Question on angular acceleration in OCR A Level further maths textbook

Question

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?

Answer 1

I find the text slightly hard to read; due to the notation jumping around. Maybe more consistency, like this, better: $$r\quad\overset{.}r\quad\overset{..}r \\ r\quad v\quad a\\\quad\\ \theta\quad\overset{.}\theta\quad\overset{..}\theta\\ \theta\quad\omega\quad\alpha$$

An equation for linear (tangential) speed is then derived and is given by $$v=r\omega.$$

This is the linear speed of circular motion.

The formula for acceleration is given by $a=v\omega.$ Since $v=r\omega,$ you can write $a=r\omega^2$ and, since $\omega = \frac{v}{r},$ you can write $$a=\frac{v^2}{r}.$$

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

Sometimes acceleration is written as $$a=r\overset{..}{\theta}$$

This “$a$” is the tangential acceleration of circular motion; let's relabel it $a_{\text{tangential}}.$ For uniform circular motion, that is, when the angular acceleration $\overset{..}{\theta}$ equals $0,$ this quantity equals $0.$

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

Since $a_{\text{centripetal}}$ does not generally equal $a_{\text{tangential}},$ the above is indeed false.

(The linear acceleration of circular motion is the vector sum of $a_{\text{centripetal}}$ and $a_{\text{tangential}}.$)

P.S. Here's a nice summary: A Tale of Three Accelerations, or The Differences between Angular, Tangential, and Centripetal Accelerations

Answer 2

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 $$ \alpha = \frac{d}{dt}\dot{\theta} $$