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2. Kinematics of the material point in orthogonal curvilinear coordinates

Curvilinear coordinates (q1, q2, q3) are a coordinate system for Euclidean

space in which the coordinate lines may be curved.

Examples: polar, rectangular, spherical, and cylindrical coordinate systems.

Lecture 2. Curvilinear coordinates 1

Theoretical Mechanics

(2.3)

We assume the existence of a vector function 𝑟 : Ω ⊂ 𝑅3 → 𝐷 ⊂ 𝑅3 such that

Lecture 2. Curvilinear coordinates 2


(2.1)

(2.2)

This function associates to each triplet 𝑞1, 𝑞2, 𝑞3 ∋ Ω an unique position

M 𝑥, 𝑦, 𝑧 ∈ R3 and vice versa.

(2.3)

Thus, the application should be is a diffeomorphism of class C2 and in order to exist

is necessary that:

3


(2.4)

(2.5)

In this case the equation of motion are

Lecture 2. Curvilinear coordinates

4


The curves:

Γ1:

Γ2:

Γ3:

are called curves of coordinates. In a similar way the surfaces of coordinates are

obtained.


5


The elementary displacement on

is:

and the length of the curve arc covered by

particle M is given by:


𝑀 𝑑𝑟 2

𝑞1

𝑞2

𝑞3

𝑑𝑟 1 𝑑𝑟 3

𝑒 1

𝑒 2

𝑒 3

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Taking into account that 𝑑𝑟 is tangent to the curve Γ1 then

Generally, we have:

where 𝐻𝑖 is the coefficient of Lamé corresponding to the curve Γ1 (i = 1,2,3).

However, for a general motion the elementary displacement on the trajectory is:

𝑑𝑟 = 𝑑𝑟 𝑖

3

𝑖=1

= 𝜕𝑟

𝜕𝑞𝑖

3

𝑖=1

𝑑𝑞𝑖 = 𝐻𝑖𝑑𝑞𝑖𝑒 𝑖

3

𝑖=1

𝑑𝑠 = 𝑑𝑟 = 𝐻𝑖𝑑𝑞𝑖2

3

𝑖=1

; 𝑑𝑠𝑖 = 𝐻𝑖𝑞𝑖 , 𝑖 = 1,2,3

(2.6)

(2.7)

(2.8)


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Velocity in curvilinear coordinates

Taking into account that and

one obtain:

where

𝑣𝑖 = 𝐻𝑖𝑞𝑖 , (𝑖 = 1,2,3)

are the components of the velocity in the frame 𝑒 1, 𝑒 2, 𝑒 3 .

or

𝑣 = 𝐻𝑖𝑞𝑖 𝑒 𝑖

3

𝑖=1

= 𝑣𝑖𝑒 𝑖

3

𝑖=1

(2.9)

(2.8)


Lecture 1. Introduction 8


Acceleration in curvilinear coordinates

Consider the components of the acceleration:

We have

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Finally, one obtain

(2.10)

Remark: If the curve of coordinate is a straight line or a circle of radius R then we

have 𝐻line = 1 and 𝐻circle = 𝑅.

Indeed, for Γ1 line we have 𝑑𝑞1 = 𝑑𝑠1


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For Γ1 circle we have 𝑑𝑠1 = 𝑅𝑑𝜃, and thus


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Particular curvilinear coordinates –cylindrical coordinates

(q1, q2, q3) :


Equations of the motion are:

Functional relations and the Jacobian:

(2.11)

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The curves:

Γ𝑟 is a line with the versor e𝑟 𝐻𝑟 = 1

Γ𝜃 is a circle with the versor e𝜃 𝐻𝜃 = 𝑟

Γ𝑧 is a line with the versor e𝑧 𝐻𝑧 = 1


The curves of coordinates (𝑟, 𝜃, 𝑧) are orthogonal.

(orthogonality condition)

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Thus, the velocity and acceleration components are:

(2.12)


𝑑𝑟 = 𝑑𝑟 𝑟 + 𝑑𝑟 𝜃 + 𝑑𝑟 𝑧 = 𝑑𝑟 𝑒 𝑟 + 𝑟𝑑𝜃 𝑒 𝜃 + 𝑑𝑧 𝑒 𝑟

𝑑𝑠 = 𝑑𝑟 = 𝑑𝑟 2 + 𝑟𝑑𝜃 2 + 𝑑𝑧 2

Displacement:

𝑑𝑠𝑟 = 𝐻𝑟𝑑𝑟 = 𝑑𝑟; 𝑑𝑠𝜃 = 𝐻𝜃𝑑𝜃 = 𝑟𝑑𝜃; 𝑑𝑠𝑧 = 𝐻𝑧𝑑𝑧 = 𝑑𝑧

𝑑𝑟 𝑟 = 𝑑𝑠𝑟𝑒 𝑟 = 𝑑𝑟𝑒 𝑟; 𝑑𝑟 𝜃 = 𝑑𝑠𝜃𝑒 𝜃 = 𝑟𝑑𝜃𝑒 𝜃; 𝑑𝑟 𝑧 = 𝑑𝑠𝑧𝑒 𝑟 = 𝑑𝑧𝑒 𝑟;

(2.13)

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(2.14)


𝑎 = 𝑎𝑟𝑒 𝑟 + 𝑎𝜃𝑒 𝜃 + 𝑎𝑧𝑒 𝑧 = 𝑟 − 𝑟𝜃 2 𝑒 𝑟 +1

𝑟

𝑑

𝑑𝑡𝑟2𝜃 𝑒 𝜃 + 𝑧 𝑒 𝑧

15


(2.15)

Remark: For z=0, the motion in polar coordinates is obtained.

Equations of the motion are:

The velocity and acceleration components are

𝑣𝑟 = 𝑟 is the radial velocity

𝑣𝜃 = 𝑟𝜃 is the transversal (circumferential) velocity

(2.16)


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Remark: It is also possible to calculate directly, using (2.6) the values for the

Lame’s coefficients in cylindrical coordinates.

(2.17)


𝐻𝑟 =𝑑𝑥

𝑑𝑟

2

+𝑑𝑦

𝑑𝑟

2

+𝑑𝑧

𝑑𝑟

2

= cos2 𝜃 + sin2 𝜃 + 02 = 1

𝐻𝜃 =𝑑𝑥

𝑑𝜃

2

+𝑑𝑦

𝑑𝜃

2

+𝑑𝑧

𝑑𝜃

2

= r2cos2 𝜃 + 𝑟2sin2 𝜃 + 02 = 𝑟

𝐻𝑧 =𝑑𝑥

𝑑𝑧

2

+𝑑𝑦

𝑑𝑧

2

+𝑑𝑧

𝑑𝑧

2

= 02 + 02 + 12 = 1

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The angular velocity describes the rate of

change of angular position, and the

instantaneous axis of rotation. In this case

(counter-clockwise rotation) the vector points up.

In this case the normal and

tangent versors and the polar

coordinate versors have the

same directions.

https://ocw.mit.edu/courses/aeronautics-and-astronautics/16-07-dynamics-fall-2009/lecture-

notes/MIT16_07F09_Lec05.pdf

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How to find the sense of the angular velocity:


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Particular curvilinear coordinates –spherical coordinates

The curves:

Γ𝑟 is a line with the versor e𝑟 𝑟 𝐻𝑟 = 1

Γ𝜃 is a circle of radius 𝑟 with the versor e𝜃 𝜃 𝐻𝜃 = 𝑟

Γ𝜙 is a circle of radius 𝑟 sin 𝜃 with the versor e𝜙 𝜙 𝐻𝜙 = 𝑟 sin 𝜃

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(2.18)


𝑑𝑟 = 𝑑𝑟 𝑟 + 𝑑𝑟 𝜃 + 𝑑𝑟 𝜙 = 𝑑𝑟 𝑒 𝑟 + 𝑟𝑑𝜃 𝑒 𝜃 + 𝑟 sin 𝜃 𝑑𝜙𝑒 𝜙

𝑑𝑠 = 𝑑𝑟 = 𝑑𝑟 2 + 𝑟𝑑𝜃 2 + 𝑟 sin 𝜃 𝑑𝜙 2

Displacement:

𝑑𝑠𝑟 = 𝐻𝑟𝑑𝑟 = 𝑑𝑟; 𝑑𝑠𝜃 = 𝐻𝜃𝑑𝜃 = 𝑟𝑑𝜃; 𝑑𝑠𝜙 = 𝐻𝜙𝑑𝜙 = 𝑟 sin 𝜃 𝑑𝜙

𝑑𝑟 𝑟 = 𝑑𝑟𝑒 𝑟; 𝑑𝑟 𝜃 = 𝑟𝑑𝜃𝑒 𝜃; 𝑑𝑟 𝜙 = 𝑟 sin 𝜃 𝑑𝜙𝑒 𝜙;

(2.19)

Velocity: 𝑣 = 𝑣𝑟𝑒 𝑟 + 𝑣𝜃𝑒 𝜃 + 𝑣𝜙𝑒 𝜙

𝑣𝑟 = 𝑟

𝑣𝜃 = 𝑟 𝜃

𝑣𝜙 = 𝑟 sin 𝜃 𝜙

24


(2.20)


Acceleration:

Other choice :

(2/16)

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Example: J. L. Meriam, L. G. Kraige, J. N. Bolton, Engineering Mechanics:

Dynamics, 8th Ed, Wiley, 2015

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Example: J. L. Meriam, L. G. Kraige, J. N. Bolton, Engineering Mechanics:

Dynamics, 8th Ed, Wiley, 2015

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