How do you go about finding the normal acceleration of a plane associated with polar co ordinates?

Question

Question

A plane, having just taken off, has a constant speed of v=94.3m/s. When $\theta=20^{o}$, the plane is climbing at an ever steepening rate of 0.17 rad/s.

What is the normal acceleration of the plane $a_n$ this instant?

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My Answer

I have no idea how to do this since this uses polar co ordinates, and I am quite new to polar co ordinates. No clue, would really appreciate if someone provide hints or a walkthrough.

Answer 1

A simplistic view of this (probably too simplistic) is to say that as the angle is changing constantly, the path of the plane is approximately (?) circular.

Then the acceleration towards the centre of the circle, i.e. normal to the path, is $$r\dot{\theta}^2=r\dot{\theta}\times\dot{\theta}.$$This is the same as linear speed $\times$ rate of increase of angle, which is $94.3\times0.17$ m/$s^2.$

However I realise that the angle $20^o$ must be somehow relevant, so in all likelihood this is wrong.