Your wife could be right about the rocket.
Whenever we say something is small physically, we need to be sure what it is small with respect to. In the case of a thread, the drag force exerted by air beats the centripetal force keeping it taut. Because centripetal forces scale with mass, a denser thread will work.
The issue facing a space elevator is different. For a rigid cable, we need the terminus to be above geostationary orbit. The height of the atmosphere is about 100km from sea level, while geostationary orbit is about 36000 km, so very little of the cable will be exposed to atmospheric drag. What net drag there is will be mostly from the prevailing winds, I suspect. The effect of this will be small but persistent. If it is never corrected, over time the elevator will drift away from vertical. Whether this effect is large enough to matter over, say, the lifetime of a civilization, I don't know.
For the other part of your question - what mass would we need to keep it taut? - I think your intuition about the rope is good. Here our concern isn't wrapping, but whether the rope will collapse. Imagine the rope is made of little beads of mass d$m$ connected by little massless cables of length d$h$. In the rotating frame, a bead at height $h$ in the cable experiences a total force $\mathrm{d}F = -\mathrm{d}m\frac{M_eG}{(h+R_e)^2}+T(h+\mathrm{d}h) - T(h) + \mathrm{d}m \, \omega^2(h+R_e) = 0$ for a system in equilibrium, where $T$ is the tension. If we call the mass per unit length of the cable $\lambda$, then we can turn this into a differential equation:
$$\frac{\mathrm{d}}{\mathrm{d}h} T(h) = \lambda \left ( \frac{M_eG}{(h+R_e)^2} - \omega ^2 (h+R_e) \right )$$
Integrating:
$$ T(h) = \lambda \left ( M_eG \left( \frac{1}{R_e} - \frac{1}{R_e+h} \right ) - \frac{1}{2}\omega^2 (h^2 + 2 h R_e) \right) +C $$
We find the integration constant $C$ we look at the base of the cable. The tension there must be sufficient to keep the whole cable in place. The force the whole cable exerts at that point is
$$T(0)=C=\int_0^L \lambda \left ( -\frac{M_eG}{(h+R_e)^2} + \omega ^2 (h+R_e) \right ) \mathrm{d}h$$
where $L$ is the total length of the cable. We can see right away that this integral becomes more and more positive as $L$ increases, so even without finishing the integral we know there is some value of $L$ that will make this base tension positive. The condition that the cable stay taut is the condition $T(h) > 0 \;\forall\; h < L$ - that is, there is tension in the cable everywhere but the very end. Plug in the value of $C$ and you will find that if $C$ is positive, this is the case.
So all we technically need is a long cable. But it might be more efficient to have a counterweight at the end.
You are correct : the diagram is unrealistic. The entire problem leaves out a lot of important information. It is a very disappointing effort from such a well-known teaching site.
Mass $m_1$ does not appear to be moving on a frictionless horizontal table - although that would be a reasonable assumption. Gravity must be assumed to act on mass $m_2$ otherwise there could be no centripetal force. And the preceding paragraph refers to $m_2$ as a counterweight. Gravity must act on $m_1$ also. Without a table to rest on, mass $m_1$ would not rotate in a perfectly horizontal circle however fast it moved. The string would always be inclined downwards by some small angle (see Conical Pendulum). For this reason alone the centripetal force would indeed be less than the counter-weight.
The tube has a rounded opening or bell which, like a pulley, allows the string to change direction smoothly. The implication is that the tube is frictionless, and that the tension is the same in both sections of the string. If static friction were included the tension in the nearly-horizontal section of the string could be greater or less than that in the vertical section, because friction could act in either direction.
The bell opens out horizontally. Perhaps the person who devised the question intended that this would ensure that the string remained horizontal when it left contact with the bell, so that mass $m_1$ is guaranteed to travel in a horizontal circle. If so, this is a naive assumption which is incorrect. The string will change direction abruptly at the rim where it is no longer supported by the bell.
Lastly, it is not stated that $m_2$ hangs freely, but that seems to be implied. Like $m_1$, counterweight $m_2$ could also be resting on an invisible horizontal surface, so that part of its weight is balanced by the normal contact force.
The length of the string is fixed, and the problem states that $m_1$ moves in a circle, therefore $m_2$ does not move vertically. It is held in equilibrium by the upward tension in the string and the downward force of gravity. It is not necessary to assume that $m_2$ rests on a horizontal surface. The fact that $m_2$ is not moving vertically (ie that it is not falling) does not mean that it is resting on a horizontal surface which stops it from falling.
Best Answer
You are missing buoyant force. Let us take a simple example when you pour some dirt in water and shake after some time more dense object settle.Why ? Because of buoyant force $$F=mg-V_{obj}d_{fluid}g$$ $$a=g-\frac{d_{fluid}}{d_{obj}}g$$ Hence the less dense object remain suspended for a long time.
In the same way, in centrifuge there is outward acceleration same a $g$(due to centrifugal force) but greater in magnitude and buoyant force due to that.Hence dense particles moves away from the axis and less dense particles towards the axis.
Although to an "inertial observer" the spinning centrifuge does not produce a "real force" it does increase the pressure within the fluid,but to an observer "in the test tube" it is same as increasing the gravitational force.