newtonian mechanics - What is the centripetal acceleration for non-uniform speed?

In NCERT physics page no. 122 example 6.7 there is an argument next to equation 6.12 i.e., $T_A -mg= \frac {mv_0^2}{L}$ which means that at the lowest point the centripetal force is equal to $\frac {mv_0^2}{L}$ which means that centripetal acceleration is $\frac {v^2}{r}$ which I think isn't true as the speed of the ball is constantly changing so we can't use the formula of $\frac {v^2}{r}$ for calculating centripetal acceleration and hence the force. I know the derivation of the formula of centripetal acceleration for uniform circular motion from Halliday Resnick and Walker and the formula is derived on the assumption that speed is constant. So I want a confirmation whether my thinking about the logic being used is wrong is correct or not?

Answer

In polar coordinates, the acceleration vector for planar motion is given by

$$\mathbf a=(\ddot r-r\dot\theta^2)\hat r+(r\ddot\theta+2\dot r\dot\theta)\hat\theta$$

If our motion is along a circle, we have $\dot r=\ddot r=0$, so our acceleration reduces to

$$\mathbf a=-r\dot\theta^2\hat r+r\ddot\theta\hat\theta$$

The centripetal acceleration is the radial component of the acceleration

$$a_c=r\dot\theta^2$$

Using $\dot\theta=v/r$ we end up with the familiar result

$$a_c=\frac{v^2}{r}$$

Notice how we didn't assume anything about the speed $v$. This expression is valid for when $v$ is not constant. We will just have a changing centripetal acceleration, and we will also have a non-zero tangential acceleration as $\ddot\theta=\dot v/r\neq 0$.