User:Egm6321.f10.team2.patterson/Homework 1

Given
The horizontal force acting on a wheel/ magnet travelling along a flexible guideway can be modeled by the following equation:

$$\begin{align} & {{c}_{0}}\left( {{y}^{1}},t \right)=-{{F}^{1}}\left\{ 1-\overline{R}u_{ss}^{2}\left( {{y}^{1}},t \right) \right\}-{{F}^{2}}u_{s}^{2}\left( {{y}^{1}},t \right)-{}^{T}\!\!\diagup\!\!{}_{R}\;+M\left\{ \left\{ 1-\overline{R}u_{ss}^{2}\left( {{y}^{1}},t \right) \right\}\left\{ u_{tt}^{1}\left( {{y}^{1}},t \right)-\overline{R}u_{stt}^{2}\left( {{y}^{1}},t \right) \right\}+u_{s}^{2}\left( {{y}^{1}},t \right)u_{tt}^{2}\left( {{y}^{1}},t \right) \right\} \\ & \\ \end{align}$$

Find
Perform a dimensional analysis on all of the terms in the given equation and provide an explanation of its physical meaning

$$\begin{align} & {{c}_{0}}\left( {{y}^{1}},t \right)=-{{F}^{1}}\left\{ 1-\overline{R}u_{ss}^{2} \right\}-{{F}^{2}}u_{s}^{2}-{}^{T}\!\!\diagup\!\!{}_{R}\;+M\left\{ \left\{ 1-\overline{R}u_{ss}^{2} \right\}\left\{ u_{tt}^{1}-\overline{R}u_{stt}^{2} \right\}+u_{s}^{2}u_{tt}^{2} \right\} \\ & \left[ {{c}_{0}}\left( {{y}^{1}},t \right) \right]=-\left[ {{F}^{1}} \right]\left\{ \left[ 1 \right]-\left[ \overline{R} \right]\left[ u_{ss}^{2} \right] \right\}-\left[ {{F}^{2}} \right]\left[ u_{s}^{2} \right]-{}^{\left[ T \right]}\!\!\diagup\!\!{}_{\left[ R \right]}\;+\left[ M \right]\left\{ \left\{ \left[ 1 \right]-\left[ \overline{R} \right]\left[ u_{ss}^{2} \right] \right\}\left\{ \left[ u_{tt}^{1} \right]-\left[ \overline{R} \right]\left[ u_{stt}^{2} \right] \right\}+\left[ u_{s}^{2} \right]\left[ u_{tt}^{2} \right] \right\} \\ \end{align}$$

Solution
The dimensional analysis of each term of the given equation is detailed below:

$$\begin{align} & \left[ {{F}^{1}} \right]=\left[ {{F}^{2}} \right]=ML{{T}^{-2}}\text{ } \\ & \left[ \overline{R} \right]=\left[ R \right]=\left[ {{u}^{1}} \right]=\left[ {{u}^{2}} \right]=L \\ & \left[ u_{ss}^{2} \right]={{L}^{-1}} \\ & \left[ u_{s}^{2} \right]=1\text{ } \\ & \left[ T \right]=M{{L}^{2}}{{T}^{-2}}\text{    } \\ & \left[ u_{tt}^{1} \right]=\left[ u_{tt}^{2} \right]=L{{T}^{-2}} \\ & \left[ u_{stt}^{2} \right]={{T}^{-2}} \\ \end{align}$$

Inserting each dimension into the equation yields the following:

$$\begin{align} & \left[ {{c}_{0}}\left( {{y}^{1}},t \right) \right]=-F\left\{ \left[ 1 \right]-L{{L}^{-1}} \right\}-F\cdot 1-{}^{M{{L}^{2}}{{T}^{-2}}}\!\!\diagup\!\!{}_{L}\;+M\left\{ \left\{ \left[ 1 \right]-L{{L}^{-1}} \right\}\left\{ L{{T}^{-2}}-L{{T}^{-2}} \right\}+1\cdot L{{T}^{-2}} \right\} \\ & \left[ {{c}_{0}}\left( {{y}^{1}},t \right) \right]=-F-F-ML{{T}^{-2}}+M\left\{ 1\left\{ L{{T}^{-2}} \right\}+L{{T}^{-2}} \right\}=-F-F-F+F+F \\ & \left[ {{c}_{0}}\left( {{y}^{1}},t \right) \right]=F \\ \end{align}$$

Each term in the given equation has the following physical explanation:
 * The axial force on the wheel, F1, which is influenced by the distance from the guideway centroid and the center of the wheel/magnet and the curvature of the guideway
 * The transverse force on the wheel projected into the axial direction, F2 multiplied by the transverse displacement u2s
 * The axial force generated from the Torque applied to the wheel (not the case for a magnet), $${}^{T}\!\!\diagup\!\!{}_{R}\;$$
 * Mass of the wheel multiplied by it's acceleration.

The sum of each term equals the horizontal force acting on a wheel/ magnet.

Given
Show why $${{c}_{3}}\left( {{Y}^{1}},t \right){{\ddot{Y}}^{1}}$$ is nonlinear with respect to Y1

Solution
A nonlinear ordinary differential equation (ODE) is as an ODE that is not linear. There are few properties of ODEs that distinguish them as linear. First, the dependent variable and all of the derivatives of the dependent variable are of a degree one. Second, nonlinear functions of the dependent variable or its derivatives aren’t allowed.1

In general, linear ODEs must satisfy the following statement:

$$F\left( \alpha u+\beta v \right)=\alpha F\left( u \right)+\beta F\left( v \right)$$

Evaluating the left hand side (LHS) of the equation for $$F\left( {{Y}^{1}},t \right)={{c}_{3}}\left( {{Y}^{1}},t \right){{\ddot{Y}}^{1}}\left( t \right)$$ yields:

$$\begin{align} & F\left( \alpha u+\beta v \right)={{c}_{3}}\left( \alpha u+\beta v,t \right){{{\ddot{Y}}}^{1}}\left( t \right) \\ & \text{                  }=M\left[ 1-\overline{R}u_{ss}^{2}\left( \alpha u+\beta v,t \right) \right]{{{\ddot{Y}}}^{1}}\left( t \right) \\ \end{align}$$

An evaluation of the right hand side (RHS) of the equation yields:

$$\begin{align} & \alpha F\left( u \right)+\beta F\left( v \right)=\alpha {{c}_{3}}\left( u,t \right){{{\ddot{Y}}}^{1}}\left( t \right)+\beta {{c}_{3}}\left( v,t \right){{{\ddot{Y}}}^{1}}\left( t \right) \\ & \text{                        }=\alpha M\left[ 1-\overline{R}u_{ss}^{2}\left( u,t \right) \right]{{{\ddot{Y}}}^{1}}\left( t \right)+\beta M\left[ 1-\overline{R}u_{ss}^{2}\left( v,t \right) \right]{{{\ddot{Y}}}^{1}}\left( t \right) \\ & \text{                        }=M\left[ \alpha \left( 1-\overline{R}u_{ss}^{2}\left( u,t \right) \right)+\beta \left( 1-\overline{R}u_{ss}^{2}\left( v,t \right) \right) \right]{{{\ddot{Y}}}^{1}}\left( t \right) \\ \end{align}$$

It is evident that the RHS does not equal the LHS and the expression is nonlinear because

$$F\left( \alpha u+\beta v \right)\ne \alpha F\left( u \right)+\beta F\left( v \right)$$

1Zill, D., and Cullen, M. Differential Equations with Boundary-Value Problems. Pacific Grove, CA: Brooks/ Cole,