向心力

The centripetal force, or center-seeking force, is the force that keeps an object moving on a curved path.

Since the object is not moving in a straight line, it must be accelerating, so naturally we must have some force in play. This force is what we are looking at.

We will only derive the formula for uniform circular motion, but the centripetal acceleration may be applied in other cases as well.

Suppose that we have a position function $\vec{r}(t)$ relative to the center of the circle, which is a vector-valued function. Let the velocity function be $\vec{v}(t)$, and let the acceleration be $\vec{a}(t)$. Let the circle have radius $R$.

Since the motion is uniform and circular, we may express it as follows: $$\vec{r}(t)=R\left<\cos(\omega t),\sin(\omega t)\right>$$ Differentiating twice, $$\vec{v}(t)=R\omega\left<-\sin(\omega t),\cos(\omega t)\right>$$ $$\vec{a}(t)=R\omega^2\left<-\cos(\omega t),-\sin(\omega t)\right>$$ First, notice that $\vec{a}(t)\parallel\vec{r}(t)$. In fact, the two point in opposite directions. This determines the direction.

Calculating the magnitude, $$|\vec{a}(t)|=\omega^2 R=\frac{v^2}{R}$$ In the last part, we used $v=\omega R$ for uniform circular motion.

So how do we use this outside uniform circular motion?

A theorem of vector-valued functions which I will not prove here states that the acceleration vector always lies in the osculating plane of the graph of the function. Moreover, it states the following: $$\vec{a}(t)=\frac{d^2s}{dt^2}\vec{T}+\kappa\left(\frac{ds}{dt}\right)^2\vec{N}$$ Here, $s$ is an arc length parameter, $\kappa$ is the curvature, $\vec{T}$ is the principal unit tangent vector, and $\vec{N}$ is the principal unit normal vector.

Notice that we have divided the acceleration into a normal component and a tangential component. It is the normal component that we seek.

Indeed, since the curvature is the reciprocal of the radius of curvature, we see that the normal component of the acceleration, or the centripetal acceleration, is the square of the speed divided by the radius of curvature.