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Report1_Team5_A2: 9/11/2012 5:00pm EST

Report 1, Team 5

R1.1 Second Total Time Derivative

Given

Given a multivariable function,$$ f(S,t) $$, where $$S$$ is a function of $$t$$.

Find Show that its second total derivative, when evaluated at $$S = Y^1(t)$$, is the equation (1) p.1a-5, which is shown below:


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$$\frac {d^2f}{dt^2} = f_{,S}(Y^1,t)\ddot Y^1+ f_{,SS}(Y^1,t)\dot {(Y^1)}^2 + 2f_{,St}(Y^1,t)\dot Y^1+ f_{,tt}(Y^1,t)$$
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Where the notations for the partial derivatives are defined as follows:


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

This solution was solved on our own.

Derivation of First total time derivative
The first total derivative of $$f(S,t)$$ can be found by using the Chain Rule for multivariable functions as shown below.


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$$ \displaystyle \frac{df}{dt}=\frac{\partial f(S,t)}{\partial S} \dot S + \frac{\partial f(S,t)}{\partial t} $$ (1.1.1)
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where,


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$$\dot S := \frac {dS}{dt}$$
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The second total derivative of $$ f(S(t),t) $$ can be obtained by taking the total derivative of equation 1.1.1 by using the Chain Rule.


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$$\frac {d^2f}{dt^2} = \frac {\partial \left(\displaystyle \frac {\partial f(S,t)}{\partial S}\dot S\right)}{\partial S} \dot S + \frac {\partial \left(\displaystyle \frac {\partial f(S,t)}{\partial S}\dot S \right)}{\partial t} + \frac {\partial \left(\displaystyle \frac {\partial f(S,t)}{\partial t} \right)}{\partial S} \dot S + \frac {\partial \left(\displaystyle \frac {\partial f(S,t)}{\partial t} \right)}{\partial t}$$ (1.1.2)
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Expanding first two terms using the Product Rule gives,


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$$\frac {d^2f}{dt^2} = \frac {\partial^2 f(S,t)}{\partial S^2} \dot S^2 + \frac {\partial f(S,t)}{\partial S} \frac {\partial \dot S}{\partial S} \dot S + \frac {\partial^2 f(S,t)}{\partial S \partial t} \dot S + \frac {\partial f(S,t)}{\partial S} \ddot S $$
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$$ + \frac {\partial^2 f(S,t)}{\partial t \partial S} \dot S + \frac {\partial^2 f(S,t)}{\partial t^2} $$ (1.1.3)
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If the function $$f$$ has continuous second partial derivatives, then


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$$ \frac {\partial^2 f(S,t)}{\partial S \partial t} = \frac {\partial^2 f(S,t)}{\partial t \partial S}$$ (1.1.4)
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and since


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$$ \frac {\displaystyle \partial \left(\frac {dS}{dt} \right)}{\partial S} = \frac {\displaystyle d \left(\frac {\partial S}{\partial S} \right)}{dt} = 0$$ (1.1.5)
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equation 1.1.3 reduces to


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$$\frac {d^2f}{dt^2} = \frac {\partial^2 f(S,t)}{\partial S^2} \dot S^2 + 2\frac {\partial^2 f(S,t)}{\partial S \partial t} \dot S + \frac {\partial f(S,t)}{\partial S} \ddot S + \frac {\partial^2 f(S,t)}{\partial t^2} $$ (1.1.6)
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Using the notation for partial derivatives defined in the Problem Statement, equation 1.1.6 can be rewritten and rearranged as


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$$\frac {d^2f}{dt^2} = f_{,S}(S,t)\ddot S + f_{,SS}(S,t)\dot S^2 + 2f_{,St}(S,t)\dot S + f_{,tt}(S,t)$$ (1.1.7)
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When $$f$$ is evaluated at $$ S = Y^1(t)$$, equation 1.1.7 becomes


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$$\frac {d^2f}{dt^2} = f_{,S}(Y^1,t)\ddot Y^1+ f_{,SS}(Y^1,t)\dot {(Y^1)}^2 + 2f_{,St}(Y^1,t)\dot Y^1+ f_{,tt}(Y^1,t)$$ (1.1.8)
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which is the same as equation (1) from p.1a-5.

R1.2 First and Second total Time Derivative with derivative of the Coriolis Force

Given

First Total Time Derivative

Second Total Time Derivative

Find

Derive the First and Second Total Time Derivative and show the similarity with the derivation of the Coriolis force

Note: (1.2.2) Second total time derivative is similar to the acceleration of a particle in a rotating frame that gives rise to the Coriolis force (undergraduate dynamics)

(1.2.1) First total time derivative is related to the material time derivative and the Reynolds Transport Theorem in Continuum Mechanics (which covers Heat, Solids, Fluids, Electromagnetics)

Solution

This solution was solved on our own.

Derivation of Second total time derivative
Chain Rule for a function with two variables is shown below.


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$$ \frac {d}{dt}f(x,y) = \frac {\partial f(x,y)}{\partial x} \frac {dx}{dt} + \frac {\partial f(x,y)}{\partial y} \frac {dy}{dt} $$ (1.2.3)
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Applying the chain rule to the function $$f(S,t)$$ gives


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$$ \frac {d}{dt}f(S,t) = \frac {\partial f(S,t)}{\partial S} \frac {dS}{dt} + \frac {\partial f(S,t)}{\partial t} \cancelto{1}{\frac {dt}{dt}} $$ (1.2.4)
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Substituting $$Y^1(t)$$ for $$S$$ gives


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$$ \frac {d}{dt}f(Y^1(t),t) = \frac {\partial f(Y^1(t),t)}{\partial S} \frac {dY^1(t)}{dt} + \frac {\partial f(Y^1(t),t)}{\partial t} $$ (1.2.5)
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Using the notation from p.1a-5,


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$$ \dot {Y^1} := \frac {dY^1(t)}{dt} $$
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equation (1.2.5) is written as


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$$ \frac {d}{dt}f(Y^1(t),t) = \frac {\partial f(Y^1(t),t)}{\partial S} \dot {Y^1(t)} + \frac {\partial f(Y^1(t),t)}{\partial t} $$ (1.2.6)
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which is the same as the Second Total Time Derivative (1.1.2). Derivation of the First Total Time Derivative (1.1.1) is shown in R1.1.

Derivation of Corilois Force
The Corilois Force is defined as
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$$F_C:=-2m\Omega \times v $$ (1.2.7)
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where $$ v $$ is the velocity of the particle in the rotating system, and $$ \Omega $$ is the angular velocity vector.

The acceleration of an object on Earth is obtained by taking the second derivative of its position, $$\vec{R}.$$

Taking the first derivative gives the velocity,
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$$ \vec{\nu} := \frac{d\vec{R}_I}{dt} = \frac{d\vec{R}_E}{dt} + \vec{\omega} \times \vec{R}_E $$ (1.2.8)
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Subscripts $$I$$ and $$E$$ indicate inertial and Earth-fixed reference frames, respectively, and $$\omega$$ is the rotation of the earth w.r.t. the inertial frame.

Acceleration is obtained by taking the derivative of $$\vec{\nu}$$.
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$$ \vec{a} := \frac{d\vec{\nu}_I}{dt} = \frac{d\vec{\nu}_E}{dt} + \vec{\omega} \times \vec{\nu}_E $$ (1.2.9)
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Substituting the expression for $$\vec{\nu_{E}}$$ from (1.1.9) into (2) gives
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$$ \vec{a} := \frac{d}{dt}\left(\frac{d\vec{R}_E}{dt} + \vec{\omega} \times \vec{R}_E\right) + \vec{\omega} \times \left(\frac{d\vec{R}_E}{dt} + \vec{\omega} \times \vec{R}_E\right) $$ (1.2.10)
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Separating out the components gives
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$$ \vec{a} := \frac{d^2\vec{R}_E}{dt^2} + \frac{d\vec{\omega}}{dt} \times \vec{R}_E + \vec{\omega} \times \frac{d\vec{R}_E}{dt} + \vec{\omega} \times \frac{d\vec{R}_E}{dt} +\vec{\omega} \times (\vec{\omega} \times \vec{R}_E) $$ (1.2.10)
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$$ \vec{a} := \frac{d^2\vec{R}_E}{dt^2} + \left(\frac{d\vec{\omega}}{dt} \times \vec{R}_E \right) + 2\left(\vec{\omega} \times \frac{d\vec{R}_E}{dt}\right) + \vec{\omega} \times (\vec{\omega} \times \vec{R}_E) $$ (1.2.11)
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Rearranging the terms in (1.2.10) gives
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$$ \vec{a} := \left(\frac{d\vec{\omega}}{dt} \times \vec{R}_E \right) + \vec{\omega} \times (\vec{\omega} \times \vec{R}_E) + 2\left(\vec{\omega} \times \frac{d\vec{R}_E}{dt}\right) + \frac{d^2\vec{R}_E}{dt^2} $$ (1.2.12)
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which looks very similar to the Second Total Time Derivative equation (1.2.2). The two equations would be virtually identical if the substitutions below were made into equation the First Total Time Derivative equation (1.2.1).
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$$ f,_S(Y^1,t) = \vec{R}_E \left(\therefore f,_{St}(Y^1,t) = \frac{d\vec{R}_E}{dt} \right) $$
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$$ Y^1 = \vec{\omega} $$
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Both of the equations have a first term that contains an "acceleration" term; a second term that contains a squared "velocity" term; a third term that has a factor 2 multiplying the product of two different time derivatives; and a fourth term which is the second time derivative of the original function.

R1.3 Dimensional Analysis

Given

“Ordinary” Differential Equation (ODE), where $$M$$ = mass of wheel/magnet


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$$ \displaystyle c_0(Y^1,t)=\underbrace{-F^1[1 - \bar R u^2_{,SS}(Y^1,t)]}_{\color{blue}{A}} - F^2 u^2_{,S} - \frac{T}{R} + M\left[(1-\bar R u^2_{,SS})(u^1_{,tt}-bar R u^2_{,S tt})+u^2_{,S} u^2_{,tt} \right ] $$ (1.3.1)
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$$ A:=F^1[1-\bar Ru^2_{,SS}] $$ (1.3.2)
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$$c_0$$ is horizontal force acting on wheel/magnet

Find

Analyze the dimension of all terms in “Ordinary” Differential Equation (ODE) (1.3.1) and provide the physical meaning

Solution

This solution was solved on our own.

We will define the following:
$$ F $$ is force

$$ S $$ is time

$$ M $$ is mass

$$ L $$ is length

First term
in which

$$F^1$$ is the axial force of the wheel/magnet
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$$[F^1]=F$$
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$$1$$ is a constant, which is non-dimensional
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$$[1]=1$$
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$$R$$ is the distance between the guideway and center of the wheel/magnet
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$$[\bar R]=L$$
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$$u^2$$ is the transversal displacement of the guideway.
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$$[u^2]=L$$
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$$u^2_{,SS}$$ is the curvature of the guideway deformed by the force of the wheel/magnet.
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$$[u^2_{,SS}]=\frac{1}{R}=\frac{1}{L}=L$$
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So

Second Term
in which $$F^2$$ is force of the wheel/magnet in the vertical direction.
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$$[F^2]=F$$
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$$u^2_{,S}$$ is the guideway slope.
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$$[u^2_{,S}]=\frac{L}{L}$$
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So

Third Term
in which

$$T$$ is the Torque of the wheel/magnet
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$$[T]= F*L$$
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$$R$$ is the radius of the magnet/wheel.
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$$[R]= L$$
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so

Fourth Term
in which

$$M$$ is the mass of the magnet/wheel.
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$$[M]=M$$
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$$u^1$$ is the axial displacement of the guideway.
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$$[u^1]=L$$
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$$u^1_{,tt}$$ is the acceleration of the wheel/magnet.
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$$[u^1_{,tt}]=\frac{L}{S^2}$$
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$$u^2_{,Stt}$$ is the angular acceleration of the wheel/magnet over the spacial displacement.
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$$[u^2_{,Stt}]=\frac{1}{S^2}$$
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so

Conclusion
Evaluating for $$c_0$$ in (1.3.1) gives $$c_0$$ as $$F$$, which is the horizontal force acting on the wheel/magnet.
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$$c_0 = -F[1-L(\frac{1}{L})]-F(\frac{L}{L})-(\frac{{}F\cdot L}{L}) + M[(1-L(\frac{1}{L}))(\frac{L}{S^2}-L(\frac{L}{L\cdot S^2}))+((\frac{L}{L})(\frac{L}{S^2}))]$$ (1.3.3)
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R1.4 Polar Coordinates

Given A point $$\xi$$ in a 3-D curvilinear coordinates $$(\xi_1,\xi_2,\xi_3)$$ and is written as $$\xi=(\xi_1,\xi_2,\xi_3)$$



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

Solution

This solution was solved on our own.



R1.5 Simplifications of Separated Equations

Given

Find

Show that

becomes

where (1.5.3) is a particular case of

Solution

This solution was solved on our own.

Using the following substitutions

we can rewrite (1.5.2) as the following.

By applying the product rule and expanding the terms, we can reduce the equation further.

To show that (1.5.2) becomes (1.5.3), we use the following substitutions in our notation

So that (1.5.9) becomes

which is a particular case of

R1.6 Nonlinear 2nd-order ODE

Given

Equation (2) from p.5-4 gives the following expansion for the equation of motion of the wheel/magnet

It is assumed that the term $$\displaystyle u^2_{,SS}(Y^1,t)$$ is linear.

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

Solution

This solution was solved on our own.

From equation 1.6.1 we get,

For an operator or a function to be linear, it has to satisfy the following condition:

This condition can be broken down into two separate conditions which have to be satisfied simultaneously,

1. The condition of homogeneity:

2. The condition of linearity

As both of these conditions have to be satisfied simultaneously, an operator or function that does not satisfy any one of the two conditions above can be proved as nonlinear.

Initially, checking the condition of homogeneity (Equation 1.6.4)

Now substituting $$ \alpha Y^1 $$ for $$ Y^1 $$.

Since the term $$\displaystyle u^2_{,SS}(Y^1,t)$$ is linear, we can write:

But if the term is to be homogenous then,

It is evident from Equations (1.6.7) and (1.6.8) that,

Thus,

So the given term $$\displaystyle c_3(Y^1,t)\ddot Y^1$$ is not homogenous with respect to $$\displaystyle Y^1$$. As it is one of the two conditions to be simultaneously satisfied for linearity, we can say that term $$\displaystyle c_3(Y^1,t)\ddot Y^1$$ is also not linear with respect to $$\displaystyle Y^1$$.

R1.7 Linearity of an Equation

Given

Find

Show that $$L_{2}(\cdot )$$ is linear in equation (1.7.1).

Solution

This solution was solved on our own.

A function can be said to be linear if and only if

In this problem, we can get:

Since the equation satisfies (1.7.2) we have proved that the equation (1.7.1) is linear.

Contributors Report Leader: Anh