Let's say we have a particle rotating in the $xy$ plane, centered at $(0,0)$, with a radius $r$ and constant speed $v$. When it's at the point $(0,r)$, the horizontal component of the velocity is $v_x=v$ and vertical component is $v_y=0$. Moreover, the direction of total force on it, the centripetal force, is along the negative $y$ axis. After an infinitesimal time has passed, the vertical velocity increases in the negative $y$ direction by an infinitesimal amount and the horizontal component decreases by an infinitesimal amount such that the vector resultant of the new components equals $v$.
Now my question is, the vertical component changed due to the centripetal force, which was in that direction. But since at that particular moment, there was no force in the horizontal direction, what caused the change in that component?
Best Answer
Your question is interesting because it shows the dangers of working with infinitesimals without a careful control of their meaning.
Basically, your question is applicable to any motion, when the trajectory in phase space reaches an extremum along one of the directions. So, for simplicity let's discuss the simple one dimensional harmonic oscillator, which, in your example exactly corresponds to the $x$- component of the uniform circular motion.
The equation of motion is $$ \ddot x(t) = - \omega^2x(t), $$ valid for any time $t$. When $x=0$, acceleration is zero and the speed along $x$ direction is maximum.
So,how it happens that there is a decrease of $\dot x$?
A naïf application of infinitesimals is misleading. In the present case $$ \dot x(t+dt) \simeq \dot x(t) + \ddot x(t) dt $$ would imply $\dot x(t+dt) = \dot x(t) $. But this is just the first order result. First order approximations are the leading term of a local analysis of the behavior of a regular function provided they do not vanish. When, like in the case of an extremum of velocity at time $t_0$ (threfore $\ddot x(t_0)=0$), the first order variation is zero, it is necessary to look for the next non-zero term in an expansion in powers of $dt$. In the present case: $$ \dot x(t_0+dt) = \dot x(t_0) + \ddot x(t_0) dt + \frac12 \dddot x(t_0) dt^2 = \dot x(t_0) - \frac12 \omega^2 \dot x(t_0) dt^2, $$ where use has been done of the equation of motion (by taking a time derivative of both sides) in order to express the third derivative of $x$ with respect to time as a function of $\dot x$.
Therefore, one can see that, at the dominant non vanishing order in $dt$, $\dot x$ is correctly varying.