User:Egm6321.f12.team7/report1

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Problem 1[edit]

We have solved this problem by ourselves.

Given[edit]

(1.1)

Find[edit]

,

where and

(1.2)

Solution[edit]

First, we get the first derivative of with respect to time,

, with

(1.3)

Then we get,

(1.4)

(1.5)

(1.6)

Bring (1.5) and (1.6) into (1.4), we can get,

(1.7)

i.e.

,

where and

(1.8)


Problem 2[edit]

We have solved this problem by ourselves.

Given[edit]

The Equation of Motion (EOM) for the meglev train is modeled by:

(2.1)

Find[edit]

  • Get the derivation of the 1st total time derivative.
  • Get the derivation of the 2nd total time derivative.
  • Show the similarity with the derivation of the Coriolis force.

Solution[edit]

* Get the derivation of the 1st total time derivative[edit]

Apply the chain rule to , we have,

(2.2)

Given that for is evaluated at in the Eq. (2-1). Then the equation can be rewritten as,

(2.3)

Define the notation as below,

(2.4)

As the result, Eq. (2-1) can be simplified to,

(2.5)

* Get the derivation of the 2nd total time derivative[edit]

Take the 2nd derivative of the Eq. (2.2) with respect to , and apply the chain rule yields. We have,

(2.6)

Define the notations as below,

(2.7)

Then the Eq. (2.6) can be rewritten as below,

(2.8)

* Show the similarity with the derivation of the Coriolis force[edit]

Define as the position vector indicating the position of the origin of the reference frame, and define is a point fixed on an object rotating with an angular velocity of with respect to inertial reference frame . And define the three mutual perpendicular unit vectors . Then, can be defined as,

(2.9)

The velocity of the object as viewed by an observer fixed to the inertial reference frame is defined as below,

(2.10)

The acceleration of the object as viewed by an observer fixed to the inertial reference frame is defined as below,

(2.11)

Substitute from the Eq. (2.10) into the Eq. (2.11),and we have,

(2.12)

Expand the Eq. (2.12) as below.

As the result, the Eq.(2.12) can be rewritten as,

(2.13)

by the definition,the expression on RHS can also be written as,

(2.14)

The Eq. (2-8) and the Eq. (2-13) are related as below,

Eq. (2-8)
Eq. (2-13)
LHS
Expression1 on RHS
Expression 2 on RHS
Expression 3 on RHS
Expression 4 on RHS

Problem 3[edit]

We have solved this problem by ourselves.

Given[edit]

The expression of is given by,

(3.1)

Find[edit]

Analyze the dimension of each term and provide the physical meaning.

Solution[edit]

*Dimensional analysis of each term[edit]

Left hand side:

(3.2)

Right hand side:

  • First term

(3.3)

(3.4)

(3.5)

(3.6)

(3.7)

(3.8)

  • Second term

(3.9)

(3.10)

(3.11)

  • Third term

(3.12)

(3.13)

(3.14)

  • Fourth term

(3.15)

(3.16)

(3.17)

(3.18)

(3.19)

(3.20)

(3.21)

(3.22)

(3.23)

*Physical meaning[edit]

is the horizontal force which act on wheel/magnet of the train. Other terms are forces from different source.

Problem 4[edit]

We have solved this problem by ourselves.

Given[edit]

The polar coordinate

Find[edit]

Draw the polar coordinate lines in a 2-D plane emanating from a point, not at the origin.

Solution[edit]

Suppose a non-origin point .

Draw the polar coordinate lines in XY coordinate system from the point , then we could derive:

(4.1)

(4.2)

According to (4.2), as changes, two sets of lines could be draw as shown in the following figure(4-1):

figure(4-1)

Problem 5[edit]

We have solved this problem by ourselves.

Given[edit]

From the lecture note, we have

(5.1)


Find[edit]

Show that equation(5.1) becomes

(5.2)


Solution[edit]

According to the Chain Rule, equation (5.1) yields

(5.3)


After simplification, it becomes

(5.4)


Obviously, When , equation(5.4) and equation(5.2) are the same.

Problem 6[edit]

We have solved this problem by ourselves.

Given[edit]

The Equation of Motion of Wheel/Magnet has four terms, which are given by,

(6.1)

The general expression of is given by,

(6.2)

From the lecture note[}, we can Consider as an operator, is linear if and only if,

(6.3)

Where and are two possible arguments for , and

The derivative operator is a linear operator, which is given by,

(6.4)

Find[edit]

is nonlinear with repect

Solution[edit]

Suppose we have,

(6.5)

For the right-hand side, we have,

(6.6)

The left-hand side is given by,

(6.7)

Since the derivative operator is linear operator, we can get,

(6.8)

So (6.7) turns into,

(6.9)

If is nonlinear with respect to , by comparision between (6.5) and (6.8) we can find that,

(6.10)


Even if is linear with respect to , on the left-hsnd side we have,

(6.11)

Obviously,

(6.12)

Problem 7[edit]

We have solved this problem by ourselves.

Given[edit]


Find[edit]

Show that is linear.

Solution[edit]

If is linear, then the following must be true:















Thus is a linear operator.

Contribution[edit]

Problem Assignments
Problem # Solved&Typed by Reviewed by
1 Goran Marjanovic, Zixiang Liu all
2 Xiaolong Wu, Lin,Hao all
3 Zhou Zhe, Rui Kong, Anup Parikh all
4 Goran Marjanovic, Zixiang Liu all
5 Xiaolong Wu, Lin,Hao all
6 Zhou Zhe, Rui Kong, Anup Parikh all
7 Goran Marjanovic, Zixiang Liu all