User:Egm6321.f12.team2/Report1

=Problem 1.1=

Given
The first total time derivative is: where

Show That
where

Total Time Derivative
Here $$\displaystyle S=Y^1(t) $$, the total time derivative of the given function can be expressed as:

Substitute $$\displaystyle \dot Y^1 $$ for $$\displaystyle \frac{dY^1}{dt} $$:

Second Total Time Derivative
Take the second total time derivative of the equation (1.3):

We solved this by ourselves

=Problem 1.2=

Given
with $$\displaystyle \dot Y^1 := \frac {dY^1(t)}{dt}$$ with $$ \displaystyle f_{,S}(Y^1,t) :=\frac{\partial f(Y^1,t)}{\partial S}$$; $$ \displaystyle f_{,St}(Y^1,t):=\frac{\partial ^2 f(Y^1,t)}{\partial S \partial t} $$

Find
Derive And show the similarity with the derivation of the Coriolis force.

Solution
The function $$\displaystyle f $$ is a function of both space,$$\displaystyle S $$ and time, $$\displaystyle t $$. Here $$\displaystyle S=Y^1(t) $$. The total time derivative of the given function can be expressed as:



Substitute

 Take the second total time derivative of the equation:

Derivation of Coriolis Force


The instantaneous position vector of a particle $$ \overrightarrow{s}(t) $$ at time $$t$$ in a rotating frame as seen from an inertial frame is given by $$ \overrightarrow{s}(t)=\overrightarrow{s}_{0} + \overrightarrow{r}(t) $$. Where $$\overrightarrow{r}(t)$$ is the position vector of the particle from the rotating frame and $$\overrightarrow{s}_{0}$$ is the position vector of the origin of the rotating coordinate axis from the inertial frame.

Here $$\overrightarrow{r}(t) = r_{1}\hat{u}_{1}+r_{2}\hat{u}_{2}+r_{3}\hat{u}_{3}$$ where, $$\hat{u}_{1}$$, $$\hat{u}_{2}$$, $$\hat{u}_{3}$$ are the unit vectors along the rotating coordinate axes.

The Velocity of the particle in the inertial frame is given by $$ \overrightarrow{v}(t) $$ where, The Acceleration of the particle in the inertial frame is given by $$ \overrightarrow{a}(t) $$ where,

Then the acceleration vector $$ \overrightarrow{a}(t) $$ can be written as,

The two differentials are evaluated individually. The first differential $$ 1 $$from Eqn. 2.0 is

Let $$\frac{dv_i}{dt}=a_i$$,then

$$ \frac{d}{dt}(v_i\hat{u_i})=a_i\hat{v_i}+v_i\frac{d{v_i}}{dt} $$ The second differential $$ 2 $$ from Equation 2.0 can be written as

Substituting both these differentials into the equation for $$a$$ will give,

In this expansion, the term corresponding to the coriolis force is $$2v_i\frac{d\hat{u_i}}{dt}$$.

This equation is of the same form as the equation for $$\frac{d^2}{dt^2} f\left(Y^1(t),t\right)$$. Here the term corresponding to the coriolis force $$2v_i\frac{d\hat{u_i}}{dt}$$ and the term $$2f_{,S t}(Y^1,t) \dot Y^1$$ in the expansion of $$\frac{d^2}{dt^2} f\left(Y^1(t),t\right)$$ are equivalent in their purpose.

We solved this by ourselves

=Problem 1.3=

Given
Where:


 * $$\displaystyle u^2_{,SS}(Y^1,t) := \frac {\partial u^2(Y^1,t)}{(\partial S)^2}$$


 * $$\displaystyle u^1_{,tt}(Y^1,t) := \frac {\partial u^1(Y^1,t)}{(\partial t)^2}$$

Equations 3.1 and 3.2 refer to Ref: VQ&O 1989 CMAME Eqs.(2.5a), and Ref: VQ&O 1989 CMAME Eqs.(2.5b), respectively.

Find
Analyze the dimension of all terms in $$c_0(Y^1,t)$$, and provide the physical meaning.

Solution
Dimensions: $$[F^1]=Force$$ $$[\bar{R}]=Length$$ $$[u^2_{,SS}(Y^1,t)]=Length^{-1}$$ $$[\bar{R} u^2_{,SS}(Y^1,t)]=1$$ $$[F^2]=Force$$ $$[u^2_{,S}]=1$$ $$[F^2u^2_{,S}]=Force$$ $$[T]=Torque$$ $$[R]=Length$$ $$[\frac TR]=Force$$ $$[u^1_{,tt}]=Length/Time^2$$ $$[u^2_{,Stt}]=Time^{-2}$$ $$[u^2_{,S}]=1$$ $$[u^2_{,tt}]=Length/Time^2$$ $$[M]=Mass$$ $$[M[(1-\bar R u^2_{,SS})(u^1_{,tt}-\bar u^2_{,Stt})+u^2_{,S}u^2_{,tt}]]=Force$$ $$[(1-\bar R u^2_{,SS})(u^1_{,tt}-\bar u^2_{,Stt})+u^2_{,S}u^2_{,tt}]=Length/Time^2$$

Physical meanings:

$$c_0(Y^1,t) = horizontal \; force \; acting \; on \; wheel/magnet$$ $$\bar{R} = adjusted \; effect \; of \; radius \; of \; wheel$$ $$R = radius \; of \; wheel$$ $$u^2_{,SS}(Y^1,t) = curvature$$ $$u^2_{,S} = gradient$$ $$M = mass $$ $$[(1-\bar R u^2_{,SS})(u^1_{,tt}-\bar u^2_{,Stt})+u^2_{,S}u^2_{,tt}]=Length/Time^2 = acceleration $$ $$[\frac TR] =Force\ due\ to\ torque\ in\ the\ horizontal\ direction.$$ $$-F^1[1-\bar R u^2_{,SS}(Y^1,t)]=Force\ due\ to\ curvature\ of\ the\ path\ in\ thr\ horizontal\ direction.$$ $$F^2u^2_{,S}(Y^1,t)=Component\ of\ F^{2} force\ acting\ along\ the\ path\ of\ motion$$ We solved this by ourselves

=Problem 1.4=

Find
Draw the polar coordinate lines, $$\displaystyle (\psi_1,\psi_2)=(r,\theta) $$ in a 2-D plane emanating from a point, not at the origin.

Solution


Polar coordinate lines, $$\displaystyle (\psi_1,\psi_2)$$, originating from point P can be expressed in 2-D polar coordinates, $$\displaystyle (r,\theta) $$ as shown in the figure above.

We solved this by ourselves.

=Problem 1.5=

Find
Show that the Given equation has the following form:

Solution
Referring to p.4-5 (1), we can simplify to:

And continue to simplify:

And finally,

We solved this by ourselves.

=Problem 1.6=

Given
In the equation of motion (EOM) of wheel/magnet, (1) p.3-3,

Find
Show $$\displaystyle c_3(Y^1,t)\ddot Y^1$$ is nonlinear regardless if $$\displaystyle u^2(Y^1,t) $$ is linear or nonlinear with respect to $$\displaystyle Y^1$$.

Solution
Preliminaries:

$$\displaystyle Y^1:[t_0,+\infty)\rightarrow \mathbb R $$ $$\displaystyle\forall f,g:[t_0,+\infty) \rightarrow \mathbb R $$ $$\displaystyle \forall \alpha,\beta \in \mathbb R $$ $$\displaystyle c_3(Y^1,t)= M[1-\bar {R} \frac{\partial^2 u^2}{\partial s^2}(Y^1,t)] $$  Analysis:

is nonlinear if and only if

1.) From VQ & Olsson 1989 (2.7b), $$\displaystyle u^2(Y^1,t) $$ is nonlinear wrt $$\displaystyle Y^1 $$

Thus,

Since $$\displaystyle \ddot u^2 $$ is contained by the partial differential equation governing $$\displaystyle u^2 $$, and is nonlinear w.r.t $$\displaystyle Y^1 $$,

Thus, A is still nonlinear.

2. Check if A is still nonlinear if $$\displaystyle u^2(Y^1,t) $$ were linear wrt $$\displaystyle Y^1 $$:

If $$\displaystyle u^2(Y^1,t) $$ were linear wrt $$\displaystyle Y^1 $$, then

Thus,

Thus, D is nonlinear in all cases.

We solved this by ourselves. Class notes Pea1.f12.sec5, and VQ & Olsson 1989 paper were used as references.

=Problem 1.7=

Given
An ODE of 2nd order with variable coefficients is read as

where $$\displaystyle a_0(x), a_1(x)$$ are variable coefficients

Find
Show that

is a linear operator.

Solution
We must show that

so,

Furthermore,

So,

And is therefore, linear.

We solved this by ourselves

=Contributing Members=