User:Egm6321.f12.team3/report1

= R*1.1 = I solved it on my own

Problem
Show That:

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

Given
$$\displaystyle \left.f(S,t) \right|_{S=Y^1(t)}=f(Y^1(t),t)$$

Solution
First Total Time Derivative:

$$\displaystyle \frac{d}{dt}f(S(t),t)= \underbrace{\frac{\partial f(S(t),t)}{\partial S} \dot S(t)}_{\color{red}{u}}+\underbrace{\frac{\partial f(s(t),t)}{\partial t}}_{\color{blue}{v}}$$

Now let us take the derivative of u and v with respect to time.

Since u is a product of two terms, we apply the product rule

Product rule states:

$$\displaystyle \frac{d(a(x)\cdot b(x))}{dx} =\frac{d(a(x))}{dx}\cdot {b(x)} + \frac{d(b(x))}{dx}\cdot {a(x)}$$

For $$ \displaystyle u =\frac{\partial f(S(t),t)}{\partial S} \dot S(t)$$

applying Chain rule to equation (1.1)

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

For $$\displaystyle v =\frac{\partial f(S(t),t)}{\partial t}$$

$$\displaystyle \frac{dv}{dt} ={\frac{\partial}{\partial t} (\frac{\partial f(S(t),t)}{\partial t} )}$$

Then the second total time derivative is:

Plugging equations (1.1) & (1.2) into (1.3) yields:

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

Combine like terms:

Since $$\displaystyle \left.f(S,t) \right|_{S=Y^1(t)}=f(Y^1(t),t)$$

equation 1.4 becomes:

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

Hence the second total time derivative is:

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

= R*1.2 =

I referred to this page for the derivation of coriolis acceleration

Given
$$\displaystyle \left. f(S,t) \right|_{S=Y^1(t)} = f(Y^1(t),t) $$

Problem
Derive the first and second time derivative of $$f$$.

Also show the similarity between the second time derivative and the Coriolis acceleration expression.

Solution
Using the chain rule,

This is the first time derivative.

The second time derivative is calculated as follows:

Coriolis Acceleration Derivation
Angular velocity of a vector is defined as $$\displaystyle \dot \vec r = \vec \omega \times \vec r $$

Consider a point in space $$ \displaystyle \vec a = a_x \hat i+ a_y \hat j+a_z\hat k $$

Velocity of the particle as seen in an inertial frame is

Where $$ \left. \frac {d\vec a}{dt}\right|_A $$ is the local velocity in the accelerating frame of reference.

Differentiating again,

The above equation is the expression showing the coriolis and centrifugal accelerations.

The similarity between the second time derivative and the above expression are listed below

$$ \displaystyle 2\dot Y \frac{\partial f}{\partial S \partial t}\cong 2 \vec \omega \times \frac{d\vec a}{dt} $$

$$ \displaystyle \frac{\partial^2 f}{\partial t^2}\cong \left.\frac{d^2\vec a}{dt^2}\right|_A $$

$$ \displaystyle \ddot Y^1\frac{\partial f}{\partial S}\cong \frac{d\vec \omega}{dt}\times \vec a $$

$$ \displaystyle (\dot Y^1)^2\frac{\partial^2 f}{\partial S^2}\cong \vec \omega\times (\vec \omega\times \vec a ) $$

= R*1.3 =

I solved it on my own

Problem
Given that $$\displaystyle c_0(Y^1,t)= \underbrace{-F^1[1- \overline{R} u^2_{,SS}(Y^1,t)]}_{\color{red}{Term 1}} - \underbrace{F^2 u^2_{,S}(Y^1,t)}_{\color{red}{Term 2}} - \underbrace{\frac{T}{R}}_{\color{red}{Term 3}}

+ \underbrace{M[(1-\overline{R} u^2_{,SS}(Y^1,t)) (u^1_{,tt} - \overline{R} u^2_{,S tt}(Y^1,t)) + u^2_{,S}(Y^1,t) u^2_{,tt}(Y^1,t)]}_{\color{red}{Term 4}}$$

where $$\displaystyle c_0(Y^1,t)$$ is the 'Force' term (negative in direction) in the Equation of Motion (EOM) of a Wheel of a High Speed Train.

The EOM is given by $$\displaystyle c_3(Y^1,t) \ddot{Y}^1 + c_2(Y^1,t) \dot{Y}^1 + c_0(Y^1,t)=0$$

Check for dimensional consistency of all terms of $$ c_0(Y^1,t)$$ and provide the physical meaning of those terms.

Solution
Let us change the Notation of the Torque Variable 'T' in the equation to 'Tr' to avoid confusion with the Time Dimension T later in our solution. Now, let us find out the dimensions of each variable. Also, since the equation represents Force, let us assume Force (F) to be one of the elementary dimensions in our solution for simplicity. From the above dimension check, we can find the dimensions of the following Partially Differentiated Terms. Now, Let us analyze the dimensions of each term on the right hand side of the equation.

Term 1: Term 2: Term 3: Term 4:

Since there are two sub-terms here, let us take it step by step. From Equations $$\displaystyle(1)-(17)$$,

Hence, the equation is consistent dimensionally.

Physical Meanings of the Terms of the Equation

Term 1: It gives the Force acting on the wheel in the Horizontal Direction.

Term 2: It gives the Force acting on the wheel in the Vertical Direction.

Term 3: It gives the Torque acting on the wheel.

Term 4: It gives Mass of the wheel times its Linear Acceleration.

= R*1.4 =

I solved it on my own

Problem
Draw the polar coordinate lines $$\displaystyle (\xi_1,\xi_2)=(r,\theta) $$ in a 2-D plane emanating from a point, not at the origin.

Solution


= R1.5 = I solved it on my own

Problem
Show that Equation (3) p.4-4 of the lecture slide from Mtg 5 becomes $$\displaystyle y''(x)+\frac{g'(x)}{g(x)}y'(x) +a_0(x)y(x)=0$$

Given
$$\displaystyle \frac{1}{g_i(\xi_i)}\frac{d}{d\xi_i}[g_i(\xi_i) \frac{d \chi_i(\xi_i)}{d \xi_i}]+ f_i(\xi_i)\chi_i(\xi_i)=0$$

Solution
To simplify equation 5.1, the following substitutions are made:

$$\displaystyle \xi_i \rightarrow x$$

$$\displaystyle \chi_i(\xi_i) \rightarrow y(x)$$

$$\displaystyle g_i(\xi_i) \rightarrow g(x)$$

$$f_i(\xi_i) \rightarrow a_0(x)$$

After making the substitutions, we arrive at this equation:

First, take the derivative of the term inside the brackets by applying the product rule:

Now we have this equation:

Replace the longhand notation for derivatives with the prime notation:

$$\displaystyle \frac{dg(x)}{dx}=g'(x)$$ $$\displaystyle \frac{d^2y(x)}{dx^2}=y''(x)$$

Divide the terms inside the bracket by $$g(x)$$

$$\displaystyle y''(x)+\frac{g'(x)}{g(x)}y'(x)+a_0(x)y(x)=0$$

= R1.6 = I solved it on my own

Given the term $$\displaystyle C_3(Y^1,t)\ddot{Y}^1$$  is nonlinear, thus making the Equation of motion (EOM) of a high speed train a nonlinear second order ODE.

Prove
Show that $$\displaystyle C_3(Y^1,t)\ddot{Y}^1$$ is nonlinear irrespective of whether $$\displaystyle u^2_,ss(Y^1,t)$$ is nonlinear or linear wrt $$\displaystyle Y^1$$.

Solution
The term $$\displaystyle C_3(Y^1,t)\ddot{Y}^1$$ can be expressed as

We know that for a fuction F to be linear, it should satisfy : $$\displaystyle \forall \alpha, \beta \in \mathbb R $$

To prove linearity, from $$(Eq.6.1)$$ and $$(Eq.6.2)$$ the left hand side $$(F(\alpha u+\beta v))$$ can be written as

expanding $$(Eq.6.3)$$ we get,

The right hand side $$(\alpha F(u) + \beta F(v))$$ can be written as

$$(Eq.6.4)$$ and $$(Eq.6.5)$$ are not equal. Hence, $$(Eq.6.1)$$ is non-linear always.

= R*1.7 = I solved it on my own

Problem
Show that second order differential operator $$ L_2(\cdot)$$ is linear.

Solution
To prove linearity, show that: $$F(\alpha u + \beta v) = \alpha F(u) + \beta F(v) \, \forall \alpha, \beta \in \mathbb R$$

Solve for the first derivative.

$$\displaystyle \frac{d (\cdot)}{dx} = \frac{d(\alpha u + \beta v)}{dx}$$

$$\displaystyle = \frac{d(\alpha u )}{dx} +\frac{d(\beta v )}{dx}$$

$$\displaystyle = \alpha\frac{d(u)}{dx} +\beta\frac{d(v)}{dx}$$

$$\displaystyle = \alpha u' + \beta v'$$

Use the first derivative to solve for the second derivative.

$$\displaystyle \frac{d^2 (\cdot)}{dx^2} = \frac{d}{dx}(\frac{d (\cdot)}{dx})$$

$$\displaystyle = \frac{d}{dx}(\alpha\frac{d(u)}{dx} +\beta\frac{d(v)}{dx})$$

$$\displaystyle = \frac{d}{dx}(\alpha\frac{d(u)}{dx})+ \frac{d}{dx}(\beta\frac{d(v)}{dx})$$

$$\displaystyle = \alpha\frac{d^2 u}{dx^2}+ \beta\frac{d^2 v}{dx^2}$$

$$\displaystyle = \alpha u + \beta v$$

Now, solve for $$L_2(\alpha u + \beta v)$$.

Hence, $$ L_2(\cdot)$$ is linear because it satisfies:

$$ \displaystyle L_2(\alpha u + \beta v)= \alpha L_2(u) + \beta L_2(v)$$

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