# 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):

 {\displaystyle {\begin{aligned}&k,n\in \mathbb {R} \\&m={\text{mass of the particle}}\\&g={\text{acceleration of gravity}}\end{aligned}}}

## Find

${\displaystyle \left(1\right)}$ Drive Equation of Motion (EOM)
${\displaystyle \left(2\right)}$ Particular case ${\displaystyle \displaystyle k=0}$ : Verify ${\displaystyle y\left(x\right)}$ is parabolla
${\displaystyle \left(3\right)}$ Consider ${\displaystyle k\neq 0}$, ${\displaystyle \displaystyle {{v}_{x}}=0}$,
${\displaystyle (3.1)}$ Find ${\displaystyle {{v}_{y}}\left(t\right)}$, ${\displaystyle \displaystyle y(t)}$ for ${\displaystyle \displaystyle m}$ = constant
${\displaystyle (3.2)}$ Find ${\displaystyle {{v}_{y}}\left(t\right)}$, ${\displaystyle \displaystyle y(t)}$ if ${\displaystyle m=m\left(t\right)}$

## 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
 ${\displaystyle \displaystyle W=mg}$
2) Inertia force
 ${\displaystyle {{F}_{a}}=ma=m\cdot {\frac {dv}{dt}}}$ for particle with constant mass
3) Air resistance which is proportional to the velocity of particle
 ${\displaystyle \displaystyle {{F}_{D}}=k{{v}^{n}}}$

Now consider the force equilibrium in both horizontal and vertical direction

a) Force Equilibrium in horizontal direction:
 ${\displaystyle \sum {{{F}_{H}}=0}}$
 ${\displaystyle m\cdot {\frac {d{{v}_{x}}}{dt}}+k{{v}^{n}}\cos \alpha =0}$, where ${\displaystyle \displaystyle {{v}_{x}}\to }$ horizontal component of velocity
 ${\displaystyle m\cdot {\frac {d{{v}_{x}}}{dt}}=-k{{v}^{n}}\cos \alpha }$ ${\displaystyle \displaystyle (Eq.1.1)}$
b) Force Equilibrium in the vertical direction:
 ${\displaystyle \sum {{{F}_{V}}=0}}$
 ${\displaystyle m\cdot {\frac {d{{v}_{y}}}{dt}}+k{{v}^{n}}\sin \alpha +mg=0}$, where ${\displaystyle \displaystyle {{v}_{y}}\to }$ vertical component of velocity
 ${\displaystyle m\cdot {\frac {d{{v}_{y}}}{dt}}=-k{{v}^{n}}\sin \alpha -mg}$ ${\displaystyle \displaystyle (Eq.1.2)}$

#### Part 2

Particular case: When ${\displaystyle \displaystyle k=0}$

Eq.1.1 reduces to

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

Integrating the above equation gives:

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

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

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

Now Eq.1.3 becomes:

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

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

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

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

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

Therefore,

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

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

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

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

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

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

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

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

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

Now Eq.1.5 becomes,

 ${\displaystyle {{v}_{t}}\left(t\right)=-gt+{{v}_{{y}_{0}}}}$

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

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

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

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

Therefore,

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

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

 ${\displaystyle y\left(x\right)=-{\frac {g}{2}}{{\left({\frac {x-{{x}_{0}}}{{v}_{{x}_{0}}}}\right)}^{2}}+{{v}_{{y}_{0}}}\left({\frac {x-{{x}_{0}}}{{v}_{{x}_{0}}}}\right)+{{y}_{0}}}$ ${\displaystyle \displaystyle (Eq.1.6)}$

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

#### Part 3

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

 {\displaystyle {\begin{aligned}m{\frac {dv_{x}}{dt}}=&-kv^{n}\cos \alpha \\m{\frac {dv_{x}}{dt}}=&-kv^{n}{\frac {v_{x}}{v}}\\m{\frac {dv_{x}}{dt}}=&-kv^{n-1}v_{x}\end{aligned}}} (1.13)

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

#### Part 3.1

Solving

 {\displaystyle {\begin{aligned}m{\frac {dv_{y}}{dt}}=-k(v_{y})^{n}-mg\end{aligned}}} (1.2)

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

 {\displaystyle {\begin{aligned}\underbrace {k(v_{y})^{n}+mg} _{M(v_{y},t)}+\underbrace {\left.m\right.^{}} _{N(v_{y},t)}{\frac {dv_{y}}{dt}}=0\end{aligned}}} (1.14)

so that

 {\displaystyle {\begin{aligned}{\frac {\partial \phi }{\partial t}}dt+{\frac {\partial \phi }{\partial v_{y}}}dv_{y}=0\end{aligned}}} (1.15)
 {\displaystyle {\begin{aligned}M(v_{y},t)={}&k(v_{y})^{n}+mg{}&=\phi _{t}\\N(v_{y},t)={}&m&=\phi _{v_{y}}\end{aligned}}} (1.16)

The second condition of exactness

 {\displaystyle {\begin{aligned}M(v_{y},t)_{v_{y}}=N(v_{y},t)_{t}\end{aligned}}} (1.17)

Which is met if ${\displaystyle m}$ is not a function of time

 {\displaystyle {\begin{aligned}M(v_{y},t)_{v_{y}}=knv_{y}^{n-1}\neq N(v_{y},t)_{t}={\frac {dm(t)}{dt}}\end{aligned}}} (1.18)

Thus, the equation is non-exact.

The equation can be made exact through the integrating factor method

 {\displaystyle {\begin{aligned}h(v_{y},t)\left(kv_{y}^{n}+mg\right)+h(v_{y},t)m{\frac {dv_{y}}{dt}}=0\end{aligned}}} (1.19)

Then expressing as a total derivative and testing for exactness

 {\displaystyle {\begin{aligned}h(v_{y},t)\left(kv_{y}^{n}+mg\right)dt+h(v_{y},t)mdv_{y}=0\end{aligned}}} (1.20)
 {\displaystyle {\begin{aligned}\phi _{t}&=h(v_{y},t)\left(kv_{y}^{n}+mg\right)\\\phi _{v_{y}}&=h(v_{y},t)m\\\phi _{t,v_{y}}&=\phi _{v_{y},t}\end{aligned}}} (1.21)
 {\displaystyle {\begin{aligned}{\frac {\partial h(v_{y},t)}{\partial v_{y}}}\left(kv_{y}^{n}+mg\right)+h(v_{y},t){\frac {\partial \left(kv_{y}^{n}+mg\right)}{\partial v_{y}}}&={\frac {\partial h(v_{y},t)}{\partial t}}m+h(v_{y},t){\frac {\partial m}{\partial t}}\end{aligned}}} (1.22)

Letting ${\displaystyle \partial h/\partial v_{y}=0}$

 {\displaystyle {\begin{aligned}{\cancelto {0}{\frac {\partial h(v_{y},t)}{\partial v_{y}}}}\left(kv_{y}^{n}+mg\right)+h(v_{y},t){\frac {\partial \left(kv_{y}^{n}+mg\right)}{\partial v_{y}}}&={\frac {\partial h(v_{y},t)}{\partial t}}m+h(v_{y},t){\cancelto {0}{\frac {\partial m}{\partial t}}}\end{aligned}}} (1.23)
 {\displaystyle {\begin{aligned}h(v_{y},t){\frac {\partial \left(kv_{y}^{n}+mg\right)}{\partial v_{y}}}&={\frac {\partial h(v_{y},t)}{\partial t}}m\end{aligned}}} (1.24)
 {\displaystyle {\begin{aligned}h(v_{y},t){\frac {nkv_{y}^{n}}{m}}&={\frac {\partial h(v_{y},t)}{\partial t}}\end{aligned}}} (1.25)
 {\displaystyle {\begin{aligned}h(v_{y},t)=\exp \left({\frac {kv_{y}^{n}}{m}}t\right)\end{aligned}}} (1.26)

Then

 {\displaystyle {\begin{aligned}\phi _{t}=\exp \left({\frac {kv_{y}^{n}}{m}}t\right)\left(kv_{y}^{n}+mg\right)\end{aligned}}} (1.27)
 {\displaystyle {\begin{aligned}\phi &=\int {\phi _{t}dt}=\int {\exp \left({\frac {kv_{y}^{n}}{m}}t\right)\left(kv_{y}^{n}+mg\right)dt}\\&\phi =\left(kv_{y}^{n}+mg\right){\frac {m}{kv_{y}^{n}}}\exp \left({\frac {kv_{y}^{n}}{m}}t\right)+f(v_{y})=k_{2}\end{aligned}}} (1.28)
 {\displaystyle {\begin{aligned}\phi _{v}=m\exp \left({\frac {kv_{y}^{n}}{m}}t\right)\end{aligned}}} (1.29)
 {\displaystyle {\begin{aligned}\phi &=\int {\phi _{v_{y}}dv_{y}}=\int {m\exp \left({\frac {kv_{y}^{n}}{m}}t\right)dv_{y}}=k_{2}\\&\phi =m\int {\exp \left({\frac {kv_{y}^{n}}{m}}t\right)dv_{y}}+f(t)=k_{2}\end{aligned}}} (1.30)

Combining

 {\displaystyle {\begin{aligned}\phi &=m\int {\exp \left({\frac {kv_{y}^{n}}{m}}t\right)dv_{y}}+\left(kv_{y}^{n}+mg\right){\frac {m}{kv_{y}^{n}}}\exp \left({\frac {kv_{y}^{n}}{m}}t\right)=k_{2}\end{aligned}}} (1.30)

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

#### Part 3.2

If ${\displaystyle m(t)\neq 0}$ the integrating factor method is complicated in equation 1.22 as the partial of ${\displaystyle m}$ with respect to time remains, complicating the expression for ${\displaystyle h}$

[Author]

# Problem 2 - Derive EOM (SC-L1-ODE)

From Meeting 13, p. 13-3

## Given

 {\displaystyle {\begin{aligned}{m}_{1}{l}^{2}{\ddot {{\theta }_{1}}}=-k{a}^{2}({\theta }_{1}-{\theta }_{2})-{m}_{1}gl{\theta }_{1}+{u}_{1}l\\\end{aligned}}} ${\displaystyle \displaystyle (2-1)}$
 {\displaystyle {\begin{aligned}{m}_{2}{l}^{2}{\ddot {{\theta }_{2}}}=-k{a}^{2}({\theta }_{2}-{\theta }_{1})-{m}_{2}gl{\theta }_{2}+{u}_{2}l\\\end{aligned}}} ${\displaystyle \displaystyle (2-2)}$
 {\displaystyle {\begin{aligned}{m}_{1},{m}_{2}\end{aligned}}} : mass of the each pendulum {\displaystyle {\begin{aligned}{\theta }_{1},{\theta }_{2}\end{aligned}}} : the angle from the vertical to the each pendulum {\displaystyle {\begin{aligned}{u}_{1},{u}_{2}\end{aligned}}} : applied forces to the each pendulum {\displaystyle {\begin{aligned}l\end{aligned}}} : length of the pendulum {\displaystyle {\begin{aligned}k\end{aligned}}} : force constant(or spring constant) {\displaystyle {\begin{aligned}g\end{aligned}}} : acceleration of gravity

## Find

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

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

 {\displaystyle {\begin{aligned}{\dot {\textbf {x}}}(t)={\textbf {A}}(t)\cdot {\textbf {x}}(t)+{\textbf {B}}(t)\cdot {\textbf {u}}(t)\end{aligned}}} ${\displaystyle \displaystyle (2-3)}$

where,

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

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

Dimension of matrix

{\displaystyle {\begin{aligned}{\dot {\textbf {x}}}\ :\ 4\times 1\end{aligned}}}

{\displaystyle {\begin{aligned}{\textbf {x}}\ :\ 4\times 1\end{aligned}}}

{\displaystyle {\begin{aligned}{\textbf {A}}\ :\ 4\times 4\end{aligned}}}

{\displaystyle {\begin{aligned}{\textbf {B}}\ :\ 4\times 2\end{aligned}}}

{\displaystyle {\begin{aligned}{\textbf {U}}\ :\ 2\times 1\end{aligned}}}

## Solve

Step 1. Derivation

Background Knowledge

1. Torque [1]

2. Hooke's Law [2]

3. Pendulum [3]

4. Moment [4]

5. Moment of inertia [5]

6. Angular acceleration [6]

Derive Using above background,

 ${\displaystyle \tau =I\cdot \alpha =\sum {\tau }(=}$ Torque of the spring force + Torque of the gravity force + Torque of the applied force ) ${\displaystyle \displaystyle (2-4)}$

where,

{\displaystyle {\begin{aligned}\tau \end{aligned}}} : torque

{\displaystyle {\begin{aligned}I=m{l}^{2}\end{aligned}}} : moment of inertia

{\displaystyle {\begin{aligned}\alpha ={\ddot {\theta }}\end{aligned}}} : angular acceleration

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

Torque of the spring force

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

 {\displaystyle {\begin{aligned}F=-kx\end{aligned}}} ${\displaystyle \displaystyle (2-5)}$

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

Therefore, Torque of the spring force is,

 {\displaystyle {\begin{aligned}{\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{aligned}}} ${\displaystyle \displaystyle (2-6)}$

Torque of the gravity force

 {\displaystyle {\begin{aligned}{\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{aligned}}} ${\displaystyle \displaystyle (2-7)}$

Torque of the applied force

 {\displaystyle {\begin{aligned}{\vec {\tau }}={\vec {r}}\times {\vec {F}}={u}_{1}\cdot l\mathrm {cos} {\theta }_{1}\approx {u}_{1}\cdot l\end{aligned}}} ${\displaystyle \displaystyle (2-8)}$

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

 ${\displaystyle {m}_{1}{l}^{2}{\ddot {{\theta }_{1}}}=-k{a}^{2}({\theta }_{1}-{\theta }_{2})-{m}_{1}gl{\theta }_{1}+{u}_{1}l}$

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

 ${\displaystyle {m}_{2}{l}^{2}{\ddot {{\theta }_{2}}}=-k{a}^{2}({\theta }_{2}-{\theta }_{1})-{m}_{2}gl{\theta }_{2}+{u}_{2}l}$

Step 2. Find A, B and U

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

 {\displaystyle {\begin{aligned}{\begin{bmatrix}\ {\dot {{\theta }_{1}}}\\\ {\ddot {{\theta }_{1}}}\\\ {\dot {{\theta }_{2}}}\\\ {\ddot {{\theta }_{2}}}\\\end{bmatrix}}=\underbrace {\begin{bmatrix}\cdots &&\\&\cdots &\\&&\cdots \end{bmatrix}} _{\mathrm {Matrix} \mathbf {A} (4\times 4)}{\begin{bmatrix}\ {\theta }_{1}\\\ {\dot {{\theta }_{1}}}\\\ {\theta }_{2}\\\ {\dot {{\theta }_{2}}}\\\end{bmatrix}}+\underbrace {\begin{bmatrix}\cdots &&\\&\cdots &\\&&\cdots \end{bmatrix}} _{\mathrm {Matrix} \mathbf {B} (4\times 2)}{\begin{bmatrix}\ {u}_{1}l\\\ {u}_{2}l\\\end{bmatrix}}\end{aligned}}} ${\displaystyle \displaystyle (2-9)}$

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

 {\displaystyle {\begin{aligned}{\ddot {{\theta }_{1}}}=\left(-{\frac {k{a}^{2}}{{m}_{1}{l}^{2}}}-{\frac {g}{l}}\right){\theta }_{1}+\left({\frac {k{a}^{2}}{{m}_{1}{l}^{2}}}\right){\theta }_{2}+{\frac {1}{{m}_{1}{l}^{2}}}{u}_{1}l\\{\ddot {{\theta }_{2}}}=\left({\frac {k{a}^{2}}{{m}_{2}{l}^{2}}}\right){\theta }_{1}+\left(-{\frac {k{a}^{2}}{{m}_{2}{l}^{2}}}-{\frac {g}{l}}\right){\theta }_{2}+{\frac {1}{{m}_{2}{l}^{2}}}{u}_{2}l\\\end{aligned}}} ${\displaystyle \displaystyle (2-10)}$

Let's put them to (2-9)

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

## Given

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

 {\displaystyle {\begin{aligned}&k={\text{spring constant}}\\&{{m}_{1}},{{m}_{2}}={\text{mass of pendulums}}\\&{{\theta }_{1}},{{\theta }_{2}}={\text{Rotation angles}}\\&{{u}_{1}},{{u}_{2}}={\text{control forces}}\end{aligned}}}

## Find

${\displaystyle \left(1\right)}$ Derive equation of motion:
 ${\displaystyle {{m}_{1}}{{l}^{2}}{{{\overset {\centerdot \centerdot }{\mathop {\theta } }}\,}_{1}}=-k{{a}^{2}}\left({{\theta }_{1}}-{{\theta }_{2}}\right)-{{m}_{1}}gl{{\theta }_{1}}+{{u}_{1}}l}$ ${\displaystyle \displaystyle (Eq.2.1)}$
 ${\displaystyle {{m}_{2}}{{l}^{2}}{{{\overset {\centerdot \centerdot }{\mathop {\theta } }}\,}_{2}}=-k{{a}^{2}}\left({{\theta }_{2}}-\theta 1\right)-{{m}_{2}}gl{{\theta }_{2}}+{{u}_{2}}l}$ ${\displaystyle \displaystyle (Eq.2.2)}$
${\displaystyle \left(2\right)}$ Write Eq.2.1 and Eq.2.2 in the form of Mtg 13 (c),page2 , of:
 ${\displaystyle {\underset {-}{\overset {\centerdot }{\mathop {x} }}}\,={\underset {-}{\mathop {A} }}\,\left(t\right){\underset {-}{\mathop {x} }}\,\left(t\right)+{\underset {-}{\mathop {B} }}\,\left(t\right){\underset {-}{\mathop {u} }}\,\left(t\right)}$ ${\displaystyle \displaystyle (Eq.2.3)}$
 Given ${\displaystyle {\underset {-}{\mathop {x} }}\,={{\left[{\begin{matrix}{{\theta }_{1}}&{\overset {\centerdot }{\mathop {{\theta }_{1}} }}\,&{{\theta }_{2}}&{\overset {\centerdot }{\mathop {{\theta }_{2}} }}\,\\\end{matrix}}\right]}^{T}}}$ and ${\displaystyle {\underset {-}{\mathop {u} }}\,={{\left[{\begin{matrix}{{u}_{1}}l&{{u}_{2}}l\\\end{matrix}}\right]}^{T}}}$

## Solution

${\displaystyle \left(1\right)}$ Derive equation of motion:
(a) Consider Free Body Diagram of left pendulum:
For small angle:
 ${\displaystyle \displaystyle l\sin {{\theta }_{1}}=l{{\theta }_{1}}}$ and ${\displaystyle \displaystyle l\cos {{\theta }_{1}}=l}$
 ${\displaystyle acceleration,{{a}_{1}}={\frac {{{d}^{2}}\left(l{{\theta }_{1}}\right)}{d{{t}^{2}}}}=l{{{\overset {\centerdot \centerdot }{\mathop {\theta } }}\,}_{1}}}$
 ${\displaystyle Inertiaforce={{m}_{1}}l{{{\overset {\centerdot \centerdot }{\mathop {\theta } }}\,}_{1}}}$
 ${\displaystyle \displaystyle Springforce=ka{{\theta }_{1}}-ka{{\theta }_{2}}=ka({{\theta }_{1}}-{{\theta }_{1}})}$

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

 ${\displaystyle \sum {{{M}_{A}}=0\Rightarrow }\left({{m}_{1}}l{{{\overset {\centerdot \centerdot }{\mathop {\theta } }}\,}_{1}}\right)\cdot l+\left(ka({{\theta }_{1}}-{{\theta }_{2}})\right)\cdot a+\left({{m}_{1}}g\right)\cdot l{{\theta }_{1}}-\left({{u}_{1}}\right)\cdot l=0}$
 ${\displaystyle \Rightarrow {{m}_{1}}{{l}^{2}}{{{\overset {\centerdot \centerdot }{\mathop {\theta } }}\,}_{1}}=-k{{a}^{2}}({{\theta }_{1}}-{{\theta }_{2}})-{{m}_{1}}gl{{\theta }_{1}}+{{u}_{1}}l}$ ${\displaystyle \displaystyle (Eq.2.1)}$
(b) Consider Free Body Diagram of right pendulum:
For small angle:
 ${\displaystyle \displaystyle l\sin {{\theta }_{2}}=l{{\theta }_{2}}}$ and ${\displaystyle \displaystyle l\cos {{\theta }_{2}}=l}$
 ${\displaystyle acceleration,{{a}_{2}}={\frac {{{d}^{2}}\left(l{{\theta }_{2}}\right)}{d{{t}^{2}}}}=l{{{\overset {\centerdot \centerdot }{\mathop {\theta } }}\,}_{2}}}$
 ${\displaystyle Inertiaforce={{m}_{2}}l{{{\overset {\centerdot \centerdot }{\mathop {\theta } }}\,}_{2}}}$
 ${\displaystyle \displaystyle Springforce=ka{{\theta }_{2}}-ka{{\theta }_{1}}=ka({{\theta }_{2}}-{{\theta }_{1}})}$

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

 ${\displaystyle \sum {{{M}_{B}}=0\Rightarrow }\left({{m}_{2}}l{{{\overset {\centerdot \centerdot }{\mathop {\theta } }}\,}_{2}}\right)\cdot l+\left(ka({{\theta }_{2}}-{{\theta }_{1}})\right)\cdot a+\left({{m}_{2}}g\right)\cdot l{{\theta }_{2}}-\left({{u}_{2}}\right)\cdot l=0}$
 ${\displaystyle \Rightarrow {{m}_{2}}{{l}^{2}}{{{\overset {\centerdot \centerdot }{\mathop {\theta } }}\,}_{2}}=-k{{a}^{2}}({{\theta }_{2}}-{{\theta }_{1}})-{{m}_{2}}gl{{\theta }_{2}}+{{u}_{2}}l}$ ${\displaystyle \displaystyle (Eq.2.2)}$
${\displaystyle \left(2\right)}$ 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,
 ${\displaystyle {{{\overset {\centerdot \centerdot }{\mathop {\theta } }}\,}_{1}}={\frac {-\left(k{{a}^{2}}+{{m}_{1}}gl\right)}{m{}_{1}{{l}^{2}}}}{{\theta }_{1}}+{\frac {k{{a}^{2}}}{m{}_{1}{{l}^{2}}}}{{\theta }_{2}}+{\frac {{{u}_{1}}l}{m{}_{1}{{l}^{2}}}}}$ ${\displaystyle \displaystyle (Eq.2.4)}$
 ${\displaystyle {{{\overset {\centerdot \centerdot }{\mathop {\theta } }}\,}_{2}}={\frac {k{{a}^{2}}}{m{}_{2}{{l}^{2}}}}{{\theta }_{1}}+{\frac {-\left(k{{a}^{2}}+{{m}_{2}}gl\right)}{m{}_{2}{{l}^{2}}}}{{\theta }_{2}}+{\frac {{{u}_{2}}l}{m{}_{2}{{l}^{2}}}}}$ ${\displaystyle \displaystyle (Eq.2.5)}$
Now Eq.2.4 and Eq.2.5 can be put in the form of Eq.2.3 as:
 ${\displaystyle \left[{\begin{matrix}{{{\overset {\centerdot }{\mathop {\theta } }}\,}_{1}}\\{{{\overset {\centerdot \centerdot }{\mathop {\theta } }}\,}_{1}}\\{{{\overset {\centerdot }{\mathop {\theta } }}\,}_{2}}\\{{{\overset {\centerdot \centerdot }{\mathop {\theta } }}\,}_{2}}\\\end{matrix}}\right]=\underbrace {\left[{\begin{matrix}0&1&0&0\\{\frac {-\left(k{{a}^{2}}+{{m}_{1}}gl\right)}{m{}_{1}{{l}^{2}}}}&0&{\frac {k{{a}^{2}}}{m{}_{1}{{l}^{2}}}}&0\\0&0&0&1\\{\frac {k{{a}^{2}}}{m{}_{2}{{l}^{2}}}}&0&{\frac {-\left(k{{a}^{2}}+{{m}_{2}}gl\right)}{m{}_{2}{{l}^{2}}}}&0\\\end{matrix}}\right]} _{{\underset {-}{\mathop {A} }}\,}\left[{\begin{matrix}{{\theta }_{1}}\\{{{\overset {\centerdot }{\mathop {\theta } }}\,}_{1}}\\{{\theta }_{2}}\\{{{\overset {\centerdot }{\mathop {\theta } }}\,}_{2}}\\\end{matrix}}\right]+\underbrace {\left[{\begin{matrix}0&0\\{\frac {1}{m{}_{1}{{l}^{2}}}}&0\\0&0\\0&{\frac {1}{m{}_{2}{{l}^{2}}}}\\\end{matrix}}\right]} _{{\underset {-}{\mathop {B} }}\,}\left[{\begin{matrix}{{u}_{1}}l\\{{u}_{2}}l\\\end{matrix}}\right]}$ ${\displaystyle \displaystyle (Eq.2.6)}$
Where:
 ${\displaystyle {\underset {-}{\mathop {A} }}\,=\left[{\begin{matrix}0&1&0&0\\{\frac {-\left(k{{a}^{2}}+{{m}_{1}}gl\right)}{m{}_{1}{{l}^{2}}}}&0&{\frac {k{{a}^{2}}}{m{}_{1}{{l}^{2}}}}&0\\0&0&0&1\\{\frac {k{{a}^{2}}}{m{}_{2}{{l}^{2}}}}&0&{\frac {-\left(k{{a}^{2}}+{{m}_{2}}gl\right)}{m{}_{2}{{l}^{2}}}}&0\\\end{matrix}}\right]}$
 ${\displaystyle {\underset {-}{\mathop {B} }}\,=\left[{\begin{matrix}0&0\\{\frac {1}{m{}_{1}{{l}^{2}}}}&0\\0&0\\0&{\frac {1}{m{}_{2}{{l}^{2}}}}\\\end{matrix}}\right]}$

## Contributing Members

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

[Author]

# Problem 3 - Derive (L1-ODE-CC)

From Meeting 14, p. 14-1

## Given

 {\displaystyle {\begin{aligned}{\dot {x}}(t)=a(t)\cdot x(t)+b(t)\cdot u(t)\end{aligned}}} ${\displaystyle \displaystyle (3-1)}$
 ${\displaystyle x(t)={e}^{a(t-{t}_{0})}\cdot x({t}_{0})+\int _{{t}_{0}}^{t}{e}^{a(t-\tau )}\cdot b\cdot u(\tau )d\tau }$ ${\displaystyle \displaystyle (3-2)}$

Derive (3-2)

## Solve

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

 {\displaystyle {\begin{aligned}{\dot {x}}(t)-a\cdot x(t)=b\cdot u(t)\end{aligned}}} (3.2)

Let's find the integrating factor first.

As the coefficient for the {\displaystyle {\begin{aligned}{\dot {x}}\end{aligned}}} is 1,

 {\displaystyle {\begin{aligned}h(x)=exp\left[\int _{}^{t}(-a)ds\right]={e}^{-at}\end{aligned}}} (3.3)

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

 {\displaystyle {\begin{aligned}{e}^{-at}{\dot {x}}(t)-{e}^{-at}a\cdot x(t)={e}^{-at}\cdot b\cdot u(t)\end{aligned}}} (3.4)
 {\displaystyle {\begin{aligned}\left[{e}^{-at}\cdot x(t)\right]'={e}^{-at}\cdot b\cdot u(t)\end{aligned}}} (3.5)

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

 {\displaystyle {\begin{aligned}\int _{{t}_{0}}^{t}\left[{e}^{-a\tau }\cdot x(\tau )\right]'=\int _{{t}_{0}}^{t}{e}^{-a\tau }\cdot b\cdot u(\tau )d\tau \end{aligned}}} (3.6)
 {\displaystyle {\begin{aligned}{e}^{-at}\cdot x(t)-{e}^{-a{t}_{0}}\cdot x({t}_{0})=\int _{{t}_{0}}^{t}{e}^{-a\tau }\cdot b\cdot u(\tau )d\tau \end{aligned}}} (3.7)

rearrange eqn(3.7),

 {\displaystyle {\begin{aligned}{e}^{-at}\cdot x(t)={e}^{-a{t}_{0}}\cdot x({t}_{0})+\int _{{t}_{0}}^{t}{e}^{-a\tau }\cdot b\cdot u(\tau )d\tau \end{aligned}}} (3.8)
 ${\displaystyle x(t)={e}^{a(t-{t}_{0})}\cdot x({t}_{0})+\int _{{t}_{0}}^{t}{e}^{a(t-\tau )}\cdot b\cdot u(\tau )d\tau }$  (3.9)

[Author]

# Problem 4 - Expand Taylor series(exponential and exponential matrix)

From Meeting 14, p. 14-2

## Given

 ${\displaystyle \displaystyle {e}^{x}=1+{\frac {x}{1!}}+{\frac {x^{2}}{2!}}+...}$ ${\displaystyle \displaystyle \ =\sum _{k=0}^{\infty }{\frac {x^{k}}{k!}}}$ ${\displaystyle \displaystyle (4-1)}$
 ${\displaystyle \displaystyle \underbrace {{e}^{\underline {A}}} _{{\underline {A}}\ :\ n\times n\ Matrix}=\underbrace {\underline {\mathit {I}}} _{n\times n\ Unit\ mat.}+{\frac {\underline {A}}{1!}}+{\frac {{\underline {A}}^{2}}{2!}}+...}$ ${\displaystyle \displaystyle \ =\sum _{k=0}^{\infty }{\frac {{\underline {A}}^{k}}{k!}}}$ ${\displaystyle \displaystyle (4-2)}$

1) Derive (4-1)

2) Derive (4-2)

## Solve

Solution of 1)

Using Taylor series [7],

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

which can be written in the more compact sigma notation [8] as

${\displaystyle \sum _{n=0}^{\infty }{\frac {f^{(n)}(a)}{n!}}\,(x-a)^{n}}$

In the particular case where a = 0, the series is also called a Maclaurin series [9]

 ${\displaystyle f(x)=f(0)+f'(0)x+{\frac {f''(0)}{2!}}x^{2}+{\frac {f^{(3)}(0)}{3!}}x^{3}+\cdots .}$ ${\displaystyle \ =\sum _{n=0}^{\infty }{\frac {f^{(n)}(0)}{n!}}\,(x)^{n}}$ ${\displaystyle \displaystyle (4-3)}$
 ${\displaystyle \displaystyle f(x):={e}^{x}}$ ${\displaystyle \displaystyle (4-4)}$
 ${\displaystyle \displaystyle f(0)=1,\ f'(x)={e}^{x},\ f'(0)=1,\ ...}$ ${\displaystyle \displaystyle (4-5)}$
 ${\displaystyle \displaystyle f(x)=1+x+{\frac {x^{2}}{2!}}+{\frac {x^{3}}{3!}}\ ...}$ ${\displaystyle \displaystyle (4-6)}$
 ${\displaystyle \therefore {e}^{x}=\sum _{k=0}^{\infty }{\frac {x^{k}}{k!}}}$

Solution of 2)

Using Taylor series and Maclaurin series

 ${\displaystyle \displaystyle f({\underline {x}}):=\underbrace {{e}^{\underline {A}}} _{{\underline {A}}\ :\ n\times n\ Matrix}}$ ${\displaystyle \displaystyle (4-7)}$

<Background Knowledge> - Exponential Matrix, [10] Identity Matrix [11]

 ${\displaystyle \displaystyle f({\underline {0}})=\underbrace {\underline {\mathit {I}}} _{n\times n\ Unit\ mat.},\ f'({\underline {x}})={e}^{\underline {A}},\ f'({\underline {0}})={\underline {\mathit {I}}},\ ...}$ ${\displaystyle \displaystyle (4-8)}$
 ${\displaystyle \displaystyle f({\underline {x}})=1+{\underline {A}}+{\frac {{\underline {A}}^{2}}{2!}}+{\frac {{\underline {A}}^{3}}{3!}}\ ...}$ ${\displaystyle \displaystyle (4-9)}$
 ${\displaystyle \therefore {e}^{\underline {A}}=\sum _{k=0}^{\infty }{\frac {{\underline {A}}^{k}}{k!}}}$

[Author] Oh, Sang Min

# Problem 5 - Generalized to SC-L1-ODE-VC

From Meeting 14, p. 14-2

## Given

 L1-ODE-CC : ${\displaystyle x(t)={e}^{a(t-{t}_{0})}\cdot x({t}_{0})+\int _{{t}_{0}}^{t}{e}^{a(t-\tau )}\cdot b\cdot u(\tau )d\tau }$ ${\displaystyle \displaystyle (5-1)}$
 L1-ODE-VC : ${\displaystyle x(t)={e}^{\int _{{t}_{0}}^{t}a(\tau )d\tau }\cdot x({t}_{0})+\int _{{t}_{0}}^{t}{e}^{\int _{\tau }^{t}a(s)ds}\cdot b\cdot u(\tau )d\tau }$ ${\displaystyle \displaystyle (5-2)}$
 SC-L1-ODE-CC : ${\displaystyle {\underline {x}}(t)=\underbrace {{e}^{{\underline {A}}(t-{t}_{0})}} _{{\underline {\phi }}(t,{t}_{0})}\cdot {\underline {x}}({t}_{0})+\int _{{t}_{0}}^{t}\underbrace {{e}^{{\underline {A}}(t-\tau )}} _{{\underline {\phi }}(t,\tau )}\cdot {\underline {B}}\cdot {\underline {U}}(\tau )d\tau }$ ${\displaystyle \displaystyle (5-3)}$

Dimension of matrix

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

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

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

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

## 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

 ${\displaystyle {\underline {x}}(t)={e}^{\int _{{t}_{0}}^{t}{\underline {A}}(\tau )d\tau }\cdot {\underline {x}}({t}_{0})+\int _{{t}_{0}}^{t}{e}^{\int _{\tau }^{t}{\underline {A}}(s)ds}\cdot {\underline {B}}\cdot {\underline {U}}(\tau )d\tau }$

Dimension of matrix

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

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

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

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

[Author] Oh, Sang Min

# Problem 6 - Obtaining SC-L1-ODE-CC with int. factor method

From Meeting 15, p. 15-1

[Author]

# Problem 7 - Application SC-L1-ODE-CC about rolling control of rocket

From Meeting 15, p. 15-1

## Given

 {\displaystyle {\begin{aligned}{\dot {\phi \,}}=\omega \end{aligned}}} ${\displaystyle \displaystyle (7-1)}$
 {\displaystyle {\begin{aligned}{\dot {\omega }}=-{\frac {1}{\tau \ }}\omega +{\frac {Q}{\tau \ }}\delta \,\end{aligned}}} ${\displaystyle \displaystyle (7-2)}$
 {\displaystyle {\begin{aligned}{\dot {\delta \ }}=u\end{aligned}}} ${\displaystyle \displaystyle (7-3)}$
 {\displaystyle {\begin{aligned}{\underline {\dot {x}}}(t)={\underline {A}}\ {\underline {x}}(t)+{\underline {B}}\ {\underline {u}}(t)\end{aligned}}} ${\displaystyle \displaystyle (7-4)}$
 ${\displaystyle \displaystyle \delta }$ = aileron angle(deflection) ${\displaystyle \displaystyle \phi }$ = roll angle ${\displaystyle \displaystyle \omega }$ = roll angular velocity ${\displaystyle \displaystyle Q}$ = aileron efflectiveness ${\displaystyle \displaystyle \tau }$ = roll time constant

## Find

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

## Solve

 ${\displaystyle {\begin{pmatrix}{\dot {\phi }}\\{\dot {\omega }}\\{\dot {\delta }}\end{pmatrix}}={\begin{pmatrix}0&1&0\\0&{\frac {-1}{\tau }}&{\frac {Q}{\tau }}\\0&0&0\end{pmatrix}}{\begin{pmatrix}\phi \\\omega \\\delta \end{pmatrix}}+{\begin{pmatrix}0\\0\\1\end{pmatrix}}u}$ $\displaystyle \underline{A}= \begin{pmatrix} 0 & 1 & 0 \\ 0 & \frac{-1}{\tau} & \frac{Q}{\tau}\\ 0 & 0 & 0 \end{pmatrix} , \ \underline{B}= \begin{pmatrix} 0\\ 0\\ 1 \end{pmatrix}$

[Author] Oh, Sang Min

# Problem 8

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

 {\displaystyle {\begin{aligned}{\frac {\partial h}{\partial x}}+{\frac {\partial h}{\partial y}}\cdot {\frac {dy}{dx}}=0\end{aligned}}} (8.3)

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

 {\displaystyle {\begin{aligned}{\frac {d}{dx}}h(x,y)={\frac {\partial h}{\partial x}}+{\frac {\partial h}{\partial y}}\cdot {\frac {dy}{dx}}=0\end{aligned}}} (8.4)

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

Hence, eqn(8.2) becomes,

 {\displaystyle {\begin{aligned}{h}_{y}\cdot P=0\end{aligned}}} (8.5)

There are two possible solutions.

1) {\displaystyle {\begin{aligned}P=y'=0\end{aligned}}}

2) {\displaystyle {\begin{aligned}{h}_{y}=0\end{aligned}}}

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

As {\displaystyle {\begin{aligned}{h}_{x}=0\end{aligned}}} and {\displaystyle {\begin{aligned}{h}_{y}=0\end{aligned}}},

 {\displaystyle {\begin{aligned}h={k}_{1}=\mathrm {constant} \end{aligned}}} (8.6)