What is the proof that the component of acceleration perpendicular to motion for all motion is $v^2/r$, where r is the radius of curvature? Also, what is the proof that the component of acceleration along the direction of motion is equal to $d^2s/dt^2$ where $s$ is the function of distance. These two questions deal specifically with non-circular curved trajectories.
1 Answer
Any trajectory, on a sufficiently small scale, can be thought of as being "instantaneously" part of a circular trajectory by dint of the fact that you can talk about the instantaneous radius of curvature of the trajectory, $r$.
When you look at the direction of the particle on a (small piece of a) circular orbit with radius of curvature $r$, then after a very short time $\delta t$, it will have changed direction - and that change is perpendicular to the original direction. A simple diagram shows this:
You see here the change in direction of the particle during a very short time interval $dt$, so that $dx = v\cdot dt$. We can use simple trigonometry to show that $$dx = v \cdot dt\\ dy = r (1-\cos\theta) \approx \frac12 r \theta^2\\ \theta = \frac{dx}{r} = \frac{v\cdot dt}{r}$$
Now the acceleration $a$ must be such that $\frac12 a \cdot (dt)^2 = dy$
Combining these, we find that
$$\frac12 r \left(\frac{v\cdot dt}{r}\right)^2 = \frac12 a (dt)^2\\ a = \frac{v^2}{r}$$
As for the second part, it follows simply from looking along the trajectory. At any instant, the velocity along the trajectory is $\frac{ds}{dt}$ , so the acceleration (the rate of change of velocity along the trajectory) is $\frac{d^2s}{dt^2}$.