Bohr quantization hypothesis To explain Rydberg formula, Bohr have assumed somewhat general hypothesis which is applicable to various classical system. As far as I know he assumed that for any classical system with periodic motion,
$$ \oint p_i dq_i = n_i h \qquad (1)$$
where $q_i$s are generalized coordinates and the integral is taken over full period of the coordinate.
I have tried to recover the quantization condition $mvr = n \hbar$ for a circular motion with constant velocity.
Pick any potential with rotational symmetry and assume that the particle is in a circular motion at radius $r$. Let $v$ be the speed of the particle. Then
$$ \oint p_1 dx = \oint m \dot{x}^2 dt = \oint mv^2 \sin^2 \omega t dt = \frac{1}{2} mv^2 T $$
where $T = \frac{2 \pi r}{v}$.
Substituting yields $ \pi r mv = n_1h $ or
$$ rmv = 2n_1 \hbar $$
This is not only contradicts to the known result $rmv=n \hbar$ but also is an absurdity by itself since, we can also perform the same calculation for $i=2$ where $q_2 = y$ and obtain $rmv = 2n_2 \hbar $.
So it is apparent that I have misunderstood the Bohr quantization hypothesis.
As appeared in the answer, if we regard the index on the right hand side are under influence of summation convention, then problem vanishes for this case since $ \oint \sum p_i \frac{dq_i}{dt} dt = \oint \sum mv_i^2 dt = mv \cdot 2\pi r$.
However, this interpretation does not fit into the equation (1) since then the index on the left hand side is dummy but that in the right hand side is not.
So how can I understand equation (1)?
 A: One may regard the Bohr-Sommerfeld quantisation as the quantisation of the action. We can see that the action is
$$S=\int dt L = \int dt \frac{p^2}{2m} = \frac{1}{2m}\int (\mathbf{p}dt)\cdot\mathbf{p} = \frac{1}{2m}\int d\mathbf{q} \cdot\mathbf{p}$$
Therefore, your quantisation condition is not quite correct. When you write 
$$\oint p_i dq_i$$
The Einstein summation must be applied (Note that we write $\oint$ instead of $\int$ here because we quantise the action of a closed orbit).
Hence, the correct form of the original Bohr-Sommerfeld quantisation is
$$\oint \mathbf{p} \cdot d \mathbf{q} = nh $$
By applying this equation to the circular orbit with a spherical symmetry potential, you should be able to obtain the right answer.
A: For the Bohr quantization makes sense, the generalized coordinates should be separable. In the example of the question, the variables $x$, $y$ are not separable. Hence weirdo results are obtained.
The right choice for the problem is $r$ and $\theta$. Then $\oint p_\theta d \theta = 2\pi L = n_\theta h$ so $L = n_\theta \hbar$ as desired.
