Trajectory of a point moving with acceleration perpendicular to velocity
(Just some basic high-school physics stuff; to assure myself I can still do some elementary things. :P Essentially, showing that if a particle moves with acceleration perpendicular to velocity, or velocity perpendicular to position, then it traces out a circle. Stop reading here if this is obvious.)
Suppose a point moves in the plane such that its acceleration is always perpendicular to its velocity, and of the same magnitude. What is its path like?
To set up notation: let’s say the point’s position at time is , its velocity is , and its acceleration is .
The result of rotating a point by 90° is . (E.g. see figure below)
So the fact that acceleration is at right angles to velocity means that , or, to write everything in terms of the velocity,
where we can get rid of by substituting the second equation (in the form ) into the first:
or in other words
By some theory about ordinary differential equations, which I don’t know (please help!) (but see the very related example you saw in high school, of simple harmonic motion), the solutions to this equation are and and any linear combination of those: the solution in general is
where and is the angle such that and . And the fact that gives . So . Note that is indeed perpendicular to as we wanted.
The actual trajectory can be got by integrating
to get and . This trajectory is a point moving on a circle centered at point and of radius , with speed or unit angular speed. Note that velocity is also perpendicular to the point’s position wrt the centre of the circle, as velocity is tangential to the circle, as it should be.
With a suitable change of coordinates (translate the origin to , then rotate the axes by , then scale everything so that ), this is the familiar paremetrization of the circle.
Note: Just as we derived from assuming that the acceleration is perpendicular to velocity, we can, by assuming that velocity is perpendicular to position, identically derive , i.e. that the point moves on a circle.