Question on angular acceleration in OCR A Level further maths textbook OCR A Level Further maths: mechanics, Year $1.$
Chapter $4$ Section $2$: Acceleration in horizontal circular motion.

$\omega$ is angular speed and is defined as $\frac{d\theta}{d t},$ sometimes written as $\overset{.}{\theta}.$
An equation for linear (tangential) speed is then derived and is given
by: $v=r\omega...$ $$$$... The formula for acceleration is given by $a=v\omega.$
Since $v=r\omega,$ you can write $a=r\omega^2$ (note that this equals
$r\overset{.}{\theta}^2$ ) and, since $\omega = \frac{v}{r},$ you can
write $a=\frac{v^2}{r}.$

One the side there is a "tip":

Sometimes acceleration is written as $a=r\overset{..}{\theta},$ where
$\overset{..}{\theta}=\frac{d^2\theta}{dt^2}.$

But surely this means that $\frac{d^2\theta}{dt^2} = \overset{.}{\theta}^2,$ which is false.
So is the "tip" wrong, or am I missing something here?
 A: I think the issue is you are combining steps from rotational and translational.
Using the definitions and following through as below, should highlight the differences you have in your steps.
$$
\omega = \frac{v_t}{r}\\
\alpha = \frac{a_t}{r}
$$
where rotational is on the left side
$$
a_t = \frac{dv_t}{dt}\\
\alpha = \frac{d\omega}{dt}
$$
so we an denote $\alpha$ using the translational acc definition.
$$
\alpha = \frac{1}{r}\frac{dv_t}{dt} = \frac{1}{r}\frac{d}{dt}r\omega = \frac{1}{r}\frac{d}{dt}r\dot{\theta}\\
$$
or directly using the rotational velocity
$$
\alpha = \frac{d}{dt}\dot{\theta}
$$
A: I find the text slightly hard to read; due to the notation jumping around. Maybe more consistency, like this, better: $$r\quad\overset{.}r\quad\overset{..}r \\
r\quad v\quad a\\\quad\\
\theta\quad\overset{.}\theta\quad\overset{..}\theta\\
\theta\quad\omega\quad\alpha$$


$\omega$ is angular speed and is defined as $\frac{d\theta}{d t},$ sometimes written as $\overset{.}{\theta}.$
An equation for linear (tangential) speed is then derived and is given
by: $v=r\omega...$


This is correct regarding circular motion.


... The formula for acceleration is given by $a=v\omega.$
Since $v=r\omega,$ you can write $a=r\omega^2$ (note that this equals
$r\overset{.}{\theta}^2$ ) and, since $\omega = \frac{v}{r},$ you can
write $a=\frac{v^2}{r}.$


This “$a$” refers to the circular motion's centripetal acceleration; let's relabel it $a_{\text{centripetal}}.$


Sometimes acceleration is written as $a=r\overset{..}{\theta},$ where
$\overset{..}{\theta}=\frac{d^2\theta}{dt^2}.$


This “$a$” refers to the circular motion's transverse acceleration, whose vector version is tangential to the circle; let's relabel it $a_{\text{transverse}}.$ For uniform circular motion, that is, when the angular acceleration $\overset{..}{\theta}$ equals $0,$ this quantity equals $0.$
In general, the (linear) acceleration of a particle in circular motion is the vector sum of its centripetal and transverse accelerations $a_{\text{centripetal}}$ and $a_{\text{transverse}}.$

But surely this means that $\frac{d^2\theta}{dt^2} =
\overset{.}{\theta}^2,$ which is false.

Indeed, $a_{\text{centripetal}}$ does not generally equal $a_{\text{transverse}},$ and $\overset{..}\theta$ does not generally equal $\overset{.}\theta^2.$
