The change in magnitude of centripetal acceleration
Your question is: I want to find out how centripetal acceleration changes over time as the tangential velocity changes over time
First i will calculate the equations of Motion for this case
\begin{align*}
&\text{The components of the position vector in polar coordinate are:}\\\\
&\vec{R}=
\begin{bmatrix}
r(t)\cos(\varphi(t)) \\
r(t)\sin(\varphi(t)) \\
\end{bmatrix}&(1)\\
&\text{because the velocity changes over time, the kreis radius $r$ change over time}\\\\
& \Rightarrow\\
&\vec{\dot{R}}=
\begin{bmatrix}
\dot{r}\cos(\varphi)-r\sin(\varphi)\dot{\varphi}) \\
\dot{r}\sin(\varphi)+r\cos(\varphi)\dot{\varphi}) \\
\end{bmatrix}&(2)\\\\
&\text{so the kinetic $T$ energy is:}\\
&T=\frac{1}{2}\,m\,\dot{R}^2=\frac{1}{2}\,m\left(\dot{r}^2+r^2\,\dot{\varphi}^2\right)\\
&\text{with euler lagrange approach we get the equations of motion :} \\\\
&\ddot{r}=\dot{\varphi}^2\,r&(3)\\
&\ddot{\varphi}=-2\,\frac{\dot{\varphi}\,\dot{r}}{r}&(4)
\end{align*}
The solutions of the EoM's with the initial conditions
$\varphi(t=0)=0\,,\dot{\varphi}(t=0)=\omega$ and
$r(t=0)=r_0\,,\dot{r}(t=0)=0$ are:
The solutions of the EoM's with the initial conditions \ $\varphi(t=0)=0\,,\dot{\varphi}(t=0)=\omega$ and
$r(t=0)=r_0\,,\dot{r}(t=0)=0$ are:
\begin{align*}
&r(t)=r_{{0}}\sqrt {\omega}{\frac {1}{\sqrt {{\frac {\omega}{1+{\omega}^{2}{
t}^{2}}}}}}
\\
&\varphi(t)=\arctan \left( \omega\,t \right) \\\\
&\Rightarrow\\
&\text{The centrifugal force:}\\
&F_z(t)=r\,\dot{\varphi}^2={\omega}^{5/2}r_{{0}} \left( 1+{\omega}^{2}{t}^{2} \right) ^{-2}{
\frac {1}{\sqrt {{\frac {\omega}{1+{\omega}^{2}{t}^{2}}}}}}\\
&\text{for $t=0$ we get $F_{z0}=r_0\,\omega^2$}\\
&\Rightarrow\\
&\text{Change of the Zentrifugal force over time:}\\\\
&\boxed{\Delta F_z=F_z(t)-F_{z0}}
\end{align*}
