Whilst learning about polar coordinates on circular motion, I had come across this question,
'A particle moves along a curve with polar equation $ r\theta = a$ (for $\theta \geq \frac{\pi}{4}$) in such a way that the transverse component of the acceleration is always zero. Show that the radial component of the acceleration is inversely proportional to $r^3$'
My current workings out:
Let the transverse component of acceleration be $\alpha_{\theta} = 2\dot{r}\dot{\theta} + \ddot{\theta}$.
Differentiating $r\theta = a$ with respect to time,
$$ \frac{d(r\theta)}{dt} = \frac{d(a)}{dt}$$ Because $a$ is a constant, $a = 0$, $$ \dot{r}\theta + \dot{\theta}r = 0 $$
Differentiating again for a second time with the product rule,
$$ \ddot{r}\theta + \dot{\theta}\dot{r} + \ddot{\theta}r + \dot{r}\dot{\theta} = 0 $$
Rearrange for $\ddot{\theta}$,
$$\ddot{\theta} = \frac{-2\dot{r}\dot{\theta} - \ddot{r}\theta}{r} $$
Making $\alpha_{\theta} = 0 $,
$$ 2\dot{r}\dot{\theta} + \ddot{\theta} = 0 $$
Substituting in $\ddot{\theta}$ leads to,
$$-\ddot{r}\theta = 0 $$
From here, I am not sure where to go. I have tried using this with the radial component of acceleration but got not much luck. Any suggestions would be appreciated!
The transverse acceleration is $$\frac 1r\frac{d}{dt}(r^2\dot{\theta})=0\implies r^2\dot{\theta}=k$$
Now$$r\theta=a\implies\theta=\frac ar\implies \dot{\theta}=-\frac{a}{r^2}\dot{r}$$
Combining these results gives $$\dot{r}=-\frac ka\implies\ddot{r}=0$$
The radial component of acceleration is $$\ddot{r}-r\dot{\theta}^2$$ $$=0-r\left(\frac{k}{r^2}\right)$$
And hence the result