User talk:Egm6321.f10.team4.Yoon

Team4, Try to use Joho writer.

http://writer.zoho.com/index.do

Even if you may need to sign up for using Equation function, Joho writer provides strong latex math code generator.

Also, refer this web page that is in the email from Dr. VC.

https://docs.google.com/Doc?docid=0AY5m1lfNbLroZGM4Mm5mZ2JfNThqdjlrZnBocw&hl=en

= HW Problem 1 - First and Second Order Derivative Derivations =

Given
 The nominal motion of the Maglev train can be modeled by the following equation


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

$$\displaystyle f(Y^1 (t),t) $$
 * }
 * }

Find
Derive the given modeling equations through first and second order derivatives.


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

Axial deformation under moving wheel/guideway $$ \frac{d}{dt} f \left(Y^1(t),t\right) = \frac{\partial f}{\partial s} \left(Y^1(t),t\right) \dot Y^1(t) + \frac{\partial f}{\partial t} \left(Y^1(t),t\right) $$


 * (Eq.1)


 * }
 * }


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

Transverse deformation of wheel/guideway $$\frac{d^2}{dt^2} f\left(Y^1(t),t\right) = f_{,s} \left(Y^1(t),t\right) \ddot Y^1 + f_{,ss} \left(Y^1(t),t\right) (\dot Y^1)^2 + 2 f_{,st} \left(Y^1(t),t\right) \dot Y^1 + f_{,tt} \left(Y^1(t),t\right) $$


 * (Eq.2)


 * }
 * }

Solution for (Eq.1)
First, equating the function to a dummy variable $$s$$ and taking the time derivative


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

$$\displaystyle Y^1(t)=s $$
 * }
 * }

\frac{d}{dt} f \left(Y^1(t),t\right) = \frac{d}{dt} f \left(s,t\right) $$

By the chain rule, the equation becomes
 * {| style="width:100%" border="0" align="left"



$$\displaystyle \frac{d}{dt} f \left(s,t\right) = \frac{\partial f}{\partial s} \underbrace {\frac{\partial s}{\partial t}}_{=\dot Y^1(t)}+ \frac{\partial f}{\partial t} \underbrace{\frac{\partial t}{\partial t}}_{=1} $$
 * (Eq.3)


 * }
 * }

(Equation 3) can be rewritten by reverting function $$f$$ back to its original form:


 * {| style="width:60%" border="0"

$$ \frac{d}{dt} f \left(Y^1(t),t\right) = \frac{\partial f}{\partial s} \left(Y^1(t),t\right) \dot Y^1(t) + \frac{\partial f}{\partial t} \left(Y^1(t),t\right) $$ (Eq.4)
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * style= |
 * }
 * }

Solution for (Eq.2)
Substituting the left side of (Eq.2) to the solution of (Eq.1) yields

\frac{d^2}{dt^2} f\left(Y^1(t),t\right) = \frac{d}{dt}\left( \frac{d}{dt} f\left(Y^1(t),t\right)\right) $$

Applying the chain rule again ahead of (Eq.2) and the dummy s

= \frac{\partial}{\partial s}\left( \frac{d}{dt} f\left(Y^1(t),t\right)\right) + \frac{\partial}{\partial t}\left( \frac{d}{dt} f\left(Y^1(t),t\right)\right) = \frac{\partial}{\partial s}\left( \frac{\partial f}{\partial s} \frac{\partial s}{\partial t}+ \frac{\partial f}{\partial t} \right) + \frac{\partial}{\partial t}\left( \frac{\partial f}{\partial s} \frac{\partial s}{\partial t} +  \frac{\partial f}{\partial t} \right) $$



=\frac{\partial^2 f}{\partial s^2} \left(\frac{\partial s}{\partial t}\right)^2 + \frac{\partial^2 f}{\partial s \partial t} \frac{\partial s}{\partial t} \underbrace{\frac{\partial t}{\partial t}}_{=1} + \frac{\partial f}{\partial s} \underbrace{\frac{\partial^2 s}{\partial s \partial t}}_{=0}\frac{\partial s}{\partial t} + \frac{\partial f}{\partial s} \frac{\partial^2 s}{\partial t^2}\underbrace{\frac{\partial t}{\partial t}}_{=1} + \frac{\partial^2 f}{\partial s\partial t}\frac{\partial s}{\partial t} + \frac{\partial^2 f}{\partial t^2} \underbrace{\frac{\partial t}{\partial t}}_{=1} $$

Collecting terms in the above equation by reduced order of y fucntion.

=\frac{\partial^2 f}{\partial s^2} \left(\frac{\partial s}{\partial t}\right)^2 + 2\frac{\partial^2 f}{\partial s \partial t} \frac{\partial s}{\partial t} + \frac{\partial f}{\partial s} \frac{\partial^2 s}{\partial t^2} +  \frac{\partial^2 f}{\partial t^2} $$

By transforming the Leivniz expression to Lagrange expression such as X,y instead of dx/dy,

\frac{d^2}{dt^2} f\left(Y^1(t),t\right) =f_{,ss} \underbrace{\left(\frac{\partial s}{\partial t}\right)^2}_{\dot {Y^1}^{2}} + 2f_{,st} \underbrace{\frac{\partial s}{\partial t}}_{\dot Y^1} + f_{,s}\underbrace{\frac{\partial^2 s}{\partial t^2}}_{\ddot Y^1} + f_{,tt} $$ Finally, we can complete the desired solution by displacing the dummy variable s to the original function.
 * {| style="width:60%" border="0"

$$ \frac{d^2}{dt^2} f\left(Y^1(t),t\right) =f_{,ss} \left(\dot Y^1\right)^2 + 2f_{,st} \dot Y^1 + f_{,s}\ddot Y^1 + f_{,tt} $$
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * style= |
 * }

Similaritry to/Difference from Derivation of Coriolis Force
in Rotating coordinate systems(i.e. the earth) To compare second order differential equation to derivation of Coriolis effect, the idea of fictitious forces is used. If the rotation of frame B can be described as a vector Ω pointed along the axis of rotation


 * $$ |\boldsymbol{\Omega} | = \frac {d \theta }{dt} = \omega (t) \, $$

then the time derivative of the three unit vectors consisting the vector Ω, defined as an observed frame with respect to frame A is


 * $$ \frac {d \mathbf{u}_j (t)}{dt} = \boldsymbol{\Omega} \times \mathbf{u}_j (t) \, $$

and


 * $$\frac {d^2 \mathbf{u}_j (t)}{dt^2}= \frac{d\boldsymbol{\Omega}}{dt} \times \mathbf{u}_j+ \boldsymbol{\Omega} \times \underbrace{\left(  \boldsymbol{\Omega} \times  \mathbf{u}_j (t) \right)}_{\frac {d \mathbf{u}_j (t)}{dt}}, $$

as is verified using the properties of the vector cross product. These derivative formulas now are applied to the relationship between acceleration in an inertial frame, and that in a coordinate frame rotating with time-varying angular velocity ω ( t ). From the previous section, where subscript A refers to the inertial frame and B to the rotating frame, setting aAB = 0 to remove any translational acceleration, and focusing on only rotational properties


 * $$ \frac {d^2 \mathbf{x}_{A}}{dt^2}=\mathbf{a}_B + 2\ \sum_{j=1}^3 v_j \ \frac{d \mathbf{u}_j}{dt} $$&ensp;$$+ \sum_{j=1}^3 x_j \ \frac{d^2 \mathbf{u}_j}{dt^2}\ ,$$
 * $$\mathbf{a}_A=\mathbf{a}_B +\ 2\ \sum_{j=1}^3 v_j \boldsymbol{\Omega} \times \mathbf{u}_j (t)\ $$&ensp;$$+\   \sum_{j=1}^3 x_j \frac{d\boldsymbol{\Omega}}{dt} \times \mathbf{u}_j  \  + \sum_{j=1}^3 x_j \boldsymbol{\Omega} \times \left(  \boldsymbol{\Omega} \times  \mathbf{u}_j (t) \right)\ $$
 * $$=\mathbf{a}_B $$&ensp;$$+ 2\ \boldsymbol{\Omega} \times\sum_{j=1}^3 v_j \mathbf{u}_j (t) \  $$&ensp;$$+ \frac{d\boldsymbol{\Omega}}{dt} \times \sum_{j=1}^3 x_j \mathbf{u}_j  \   $$&ensp;$$+\ \boldsymbol{\Omega} \times \left(  \boldsymbol{\Omega} \times \sum_{j=1}^3 x_j  \mathbf{u}_j (t) \right)\ .$$

Collecting terms,


 * $$\mathbf{a}_A=\mathbf{a}_B + 2\ \boldsymbol{\Omega} \times\mathbf{v}_B\ $$&ensp;$$+ \frac{d\boldsymbol{\Omega}}{dt} \times \mathbf{x}_B  \  + \boldsymbol{\Omega} \times \left(  \boldsymbol{\Omega} \times \mathbf{x}_B \right)\ .$$

Technically, the aceleration aA in the inertial frame A is not exact with the acceleration aB seen by observers in the rotational frame B but has several additional geometric acceleration terms associated with the rotation of B. Rearranging the above equation by the acceleration B

\mathbf{a}_{B} = \mathbf{a}_A - 2 \boldsymbol\Omega \times \mathbf{v}_{B} - \boldsymbol\Omega \times (\boldsymbol\Omega \times  \mathbf{x}_B )  - \frac{d \boldsymbol\Omega}{dt} \times \mathbf{x}_B \. $$

The force upon the object in the B coordinate system with respect to observers in the rotating frame A is FB = m aB. If their observations are to result in the correct force on the object when using Newton's laws, they must consider that the additional force Ffict is present, so the end result is FB = FA + Ffict. Thus, the fictitious force used by observers in B to get the correct behavior of the object from Newton's laws equals:



\mathbf{F}_{\mathrm{fict}} = \underbrace{- 2 m \boldsymbol\Omega \times \mathbf{v}_{B}}_{Coriolis force} \underbrace{- m \boldsymbol\Omega \times (\boldsymbol\Omega \times \mathbf{x}_B )}_{centrifugal force} $$&ensp;$$ \underbrace{\ - m \frac{d \boldsymbol\Omega }{dt} \times \mathbf{x}_B}_{Euler force=0}  \. $$

When the rate of rotation doesn't change, as is typically the case for a planet, the Euler force is zero.

= HW1.problem 6 =

Given

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

$$ \left(1-x^2 \right)\cdot y''-2x\cdot y'+2y=0$$
 * style="width:95%" |
 * style="width:95%" |
 * (1)
 * }

Verify

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

$$L_{2} \left( {y}^{1}_{H}\right)=L_{2} \left( {y}^{2}_{H}\right)=0$$
 * style="width:95%" |
 * style="width:95%" |
 * (2)
 * }

Solution

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

To find $$L_{2} \left( {y}^{1}_{H}\right)=0$$
 * style="width:95%" |
 * style="width:95%" |

Differntiating the first given solution, and differentiating again to find

$${y}^{1}_{H}\left(x \right)=x$$

$$\frac{dy}{dx}=1,$$

$$ \frac{d^2y}{dx^2}=0$$

These can then be substituted into the given differential equation (1) to find

$$ L_{2} \left( {y}^{1}_{H}\right)=\left(1-x^2 \right)\cdot \left( 0 \right)-2x\left( 1 \right)+2x=0$$

To find $$L_{2} \left( {y}^{2}_{H}\right)=0$$

Repeatly differentiated the second solution to find

$${y}^{2}_{H}\left(x \right)= \frac{x}{2}log \left( \frac{1+x}{1-x} \right)-1$$

$$\frac{dy}{dx}= \frac{1}{2}log \left( \frac{1+x}{1-x} \right)+ \left( \frac{1}{1+x}+ \frac{1}{1-x}\right)=\frac{1}{2}log \left( \frac{1+x}{1-x} \right)+ \left(  \frac{x}{1-x^2}\right),  $$

$$\frac{d^2y}{dx^2}= \frac{1}{2} \left( \frac{1}{1+x}+ \frac{1}{1-x} \right)+ \frac{1\cdot \left( 1-x^2\right)-x\cdot \left( -2x\right)}{\left( 1-x^2 \right)^2}=\frac{1}{ \left(1-{x}^{2} \right)}+ \frac{1+x^2}{\left( 1-x^2 \right)^2}$$

Again, substituted into the given differential equation (1) to find

$$L_{2} \left( {y}^{2}_{H}\right)=\left(1-x^2 \right)\cdot \left( \frac{1}{ \left(1-{x}^{2} \right)}+ \frac{1+x^2}{\left( 1-x^2 \right)^2} \right)-2x\left( \frac{1}{2}log \left( \frac{1+x}{1-x} \right)+ \left( \frac{x}{1-x^2}\right) \right)+2\left(\frac{x}{2}log \left( \frac{1+x}{1-x} \right)-1\right)=1+\frac{1-x^2}{1-x^2}-2=0$$

Thus,
 * }


 * {| style="width:20%" border="0"

$$L_{2} \left( {y}^{1}_{H}\right)=L_{2} \left( {y}^{2}_{H}\right)=0$$
 * style="width:80%; padding:10px; border:2px solid #8888aa" |
 * style="width:80%; padding:10px; border:2px solid #8888aa" |


 * style= |
 * }

==(Referred from Wikipedia)Mathematical derivation of fictitious forces for understanding of Coriolis effect ["Fictitious force" http://en.wikipedia.org/w/index.php?title=Fictitious_force&action=submit] ==



General derivation
Many problems require use of noninertial reference frames, for example, those involving satellites and particle accelerators. Figure 2 shows a particle with mass m and position vector xA(t) in a particular inertial frame A. Consider a non-inertial frame B whose origin relative to the inertial one is given by XAB(t). Let the position of the particle in frame B be xB(t). What is the force on the particle as expressed in the coordinate system of frame B?

To answer this question, let the coordinate axis in B be represented by unit vectors uj with j any of { 1, 2, 3 } for the three coordinate axes. Then


 * $$ \mathbf{x}_{B} = \sum_{j=1}^3 x_j\ \mathbf{u}_j \ . $$

The interpretation of this equation is that xB is the vector displacement of the particle as expressed in terms of the coordinates in frame B at time t. From frame A the particle is located at:


 * $$\mathbf{x}_A =\mathbf{X}_{AB} + \sum_{j=1}^3 x_j\ \mathbf{u}_j \ . $$

As an aside, the unit vectors { uj } cannot change magnitude, so derivatives of these vectors express only rotation of the coordinate system B. On the other hand, vector XAB simply locates the origin of frame B relative to frame A, and so cannot include rotation of frame B.

Taking a time derivative, the velocity of the particle is:


 * $$ \frac {d \mathbf{x}_{A}}{dt} =\frac{d \mathbf{X}_{AB}}{dt}+ \underbrace{\sum_{j=1}^3 \frac{dx_j}{dt} \mathbf{u}_j}_{\mathbf{v}_B} + \sum_{j=1}^3 x_j\ \frac{d \mathbf{u}_j}{dt} \ . $$

The second term summation is the velocity of the particle, say vB as measured in frame B. That is:


 * $$ \frac {d \mathbf{x}_{A}}{dt} =\mathbf{v}_{AB}+ \mathbf{v}_B + \sum_{j=1}^3 x_j \ \frac{d \mathbf{u}_j}{dt} \ . $$

The interpretation of this equation is that the velocity of the particle seen by observers in frame A consists of what observers in frame B call the velocity, namely vB, plus two extra terms related to the rate of change of the frame-B coordinate axes. One of these is simply the velocity of the moving origin vAB. The other is a contribution to velocity due to the fact that different locations in the non-inertial frame have different apparent velocities due to rotation of the frame; a point seen from a rotating frame has a rotational component of velocity that is greater the further the point is from the origin.

To find the acceleration, another time differentiation provides:


 * $$ \frac {d^2 \mathbf{x}_{A}}{dt^2} = \mathbf{a}_{AB}+\frac {d\mathbf{v}_B}{dt} + \sum_{j=1}^3 \frac {dx_j}{dt} \ \frac{d \mathbf{u}_j}{dt} + \sum_{j=1}^3 x_j \ \frac{d^2 \mathbf{u}_j}{dt^2}\ . $$

Using the same formula already used for the time derivative of xB, the velocity derivative on the right is:


 * $$\frac {d\mathbf{v}_B}{dt} =\sum_{j=1}^3 \frac{d v_j}{dt} \ \mathbf{u}_j+ \sum_{j=1}^3 v_j \ \frac{d \mathbf{u}_j}{dt} =\mathbf{a}_B + \sum_{j=1}^3 v_j \ \frac{d \mathbf{u}_j}{dt} \ . $$

Consequently,
 * $$ \frac {d^2 \mathbf{x}_{A}}{dt^2}=\mathbf{a}_{AB}+\mathbf{a}_B + 2\ \sum_{j=1}^3 v_j \ \frac{d \mathbf{u}_j}{dt} + \sum_{j=1}^3 x_j \ \frac{d^2 \mathbf{u}_j}{dt^2}\ . $$&ensp;&ensp;&ensp;(Eq. 1)

The interpretation of this equation is as follows: the acceleration of the particle in frame A consists of what observers in frame B call the particle acceleration aB, but in addition there are three acceleration terms related to the movement of the frame-B coordinate axes: one term related to the acceleration of the origin of frame B, namely aAB, and two terms related to rotation of frame B. Consequently, observers in B will see the particle motion as possessing "extra" acceleration, which they will attribute to "forces" acting on the particle, but which observers in A say are "fictitious" forces arising simply because observers in B do not recognize the non-inertial nature of frame B.

The factor of two in the Coriolis force arises from two equal contributions: (i) the apparent change of an inertially constant velocity with time because rotation makes the direction of the velocity seem to change (a dvB / dt term) and (ii) an apparent change in the velocity of an object when its position changes, putting it nearer to or further from the axis of rotation (the change in Σ xj $$\stackrel{\frac{d}{dt}}{}$$uj due to change in x j ).

To put matters in terms of forces, the accelerations are multiplied by the particle mass:


 * $$\mathbf{F}_A = \mathbf{F}_B + m\ \mathbf{a}_{AB}+ 2m\ \sum_{j=1}^3 v_j \ \frac{d \mathbf{u}_j}{dt} + m\ \sum_{j=1}^3 x_j \ \frac{d^2 \mathbf{u}_j}{dt^2}\ . $$

The force observed in frame B, FB = m aB is related to the actual force on the particle, FA, by:


 * $$\mathbf{F}_B = \mathbf{F}_A + \mathbf{F}_{\mbox{fictitious}} \ . $$

where:


 * $$ \mathbf{F}_{\mbox{fictitious}} =-m\ \mathbf{a}_{AB} -2m\ \sum_{j=1}^3 v_j \ \frac{d \mathbf{u}_j}{dt} - m\ \sum_{j=1}^3 x_j \ \frac{d^2 \mathbf{u}_j}{dt^2}\ . $$

Thus, we can solve problems in frame B by assuming that Newton's second law holds (with respect to quantities in that frame) and treating Ffictitious as an additional force.

Below are a number of examples applying this result for fictitious forces. More examples can be found in the article on centrifugal force.

Rotating coordinate systems
A common situation in which noninertial reference frames are useful is when the reference frame is rotating. Because such rotational motion is non-inertial, due to the acceleration present in any rotational motion, a fictitious force can always be invoked by using a rotational frame of reference. Despite this complication, the use of fictitious forces often simplifies the calculations involved.

To derive expressions for the fictitious forces, derivatives are needed for the apparent time rate of change of vectors that take into account time-variation of the coordinate axes. If the rotation of frame B is represented by a vector Ω pointed along the axis of rotation with orientation given by the right-hand rule, and with magnitude given by


 * $$ |\boldsymbol{\Omega} | = \frac {d \theta }{dt} = \omega (t) \, $$

then the time derivative of any of the three unit vectors describing frame B is:


 * $$ \frac {d \mathbf{u}_j (t)}{dt} = \boldsymbol{\Omega} \times \mathbf{u}_j (t) \, $$

and


 * $$\frac {d^2 \mathbf{u}_j (t)}{dt^2}= \frac{d\boldsymbol{\Omega}}{dt} \times \mathbf{u}_j +\boldsymbol{\Omega} \times \frac{d \mathbf{u}_j (t)}{dt} $$&ensp;$$= \frac{d\boldsymbol{\Omega}}{dt} \times \mathbf{u}_j+ \boldsymbol{\Omega} \times \left(  \boldsymbol{\Omega} \times  \mathbf{u}_j (t) \right)\, $$

as is verified using the properties of the vector cross product. These derivative formulas now are applied to the relationship between acceleration in an inertial frame, and that in a coordinate frame rotating with time-varying angular velocity ω ( t ). From the previous section, where subscript A refers to the inertial frame and B to the rotating frame, setting aAB = 0 to remove any translational acceleration, and focusing on only rotational properties (see Eq. 1):


 * $$ \frac {d^2 \mathbf{x}_{A}}{dt^2}=\mathbf{a}_B + 2\ \sum_{j=1}^3 v_j \ \frac{d \mathbf{u}_j}{dt} $$&ensp;$$+ \sum_{j=1}^3 x_j \ \frac{d^2 \mathbf{u}_j}{dt^2}\ ,$$
 * $$\mathbf{a}_A=\mathbf{a}_B +\ 2\ \sum_{j=1}^3 v_j \boldsymbol{\Omega} \times \mathbf{u}_j (t)\ $$&ensp;$$+\   \sum_{j=1}^3 x_j \frac{d\boldsymbol{\Omega}}{dt} \times \mathbf{u}_j  \  + \sum_{j=1}^3 x_j \boldsymbol{\Omega} \times \left(  \boldsymbol{\Omega} \times  \mathbf{u}_j (t) \right)\ $$
 * $$=\mathbf{a}_B $$&ensp;$$+ 2\ \boldsymbol{\Omega} \times\sum_{j=1}^3 v_j \mathbf{u}_j (t) \  $$&ensp;$$+ \frac{d\boldsymbol{\Omega}}{dt} \times \sum_{j=1}^3 x_j \mathbf{u}_j  \   $$&ensp;$$+\ \boldsymbol{\Omega} \times \left(  \boldsymbol{\Omega} \times \sum_{j=1}^3 x_j  \mathbf{u}_j (t) \right)\ .$$

Collecting terms, the result is the so-called acceleration transformation formula:


 * $$\mathbf{a}_A=\mathbf{a}_B + 2\ \boldsymbol{\Omega} \times\mathbf{v}_B\ $$&ensp;$$+ \frac{d\boldsymbol{\Omega}}{dt} \times \mathbf{x}_B  \  + \boldsymbol{\Omega} \times \left(  \boldsymbol{\Omega} \times \mathbf{x}_B \right)\ .$$

The physical acceleration aA due to what observers in the inertial frame A call real external forces on the object is, therefore, not simply the acceleration aB seen by observers in the rotational frame B, but has several additional geometric acceleration terms associated with the rotation of B. As seen in the rotational frame, the acceleration aB of the particle is given by rearrangement of the above equation as:

\mathbf{a}_{B} = \mathbf{a}_A - 2 \boldsymbol\Omega \times \mathbf{v}_{B} - \boldsymbol\Omega \times (\boldsymbol\Omega \times  \mathbf{x}_B )  - \frac{d \boldsymbol\Omega}{dt} \times \mathbf{x}_B \. $$

The net force upon the object according to observers in the rotating frame is FB = m aB. If their observations are to result in the correct force on the object when using Newton's laws, they must consider that the additional force Ffict is present, so the end result is FB = FA + Ffict. Thus, the fictitious force used by observers in B to get the correct behavior of the object from Newton's laws equals:



\mathbf{F}_{\mathrm{fict}} = - 2 m \boldsymbol\Omega \times \mathbf{v}_{B} - m \boldsymbol\Omega  \times (\boldsymbol\Omega \times \mathbf{x}_B ) $$&ensp;$$\ - m \frac{d \boldsymbol\Omega  }{dt} \times \mathbf{x}_B \. $$

Here, the first term is the Coriolis force, the second term is the centrifugal force, and the third term is the Euler force. When the rate of rotation doesn't change, as is typically the case for a planet, the Euler force is zero.

Reference http://en.wikipedia.org/w/index.php?title=Fictitious_force&action=submit

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