Suppose I am enumerating reasons not to fly. Is it then correct to write/say:


  Firstly, I prefer the train because I can see the landscape. Secondly, I have control over my luggage, and thirdly, it is better for the environment.


Or is it rather:


  First, I prefer the train because I can see the landscape. Second, I have control over my luggage, and third, it is better for the environment.


I thought the first should be correct, but I find the second one in written texts. Which one is correct?


Question: Shape of rotating rope (lasso problem?)
Answer: It's an interesting problem. I've tried to study it neglecting the air friction, which is likely to be nonnegligible. So let's parametrize the rope of length $L$ by the vector $(z(l), r(l))$, with $l$ being the curviligne coordinate along the rope and $(z,r)$ the cylindrical coordinates of the rope (I've omitted $\theta$, which can be assumed to be constant in the rotating frame [see my comment below this post for a justification of this omission]). 
In the rotating frame, the potential energy of the rope is
$$E_p=\int_0^L \mu dl\left(gz-\frac12\Omega^2r^2\right)$$
and the definition of the curviligne integral gives the constraint $r&#39;^2+l&#39;^2=1$. So the problem can be translated into minimizing the following quantity :
$$\begin{align}
 \int_0^Ldl \mathcal L(z,r;z&#39;,r&#39;;l) &amp; &amp;
 \text{with } \mathcal L(z,r;z&#39;,r&#39;;l)=\mu gz - \tfrac12\mu\Omega^2r^2 - \lambda(z&#39;^2+r&#39;^2).
\end{align}$$
One should not forget (as I did in a first tentative answer) that the Lagrange multiplier $\lambda$ depends on $l$/
It is a standard Euler-Lagrange problem and its solution is given by
$$\begin{align}
 \frac{\partial\mathcal L}{\partial z}-\frac{d}{dl}\frac{\partial\mathcal L}{\partial z}&amp;=0&amp;
&amp;\text{and}&amp;
 \frac{\partial\mathcal L}{\partial r}-\frac{d}{dl}\frac{\partial\mathcal L}{\partial r}&amp;=0&amp;
\end{align}$$
The equation in $z$ becomes
$$\begin{align}
\mu g&amp;=-2\frac{d}{dl}(\lambda z&#39;)&amp;
\lambda z&#39;&amp;= -\tfrac12 \mu g(l-l_0),
\end{align}$$
where $l_0$ is an integration constant. For the optimal solution $z&#39;\le 0$, since replacing $z&#39; by -|z&#39;|$ can only decrease $\mathcal L$, ans we have $z&#39;=-\sqrt{1-r&#39;^2}$. One has therefore
$$\lambda=\frac{\mu g(l-l_0)}{2\sqrt{1-r&#39;^2}}$$.
The equation in $r$ is then
$$-\mu\Omega^2r=-2\frac{d}{dl}(\lambda r&#39;) =-\frac{d}{dl}\frac{\mu g(l-l_0)r&#39;}{\sqrt{1-r&#39;^2}}$$
We have then the following equation to integrate
$$0=\frac{\Omega^2}{g}r(1-r&#39;^2)^{3/2}-r&#39;(1-r&#39;^2)-(l-l_0)r&#39;&#39;.$$
I don't know how to integrate it analytically, but when $l-l_0&lt;0$, it seems that we can have oscillations.

Edited to add :

If one looks at the signs of $r&#39;$ and $r&#39;&#39;$ as function of $r$ and $r&#39;$ for a fixed $l$, one can quickly draw some arrows in phase space (i.e.) and see that the flow goes "in circles" when $l&lt;l_0$ and tends asymptotically to the curve $\frac{\Omega^2}{g}r=\frac{r&#39;}{\sqrt{1-r&#39;^2}}$ when $l&gt;l0$. So one can expect a finite number of oscillations of the rope. But the whole question is then how $l_0$ relates to the parameters of the system.

Edit: In the limit of almost vertical rope ($r'\ll1$), the equation becomes
$$0=\frac{\Omega^2}{g}r-r&#39;-(l-l_0)r&#39;&#39;$$
which is simpler. And indeed, Wolfram Alpha has an analytical solution using modified Bessel functions.