= Solution
{c}
All time <derivatives> in this part are taken relative to the <rotating reference frame>, so its <basis> is held fixed during <differentiation>. Put $f(\phi)=p(\phi)\sin\phi$ and $q(\phi)=p(\phi)\cos\phi$. The <chain rule> and <product rule> give
$$
\boxed{\ddot{\mathbf r}=\ddot\phi(f'\mathbf e_1-q'\mathbf e_3)+\dot\phi^2(f''\mathbf e_1-q''\mathbf e_3).}
$$
Explicitly,
$$
\begin{aligned}
f'&=p'\sin\phi+p\cos\phi,&
f''&=p''\sin\phi+2p'\cos\phi-p\sin\phi,\\
q'&=p'\cos\phi-p\sin\phi,&
q''&=p''\cos\phi-2p'\sin\phi-p\cos\phi.
\end{aligned}
$$
The separate <Coriolis force> and <centrifugal force> terms in the stated <equation> account for the motion of the <basis> relative to an inertial observer.
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