displaystyle \frac{\partial}{\partial t}\frac{\partial f(Y^1(t),t)}{\partial t}=\frac{\partial}{\partial S}\frac{\partial S}{\partial t}\left [\frac{\partial f}{\partial S}\frac{\partial S}{\partial t}\right]=\frac{\partial}{\partial S}\dot Y^1\left[\frac{\partial f}{\partial S}\dot Y^1\right]=\frac{\partial^2 f(Y^1(t),t)}{\partial^2 S}(\dot Y^1)^2 $$

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

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

 Solution by Egm6321.f11.team1.colsonfe 21:15, 7 September 2011 (UTC)

= R1.2 Coriolis Force Derivation =

Find
Derive the following two equations:

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

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

Show similarity between the above derivations and derivation of Coriolis Force.

Basic Equations
Note the following notation of using $$S_0$$ for inertial (non-moving) and $$S$$ for rotating frames.

Note that I define the position vector $$\mathbf r =x \hat{\mathbf x}+y \hat{\mathbf y}+z \hat{\mathbf z}$$.

Newton's Second Law:

$$m\left(\frac{d^2\mathbf{r}}{dt^2}\right)_{S_0}=\mathbf{F}$$

Time Derivative in a Rotating Frame: 


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\left(\frac{d\mathbf{Q}}{dt}\right)_{S_0}=\left(\frac{d\mathbf{Q}}{dt}\right)_S+\mathbf\Omega\times\mathbf Q $$ $$ Note this identity relates the derivative of any one vector $$\mathbf Q$$ as measured in the inertial frame $$S_0$$ to the corresponding derivative in the rotating frame $$S$$.
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Equation\;1.2.1)
 * }
 * }

Note that $$\mathbf\Omega$$ is the angular velocity of the rotating frame.

Solution
Derive Equation 1.1.1:

Via application of chain rule we can begin with

$$\frac{df}{dt}=\frac{\partial f}{\partial S}\frac{\partial S}{\partial t}+\frac{\partial f}{\partial t}\frac{\partial t}{\partial t} $$

$$ \frac{df}{dt}=\frac{\partial f}{\partial S}\dot Y^1+\frac{\partial f}{\partial t}\cdot 1 $$

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

$$f$$ Derivations Summary

Coriolis Derivation

Let's use Equation 1.2.1 to help us express $$\mathbf r$$ in terms of derivatives.

$$\left(\frac{d\mathbf{r}}{dt}\right)_{S_0}=\left(\frac{d\mathbf{r}}{dt}\right)_S+\mathbf\Omega\times\mathbf r$$

Differentiating a second time we find

$$\left(\frac{d^2\mathbf{r}}{dt^2}\right)_{S_0}=\left(\frac{d}{dt}\right)_{S_0}\left(\frac{d\mathbf{r}}{dt}\right)_{S_0}$$

$$\left(\frac{d^2\mathbf{r}}{dt^2}\right)_{S_0}=\left(\frac{d}{dt}\right)_{S_0}\left[\left(\frac{d\mathbf{r}}{dt}\right)_{S}+\mathbf\Omega\times\mathbf r\right]$$

Thus again applying our rotating frame equation we find

$$\left(\frac{d^2\mathbf{r}}{dt^2}\right)_{S_0}=\left(\frac{d}{dt}\right)_{S}\left[\left(\frac{d\mathbf{r}}{dt}\right)_{S}+\mathbf\Omega\times\mathbf r\right]+\mathbf\Omega\times\left[\left(\frac{d\mathbf{r}}{dt}\right)_{S}+\mathbf\Omega\times\mathbf r\right]$$

Once expanded this messy result can be rewritten using dot notation:

$$\left(\frac{d^2\mathbf{r}}{dt^2}\right)_{S_0}=\mathbf{\ddot r}+2\mathbf\Omega\times\mathbf{\dot r}+\mathbf\Omega\times(\mathbf\Omega\times\mathbf r)$$

Now substitute the result into Newton's Second Law

$$m\mathbf{\ddot r}=\mathbf F +2m\mathbf{\dot r}\times\mathbf{\Omega}+m(\mathbf\Omega\times\mathbf r)\times\mathbf\Omega$$

$$\mathbf F$$ denotes the sum of all the forces as identified in any inertial frame. However, there are two additional terms on the right side of the equation. The final term is the centrifugal force. The middle term is what we are interested in here, this is the Coriolis force.

$$\mathbf F_{cor} = 2m\mathbf{\dot r}\times\mathbf\Omega$$

Coriolis Derivations Summary

Qualitatively we can visualize the parallelism between the first derivatives and second derivatives by noting the difference between the two tables presented above. We can do a term by term comparison of the first derivative of $$f(Y^1(t),t)$$ and velocity and the second derivative of $$f(Y^1(t),t)$$ and acceleration.

 Solution by Egm6321.f11.team1.colsonfe 02:17, 12 September 2011 (UTC)

= R1.3 Analyze the dimension of all terms in (2) p.3-3 and provide the physical meaning = Draft Solution: In general, the terms for this equation govern the vehicle motion with the electromagnetic magnet and constant air gap on a flexible beam. $$c_3(Y^1,t)$$ is the force in Newtons along nominal motion in $$X^1$$ axis. The two terms, $$F^1$$ and $$F^2$$ are the general components of the applied force in Newtons. Terms of u with the form $$u^2$$ represent displacements of the material point S on the centroidal line. $$\bar{R}$$ is the distance from the beam line of centroids to the wheel center of mass.

See general link for figure from journal article. http://en.wikiversity.org/wiki/File:EGM6341.s11.TEAM1.WILKS_EC1.svg

Journal article: http://clesm.mae.ufl.edu/~vql/pdf/cmame.1989.vuquoc.olsson.pdf

= R1.4 Temporarily Unavailable =

= R1.5 =

Given
Consider the variables seperated in 3_D curvilinear coordinates $$X\left(\xi\right)=\left(X(\xi_1),X(\xi_2),X(\xi_3)\right)$$ the separated equations for $$\displaystyle i=1,2,3$$ is

Find
These seperated equations can be expressed by

$$\displaystyle y''+a_1(x)y'+a_0(x)y=0$$

in which,

$$a_1(x)=\frac{g'(x)}{g(x)}$$

Solution
According to the differential of product, $$(Eq.5.1)$$ is

$$\frac{1}{g_i(\xi_i)}\left[\frac{dg_i(\xi_i)}{d\xi_i}\frac{dx_i(\xi_i)}{d\xi_i}+g_i(\xi_i)\frac{d^2x_i(\xi_i)}{d\xi_i^2}\right]+f_i(\xi_i)x_i(\xi_i)=0$$

For $$\displaystyle i=1,2,3$$ , set $$\displaystyle x$$ is the varible $$\displaystyle \xi_i$$, $$\displaystyle y$$ is its value $$\displaystyle x_i$$.

As for a specific $$\displaystyle i=1,2,3$$ , $$(Eq.5.2)$$ can be written as

$$\displaystyle a_2(x)y''+a_1(x)y'+a_0(x)y=0$$

in which,

$$\displaystyle a_2(x)=1;$$ $$\displaystyle a_1(x)=\frac{g'(x)}{g(x)}$$

= R1.6 =

Given
In the Equation of Motion, the first term is

Find
Prove $$\displaystyle C_3(Y^1,t)\ddot{Y}^1$$  is nonlinear respect to $$\displaystyle Y^1$$.

Solution
If a function $$\displaystyle F(\cdot)$$ is linear, it must be satisfied by the following for any arbitrary number $$\displaystyle \alpha,\beta$$ :

As for the first term of EOM, we set $$\displaystyle F(\cdot)=C_3(\cdot,t)\ddot{(\cdot)}$$

Therefore, the left side of $$(Eq.6.2)$$ can be written as $$\displaystyle F(\alpha u+\beta v)=M\left[1-\overline{R}u^2_,ss(\alpha u+\beta v,t)\right]\frac{d^2(\alpha u+\beta v)}{dt^2}$$.

The right hand side of $$(Eq.6.2)$$ is

$$\displaystyle \alpha F(u)+\beta F(v)=\alpha M[1-\overline{R}u^2_,ss(u,t)]\frac{d^2 u}{dt^2}+\beta M[1-\overline{R}u^2_,ss(v,t)]\frac{d^2 v}{dt^2}$$.

It is obvious that $$\displaystyle F(\alpha u+\beta v)\neq \alpha F(u)+\beta F(v)$$, $$\displaystyle \forall \alpha ,\beta \in \mathbb R$$ In all, $$ (Eq.6.1) $$ is nonlinear with respect to $$\displaystyle Y^1$$.

= R1.7 Proof of linearity =

Find
Show that equation $$(Eq.7.1)$$ is linear

Solution
Proof of Linearity:

In order for equation $$(Eq.7.1)$$ to be linear, the equation:

$$L_2(\alpha u + \beta v) = \alpha L_2(u) + \beta L_2(v)$$  $$\forall u, v,$$   and    $$\forall \alpha, \beta \in \Re$$

 $$(Eq.7.2)$$ must be satisfied. This is known as the superposition principle.

Where $$u$$ and $$v$$ are functions of $$x$$, and $$\alpha$$ and $$\beta$$ are scalars. Next, we define the equation $$L_2(\alpha u + \beta v)$$ as:

$$L_2(\alpha u + \beta v) = \frac{d^2(\alpha u + \beta v)}{dx^2} + a_1 x \frac{d(\alpha u + \beta v)}{dx} + a_o x(\alpha u + \beta v)$$

Grouping $$\alpha $$ and $$\beta $$ terms on both side of the equation yields:

$$\alpha L_2(u) + \beta L_2(v) = \alpha[\frac{d^2 u}{dx^2} + a_1 x\frac{d u}{dx} + a_0 x u] + \beta [\frac{d^2 v}{dx^2} + a_1 x\frac{d v}{dx} + a_0 x v]$$

Substitution of $$u$$ and $$v$$ into $$(Eq.7.1)$$ on the left hand side yields: $$\alpha[\frac{d^2 u}{dx^2} + a_1 x\frac{d u}{dx} + a_0 x u] + \beta [\frac{d^2 v}{dx^2} + a_1 x\frac{d v}{dx} + a_0 x v] = \alpha[\frac{d^2 u}{dx^2} + a_1 x\frac{d u}{dx} + a_0 x u] + \beta [\frac{d^2 v}{dx^2} + a_1 x\frac{d v}{dx} + a_0 x v]$$

Since the superposition principle satisfied, the function is linear.

References: ,

Problem 3: | Team 4 - Correctly defined differential operator, clear write up.

Problem 4: | Team 6 - Correctly defined differential operator, complete solution.

This solution was created by Ben Neri, Checked by XXXX, and compiled into the complete report by XXXX.

= R1.8 The homogeneous solution is y = $$c_1Y^1_H(x) + c_2Y^2_H(x)$$. Find the constants from the initial and boundary values =

(1) Boundary conditions: y(a) = $$\alpha$$ and y(b) = $$\beta$$ (2) Initial conditions: y(a) = $$\alpha$$ and y'(a) = $$\beta$$

Draft Solution: The constants $$C_1$$ and $$C_2$$ can be solved by using a system of equations. The boundary conditions give two equations with two unknowns. By solving for one constant in terms of the other then a simple substituion and elimination can solve for one of the constants. The boundary conditions can then be used to find the second constant. Similarly, the initial conditions can be solved with two equations and two unknowns.

(a) $$C_1Y^1_H(a) + c_2Y^2_H(a) = \alpha$$ (b) $$C_1Y^1_H(b) + c_2Y^2_H(b) = \beta$$

By substitution and algebraic manipulation, $$C_1$$ and $$C_2$$ can be found for the boundary conditions. The similar system of two equations and two unknowns can be applied to the initial conditions.

(a) $$C_1Y^1_H(a) + c_2Y^2_H(a) = \alpha$$ (b) $$C_1Y^1_H(b)' + c_2Y^2_H(b)' = \beta$$

Upon group consensus, the details of this solution can be epanded to show the specific derivations and constant values. (This just becomes an algebra problem).

The references are King (2003) and the lecture notes.

= References =