How can a mass $m_1$ lift a heavier mass $m_2=2m_1$? 
In below diagram, in what angle $\theta$ should we drop $m_1$ so that, right at the moment of dropping, $m_2$ would lift upwards? Assume $m_2=2m_1$.

This is in my homework and I don't even understand the question. I think the question is problematic. Is it not? I mean, $m_1$ is lighter than $m_2$. How on earth could it be used to lift $m_2$? Any tip is appreciated.

[EDIT] My teacher concurred that this was a wrong question. Thank you for your kind answers.
 A: So if I were reading this, I would interpret this as saying that $m_2$ lifts upwards right as $m_1$ is at the bottom of its swing, rather than right as it is released. When it is released, it has no centrifugal force to back up its weight  in tensioning the string and thus lift up the other mass: that comes as it starts moving faster.
Basics you should know
The first thing that I'm going to say to you as a physics student is that these pulleys are places where the conservation of momentum is being violated. Or, if you prefer, Newton's third law applies but one of the forces is being applied outside of the system as you understand it, it's being applied to the Earth or so, which you are then assuming is not moving. Same difference, whichever explanation you prefer. They are able to exert an upward force on the string, which allows them to potentially pull up a mass.
But, force balance is different than Newton's third law. Probably the biggest problem for students is mixing up these two, so let me be clear. If there is something out there which is not accelerating, for example me sitting on a chair, then that means that the forces on that thing are balanced: so I am being pulled down by gravity, but pushed up by my chair. And my chair is being pushed down by me (that is the Third Law!) and pushed down by its own gravity, but pushed up by the floor. Because it's not accelerating either, it remains at rest. Force balance is not a law: it is a way things happen to be right now that helps you to analyze a problem. Maybe I in my chair am not in a state of force balance, maybe I am sitting in my chair but it is tipping over and I am about to hit the ground, and we are both accelerating towards this unpleasant end. Or maybe the floor is not strong enough to hold the both of us and we are beginning to plummet through it, to the floor below.
You can still steal this idea of force balance even for a pulley, and that is the idea of tension. The idea is that the pulley offers no friction for rotation, it cannot resist a torque. So there is torque-balance. This means that the magnitude of the force of the rope coming into one side, is the same as the magnitude of the force of the rope on the other side, those two forces are equal and “opposite” not in direction but orientation around the wheel. The pulley preserves tension by being frictionless.
How to simplify the problem
The proper set of equations of motion is likely to be very hard here, and if I were studying this system myself I would use a branch of mathematics which is probably too advanced for you at this stage of your education, called Lagrangian mechanics or “variational calculus,” and even then I would probably make a simplifying assumption that the rope between $m_1$ and its pulley stays perfectly straight, and also that the pulleys are all incredibly small and the rope is massless.
To simplify it to the point where you can analyze it, probably we should put $m_2$ on a scale and ask when that scale says that $m_2$ weighs exactly zero. The basic problem with $m_2$ becoming airborne is that then the distance between $m_1$ and its pulley is changing and that looks like a big mess of questions to answer. But, if we imagine doing the experiment over and over slowly increasing this initial $\theta$ and then looking at that scale, then after this critical value of $\theta$ where $m_2$ goes airborne, the scale must read zero as the tension in the rope pulls the mass off of the scale. So I imagine it continuously decreases to zero and I can get this critical angle by looking at the very last experiment where $m_2$ did not go airborne.
This simplification, combined with the earlier one about small pulleys, means that $m_1$ is roughly in a circular motion about a fixed center. The constraint of a point $(x,y)$ forced to move on a circle with a dynamical angle $\theta(t)$ measured in radians is,
$$x(t) = x_0 + R \sin\theta(t),\\
y(t) = y_0 - R\cos\theta(t).$$
This is true no matter how the forces are provided, and some calculus gives the velocity and acceleration in terms of the time derivatives $\dot\theta,\ddot\theta$ of that angle:$$
v_x = R\dot\theta \cos\theta,\\
v_y = R\dot\theta \sin\theta,\\
a_x = R\ddot\theta \cos\theta - R\dot\theta^2 \sin\theta,\\
a_y = R\ddot\theta \sin\theta + R\dot\theta^2 \cos\theta.$$ This has a very nice geometric interpretation, the vector $\mathbf a$ has one component parallel to the velocity with magnitude $R\ddot\theta$ and one component perpendicular to that, parallel instead to the displacement from the center point, with magnitude $R\dot \theta^2,$ which we could write as $v^2/R.$
Now, at the very moment of $m_1$ hitting its lowest point, call this $t_\text{min}$ such that $\theta(t_\text{min})=0,$ we do not have any forces to lie perpendicular to its motion and $\ddot\theta=0.$ The only forces are the tension of the string holding the mass up, and gravity pulling it down: and while the mass is still accelerating so we cannot claim force-balance, we know exactly how much it is accelerating by, and we know the forces on it:
$$
a_y(t_\text{min})= v^2/R = \sum F_i/m_1 = (T - m_1 g)/m_1.$$
Substituting $T=2m_1 g$ to make the scale read zero, we get that $v^2/R = g.$
From here you should have everything you need, if you remember that $v^2$ is connected to energy concerns. Good luck.
A: You should draw a free-body diagram for the swinging mass, $m_1$. Then you will see that the tension in the rope is a function of $\theta$ if the mass is released at some angle $\theta_0 > 0$. You should be very careful to note that the radial acceleration is not zero, but is a function of speed.
You can calculate the speed of $m_1$ as a function of angle, also, using conservation of total mechanical energy.
Those two calculations will let you construct a solution to the question.
