# Path traced out by a point

While studying uniform circular motion at school, one of my friends asked a question:

"How do I prove that the path traced out by a particle such that an applied force of constant magnitude acts on it perpendicular to its velocity is a circle?" Our physics teacher said it was not exactly a very simple thing to prove.

I really wish to know how one can prove it.Thank you!

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How easy this can prove depends on your level of mathematics. Do you understand analytic geometry or even calculus? – Siyuan Ren Jun 22 '12 at 13:35
You'll come away with so much more if spend the time to think more deeply about the problem and earn the knowledge of how to prove it. – Alfred Centauri Jun 22 '12 at 13:40
It is an extremely simple thing to prove! You should not listen to your teacher. – Ron Maimon Jun 23 '12 at 8:37
Thank you.I did prove it eventually,I was doing something extremely stupid. :) I am least bothered about how tough something is as long as I have the background to attack it. – user10060 Jun 23 '12 at 8:49

One can prove it in a more-or-less elementary way by solving a pair of simultaneous differential equations. In two dimensions, a vector that is perpendicular to a velocity $$\left(\begin{matrix}u(t)\cr v(t)\end{matrix}\right)\quad\mathrm{is}\quad\left(\begin{matrix}-v(t)\cr u(t)\end{matrix}\right).$$ The acceleration, the time derivative of the velocity, is proportional to this vector, so we have the two differential equations $$\left(\begin{matrix}\dot u(t)\cr \dot v(t)\end{matrix}\right)=\lambda\left(\begin{matrix}-v(t)\cr u(t)\end{matrix}\right).$$ If $\lambda$ is negative, the circle goes "the opposite way". If $\lambda$ is zero, the circle is a straight line. If you know differentiation and how to solve differential equations, you should be able to solve this pair of equations, and then integrate it to obtain the way in which the position changes over time. If you don't, then it may be better to be patient and wait until you come across it in the course of your studies. Learning calculus on your own to the level needed to solve this differential equation is possible, however.

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Try looking for uniform circular motion in google. It is not hard to prove it if you know something about vectors and what taking a derivative of vector function means. Force is a vector it is proportional to acceleration. Acceleration is change in velocity(remember a vector) divided by time(really shot period of time). Try to draw a circle yourself and some velocity, acceleration and position vectors

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If you don't want or know how to solve a pair of simultaneous differential equations, try this more elementary approach using complex numbers and ordinary time derivatives.

Consider the arbitrary path, with parameter t, in the complex plane:

$r(t)e^{i\theta(t)}$

The "velocity" is the time derivative:

$[\frac{dr}{dt} + ir(t)\frac{d\theta}{dt}] e^{i\theta(t)}$

The "acceleration" is the 2nd time derivative:

$\{[\frac{d^2r}{dt^2} - r(t)(\frac{d\theta}{dt})^2] + i[2(\frac{dr}{dt}\frac{d\theta}{dt}) + r(t)\frac{d^2\theta}{dt^2}]\}e^{i\theta(t)}$

We require that the "acceleration" be orthogonal to the "velocity". In polar representation, this just means that the ratio of the two complex numbers is an imaginary number (multiplication by $i$ is a rotation of 90 degrees in the complex plane).

If you stare at the two derivatives a bit, you see that this condition only holds if $r$ and $\frac{d\theta}{dt}$ are constants:

$r(t) = R$

$\frac{d\theta}{dt} = \omega$

Then the "velocity" is just:

$iR\omega e^{i\omega t}$

and the "acceleration" is just:

$-R\omega^2e^{i\omega t}$

So, the "velocity" and "acceleration" are indeed orthogonal. The path then is just:

$Re^{i\omega t}$

This is just a circle in the complex plane.

Of course, you could have done this with polar coordinates in the x-y plane but polar complex numbers make the derivatives so much easier!

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I think you can prove it you can prove that the acceleration vector $\vec{a}$ is decomposed into two components $$\vec{a} = \dot{v}\, \hat{e} + \frac{v^2}{r} \hat{n}$$ one tangential to motion along the unit direction vector $\hat{e}$ and one perpendicular to along the unit direction $\hat{n}$, with tangential speed $v$, change in speed $\dot v$ and path radius of curvature $r$.

So if the path was not a circle, there would be a component of $\vec{a}=\frac{\vec{F}}{m}$ along the tangential vector $\hat{e}$ changing the speed $v$ as to moves along. The hole thing hinges on the fact that speed does not change in uniform circular motion.

The above comes from math called Differential Geometry along a curve, and it is related to accelerations here.

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