User:Egm6321.f11.team1/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)

This solution was created by Fenner Colson, Checked by Ben Neri, and compiled into the complete report by Zexi Zheng.

= 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)

This solution was created by Fenner Colson, Checked by Ben Neri, and compiled into the complete report by Zexi Zheng.

= R1.3 Analyze the dimension of all terms in (2) p.3-3 and provide the physical meaning = R*1.3: Analyze the dimension of all terms in (2) p.3-3 and provide the physical meaning.

In general, the terms for this equation govern the vehicle motion with the electromagnetic magnet and constant air gap on a flexible beam. $$c_0(Y^1,t) := -F^1[1-\bar{R}u^2_{\text{,SS}}(Y^1,t)]-F^2u^2_{\text{,S}}(Y^1,t)-T/R + M[[1-\bar{R}u^2_{\text{,SS}}(Y^1,t)][u^1_{\text{,tt}}(Y^1,t)-\bar{R}u^2_{\text{,Stt}}(Y^1,t)]$$ $$+ u^2_{\text{,S}}(Y^1,t)u^2_{\text{,tt}}(Y^1,t)]$$

The term for $$c_0$$ is derived from the dynamics of the nominal motion, which have component terms that depend on structural displacement - the beam deformations. This derivation contrasts the previous a priori assumptions about beam rigidity in describing nominal motion. $$c_3(Y^1,t)\ddot{Y}^1+c_2(\dot{Y}^1)^2 + c_1(Y^1,t)\dot{Y}^1+c_0(Y^1,t) = 0$$ $$c_3(Y^1,t)$$ is the mass, and $$c_0(Y^1,t)$$ is the force in Newtons along motion for the equation. The terms with $$\dot{Y}^1$$ describe convection. 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. In terms of structural deformation, u is indicative of the flexible dynamics related to the beam's structural deformation. This contrasts the ideal form of displacement of nominal motion if the beam were theoretically undeformed. The value $$\bar{R}$$ is the distance from the beam line of centroids to the wheel center of mass. The partial derivative of u with respect to S is $$u^2_{\text{,S}} (Y^1,t)$$. Similarly, the 2nd partial derivative of u with respect to S is given by $$u^2_{\text{,SS}}(Y^1,t)$$. The 2nd partial derivative of u with respect to t is $$u^2_{\text{,tt}}(Y^1,t)$$. The partial derivative of the latter term with respect to S is $$u^2_{\text{,Stt}}(Y^1,t)$$.

References for R1.3 Vu-Quoc, L. and Olsson, M., "A computational procedure for interaction of high-speed vehicles on flexible structures without assuming known vehicle nominal motion, Computer Methods in Applied Mechanics and Engineering, Vol.76, pp.207-244, 1989. http://clesm.mae.ufl.edu/~vql/pdf/cmame.1989.vuquoc.olsson.pdf Vu-Quoc, L. Class Lecture: Principles of Engineering Analysis. University of Florida, Gainesville, FL, 25 Aug 2011. See general link for figure from journal article. http://en.wikiversity.org/wiki/File:EGM6341.s11.TEAM1.WILKS_EC1.svg

This solution was created by Manuel Steele, Checked by Ben Neri, and compiled into the complete report by Zexi Zheng.

= R1.4 Draw the polar coordinate lines=

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

Solution


This solution was created by Yi Zhao, Checked by Ben Neri, and compiled into the complete report by Zexi Zheng.

= 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,

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

This solution was created by Jing Pan, Checked by Ben Neri, and compiled into the complete report by Zexi Zheng.

= 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 $$\displaystyle Y^1$$.

This solution was created by Jing Pan, Checked by Ben Neri, and compiled into the complete report by Zexi Zheng.

= 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.

Link To original solution:

http://en.wikiversity.org/w/index.php?title=User:EGM6321.F11.TEAM1.R.1.7.Playpen

This solution was created by Ben Neri, Checked by Fenner Colson, and compiled into the complete report by Zexi Zheng.

= 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) = $$\beta$$

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) = $$\beta$$

Boundary Conditions: 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 substitution 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 derivation begins here. From (a) we have $$C_1Y^1_H(a) = \alpha - C_2Y^2_H(a)$$ $$C_1 = (\alpha - C_2Y^2_H(a))/Y^1_H(a)$$ From (b), we get $$C_1Y^1_H(b) = \beta - c_2Y^2_H(b)$$ $$C_1 = (\beta - c_2Y^2_H(b))/Y^1_H(b)$$ The solutions of the 1st constant from (a) and (b) can be equated and solved for the unknown. $$(\alpha - C_2Y^2_H(a))/Y^1_H(a) = (\beta - c_2Y^2_H(b))/Y^1_H(b)$$ After algebraic manipulation, the 2nd constant is $$C_2 = [\beta - \alpha(Y^1_H(b)/Y^2_H(a))/[Y^2_H(b) - (Y^1_H(b)Y^2_H(a)/Y^1_H(a)]$$ The 1st constant can now be solved. $$C_1 = \beta - C_2(Y^2_H(b)/Y^1_H(b))$$ $$C_1 = [\beta/Y^1_H(b) - \alpha/Y^2_H(a)]/[1 - (Y^1_H(b)Y^2_H(a)/Y^2_H(b)Y^1_H(a))]$$  Initial 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) $$C_1Y^1_H(b)' + C_2Y^2_H(b)' = \beta$$ Solve for $$C_1$$ then substitute to get $$C_2$$.  $$C_1 = (\alpha - C_2Y^2_H(a))/Y^1_H(a)$$ Next, we derive a solution for (b). $$C_2Y^2_H(b)' = C_1Y^1_H(b)' - \beta$$ $$C_2 = (C_1Y^1_H(b)')/Y^2_H(b)' - \beta/Y^2_H(b)'$$ Substitute $$C_1$$ into the equation.  Then we have $$C_2 = (\alpha(Y^1_H(b)'/Y^2_H(b)')) - (\alpha(\beta/Y^2_H(b)')) - (C_2Y^2_H(a)Y^1_H(b)'/Y^1_H(a)Y^2_H(b)') + (C_2\beta(Y^2_H(a)/(Y^1_H(a)Y^2_H(b)'))$$ $$ C_2 = \alpha[(Y^1_H(b)'/Y^2_H(b)') - (\beta/Y^2_H(b)')]/[1+(Y^2_H(a)Y^1_H(b)'/Y^1_H(a)Y^2_H(b)'+(\beta)(Y^2_H(a)/Y^1_H(a)Y^2_H(b))]$$ Further manipulation gives the solution for $$C_2$$. $$C_2 = \alpha[(Y^1_H(b)'/Y^2_H(b)')-(\beta/Y^2_H(b)')]/[1+(Y^2_H(a)Y^1_H(b)'/Y^1_H(a)Y^2_H(b)')+\beta(Y^2_H(a)/Y^1_H(a)Y^2_H(b))]$$ Which gives $$C_1$$ as $$C_1 = \alpha - (Y^2_H(a)/Y^1_H(a))[\alpha[(Y^1_H(b)'/Y^2_H(b)')-(\beta/Y^2_H(b)')]/[1+(Y^2_H(a)Y^1_H(b)'/Y^1_H(a)Y^2_H(b)')+\beta(Y^2_H(a)/Y^1_H(a)Y^2_H(b))]]$$ References for 1.8 A.C. King, J. Billingham, S.R. Otto, Differential equations:  Linear, nonlinear, ordinary, partial, Cambridge University Press, 2003. R. Ellis, D. Gulick, Calculus with Analytic Geometry, 3rd ed., Harcourt Brace Jovanovich, Publishers, 1986, pp. 936 - 962. M. Olia, D.E. Stevens, FE Fundamentals of Engineering Exam, Barron's Educational Series, Inc., 2008, pp. 17-52.

This solution was created by Manuel Steele, Checked by Ben Neri, and compiled into the complete report by Zexi Zheng.

= References =