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 11The velocity and acceleration of a particle in intrinsic coordinates are given by 𝐯=v𝐓,v=|dsdt|,𝐚=dvdt𝐓+v2ρ𝐍, 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 planeMathworldPlanetmath, even if the motion is spatial.

dvdt=a,v2ρ=b,0t<, (1)

where a,b>0 are known constants, ρ=ds/dθ the path’s radius of curvatureMathworldPlanetmath 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)=v0, θ(0)=0 (θ˙(0) will depend on v0 and ρ(0)). Then, from the equation (1)1 22So 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)=v0+at, and from (1)2 we have (by the chain rule and the definition of ρ) v2/ρ=vθ˙=b, where we perform an integration to get θ(t) 33Here, and in what it follows, abuse of notation is not confused., i.e.

θ(t)=0θ(t)dθ=b0tdtv0+at=balog(1+atv0). (2)

We now introduce parametric Cartesian coordinatesMathworldPlanetmath (x(t),y(t)), as we must deal with dx=dscosθ and dy=dssinθ in order to find out the path. Thus, we have

dx=(v0+at)cos{balog(1+atv0)}dt,dy=(v0+at)sin{balog(1+atv0)}dt. (3)

SymmetryMathworldPlanetmath 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 eiu=cosu+isinu. That is,


and by integrating,


or 44We multiply the numerator and the denominator by the complex conjugateMathworldPlanetmath 2-ib/a, and 1ib/a=1.

z(t)=v02a(2-iba)(ba)2+4{(1+atv0)2eibalog(1+atv0)-1}. (4)

Before separating real and imaginary partsMathworldPlanetmath, is advisable to make an isomorphic conformal mappingMathworldPlanetmathPlanetmath zζ over , i.e. x(t)+iy(t)ξ(τ)+iη(τ), which consists of a nondimensional process about parameters and the involved coordinates. That is,

κ:=ba,τ:=1+atv0,ζ:=azv02,tτ,t=0τ=1. (5)

Then, by making use of Euler’s formula, by introducing (5) into (4) and after a little algebraPlanetmathPlanetmath in order to separate real and imaginary parts, we obtain

(τ) =ξ(τ)+iη(τ)
=1κ2+4({τ2[2cos(κlogτ)+κsin(κlogτ)]-2}+i{τ2[2sin(κlogτ)-κcos(κlogτ)]+κ}),   (6)




η(τ)-κκ2+4=τ2κ2+4{2sin(κlogτ)-κcos(κlogτ)}.       (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


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