Check the sign of $v_x$. If it is less than zero, your direction of motion is $\displaystyle{180^\circ+\arctan\left(\frac{v_y}{v_x}\right)}$. If it is greater than zero, then $\displaystyle{\arctan\left(\frac{v_y}{v_x}\right)}$ works. If it is zero, you're at $\pm 90^\circ$, and the sign of $v_y$ tells you which. If you care about being between $0$ and $360^\circ$ or some other particular range, you can just test the result and add or subtract $360^\circ$ as appropriate.
See this Wikipedia article on a 2-argument arctangent function that exists exactly for this purpose.
You say that angle $0$ is "up" for your purposes. If by this you mean the positive $y$ direction, then you should subtract $90^\circ$ from the result obtained from the above, before testing to see if the angle is in the correct range. The reason is that standard positions of angles, in accordance with the arctangent computations you mentioned, measure the angles counterclockwise from the positive $x$-axis. If you instead measure counterclockwise from the positive $y$-axis, then your starting point is shifted up $90^\circ$, so your angles will correspondingly be shifted down $90^\circ$. (Or you could just apply the above with $v_y$ in place of $v_x$ and $-v_x$ in place of $v_y$, but this might be a bit confusing.)
First of all, you were given the weights, which already are forces; you do not multiply by acc due to gravity. Anyway, the following is what I think is a standard method of attack in physics:
The net torque $\tau$ on the lever is $(1000) (4) - (10) (12) = 3880 \,\text{ft} \,\text{lb}$. We find the angular acceleration by dividing the torque by the moment of inertia $I$, which is split into two components: $I_w$ (due to the weights on the ends), and $I_b$ (the uniform load)
$$I_w = [(1000) (4^2) + (10) (12^2)]/32.2 \approx 541.6 \, \text{ft}\, \text{lb} \, \text{sec}^2 $$
$$I_b = \left[\frac{1}{12} (97) (16^2) + (97) (4^2)\right]/32.2 \approx 112.5 \, \text{ft}\, \text{lb} \, \text{sec}^2$$
(The second term is an addition due to the parallel axis theorem; the lever has an off-center fulcrum. Also, the factor of $32.2$ is acc due to gravity.) The angular acceleration of the lever is then
$$\alpha = \frac{\tau}{I} \approx 5.93/\text{sec}^2 $$
The linear acceleration of the right end is then the length of the right arm times the angular acceleration, or about $(5.93)(12) \approx 71.2 \,\text{ft}/\text{sec}^2 $.
Best Answer
Your calculation is fully correct. As per request, here is a slight variant, with details.
Let $r$ be the radius of the ball bearing, and let $R$ be the radius of the "atom" if the ball-bearing was the nucleus. Then $$\frac{R}{r}=\frac{0.529\times 10^{-10}}{1.2\times 10^{-15}}.$$ This expresses the fact we are looking at a scaled version of the atom/nucleus. But we know $r$, so we can find $R$. We get $$R=r\frac{0.529\times 10^{-10}}{1.2\times 10^{-15}}=(1.5\times 10^{-3})\frac{0.529\times 10^{-10}}{1.2\times 10^{-15}}.$$ For the calculation, I would first find $1.5\frac{0.529}{1.2}$, which is $0.66125$, and then deal with the power of $10$, which is $-3+(-10)-(-15)$, which is $2$. So the answer, in standard scientific notation, is $6.6125\times 10^1$ (metres), though $66.125$ seems more sensible. Many Physics people would correctly complain that we should only give the answer to $2$ significant figures, since two of the important numbers were only specified to that precision.