Does the Bohr model violate the uncertainty principle?
The Heisenberg uncertainty principle states that

$$\Delta x \Delta p \geq \frac{\hbar}{2}$$

When you think about it, the whole of classical mechanics violates this principle. Think of a particle with mass $m$, a defined position $x_0$ at time $0$, a defined velocity $v_0$ at time $0$, and a constant acceleration $a$. The laws of kinematics state that at any time $t$, the position $x_t = x_0 + v_0t + \frac{1}{2}at^2$, and the momentum $p_t = m(v_0 + at)$, with zero uncertainty in both quantities, i.e. $\Delta x \Delta p = 0$. That works in most real-life cases because on a macroscopic level, quantum effects are negligible.

The Bohr model was derived via the classical assumption that the electron orbits the nucleus, much like how the Earth orbits the Sun, except that the origin of the centripetal force was electromagnetic instead of gravitational. It then went on to derive the energy levels of the electron in a hydrogen atom using the laws of uniform circular motion and electrostatics. This means that it is theoretically possible to know the position and momentum of the electron simultaneously with no uncertainty - a violation of the uncertainty principle.