User:EGM6321.f10.team6.cook/hw3

=Motion of a Particle =

Part 1
Derive the equations of motion (EOM) which are a system of coupled nonlinear first order ordinary differential equations.

Part 2
Particular case where $$\begin{align}k=0\end{align}$$, verify $$\begin{align}y(x)\end{align}$$ is a parabola

Part 3
Consider $$k\ne0$$, $$\begin{align}v_{xo}=0\end{align}$$, resulting in

Part 3.1
Find $$\begin{align}V_y(t),y(t)\end{align}$$ for $$\begin{align}m=const\end{align}$$

Part 3.2
Find $$\begin{align}V_y(t),y(t)\end{align}$$ for $$\begin{align}m=m(t)\end{align}$$

Part 1
Starting with Newton's second law expressed in terms of the change in momentum

recognizing there are two forces acting on the particle, namely drag which acts in the negative tangential direction and gravity which acts in the negative y-axis

using the given values for the forces,

where $$\hat{\textbf{i}}_t$$ is the tangential unit vector and $$\hat{\textbf{j}}$$ is the unit vector aligned with the gravitational field.

Utilizing $$\alpha$$ to project the two forces in the $$\hat{\textbf{i}}- \hat{\textbf{j}}$$ plane, where $$\hat{\textbf{i}}, \hat{\textbf{j}}$$ form a right handed Cartesian coordinate system with an outward positive third dimension.

with the particle having a constant mass the EOM are found

Part 2
with $$\begin{align}k=0\end{align}$$

so that

which has the solution by integration with respect to time

integrating the equations once again

time can be expressed as a function of x explicitly, thus y is then a function of x

Hence, y(x) is a parabola.

Part 3
With $$\begin{align}v_{xo}=0\end{align}$$, and $$\begin{align}\cos\alpha=\frac{v_x}{v}\end{align}$$ (from the geometry)

Which presents an interesting fact, once $$\begin{align}v_x\end{align}$$ is equal to zero it remains zero as its derivative is also zero (except when $$\begin{align}v=0\end{align}$$ and $$\begin{align}n-1<0\end{align}$$). Thus, if $$\begin{align}v_{xo}=0\end{align}$$ the $$\begin{align}x\end{align}$$ 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 for any $$n$$, by making it exact though the integrating factor method an expression was found.

$$\begin{align}n=0\end{align}$$
When n=0

Applying the boundary condition $$\begin{align}v_y(t=0)=v_{yo}\end{align}$$

$$\begin{align}n=1\end{align}$$
When n=1

Where finding an explicit solution fails.

Another approach is available for this particular problem when n=1. First re-expressing equation 1.2

This equation is a particular Chini equation. When n=1

which is simply a linear first order equation with the solution

The equation is separable

Applying boundary conditions $$\begin{align}v_y(t=0)=v_{yo}\end{align}$$

$$\begin{align}n=2\end{align}$$
When n=2

Where the imaginary error function appears.

Again looking at the Chini equation, but with n=2

Which is the general Riccati equation

$$\begin{align}n=3\end{align}$$
With n=3 the resulting equation is a particular case of the Abel differential equation. At this point the solutions are assumed sufficiently complex to effectively not exist.

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

The approach taken here is to look at the particular form of the Chini equation, equation 1.2, and how it reduces to a particular equation for various values of $$\begin{align}n\end{align}$$

Solving for the particular case where n=2, which is of physical interest as the drag is proportional to the velocity squared in reality.

The solution to which may be found in the scanned derivation below. As the problem consists of an unspecified function $$\begin{align}m(t)\end{align}$$ integral expressions remain in the solution. The solution procedure is to:


 * 1) Assume a known particular solution $$\begin{align}y_1(t)\end{align}$$
 * 2) Let $$\begin{align}w=y(t)-y_1(t)\end{align}$$, which yields a new ODE
 * 3) Let $$\begin{align}u=w^{-1}\end{align}$$ to reduce the order of the ODE
 * 4) Test the exactness, which fails
 * 5) Introduce an integrating factor and require the equation to be exact
 * 6) Assume the integrating factor is a function of time only
 * 7) Find the integrating factor by solving the linear ODE
 * 8) Perform partial integration of the reduced ODE with the integrating factors and combine to find $$\begin{align}\phi\end{align}$$
 * 9) Solve for $$\begin{align}u\end{align}$$
 * 10) Back substitute back to $$\begin{align}v_y\end{align}$$
 * 11) Integrate the expression to get vertical displacement as a function of time
 * 12) Simplify the expression

Then to solve one must find a particular solution for a given $$\begin{align}m(t)\end{align}$$. Once that is done the solution is purely mechanical.



Reference List of Solutions
A convenient list of solved first order ordinary differential equations may be found here, however, the solutions are presented without derivation.

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
Newton's second law of rotation

Background : small angle approximation
small angle approximation

Background : Hooke's Law
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 Statement
In the field of control engineering a linear system of coupled equations with variable coefficients may be represented as :

For the particular case of a single linear first order ODE equation 3.1 reduces to :

If the equation is time invariant (constant coefficient) then 3.2 further reduces to:

The solution to this equation may be found by applying the Euler Integrating Factor Method as detailed in lecture 10 Fall 2010. The solution is given as :

Required
Prove that equation 3.4 is the result of applying the Euler Integrating Factor Method to equation 3.3.

Solution
The terms in equation 3.3 may be re-written in the following form:

It is clear that this equation complies with the first condition of exactness.

So to make this equation exact we will apply the Euler Integrating Factor Method.

It has been previously shown that in order to meet the second condition of exactness we can multiply equation 3.5 by a function h to obtain the following:

Now the second condition of exactness may be applied, where x is the dependent variable and t is the independent variable.

Now for the special case where $${{h}_{x}}$$ equals zero we obtain the following:

Which may be rearranged to obtain:

We can now solve for h as a function of t.

Rewriting equation 3.6 we obtain:

We should recognize equation 3.12 as the derivative of (hx):

Therefore, to find the solution we integrate both sides in the interval $${{t}_{o}}\le \tau \le t$$.

Equation 3.14 yields:

Rearranging terms in equation 3/15 we obtain:

Dividing equation 3.16 by $${{e}^{-at}}$$, we obtain:

After some very simple manipulation we obtain:

Which, after applying the properties of exponents, can be rewritten as:

Given
$$\begin{align}f(x)=\exp (x)\end{align}$$ and $$\begin{align}{{x}_{o}}=0\end{align}$$

Find
Taylor Series expansion of f(x) at $$\begin{align}{{x}_{o}}=0\end{align}$$

Solution
Taylor Series expansion of function f: $$f(x) = f(a)+\frac {f'(a)}{1!} (x-a)+ \frac{f''(a)}{2!} (x-a)^2+\frac{f^{(3)}(a)}{3!}(x-a)^3+ \cdots. $$

For $$\begin{align}f(x)=\exp (x)\end{align}$$,

$$\begin{align}{{f}^{(n)}}(x)=f(x)=\exp (x)\end{align}$$

and at $$\begin{align}{{x}_{o}}=0\end{align}$$,

$$\begin{align}{{f}^{(n)}}({{x}_{0}})=1\end{align}$$

therefore,

$$\exp (x)=\exp ({{x}_{o}})+\exp ({{x}_{o}})(x-{{x}_{o}})+{{\exp }^{(2)}}({{x}_{0}})\frac{2!}+...$$

becomes

$$\exp (x)=1+\frac{1!}+\frac{2!}+\frac{3!}+...=\sum\limits_{k=0}^{n}{\frac{k!}}$$

= Genaralization of SC-L1-ODE-CC, from scalar to vector =

Given
and the solution of this L1-ODE-CC of the form

Find
The solution in vector form for system of linear ODE with constant coefficients

Solution
The equation 5.2 can be re-arranged in the following form:

It is clear that this equation complies with the first condition of exactness.

So to make this equation exact we will apply the Euler Integrating Factor Method.

It has been previously shown that in order to meet the second condition of exactness we can multiply equation 5.3 by a function h to obtain the following:

Now the second condition of exactness may be applied, where x is the dependent variable and t is the independent variable.

Now for the special case where $${{h}_{x}}$$ equals zero we obtain the following:

Which may be rearranged to obtain:

We can now solve for h as a function of t.

Rewriting equation 5.3 we obtain:

We should recognize equation 5.10 as the derivative of (hx):

Therefore, to find the solution we integrate both sides in the interval $${{t}_{o}}\le \tau \le t$$.

Equation 5.12 yields:

Rearranging terms in equation 5.13 we obtain:

Dividing equation 5.14 by $${{e}^{-At}}$$, we obtain:

After some very simple manipulation we obtain:

Which, after applying the properties of exponents, can be rewritten as:

where the dimension of matrices are:

$$\begin{align} & {{X}_{n\times 1}} \\ & {{U}_{n\times 1}} \\ & {{A}_{n\times n}} \\ & {{B}_{n\times n}} \\ \end{align}$$

Given
and the following properties of the of the transition matrix $$\phi $$

Where $$I$$ is the identity matrix

Find
Equation 6.4 using equations 6.1, 6.2 and 6.3

Solution
Substituting $$x\left( t \right)$$ in the equation 6.1 by $$\underline{\phi }\left( t,{{t}_{0}} \right)$$ to obtain the differential equation of the state transition matrix, we get:

Re-arranging equation 6.5 in the following form:

It is clear that this equation complies with the first condition of exactness.

So to make this equation exact we will apply the Euler Integrating Factor Method.

It has been previously shown that in order to meet the second condition of exactness we can multiply equation 6.6 by a function h to obtain the following:

Now the second condition of exactness may be applied, where x is the dependent variable and t is the independent variable.

Now for the special case where $${{h}_{x}}$$ equals zero we obtain the following:

Which may be rearranged to obtain:

We can now solve for h as a function of t.

Substituting 6.12 in the equation 6.6 we get

We should recognize equation 6.13 as the derivative of $$\left( h\phi \right)$$, by using the relation given in equation 6.2:

Therefore, to find the solution we integrate both sides in the interval $${{t}_{o}}\le \tau \le t$$.

Equation 6.15 yields:

Re-arranging terms in equation 6.16 we obtain:

Substituting $$\underline{\phi }\left( t,{{t}_{0}} \right)$$ in the equation 6.17 by $$x\left( t \right)$$ we obtain the differential equation in the form of equation 6.4:

Given
System of coupled Linear 1st order ODE constant coefficient


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$$\begin{align} \underline{\dot{\textbf{x}}}=\underline{\textbf{A}}\cdot \underline{\textbf{x}}(t) + \underline{\textbf{B}}\cdot \underline{\textbf{u}}(t) \end{align}$$

(7.1)
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Consider a Rocket Prevent rolling by activating ailerons


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$$\begin{align} &{\delta} = {\textrm{aileron{~} angle(deflection)}} \\ &{\phi} = {\textrm{roll{~} angle}} \\ &{\omega} ={\textrm{roll{~} angular {~} velocity}} \\ &Q = {\textrm{aileron{~}effectiveness}} \\ &{\tau} = {\textrm{roll-time{~} constant}} \\ \end{align} $$


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$$\begin{align} &\dot{\phi} = {\omega}\\ \end{align}$$

(7.2)
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$$\begin{align} &\dot{\omega} = - {\frac{1}{\tau}}{\omega} + {\frac{Q}{\tau}}{\delta}\\ \end{align}$$

(7.3)
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$$\begin{align} &\dot{\delta} = u \left({\textrm{commmand{~} signal{~} to {~} aileon{~} actuators{~} ={~} control}}\right)\\ \end{align}$$

(7.4)
 * 
 * }

Required
Put (7.2) - (7.4) in form of (7.1)

Solution
From(7.1), $$\begin{align} \dot{\phi},\dot{\omega},\dot{\delta}\end{align}$$ as regard as $$\begin{align} \dot{\textrm{x}}\end{align}$$, $$\begin{align} {\phi},{\omega},{\delta}\end{align}$$ as regard as $$\begin{align} \textrm{x(t)}\end{align}$$, put those three equations can be expressed as matrix form(7.5)


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

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

(7.5)
 * 
 * }

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

Problem Statement
An arbitrary exact second order nonlinear ordinary differential equation (N2_ODE) was generated in Lecture 17 Fall 2010. The equation was given as:

Required
Prove that equation 9.1 meets the second condition for exactness by applying the following conditions for exactness given in Lecture 15 Fall 2010 :

Solution
The first condition of exactness has already been shown to be met for equation 9.1, however, to prove that the equation is exact both equation 9.2 and 9.3 must also hold true.

Second Condition of Exactness Type 1
We will begin be applying equation 9.2, but first we will re-write equation 9.1 for clarity:

Starting with the left hand side of equation 9.2 we will examine each term independently. We begin with:

Combining all terms from the left hand side of 9.2 we obtain:

Now we will examine each term of the right hand side of equation 9.2.

Combining all terms from the right hand side of equation 9.2:

We can simplify the above to obtain:

Comparing equation 9.17 with equation 9.32 we can see that equation does meet the condition for exactness specified by equation 9.2.

Second Condition of Exactness Type 2
We will now apply equation 9.3 to equation 9.1 to test for exactness.

Starting with the left hand side of equation 9.3 we examine each term independently.

Combining all terms from the left hand side of 9.3 we obtain:

Now we will examine the right hand side of equation 9.3.

Comparing equation 9.45 with equation 9.49 we can see that equation does meet the condition for exactness specified by equation 9.3.

Since both equation 9.2 and 9.3 are valid when applied to equation 9.1, equation 9.1 is exact.

Given

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$$\begin{align} (6xy^2) y' + 2y^3 = h_{x} y' + h_{x} \end{align}$$

(10.1)
 * 
 * }

Required
Find $$\begin{align} h(x,y) \end{align}$$ and $$\begin{align} {\phi}(x,y,y') \end{align}$$

Solution
1) 1st choice, assume $$\begin{align} & {{h}_{x}}=2{{y}^{3}} \\  & {{h}_{y}}=6x{{y}^{2}} \\ \end{align}$$.

then

$$\begin{align} h(x,y)=2x{{y}^{3}}+{{k}_{1}} \end{align}$$, with k1 being integration constant.

Hence

$$\begin{align} & \overline{\phi }=h(x,y)+\int{f(x,y,p)dp}=2x{{y}^{3}}+{{k}_{1}}+3{{p}^{5}}\cos ({{x}^{2}})={{k}_{2}} \\ & \phi =2x{{y}^{3}}+3{{p}^{5}}\cos ({{x}^{2}})=k \\ \end{align}$$, with k1,k1 and k being integration constants.

2) 2nd choice, assume

$$\begin{align}{{h}_{y}}y'=2{{y}^{3}}\end{align}$$, and this implicitly assumes that

$$\begin{align}{{h}_{x}}=(6x{{y}^{2}})y'\end{align}$$.

We know that $$\begin{align}h=h(x,y)\end{align}$$, so it is not possible to have y' term in hx. Hence the story ends here, and 1st choice is the only choice.

Note
If we carry on the derivation $$\begin{align} & Assume\\ & {{h}_{y}}y'=2{{y}^{3}} \\ & hydy=2{{y}^{3}}dx \\ & \int{hydy=\int{2{{y}^{3}}dx}} \\ & \Rightarrow h(x,y)+\alpha (x)=2x{{y}^{3}}+{{C}_{1}} \\ & {{h}_{x}}=2{{y}^{3}}-\alpha '(x) \\ \end{align}$$

where $$\begin{align}\alpha (x)\end{align}$$ is a function of x only.

However, this assumption $$\begin{align}{{h}_{y}}y'=2{{y}^{3}}\end{align}$$ implicitly assumes that

$$\begin{align}{{h}_{x}}=(6x{{y}^{2}})y'\end{align}$$

Comparing both $$\begin{align}{{h}_{x}}\end{align}$$, this is a dead end, as explained in the solution part.

=Contributing Team Members=
 * EGM6321.f10.team6.cook Solved problems 1, reviewed all problems, Figures 1,2, and 7
 * Egm6321.f10.team6.csquared Solved problems 3,8, reviewed all problems
 * Egm6321.f10.team6.lee Solved problems 2,3,8, reviewed all problems
 * EGM6321.f10.team6.yoshioka Solved problems 6, reviewed all problems
 * EGM6321.f10.team6.Pengxiang Solved problems 4,5,7,10, reviewed all problems
 * EGM6321.f10.team6.Kim Solved problems 7,10, reviewed all problems

{| class="prettytable" style="margin: 1em auto 1em auto;"
 * colspan="3" | Problem Assignments


 * Problem #
 * Solved by
 * Reviewed by
 * Reviewed by


 * 1
 * cook
 * all
 * all


 * 2
 * Lee
 * all
 * all


 * 3
 * Cassano, Lee
 * all
 * all


 * 4
 * Pengxiang
 * all
 * all


 * 5
 * Pengxiang
 * all
 * 6
 * Yoshioka
 * all
 * Yoshioka
 * all


 * 7
 * Kim, Pengxiang
 * all
 * all


 * 8
 * Cassano, Lee
 * all
 * all


 * 9
 * Cassano
 * all
 * all


 * 10
 * Kim, Pengxiang
 * all
 * all


 * -}