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Problem 1 - Derive EOM (SC-N1-ODE)Edit

From Meeting 13, p. 13-1 ~ p. 13-2


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



  Drive Equation of Motion (EOM)
  Particular case   : Verify   is parabolla
  Consider  ,  ,
  Find  ,   for   = constant
  Find  ,   if  


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
2) Inertia force
  for particle with constant mass
3) Air resistance which is proportional to the velocity of particle

Now consider the force equilibrium in both horizontal and vertical direction

a) Force Equilibrium in horizontal direction:
where   horizontal component of velocity



b) Force Equilibrium in the vertical direction:


where   vertical component of velocity



Part 2

Particular case: When  

Eq.1.1 reduces to


Integrating the above equation gives:



Apply 'initial condition' to determine integration constant, 


Now Eq.1.3 becomes:


Integrate the above equation to obtain , 


Then 'Initial condition' is applied to determine , 




Now  , can be expressed in terms of  ,



Similarly When  , Eq.1.2 reduces to


Integrate the above equation to evaluate,  



Apply 'initial condition' to obtain  


Now Eq.1.5 becomes,


Then integrate the above equation to determine,  


  is determined using 'initial condition' as:




Now Substitute Eq.1.4 for   in the above equation;



Eq.1.6 is in the form of a parabolic equation. Therefore   is parabola.

Part 3

With  , and   (from the geometry)



Which presents an interesting fact, once   is equal to zero it remains zero as its derivative is also zero (except when   and  ). Thus, if   the   velocity remains zero and the equations of motion reduce to equation 1.2.

Part 3.1Edit




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   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




























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

Part 3.2Edit

If   the integrating factor method is complicated in equation 1.22 as the partial of   with respect to time remains, complicating the expression for  

Author and Proof-readerEdit



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

From Meeting 13, p. 13-3






  : mass of the each pendulum

  : the angle from the vertical to the each pendulum

  : applied forces to the each pendulum

  : length of the pendulum

  : force constant(or spring constant)

  : acceleration of gravity


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

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






Dimension of matrix







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,

  Torque of the spring force + Torque of the gravity force + Torque of the applied force )



  : torque

  : moment of inertia

  : angular acceleration

Therefore, left hand side is  

Torque of the spring force

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



  : restoring force
  : spring constant
  : displacement from the equilibrium position (in this case, x = a)

Therefore, Torque of the spring force is,



Torque of the gravity force



Torque of the applied force



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


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


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)



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



  Derive equation of motion:




  Write Eq.2.1 and Eq.2.2 in the form of Mtg 13 (c),page2 , of:





  Derive equation of motion:
(a) Consider Free Body Diagram of left pendulum:
For small angle:

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




(b) Consider Free Body Diagram of right pendulum:
For small angle:

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




  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,




Now Eq.2.4 and Eq.2.5 can be put in the form of Eq.2.3 as:




Contributing MembersEdit

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

Author and Proof-readerEdit



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

From Meeting 14, p. 14-1







Derive (3-2)


We can rearrange the eqn(3.1). As it is not time variable problem, let  



Let's find the integrating factor first.

As the coefficient for the   is 1,



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





let's integrate for the interval  





rearrange eqn(3.7),





Author and Proof-readerEdit



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

From Meeting 14, p. 14-2







1) Derive (4-1)

2) Derive (4-2)


Solution of 1)

Using Taylor series [7],


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


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










Solution of 2)

Using Taylor series and Maclaurin series



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






Author and Proof-readerEdit

[Author] Oh, Sang Min


Problem 5 - Generalized to SC-L1-ODE-VCEdit

From Meeting 14, p. 14-2











Dimension of matrix






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


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)



Dimension of matrix





Author and Proof-readerEdit

[Author] Oh, Sang Min


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

From Meeting 15, p. 15-1




Author and Proof-readerEdit



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

From Meeting 15, p. 15-1










  = aileron angle(deflection)

  = roll angle

  = roll angular velocity

  = aileron efflectiveness

  = roll time constant


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



Failed to parse (syntax error): {\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 and Proof-readerEdit

[Author] Oh, Sang Min


Problem 8Edit

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  , we know that   only.
It means  

Hence, eqn(8.2) becomes,



There are two possible solutions.



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

As   and  ,



Problem 9Edit

Problem 10Edit


  1. Torque(wikipedia)
  2. Hooke's Law(wikipedia)
  3. Pendulum(wikipedia)
  4. Moment(wikipedia)
  5. Moment of inertia(wikipedia)
  6. Angular acceleration(wikipedia)
  7. Taylor series(wikipedia)
  8. sigma notation(wikipedia)
  9. Maclaurin series(
  10. Exponential Matrix(wikipedia)
  11. Identity Matrix(wikipedia)
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