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Valentin Fadeev

Back to school

Saturday, 15 Jan 2011, 18:52
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Edited by Valentin Fadeev, Sunday, 18 Sept 2011, 23:24

This is problem 1.18 from JS (M821 course book). The question is to investigate the motion of a bead that slides on a smooth parabolic wire rotating with constant angular velocity omega about a vertical axis. x is the distance from the axis of rotation.

To simplyfy the calculations we can choose the scale so that equation of the parabola is y equals x squared divided by two .

Yes, I know as a grown up man I should write out the expression of the kinetic energy:

equation sequence cap t equals m times v squared divided by two plus m times omega squared times x squared divided by two equals m divided by two times open d times s divided by d times t close squared plus m times omega squared times x squared divided by two

where v is the tangential velocity of the bead directed along the wire. Then define potential energy as usual:

equation sequence cap v equals negative integral f of y d y equals integral m times g d y equals m times g times y equals m times g times x squared divided by two

bearing in mind that the force of gravity acts in the direction opposite to that of cap y axis, hence changing the sign.

Construct the Lagrangian:

equation sequence script cap l equals cap t minus cap v equals m divided by two times open d times s divided by d times t close squared plus m times omega squared times x squared divided by two minus m times g times x squared divided by two

Define the action:

cap s equals integral script cap l d t

Use variational principle

delta times cap s equals zero

to obtain the Euler-Lagrange equation:

d divided by d times t times open normal partial differential times script cap l divided by normal partial differential times x dot above close minus normal partial differential times script cap l divided by normal partial differential times x equals zero

Calculating separate terms:

equation sequence open d times s divided by d times t close squared equals x dot above squared plus y dot above squared equals x dot above squared times open one plus x squared close

equation left hand side normal partial differential times script cap l divided by normal partial differential times x dot above equals right hand side two times x dot above times open one plus x squared close

equation left hand side d divided by d times t times open normal partial differential times script cap l divided by normal partial differential times x dot above close equals right hand side two times x double dot times open one plus x squared close plus four times x times x dot above squared

equation left hand side normal partial differential times script cap l divided by normal partial differential times x equals right hand side two times x squared times x dot above squared plus two times omega squared times x minus two times g times x

Finally obtain the equation of motion:

x double dot times open one plus x squared close plus open g minus omega squared plus x dot above squared close times x equals zero

However, I am tempted to use an alternative method that I learned on my secondary school physics lessons. It is based on the direct application of Newtons laws and projecting vector equation equation left hand side m times a right arrow equals right hand side n ary summation over cap f right arrow on coordinate axes. Writing out the second law we have to bear in mind that the bead is acted upon by gravitational force and the force of normal reaction which arises due to Newton's third law and acts along the normal to the wire.

graph

Hence Newton's second law is expressed as follows:

equation left hand side m times a right arrow equals right hand side m times g right arrow plus cap n right arrow

Acceleration is split into the tangential and centripetal parts:

equation left hand side a right arrow equals right hand side a right arrow sub tau plus a right arrow sub c

Projecting the equation on the vertcal axis we obtain:

equation left hand side m times a sub tau times sine of alpha plus m times g equals right hand side cap n times cosine of alpha

where alpha is the angle at which the tangent crosses the horizontal axis (hence tangent of alpha equals y dot above divided by x dot above )

Then project on the horizontal axis:

equation left hand side m times a sub c equals right hand side cap n times sine of alpha plus m times a sub tau times cosine of alpha

Eliminate cap n :

cap n equals m divided by sine of alpha times open a sub c minus a sub tau times cosine of alpha close

a sub tau times open sine of alpha plus cosine squared of alpha divided by sine of alpha close minus a sub c divided by tangent of alpha plus m times g equals zero

a sub tau divided by sine of alpha minus a sub c divided by tangent of alpha plus m times g equals zero

Now calculate the tangential acceleration:

equation sequence a sub tau equals d divided by d times t times open d times s divided by d times t close equals d divided by d times t times open Square root of x dot above squared plus y dot above squared close equals d divided by d times t times open x dot above times Square root of one plus x squared close equals x double dot times open one plus x squared close plus x times x two dot above divided by Square root of one plus x squared

and the centripetal part:

equation sequence a sub c equals omega squared times cap r equals omega squared times x

equation sequence sine of alpha equals one divided by Square root of one plus one divided by open y dot above divided by x dot above close squared equals y dot above divided by Square root of x dot above squared plus y dot above squared equals x times x dot above divided by x dot above times Square root of one plus x squared equals x divided by Square root of one plus x squared

equation sequence tangent of alpha equals y dot above divided by x dot above equals x

Plug the above results into the equation:

x double dot times open one plus x squared close plus x times x two dot above divided by Square root of one plus x squared times Square root of one plus x squared divided by x minus omega squared times x divided by x plus g equals zero

to obtain the same result

x double dot times open one plus x squared close plus open g minus omega squared plus x dot above squared close times x equals zero

Further analysis on the phase plane shows that when the wire rotates not very fast ( omega less than Square root of g ) the bead oscillates around the origin in the vertical plane. If omega greater than or equals Square root of g the bead is moving along the wire away from the origin, it's velocity tending asymptotically to value Square root of omega squared minus g

Tags: m821, dynamical systems
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Valentin Fadeev

Descending to chaos

Tuesday, 4 Jan 2011, 05:54
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Edited by Valentin Fadeev, Sunday, 18 Sept 2011, 23:24

The first (unguided) steps in dynamical systems. The task is to prove that the phase paths of the following system:

x double dot plus epsilon times absolute value of x times s times g times n reverse solidus full stop x dot above plus x equals zero

are isochronous spirals, that is, every circuit of every path around the origin takes the same time. It looks a bit scary, because it involves not one but two "bad" functions. Thle easiest way to deal with these is to break the picture in quadrants and examine each part separately.

equation left hand side y dot above equals right hand side negative epsilon times absolute value of x times s times g times n times y minus x

x dot above equals y

equation left hand side d times y divided by d times x equals right hand side negative epsilon times absolute value of x times s times g times n times y divided by y minus x divided by y

First quadrant: absolute value of x equals x , s times g times n times y equals one

open one plus epsilon close times x times d times x plus y times d times y equals zero

open one plus epsilon close times x squared plus y squared equals cap c

Second quadrant: equation left hand side absolute value of x equals right hand side negative x , s times g times n times y equals one

open one minus epsilon close times x times d times x plus y times d times y equals zero

open one minus epsilon close times x squared plus y squared equals cap c

Third quadrant: equation left hand side absolute value of x equals right hand side negative x , equation left hand side s times g times n times y equals right hand side negative one , the same solution as for the
first quadrant.

Fourth quadrant: absolute value of x equals x , equation left hand side s times g times n times y equals right hand side negative one , the same solution as for the
second quadrant.

Introduce polar coordinates.

I,III:

open one plus epsilon close times rho sub one squared times cosine squared of phi plus rho sub one squared times sine squared of phi equals cap c

equation sequence rho sub one squared equals cap c divided by open one plus epsilon close times cosine squared of phi plus sine squared of phi equals cap c divided by one plus epsilon times cosine squared of phi

II,IV:

open one minus epsilon close times rho sub two squared times cosine squared of phi plus rho sub two squared times sine squared of phi equals cap c

equation sequence rho sub two squared equals cap c divided by open one minus epsilon close times cosine squared of phi plus sine squared of phi equals cap c divided by one minus epsilon times cosine squared of phi

Let cap c be a closed circuit enclosing the origin. The elapsed time
is given by the following formula:

equation sequence t equals integral over open cap c close d t equals integral over open cap c close one divided by d times x divided by d times t d x equals integral over open cap c close d times x divided by x dot above equals integral over open cap c close d times x divided by y

As the functions are periodic, it is sufficient to give proof for one loop. Due to the axial symmetry of the trajectory we can calculate transit time only in the first and the second quadrants and then double the result. If we integrate counter-clockwise, then we would follow the trajectory in a reversed direction. Therefore, we need to put the negative signe before the integrals.(oh, how long it took me to realize this..) In polar coordinates this expression has the following form:

equation sequence t sub one equals negative integral over zero under pi divided by two rho sub one super prime times cosine of phi minus rho sub one times sine of phi divided by rho sub one times sine of phi d phi equals integral over zero under pi divided by two open one minus rho sub one super prime times cosine of phi divided by rho sub one times sine of phi close d phi

equation left hand side t sub two equals right hand side integral over pi divided by two under pi open one minus rho sub two super prime times cosine of phi divided by rho sub two times sine of phi close d phi

t equals two times open integral over zero under pi divided by two open one minus rho sub one super prime times cosine of phi divided by rho sub one times sine of phi close d phi plus integral over pi divided by two under pi open one minus rho sub two super prime times cosine of phi divided by rho sub two times sine of phi close d phi close

Differentiating the expression for rho sub one by phi we obtain:

equation left hand side two times rho sub one times rho sub one super prime equals right hand side negative cap c times epsilon times two times cosine of phi of negative sine of phi divided by open one plus epsilon times cosine squared of phi close squared

equation left hand side rho sub one times rho sub one super prime equals right hand side cap c times epsilon times sine of phi times cosine of phi divided by open one plus epsilon times cosine squared of phi close squared

equation sequence rho sub one super prime divided by rho sub one equals rho sub one times rho sub one super prime divided by rho sub one squared equals cap c times epsilon times sine of phi times cosine of phi divided by open one plus epsilon times cosine squared of phi close squared times one plus epsilon times cosine squared of phi divided by cap c equals epsilon times sine of phi times cosine of phi divided by one plus epsilon times cosine squared of phi

A-ha, that's where we get the independence of the final result on the path choice: the arbitrary constant gets cancelled out.

equation sequence one minus rho sub one super prime times cosine of phi divided by rho sub one times sine of phi equals one minus epsilon times sine of phi times cosine squared of phi divided by open one plus epsilon times cosine squared of phi close times sine of phi equals one minus epsilon times cosine squared of phi divided by one plus epsilon times cosine squared of phi equals one divided by one plus epsilon times cosine squared of phi

equation left hand side cap i sub one equals right hand side integral over zero under pi divided by two d times phi divided by one plus epsilon times cosine squared of phi

Let u equals tangent of phi , then equation left hand side d times phi equals right hand side d times u divided by one plus u squared , cosine squared of phi equals one divided by one plus u squared

equation sequence cap i sub one equals integral over zero under normal infinity d times u divided by open one plus epsilon divided by one plus u squared close times open one plus u squared close equals integral over zero under normal infinity d times u divided by sum with, 3 , summands one plus epsilon plus u squared equals one divided by one plus epsilon times integral over zero under normal infinity d times u divided by one plus open u divided by Square root of one plus epsilon close squared equals equals one divided by Square root of one plus epsilon times integral over zero under normal infinity d of u divided by Square root of one plus epsilon divided by one plus open u divided by Square root of one plus epsilon close squared equals one divided by Square root of one plus epsilon times arc tangent of u divided by Square root of one plus epsilon vertical line sub zero super normal infinity equals pi divided by two times one divided by Square root of one plus epsilon

equation left hand side two times rho sub two times rho sub two super prime equals right hand side negative cap c times epsilon of negative two times cosine of phi times open negative sine of phi close divided by open one minus epsilon times cosine squared of phi close squared

equation left hand side rho sub two times rho sub two super prime equals right hand side negative cap c times epsilon times sine of phi times cosine of phi divided by open one minus epsilon times cosine squared of phi close squared

equation sequence rho sub two super prime divided by rho sub two equals rho sub two times rho sub two super prime divided by rho sub two squared equals negative cap c times epsilon times sine of phi times cosine of phi divided by open one minus epsilon times cosine squared of phi close squared times one minus epsilon times cosine squared of phi divided by cap c equals negative epsilon times sine of phi times cosine of phi divided by one minus epsilon times cosine squared of phi

equation sequence one minus rho sub two super prime times cosine of phi divided by rho sub two times sine of phi equals one plus epsilon times sine of phi times cosine squared of phi divided by open one minus epsilon times cosine squared of phi close times sine of phi equals one divided by one minus epsilon times cosine squared of phi

equation sequence cap i sub two equals integral over pi divided by two under pi d times phi divided by one minus epsilon times cosine squared of phi equals integral over pi divided by two under pi d times open phi minus pi divided by two close divided by one minus epsilon times sine squared of phi minus pi divided by two equals integral over zero under pi divided by two d times psi divided by one minus epsilon times sine squared of psi d psi equals equals integral over zero under normal infinity d times u divided by open one plus u squared close times open one minus epsilon times open one minus one divided by one plus u squared close close equals integral over zero under normal infinity d times u divided by one plus open one minus epsilon close times u squared equals one divided by Square root of one minus epsilon times integral over zero under normal infinity d of u times Square root of one minus epsilon divided by one plus open u times Square root of one minus epsilon close squared equals equals one divided by Square root of one minus epsilon times arc tangent of u times Square root of one minus epsilon vertical line sub zero super normal infinity equals pi divided by two times one divided by Square root of one minus epsilon

I made some false starts on cap i sub two getting a negative answer which is impossible for a strictly positive integrand. Still not quite sure where I went wrong. Normally these things happen when a singularity sneaks in inside the domain of integration after variable change. Anyway, I decided to cheat and shift the scale.

Hence the total transit time equals:

equation sequence t equals two times open pi divided by two times one divided by Square root of one plus epsilon plus pi divided by two times one divided by Square root of one minus epsilon close equals pi times open one divided by Square root of one plus epsilon plus one divided by Square root of one minus epsilon close

Tags: m821, dynamical systems
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