Problem: Given the following position vector of an object in space: $$R(t)=(a\cos(\omega t),a\sin(\omega t),\omega^2 t)$$ Find $\omega\neq0$ such that the sum of the tangential and normal components of acceleration equal half its speed.
I computed the tangential $A_T$ and the normal component of the acceleration $A_N$ and got the following expressions $$A_T=a\omega\sqrt{\omega^2+4}$$ $$A_N=\omega\sqrt{\frac{a^4\omega^2+2a^2\omega^2t^2+4a^4+8a^2t^2-4t^2}{a^2+2t^2}}$$ which is, unfortunately, leading to an extremely big polynomial equation that I am unable to resolve. I am guessing that the answer is much simpler, so please correct me if I am making a mistake.
