spiral motion of a particle

spiral motion of a particle

The purpose of this entry is to illustrate how calculus and another branches about elementary applied mathematics are useful to solve problems related to kinematics of a particle. To

continuation

, we state and discuss an interesting problem which is not so easy for the novices in those topics. Let us consider the plane motion of a particle subjected under the conditions over its tangential and normal

acceleration

components

1

1

The

velocity

and acceleration of a particle in intrinsic

coordinates

are given by

𝐯

=

v

⁢

𝐓

,

v

=

|

d

⁢

s

d

⁢

t

|

,

𝐚

=

d

⁢

v

d

⁢

t

⁢

𝐓

+

v

2

ρ

⁢

𝐍

,

being the acceleration expressed in its components tangential and normal and referred to a Frenet-Serret

local basis

{

𝐓

,

𝐍

}

, moving along with the particle at the

osculating plane

, even if the motion is spatial.

d

⁢

v

d

⁢

t

=

a

,

v

2

ρ

=

b

,

0

≤

t

<

∞

,

(1)

where

a

,

b

>

0

are known constants,

ρ

=

d

⁢

s

/

d

⁢

θ

the path’s

radius of curvature

and

θ

the path’s

slope angle

, at an arbitrary place occuped by the particle in its motion. At

t

=

0

, we impose intrinsic initial conditions

s

⁢

(

0

)

=

0

,

v

⁢

(

0

)

≡

s

˙

⁢

(

0

)

=

v

0

,

θ

⁢

(

0

)

=

0

(

θ

˙

⁢

(

0

)

will depend on

v

0

and

ρ

⁢

(

0

)

). Then, from the equation

(

1

)

1

2

2

So suggestive designation is useful when we are dealing with a set of equations which are indicated with a single number.

, one

integrates

to get

v

=

v

⁢

(

t

)

=

v

0

+

a

⁢

t

, and from

(

1

)

2

we have (by the chain rule and the

definition

of

ρ

)

v

2

/

ρ

=

v

⁢

θ

˙

=

b

, where we perform an integration to get

θ

⁢

(

t

)

3

3

Here, and in what it follows, abuse of notation is not confused.

, i.e.

θ

⁢

(

t

)

=

∫

0

θ

⁢

(

t

)

d

θ

=

b

⁢

∫

0

t

d

⁢

t

v

0

+

a

⁢

t

=

b

a

⁢

log

⁡

(

1

+

a

⁢

t

v

0

)

.

(2)

We now introduce parametric

Cartesian coordinates

(

x

⁢

(

t

)

,

y

⁢

(

t

)

)

, as we must deal with

d

⁢

x

=

d

⁢

s

⁢

cos

⁡

θ

and

d

⁢

y

=

d

⁢

s

⁢

sin

⁡

θ

in order to find out the path. Thus, we have

d

⁢

x

=

(

v

0

+

a

⁢

t

)

⁢

cos

⁡

{

b

a

⁢

log

⁡

(

1

+

a

⁢

t

v

0

)

}

⁢

d

⁢

t

,

d

⁢

y

=

(

v

0

+

a

⁢

t

)

⁢

sin

⁡

{

b

a

⁢

log

⁡

(

1

+

a

⁢

t

v

0

)

}

⁢

d

⁢

t

.

(3)

Symmetry

of (3) is evident and its integration is easy, if we pass to

z

-plane, i.e.

(

x

⁢

(

t

)

,

y

⁢

(

t

)

)

↦

z

⁢

(

t

)

, as we may take advantage from Euler’s formula

e

i

⁢

u

=

cos

⁡

u

+

i

⁢

sin

⁡

u

. That is,

d

⁢

z

=

(

v

0

+

a

⁢

t

)

⁢

e

i

⁢

b

a

⁢

log

⁡

(

1

+

a

⁢

t

v

0

)

⁢

d

⁢

t

=

v

0

⁢

e

log

⁡

(

1

+

a

⁢

t

v

0

)

⁢

e

log

(

1

+

a

⁢

t

v

0

)

i

⁢

b

a

⁢

d

⁢

t

,

and by integrating,

∫

0

z

⁢

(

t

)

d

z

=

v

0

⁢

∫

0

t

(

1

+

a

⁢

t

v

0

)

1

+

i

⁢

b

a

⁢

d

t

=

v

0

2

a

⁢

(

1

+

a

⁢

t

v

0

)

2

+

i

⁢

b

a

|

0

t

2

+

i

⁢

b

a

,

or

4

4

We multiply the numerator and the denominator by the

complex conjugate

2

-

i

⁢

b

/

a

, and

1

i

⁢

b

/

a

=

1

.

z

⁢

(

t

)

=

v

0

2

a

⁢

(

2

-

i

⁢

b

a

)

(

b

a

)

2

+

4

⁢

{

(

1

+

a

⁢

t

v

0

)

2

⁢

e

i

⁢

b

a

⁢

log

⁡

(

1

+

a

⁢

t

v

0

)

-

1

}

.

(4)

Before separating real and

imaginary parts

, is advisable to make an

isomorphic

conformal mapping

z

→

ζ

over

ℂ

, i.e.

x

⁢

(

t

)

+

i

⁢

y

⁢

(

t

)

↦

ξ

⁢

(

τ

)

+

i

⁢

η

⁢

(

τ

)

, which consists of a nondimensional process about

parameters

and the involved coordinates. That is,

κ

:=

b

a

,

τ

:=

1

+

a

⁢

t

v

0

,

ζ

:=

a

⁢

z

v

0

2

,

t

↦

τ

,

t

=

0

↦

τ

=

1

.

(5)

Then, by making use of Euler’s formula, by introducing (5) into (4) and after a little

algebra

in order to separate real and imaginary parts, we obtain

(

τ

)

=

ξ

⁢

(

τ

)

+

i

⁢

η

⁢

(

τ

)

=

1

κ

2

+

4

⁢

(

{

τ

2

⁢

[

2

⁢

cos

⁡

(

κ

⁢

log

⁡

τ

)

+

κ

⁢

sin

⁡

(

κ

⁢

log

⁡

τ

)

]

-

2

}

+

i

⁢

{

τ

2

⁢

[

2

⁢

sin

⁡

(

κ

⁢

log

⁡

τ

)

-

κ

⁢

cos

⁡

(

κ

⁢

log

⁡

τ

)

]

+

κ

}

)

,

⁢

(6)

whence,

ξ

⁢

(

τ

)

+

2

κ

2

+

4

=

τ

2

κ

2

+

4

⁢

{

2

⁢

cos

⁡

(

κ

⁢

log

⁡

τ

)

+

κ

⁢

sin

⁡

(

κ

⁢

log

⁡

τ

)

}

,

(7)

and

η

(

τ

)

-

κ

κ

2

+

4

=

τ

2

κ

2

+

4

{

2

sin

(

κ

log

τ

)

-

κ

cos

(

κ

log

τ

)

}

.

(8)

These are the nondimensional

parametric equations

of the path that the particle follows. It is evident the symmetry involved in (7) and (8).

Squaring

both, adding it and simplfying, we obtain the path’s equation

(

ξ

+

2

κ

2

+

4

)

2

+

(

η

-

κ

κ

2

+

4

)

2

=

τ

4

κ

2

+

4

,

(9)

which correponds to a family of spirals of parameter

κ

.

Title

spiral motion of a particle

Canonical

name

SpiralMotionOfAParticle

Date of creation

2013-03-22 17:23:42

Last modified on

2013-03-22 17:23:42

Owner

perucho (2192)

Last modified by

perucho (2192)

Numerical id

15

Author

perucho (2192)

Entry type

Topic

Classification

msc 70B05

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