In a horizontal plane with no gravity, it should not be hard to prove that the polar coordinates of the mass are related by
$m \ddot{r} - m r \dot{\theta}^2 = 0$.
Being $\theta(t)$ prescribed, s.t. $\dot{\theta} = \Omega$, the equation for the only degree of freedom $r(t)$ reads
$\ddot{r} = \Omega^2 r$,
whose solution has the form
$r(t) = A e^{-\Omega t} + B e^{\Omega t}$.
Given the initial conditions about the radial position and velocity of the mass, you can easily determine the integration constants $A$ and $B$. As an example, if $\dot{r}(0) = 0$ and $r(0) = r_0$, the solution is
$r(t) = \dfrac{r_0}{2} \left[ e^{-\Omega t} + e^{\Omega t} \right]$.
If you want the description of the motion in Cartesian coordinates, you need to use the transformations $x = r \cos \theta$, $y = r \sin \theta$.
To give a clear answer to your question: no, the mass doesn't move on a vertical straight line.
Note 1. If the gravity can't be neglected, the dynamical equation becomes $m \ddot{r} - m r \dot{\theta}^2 = - m g \sin \theta$, that has no closed form solution.
Note 2. In the computation, I didn't care about how the motion of the tube begins. You told us that its angular velocity is $\Omega$ and constant, and this can be used as a prescribed condition.
