Particle moving in polar coordinates

[Disserate]

Equations given:

r=A\theta

\theta=\frac{1}{2}\alphat^{2}

A=\frac{1}{\pi} meters per radian

\alpha is a given constant


Asks to show that radial acceleration is zero when \theta=\frac{1}{\sqrt{2}} radians.

I have tried rearranging, plugging in, and deriving to try to solve this problem to no avail. I do not know exactly how to go about doing this. I do desire an answer, but even more do I desire an explanation on how to do this. Also, i apologize for not using the template, but I did not like it very much.

 

[ehild]

Both velocity and acceleration are vectors, having magnitude and direction. The velocity is the time derivative of the position vector and the acceleration is the time derivative of the velocity, or second derivative of the position vector.
You certainly know that in case of uniform circular motion there is centripetal acceleration which is of radial direction, parallel with the position vector, but points towards the centre. But this problem is totally different. It is a spiral, the distance from the centre increases with the turning angle, so there is an extra radial acceleration because of the increase of r.

Do you know how you can write a vector with its polar coordinates, using the polar unit vectors e_r and e_θ? You can read this, especially the part about the derivatives of the position vector r

http://mathworld.wolfram.com/PolarCoordinates.html

The position vector is simply \vec r=r \vec {e_\theta}, the velocity is \vec v=\dot{\vec r} and the acceleration is \vec a=\dot{\vec v}

From mathworld, (or from your notes,)you can learn that the acceleration is
\vec a=(\ddot{r} -r\dot \theta^2)\vec e_r+(2 \dot r \dot \theta+r \ddot {\theta})\vec e_{\theta}

The radial component of acceleration is a_r=(\ddot{r} -r\dot \theta^2)

Substitute the given functions for r and theta and do the derivations.