User talk:Egm6321.f10.team5.oh/hw3

= Problem 1 - Derive EOM (SC-N1-ODE) = From Meeting 13, p. 13-1 ~ p. 13-2

Given
Figure shows the Trajectory of a projectile (ex:Rocket):

Find

 * $$\left( 1 \right)$$ Drive Equation of Motion (EOM)
 * $$\left( 2 \right)$$ Particular case $$\displaystyle k = 0 $$ : Verify $$y\left( x \right)$$ is parabolla
 * $$\left( 3 \right)$$ Consider $$k\ne 0$$, $$\displaystyle {{v}_{x}}=0$$,


 * $$ (3.1) $$ Find $${{v}_{y}}\left( t \right)$$, $$\displaystyle y(t)$$ for $$ \displaystyle m$$ = constant
 * $$ (3.2) $$ Find $${{v}_{y}}\left( t \right)$$, $$\displaystyle y(t)$$ if $$m=m\left( t \right)$$

Solve
Part 1 Consider the trajectory of a projectile (ex. Rocket)


 * Various forces acting on the projectile at time 't' are:


 * 1) Weight of the projectile


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


 * $$\displaystyle W=mg$$
 * }
 * }
 * }


 * 2) Inertia force


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


 * $${{F}_{a}}=ma=m\cdot \frac{dv}{dt}$$ for particle with constant mass
 * $${{F}_{a}}=ma=m\cdot \frac{dv}{dt}$$ for particle with constant mass


 * }
 * }


 * 3) Air resistance which is proportional to the velocity of particle


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


 * $$\displaystyle {{F}_{D}}=k{{v}^{n}}$$
 * }
 * }
 * }

Now consider the force equilibrium in both horizontal and vertical direction


 * a) Force Equilibrium in horizontal direction:


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


 * $$\sum{{{F}_{H}}=0}$$
 * }
 * }
 * }


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


 * $$m\cdot \frac{d{{v}_{x}}}{dt}+k{{v}^{n}}\cos \alpha =0$$,
 * $$m\cdot \frac{d{{v}_{x}}}{dt}+k{{v}^{n}}\cos \alpha =0$$,


 * where $$\displaystyle {{v}_{x}}\to $$ horizontal component of velocity


 * }
 * }


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

$$m\cdot \frac{d{{v}_{x}}}{dt}=-k{{v}^{n}}\cos \alpha $$ $$
 * style="width:2%; padding:1px; border:2px solid #FF0000" |
 * style="width:2%; padding:1px; border:2px solid #FF0000" |
 * $$\displaystyle (Eq. 1.1)
 * }
 * }


 * b) Force Equilibrium in the vertical direction:


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


 * $$\sum{{{F}_{V}}=0}$$
 * }
 * }
 * }


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

$$m\cdot \frac{d{{v}_{y}}}{dt}+k{{v}^{n}}\sin \alpha +mg=0$$,


 * where $$\displaystyle {{v}_{y}}\to $$ vertical component of velocity


 * }
 * }


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

$$m\cdot \frac{d{{v}_{y}}}{dt}=-k{{v}^{n}}\sin \alpha -mg$$ $$
 * style="width:2%; padding:1px; border:2px solid #FF0000" |
 * style="width:2%; padding:1px; border:2px solid #FF0000" |
 * $$\displaystyle (Eq. 1.2)
 * }
 * }

Part 2 


 * Particular case: When $$\displaystyle k=0$$

Eq.1.1 reduces to


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


 * $$m\frac{d{{v}_{x}}}{dt}=0\Rightarrow \frac{d{{v}_{x}}}{dt}=0$$
 * }
 * }
 * }

Integrating the above equation gives:


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

$$
 * $${{v}_{x}}\left( t \right)={{c}_{1}}$$
 * $$\displaystyle (Eq. 1.3)
 * $$\displaystyle (Eq. 1.3)
 * }
 * }

Apply 'initial condition' to determine integration constant,$$\displaystyle{{c}_{1}}$$


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


 * $${{v}_{x}}\left( t=0 \right)={{v}_}={{c}_{1}}$$
 * }
 * }
 * }

Now Eq.1.3 becomes:
 * {| style="width:100%" border="0"


 * $${{v}_{x}}\left( t \right)={{v}_}$$
 * }
 * }
 * }

Integrate the above equation to obtain ,$$\displaystyle x$$


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


 * $$x\left( t \right)={{v}_}t+{{c}_{2}}$$
 * }
 * }
 * }

Then 'Initial condition' is applied to determine ,$$\displaystyle{{c}_{2}}$$


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


 * $$x\left( t=0 \right)={{v}_}\times 0+{{c}_{2}}\Rightarrow {{c}_{2}}={{x}_{0}}$$
 * }
 * }
 * }

Therefore,


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


 * $$x\left( t \right)={{v}_}t+{{x}_{0}}$$
 * }
 * }
 * }

Now $$\displaystyle t$$, can be expressed in terms of $$\displaystyle x$$,


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

$$
 * $$t=\frac{x-{{x}_{0}}}$$
 * $$\displaystyle (Eq. 1.4)
 * $$\displaystyle (Eq. 1.4)
 * }
 * }

Similarly When $$\displaystyle k=0$$, Eq.1.2 reduces to


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


 * $$m\frac{d{{v}_{y}}}{dt}=-mg\Rightarrow \frac{d{{v}_{y}}}{dt}=-g$$
 * }
 * }
 * }

Integrate the above equation to evaluate, $$\displaystyle {{v}_{y}}$$


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

$$
 * $${{v}_{y}}\left( t \right)=-gt+{{c}_{3}}$$
 * $$\displaystyle (Eq. 1.5)
 * $$\displaystyle (Eq. 1.5)
 * }
 * }

Apply 'initial condition' to obtain $$\displaystyle {{c}_{3}}$$


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


 * $${{v}_{y}}\left( t=0 \right)={{v}_}={{c}_{3}}$$
 * }
 * }
 * }

Now Eq.1.5 becomes,


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


 * $${{v}_{t}}\left( t \right)=-gt+{{v}_}$$
 * }
 * }
 * }

Then integrate the above equation to determine, $$\displaystyle y $$


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


 * $$y\left( t \right)=\frac{-g{{t}^{2}}}{2}+{{v}_}t+{{c}_{4}}$$
 * }
 * }
 * }

$$\displaystyle {{c}_{4}}$$ is determined using 'initial condition' as:


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


 * $$y\left( t=0 \right)=\frac{-g\times 0}{2}+{{v}_}\times 0+{{c}_{4}}\Rightarrow {{c}_{4}}={{y}_{0}}$$
 * }
 * }
 * }

Therefore,


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


 * $$y\left( t \right)=\frac{-g{{t}^{2}}}{2}+{{v}_}t+{{y}_{0}}$$
 * }
 * }
 * }

Now Substitute Eq.1.4 for $$\displaystyle t$$ in the above equation;


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

$$y\left( x \right)=-\frac{g}{2}{{\left( \frac{x-{{x}_{0}}} \right)}^{2}}+{{v}_}\left( \frac{x-{{x}_{0}}} \right)+{{y}_{0}}$$ $$
 * style="width:2%; padding:1px; border:2px solid #FF0000" |
 * style="width:2%; padding:1px; border:2px solid #FF0000" |
 * $$\displaystyle (Eq. 1.6)
 * }
 * }

Eq.1.6 is in the form of a parabolic equation. Therefore $$\displaystyle y\left( x \right)$$ is parabola.

Part 3

With $$v_{xo}=0$$, and $$\cos\alpha=\frac{v_x}{v}$$ (from the geometry)

Which presents an interesting fact, once $$v_x$$ is equal to zero it remains zero as its derivative is also zero (except when $$v=0$$ and $$n-1<0$$). Thus, if $$v_{xo}=0$$ the $$x$$ velocity remains zero and the equations of motion reduce to equation 1.2.

Part 3.1
Solving

To show exactness, we first put into a form that shows the first condition of exactness is met

so that

The second condition of exactness

Which is met if $$m$$ is not a function of time

Thus, the equation is non-exact.

The equation can be made exact through the integrating factor method

Then expressing as a total derivative and testing for exactness

Letting $$\partial h / \partial v_y = 0$$

Then

Combining

Though the equation is not integrable, by making it exact though the integrating factor method an expression was found.

Part 3.2
If $$m(t)\ne0$$ the integrating factor method is complicated in equation 1.22 as the partial of $$m$$ with respect to time remains, complicating the expression for $$h$$

Author and Proof-reader
[Author]

[Proof-reader]

= Problem 2 - Derive EOM (SC-L1-ODE) = From Meeting 13, p. 13-3

Given

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


 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * style="width:92%; padding:10px; border:2px solid #8888aa" |

$$\begin{align} {m}_{1}, {m}_{2} \end{align}$$ : mass of the each pendulum

$$\begin{align} {\theta}_{1}, {\theta}_{2} \end{align}$$ : the angle from the vertical to the each pendulum

$$\begin{align} {u}_{1}, {u}_{2} \end{align}$$ : applied forces to the each pendulum

$$\begin{align} l \end{align}$$ : length of the pendulum

$$\begin{align} k \end{align}$$ : force constant(or spring constant)

$$\begin{align} g \end{align}$$ : acceleration of gravity


 * }
 * }
 * }

Find
1. Derive (2-1) and (2-2)

2. Write (2.1) and (2-2) in form of (2-3)

where,

$$\begin{align} \textbf{x} := {\left [ {\theta}_{1}, \dot{{\theta}_{1}}, {\theta}_{2},\dot{{\theta}_{2}}\right]}^{T} \end{align}$$

$$\begin{align} \textbf{u} := {\left[ {u}_{1}l, {u}_{2}l\right]}^{T} \end{align}$$

Dimension of matrix

$$\begin{align} \dot{\textbf{x}} \ : \ 4 \times 1 \end{align}$$

$$\begin{align} \textbf{x} \ : \ 4 \times 1 \end{align}$$

$$\begin{align} \textbf{A} \ : \ 4 \times 4 \end{align}$$

$$\begin{align} \textbf{B} \ : \ 4 \times 2 \end{align}$$

$$\begin{align} \textbf{U} \ : \ 2 \times 1 \end{align}$$

Solve
'''Step 1. Derivation '''

 Background Knowledge 

1. Torque 

2. Hooke's Law 

3. Pendulum 

4. Moment 

5. Moment of inertia 

6. Angular acceleration 

 Derive Using above background, 

where,

$$\begin{align} \tau \end{align}$$ : torque

$$\begin{align} I=m{l}^{2}\end{align}$$ : moment of inertia

$$\begin{align} \alpha=\ddot{\theta} \end{align}$$ : angular acceleration

Therefore, left hand side is $$\ - > \ \begin{align} m {l}^{2} \ddot{\theta} \end{align}$$

Torque of the spring force

From the backgroud (Hooke's Law(wikipedia)),

where, $$\begin{align} F \end{align}$$ : restoring force $$\begin{align} k\end{align}$$ : spring constant $$\begin{align} x \end{align}$$ : displacement from the equilibrium position (in this case, x = a)

Therefore, Torque of the spring force is,
 * {| style="width:100%" border="0"

$$\begin{align} \vec{\tau}=\vec{r}\times\vec{F} =-k {a}^{2}(\mathrm{sin}{\theta}_{1}-\mathrm{sin}{\theta}_{2})\approx -k{a}^{2}({\theta}_{1}-{\theta}_{2})\\ \end{align}$$ $$
 * style="width:95%" |
 * style="width:95%" |
 * $$\displaystyle (2-6)
 * }

Torque of the gravity force
 * {| style="width:100%" border="0"

$$\begin{align} \vec{\tau}=\vec{r}\times\vec{F} = -{m}_{1}g\cdot l\mathrm{sin}{\theta}_{1}\approx -{m}_{1}g\cdot l{\theta}_{1} \end{align}$$ $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right;">$$\displaystyle (2-7)
 * }

Torque of the applied force
 * {| style="width:100%" border="0"

$$\begin{align} \vec{\tau}=\vec{r}\times\vec{F} = {u}_{1}\cdot l \mathrm{cos}{\theta}_{1} \approx {u}_{1}\cdot l \end{align}$$ $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right;">$$\displaystyle (2-8)
 * }

Using (2-4) ~ (2-8)


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

$$ {m}_{1}{l}^{2}\ddot{{\theta}_{1}}=-k{a}^{2}({\theta}_{1}-{\theta}_{2})-{m}_{1}gl{\theta}_{1}+{u}_{1}l $$
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * }
 * }
 * }

(2-2) can be verified with same procedure.


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

$$ {m}_{2}{l}^{2}\ddot{{\theta}_{2}}=-k{a}^{2}({\theta}_{2}-{\theta}_{1})-{m}_{2}gl{\theta}_{2}+{u}_{2}l $$
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * }
 * }
 * }

'''Step 2. Find A, B and U'''

Let's make a equation of a matrix with the information that we already have.

rearrange the derived equations (2-1) and (2-2),

Let's put them to (2-9)


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


 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * style="width:92%; padding:10px; border:2px solid #8888aa" |

$$ \begin{bmatrix} \ \dot{{\theta}_{1}} \\ \ \ddot{{\theta}_{1}} \\ \ \dot{{\theta}_{2}} \\ \ \ddot{{\theta}_{2}} \\ \end{bmatrix} = \underbrace{ \begin{bmatrix} \ 0 & 0 & 0 & 0 \\ \ -\frac{k{a}^{2}}{{m}_{1}{l}^{2}}-\frac{g}{l} & 0 & \frac{k{a}^{2}}{{m}_{1}{l}^{2}} & 0 \\ \ 0 & 0 & 0 & 0 \\ \ \frac {k{a}^{2}}{{m}_{2}{l}^{2}}& 0 & -\frac{k{a}^{2}}{{m}_{2}{l}^{2}}-\frac{g}{l} & 0 \\ \end{bmatrix} }_{\mathrm{Matrix} \mathbf{A} (4\times4)} \begin{bmatrix} \ {\theta}_{1} \\ \ \dot{{\theta}_{1}} \\ \ {\theta}_{2} \\ \ \dot{{\theta}_{2}} \\ \end{bmatrix} + \underbrace{ \begin{bmatrix} \ 0 & 0 \\ \ \frac{1}{{m}_{1}{l}^{2}} & 0 \\ \ 0 & 0 \\ \ 0 & \frac{1}{{m}_{2}{l}^{2}} \\ \end{bmatrix} }_{\mathrm{Matrix} \mathbf{B} (4\times2)} \begin{bmatrix} \ {u}_{1}l \\ \ {u}_{2}l \\ \end{bmatrix}

$$


 * }
 * }
 * }

Given
Shown in figure are the two Pendulums connected by a spring:

Find

 * $$\left( 1 \right)$$ Derive equation of motion:


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

$$
 * $${{m}_{1}}{{l}^{2}}{{\overset{\centerdot \centerdot }{\mathop{\theta }}\,}_{1}}=-k{{a}^{2}}\left( {{\theta }_{1}}-{{\theta }_{2}} \right)-{{m}_{1}}gl{{\theta }_{1}}+{{u}_{1}}l$$
 * <p style="text-align:right;">$$\displaystyle (Eq. 2.1)
 * <p style="text-align:right;">$$\displaystyle (Eq. 2.1)


 * }
 * }


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

$$
 * $${{m}_{2}}{{l}^{2}}{{\overset{\centerdot \centerdot }{\mathop{\theta }}\,}_{2}}=-k{{a}^{2}}\left( {{\theta }_{2}}-\theta 1 \right)-{{m}_{2}}gl{{\theta }_{2}}+{{u}_{2}}l$$
 * <p style="text-align:right;">$$\displaystyle (Eq. 2.2)
 * <p style="text-align:right;">$$\displaystyle (Eq. 2.2)
 * }
 * }


 * $$\left( 2 \right)$$ Write Eq.2.1 and Eq.2.2 in the form of [[media:2010_09_21_14_56_48.djvu|Mtg 13 (c),page2 ]], of:


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

$$
 * $$\underset{-}{\overset{\centerdot }{\mathop{x}}}\,=\underset{-}{\mathop{A}}\,\left( t \right)\underset{-}{\mathop{x}}\,\left( t \right)+\underset{-}{\mathop{B}}\,\left( t \right)\underset{-}{\mathop{u}}\,\left( t \right)$$
 * <p style="text-align:right;">$$\displaystyle (Eq. 2.3)
 * <p style="text-align:right;">$$\displaystyle (Eq. 2.3)
 * }
 * }


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

$$\underset{-}{\mathop{x}}\,={{\left[ \begin{matrix} {{\theta }_{1}} & \overset{\centerdot }{\mathop{{{\theta }_{1}}}}\, & {{\theta }_{2}} & \overset{\centerdot }{\mathop{{{\theta }_{2}}}}\, \\ \end{matrix} \right]}^{T}}$$ and $$\underset{-}{\mathop{u}}\,={{\left[ \begin{matrix} {{u}_{1}}l & {{u}_{2}}l \\ \end{matrix} \right]}^{T}}$$
 * Given
 * Given
 * }
 * }

Solution

 * $$\left( 1 \right)$$ Derive equation of motion:
 * (a) Consider Free Body Diagram of left pendulum:




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


 * $$\displaystyle l\sin {{\theta }_{1}}=l{{\theta }_{1}}$$ and $$\displaystyle l\cos {{\theta }_{1}}=l$$
 * }
 * }
 * }


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


 * $$acceleration, {{a}_{1}}=\frac{{{d}^{2}}\left( l{{\theta }_{1}} \right)}{d{{t}^{2}}}=l{{\overset{\centerdot \centerdot }{\mathop{\theta }}\,}_{1}}$$
 * }
 * }
 * }


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


 * $$ Inertia force ={{m}_{1}}l{{\overset{\centerdot \centerdot }{\mathop{\theta }}\,}_{1}}$$
 * }
 * }
 * }


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


 * $$\displaystyle Spring force =ka{{\theta }_{1}}-ka{{\theta }_{2}}=ka({{\theta }_{1}}-{{\theta }_{1}})$$
 * }
 * }
 * }

Now using D'Alembert's_principle, sum of the moments about pivot(A)is equal to zero


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


 * $$\sum{{{M}_{A}}=0\Rightarrow }\left( {{m}_{1}}l{{\overset{\centerdot \centerdot }{\mathop{\theta }}\,}_{1}} \right)\cdot l+\left( ka({{\theta }_{1}}-{{\theta }_{2}}) \right)\cdot a+\left( {{m}_{1}}g \right)\cdot l{{\theta }_{1}}-\left( {{u}_{1}} \right)\cdot l=0$$
 * }
 * }
 * }


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

$$\Rightarrow {{m}_{1}}{{l}^{2}}{{\overset{\centerdot \centerdot }{\mathop{\theta }}\,}_{1}}=-k{{a}^{2}}({{\theta }_{1}}-{{\theta }_{2}})-{{m}_{1}}gl{{\theta }_{1}}+{{u}_{1}}l$$ $$
 * style="width:2%; padding:1px; border:2px solid #FF0000" |
 * style="width:2%; padding:1px; border:2px solid #FF0000" |
 * <p style="text-align:right;">$$\displaystyle (Eq. 2.1)
 * }
 * }


 * (b) Consider Free Body Diagram of right pendulum:




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


 * $$\displaystyle l\sin {{\theta }_{2}}=l{{\theta }_{2}}$$ and $$\displaystyle l\cos {{\theta }_{2}}=l$$
 * }
 * }
 * }


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


 * $$acceleration, {{a}_{2}}=\frac{{{d}^{2}}\left( l{{\theta }_{2}} \right)}{d{{t}^{2}}}=l{{\overset{\centerdot \centerdot }{\mathop{\theta }}\,}_{2}}$$
 * }
 * }
 * }


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


 * $$ Inertia force ={{m}_{2}}l{{\overset{\centerdot \centerdot }{\mathop{\theta }}\,}_{2}}$$
 * }
 * }
 * }


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


 * $$\displaystyle Spring force =ka{{\theta }_{2}}-ka{{\theta }_{1}}=ka({{\theta }_{2}}-{{\theta }_{1}})$$
 * }
 * }
 * }

Using D'Alembert's_principle, sum of the moments about pivot(B)is equal to zero


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


 * $$\sum{{{M}_{B}}=0\Rightarrow }\left( {{m}_{2}}l{{\overset{\centerdot \centerdot }{\mathop{\theta }}\,}_{2}} \right)\cdot l+\left( ka({{\theta }_{2}}-{{\theta }_{1}}) \right)\cdot a+\left( {{m}_{2}}g \right)\cdot l{{\theta }_{2}}-\left( {{u}_{2}} \right)\cdot l=0$$
 * }
 * }
 * }


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

$$\Rightarrow {{m}_{2}}{{l}^{2}}{{\overset{\centerdot \centerdot }{\mathop{\theta }}\,}_{2}}=-k{{a}^{2}}({{\theta }_{2}}-{{\theta }_{1}})-{{m}_{2}}gl{{\theta }_{2}}+{{u}_{2}}l$$ $$
 * style="width:2%; padding:1px; border:2px solid #FF0000" |
 * style="width:2%; padding:1px; border:2px solid #FF0000" |
 * <p style="text-align:right;">$$\displaystyle (Eq. 2.2)
 * }
 * }


 * $$\left( 2 \right)$$ Write Eq.2.1 and Eq.2.2 in the form of Eq.2.3(system of coupled equation):


 * Eq.2.1 can be rearranged as,


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

$$
 * $${{\overset{\centerdot \centerdot }{\mathop{\theta }}\,}_{1}}=\frac{-\left( k{{a}^{2}}+{{m}_{1}}gl \right)}{m{}_{1}{{l}^{2}}}{{\theta }_{1}}+\frac{k{{a}^{2}}}{m{}_{1}{{l}^{2}}}{{\theta }_{2}}+\frac{{{u}_{1}}l}{m{}_{1}{{l}^{2}}}$$
 * <p style="text-align:right;">$$\displaystyle (Eq. 2.4)
 * <p style="text-align:right;">$$\displaystyle (Eq. 2.4)
 * }
 * }


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

$$
 * $${{\overset{\centerdot \centerdot }{\mathop{\theta }}\,}_{2}}=\frac{k{{a}^{2}}}{m{}_{2}{{l}^{2}}}{{\theta }_{1}}+\frac{-\left( k{{a}^{2}}+{{m}_{2}}gl \right)}{m{}_{2}{{l}^{2}}}{{\theta }_{2}}+\frac{{{u}_{2}}l}{m{}_{2}{{l}^{2}}}$$
 * <p style="text-align:right;">$$\displaystyle (Eq. 2.5)
 * <p style="text-align:right;">$$\displaystyle (Eq. 2.5)
 * }
 * Now Eq.2.4 and Eq.2.5 can be put in the form of Eq.2.3 as:
 * Now Eq.2.4 and Eq.2.5 can be put in the form of Eq.2.3 as:


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


 * style="width:2%; padding:1px; border:2px solid #FF0000" |
 * style="width:2%; padding:1px; border:2px solid #FF0000" |

$$\left[ \begin{matrix} {{\overset{\centerdot }{\mathop{\theta }}\,}_{1}} \\ {{\overset{\centerdot \centerdot }{\mathop{\theta }}\,}_{1}} \\ {{\overset{\centerdot }{\mathop{\theta }}\,}_{2}} \\ {{\overset{\centerdot \centerdot }{\mathop{\theta }}\,}_{2}} \\ \end{matrix} \right]=\underbrace{\left[ \begin{matrix} 0 & 1 & 0 & 0 \\   \frac{-\left( k{{a}^{2}}+{{m}_{1}}gl \right)}{m{}_{1}{{l}^{2}}} & 0 & \frac{k{{a}^{2}}}{m{}_{1}{{l}^{2}}} & 0  \\ 0 & 0 & 0 & 1 \\   \frac{k{{a}^{2}}}{m{}_{2}{{l}^{2}}} & 0 & \frac{-\left( k{{a}^{2}}+{{m}_{2}}gl \right)}{m{}_{2}{{l}^{2}}} & 0  \\ \end{matrix} \right]}_{\underset{-}{\mathop{A}}\,}\left[ \begin{matrix} {{\theta }_{1}} \\ {{\overset{\centerdot }{\mathop{\theta }}\,}_{1}} \\ {{\theta }_{2}} \\ {{\overset{\centerdot }{\mathop{\theta }}\,}_{2}} \\ \end{matrix} \right]+\underbrace{\left[ \begin{matrix} 0 & 0 \\   \frac{1}{m{}_{1}{{l}^{2}}} & 0  \\ 0 & 0 \\   0 & \frac{1}{m{}_{2}{{l}^{2}}}  \\ \end{matrix} \right]}_{\underset{-}{\mathop{B}}\,}\left[ \begin{matrix} {{u}_{1}}l \\ {{u}_{2}}l \\ \end{matrix} \right]$$

$$
 * <p style="text-align:right;">$$\displaystyle (Eq. 2.6)
 * }
 * }


 * Where:


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

0 & 1 & 0 & 0 \\   \frac{-\left( k{{a}^{2}}+{{m}_{1}}gl \right)}{m{}_{1}{{l}^{2}}} & 0 & \frac{k{{a}^{2}}}{m{}_{1}{{l}^{2}}} & 0  \\ 0 & 0 & 0 & 1 \\   \frac{k{{a}^{2}}}{m{}_{2}{{l}^{2}}} & 0 & \frac{-\left( k{{a}^{2}}+{{m}_{2}}gl \right)}{m{}_{2}{{l}^{2}}} & 0  \\ \end{matrix} \right]$$
 * $$\underset{-}{\mathop{A}}\,=\left[ \begin{matrix}
 * $$\underset{-}{\mathop{A}}\,=\left[ \begin{matrix}


 * }
 * }


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

0 & 0 \\   \frac{1}{m{}_{1}{{l}^{2}}} & 0  \\ 0 & 0 \\   0 & \frac{1}{m{}_{2}{{l}^{2}}}  \\ \end{matrix} \right]$$
 * $$\underset{-}{\mathop{B}}\,=\left[ \begin{matrix}
 * $$\underset{-}{\mathop{B}}\,=\left[ \begin{matrix}
 * }
 * }

Contributing Members
Solved and posted by Egm6321.f10.team3.Sudheesh 15:39, 4 October 2010 (UTC)

Author and Proof-reader
[Author]

[Proof-reader]

= Problem 3 - Derive (L1-ODE-CC) = From Meeting 14, p. 14-1

Find
Derive (3-2)

Solve
We can rearrange the eqn(3.1). As it is not time variable problem, let $$\begin{align}a(t)=a, b(t)=b\end{align}$$

Let's find the integrating factor first.

As the coefficient for the $$\begin{align}\dot{x}\end{align}$$ is 1,

Multiply the integrating factor to eqn(3.2) on both side.

let's integrate for the interval $$\begin{align} \left[{t}_{0},t\right]\end{align}$$

rearrange eqn(3.7),

Author and Proof-reader
[Author]

[Proof-reader]

= Problem 4 - Expand Taylor series(exponential and exponential matrix) = From Meeting 14, p. 14-2

Given

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

$$\displaystyle {e}^{x} = 1 + \frac{x}{1!} + \frac{x^2}{2!} + ... $$ $$\displaystyle \ = \sum_{k=0}^{\infty} \frac{x^k}{k!} $$ <p style="text-align:right;">$$\displaystyle (4-1) $$
 * }
 * }


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

$$\displaystyle \underbrace{{e}^{\underline{A}}}_{\underline{A} \ : \ n \times n \ Matrix} = \underbrace{\underline{\mathit I }}_{n \times n \ Unit \ mat.} + \frac{\underline{A}}{1!} + \frac{\underline{A}^2}{2!} + ... $$ $$\displaystyle \ = \sum_{k=0}^{\infty} \frac{\underline{A}^k}{k!} $$ <p style="text-align:right;">$$\displaystyle (4-2) $$
 * }
 * }

Find
1) Derive (4-1) 

2) Derive (4-2) 

Solve
Solution of 1)

Using Taylor series ,

$$f(a)+\frac {f'(a)}{1!} (x-a)+ \frac{f''(a)}{2!} (x-a)^2+\frac{f^{(3)}(a)}{3!}(x-a)^3+ \cdots. $$

which can be written in the more compact sigma notation as


 * $$ \sum_{n=0} ^ {\infin } \frac {f^{(n)}(a)}{n!} \, (x-a)^{n} $$

In the particular case where a = 0, the series is also called a Maclaurin series
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$$f(x) = f(0)+f'(0)x + \frac{f''(0)}{2!}x^2+\frac{f^{(3)}(0)}{3!}x^3+ \cdots .$$ $$ \ = \sum_{n=0} ^ {\infin } \frac {f^{(n)}(0)}{n!} \, (x)^{n}$$ <p style="text-align:right;">$$\displaystyle (4-3) $$
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$$\displaystyle f(x) := {e}^{x} $$ <p style="text-align:right;">$$\displaystyle (4-4) $$
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$$\displaystyle f(0) = 1, \ f'(x) = {e}^{x}, \ f'(0) = 1, \ ... $$ <p style="text-align:right;">$$\displaystyle (4-5) $$
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$$\displaystyle f(x) = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} \ ... $$ <p style="text-align:right;">$$\displaystyle (4-6) $$
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$$\therefore {e}^{x} = \sum_{k=0}^{\infty} \frac{x^k}{k!} $$
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Solution of 2)

Using Taylor series and Maclaurin series


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$$\displaystyle f(\underline{x}) :=  \underbrace{{e}^{\underline{A}}}_{\underline{A} \ : \ n \times n \ Matrix} $$ <p style="text-align:right;">$$\displaystyle (4-7) $$
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<Background Knowledge> - Exponential Matrix, Identity Matrix


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$$\displaystyle f(\underline{0}) = \underbrace{\underline{\mathit  I }}_{n \times n \ Unit \ mat.}, \ f'(\underline{x}) = {e}^{\underline{A}}, \ f'(\underline{0}) = \underline{\mathit  I }, \ ... $$ <p style="text-align:right;">$$\displaystyle (4-8) $$
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$$\displaystyle f(\underline{x}) = 1 + \underline{A} + \frac{\underline{A}^2}{2!} + \frac{\underline{A}^3}{3!} \ ... $$ <p style="text-align:right;">$$\displaystyle (4-9) $$
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$$\therefore {e}^{\underline{A}} = \sum_{k=0}^{\infty} \frac{\underline{A}^k}{k!} $$
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Author and Proof-reader
[Author] Oh, Sang Min

[Proof-reader]

= Problem 5 - Generalized to SC-L1-ODE-VC = From Meeting 14, p. 14-2

Given
Dimension of matrix

$$\displaystyle \underline{x} \ : \ n \times \ 1 $$

$$\displaystyle \underline{A} \ : \ n \times \ n $$

$$\displaystyle \underline{B} \ : \ n \times \ m $$

$$\displaystyle \underline{U} \ : \ m \times \ 1 $$

Find
Generalized (5-3) to SC-L1-ODE-VC

Solve
 SC-L1-ODE-CC can be generalized to SC-L1-ODE-CC as same as L1-ODE-CC is generalized to L1-ODE-VC 

Using (5-1) ~ (5-3)

 SC-L1-ODE-VC 


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$$ \underline{x}(t)= {e}^{\int_{{t}_{0}}^{t} \underline{A} (\tau) d\tau}\cdot \underline{x}({t}_{0}) + \int_{{t}_{0}}^{t}{e}^{\int_{\tau}^{t} \underline{A} (s) ds} \cdot \underline{B}\cdot \underline{U}(\tau)d\tau $$
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Dimension of matrix

$$\displaystyle \underline{x} \ : \ n \times \ 1 $$

$$\displaystyle \underline{A} \ : \ n \times \ n $$

$$\displaystyle \underline{B} \ : \ n \times \ m $$

$$\displaystyle \underline{U} \ : \ m \times \ 1 $$

Author and Proof-reader
[Author] Oh, Sang Min

[Proof-reader]

= Problem 6 - Obtaining SC-L1-ODE-CC with int. factor method = From Meeting 15, p. 15-1

Author and Proof-reader
[Author]

[Proof-reader]

= Problem 7 - Application SC-L1-ODE-CC about rolling control of rocket = From Meeting 15, p. 15-1

Given

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$$\displaystyle \delta $$  =   aileron angle(deflection)

$$\displaystyle \phi $$ =   roll angle

$$\displaystyle \omega $$ =  roll angular velocity

$$\displaystyle Q $$ =   aileron efflectiveness

$$\displaystyle \tau $$ = roll time constant


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Find
Put (7-1) ~ (7-3) in form of (7-4)

Solve

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$$
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\begin{pmatrix} \dot\phi\\ \dot\omega\\ \dot\delta \end{pmatrix} = \begin{pmatrix} 0 & 1 & 0 \\ 0 & \frac{-1}{\tau} & \frac{Q}{\tau}\\ 0 & 0 & 0 \end{pmatrix}

\begin{pmatrix} \phi\\ \omega\\ \delta \end{pmatrix} + \begin{pmatrix} 0\\ 0\\ 1 \end{pmatrix} u $$

$$ \underline{A}= \begin{pmatrix} 0 & 1 & 0 \\ 0 & \frac{-1}{\tau} & \frac{Q}{\tau}\\ 0 & 0 & 0 \end{pmatrix}

, \ \underline{B}= \begin{pmatrix} 0\\ 0\\ 1 \end{pmatrix}

$$
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Author and Proof-reader
[Author] Oh, Sang Min

[Proof-reader]

= Problem 8 =

We can rewrite the eqn(8.1) as (8.2).

We are familiar with this equation, as we learned already. Total derivative - Egm6321.f10_HW1_prob#1_team6

As $$\begin{align} \frac{d}{dx}h(x,y)=0 \end{align}$$, we know that $$\begin{align} h(x,y)=f(y) \end{align}$$ only. It means $$\begin{align} {h}_{x}=0 \end{align}$$

Hence, eqn(8.2) becomes,

There are two possible solutions.

1) $$\begin{align} P=y'=0 \end{align}$$

2) $$\begin{align} {h}_{y}=0 \end{align}$$

If 1) were satisfied, whole problems became zero, which is trivial. We can conclude that 2) is the solution.

As $$\begin{align} {h}_{x}=0 \end{align}$$ and $$\begin{align} {h}_{y}=0 \end{align}$$,

= Problem 9 =

= Problem 10 =

= References =