User:Egm6321.f10.team6.lee/HW3

= Problem 2. Application - duoble pendulum =

Given
where,

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

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

$$\begin{align} {u}_{1}, {u}_{2} \end{align}$$ : applied forces respectively

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

$$\begin{align} k \end{align}$$ : spring constant

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

Required
1. Derive the equation (2.1)

2. Write eqn(2.1) in the form of eqn(2.2)

where,

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

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

Background : Newton's 2nd law for Rotation
(will be updated)

Background : small angle approximation
(will be updated) small angle approximation

Background : Hooke's Law
(will be updated) Hooke's Law

Verification
From the above,

where,

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

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

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

So, let's find all of the applied torques at the pendulum.


 * at pendulum 1,

we can calculate all the applied torques at the pendulum 1.

(positive for the CCW, negative for the CW)

torque by the spring force

from the Hooke's Law, we know that

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

Hence,

torque by the gravity force

This term would be negative as it is CCW direction. torque by the applied force

This term would be positive as it is CW direction.

Let's combine from eqn(2.3) to eqn(2.7) then,

in the same way, we can apply Newton's second law for rotation at pendulum 2. and we would get,

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

rearrange the derived equations (2.2),

Let's put them to the eqn(2.8)

= Problem 3. The solution of the state equation =

Given
For linear system of the control engineering,

Required

 * slove the given eqn(3.1) when it comes to the L1_ODE_CC : a & b are constants

Solution
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),

= Problem 8. solution of h(x,y) =

Given
While we solve the N2_ODE eqn(8.1)

we can get eqn(8.2) using definition of $$\begin{align} g(x,y,y') \end{align}$$

where $$\begin{align} P:=y' \end{align}$$

Required

 * without assuming h=const. find the solution of eqn(8.2)

Solution
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 $$\begin{align} {h}_{y}=0 \end{align}$$,