calculus and analytical geometry
Here's one method that can be used to derive the equations the equation for centripetal acceleration. Imagine an object in uniform circular motion — something like a person standing still on a rotating platform. They stay at the same distance from the point of rotation (r = constant) but travel around it at a constant rate (ω = constant).
Represent their position vector (r) using cartesian coordinates (x, y). Take the derivative of these coordinates to get the velocity components (vx, vy) of the person's velocity vector (v). Then take the derivative of the velocity components to get the acceleration components (ax, ay) and acceleration vector (a). Do it something like this…
|r||⇒||x =||+r cos(ωt)||y =||+r sin(ωt)|
|v =||dr||⇒||vx =||−rω sin(ωt)||vy =||+rω cos(ωt)|
|a =||dv||⇒||ax =||−rω2 cos(ωt)||ay =||−rω2 sin(ωt)|
We will now analyze these results to determine the magnitudes and relative directions of the three kinematic vectors.
Use the pythagorean theorem to get the magnitudes of position, velocity, and acceleration in polar coordinates.
|r2 = x2 + y2
r2 = [+r cos(ωt)]2 + [+r sin(ωt)]2
0r = r
The person stays a constant distance from the center of rotation. That's what it means to follow a circular path. A circle is the locus of points equidistant from a point.
Use the same process for velocity.
|v2 = vx2 + vy2
v2 = [−rω sin(ωt)]2 + [+rω cos(ωt)]2
0v = rω
The translational and angular velocities are related by the expected equation.
Repeat for acceleration.
|a2 = ax2 + ay2
a2 = [−rω2 cos(ωt)]2 + [−rω2 sin(ωt)]2
0a = rω2
Speed has a constant value, but direction is changing. There is an acceleration and that's its equation. If you don't like angular quantities, you can use algebra to state centripetal acceleration in terms of tangential velocity. Do that and you get…
Compare the trig functions of the components to determine relative directions. Start with position and acceleration, since that's the easier pair to compare. The trig functions match, but the signs are opposite. This means that whatever direction the position vector points, the acceleration vector points the opposite way. Since the position vector always points out and away from the center of rotation, the acceleration vector always points in towards the center. In other words, the acceleration is centripetal.
|x =||+r cos(ωt)||y =||+r sin(ωt)|
|ax =||−rω2 cos(ωt)||ay =||−rω2 sin(ωt)|
|ax =||−xω2||ay =||−yω2|
|a = −rω2|
Less obvious is what to do with the velocity. The sine and cosine functions are full of symmetries by themselves and with each other. There are all kinds of ways they can be flipped and shifted so that one function becomes the other. By flip I mean a change in sign or a reflection and by shift I mean the addition of a phase angle or a translation. For example…
These identities can be used to show that the velocity vector is 90° ahead of the position vector during uniform circular motion; that is to say, tangent to the circular path in the direction of motion. Maybe you learned it somewhere, but if not I'm telling you now, a tangent and a radius are perpendicular.
|x =||+r cos(ωt)||y =||+r sin(ωt)|
|vx =||−rω sin(ωt)||vy =||+rω cos(ωt)|
|vx =||+rω cos(ωt + 90°)||vy =||+rω sin(ωt + 90°)|
|There is no nice way to write this mathematically|
Since centripetal force and radius are 180° apart, centripetal force and velocity are 90° apart. This is another proof that can be done analytically, but I don't want to do it. Just refer back to the graphic near the start of this discussion.
rotating reference frame