Let's say a particle moves in plane with curvature equal to $\kappa(t) = 2t$, with constant speed of $\|v(t)\| = 5$, such that $v(0) = 5\textbf{i}$, and the particle never goes to the left of the $y$-axis. The angle that the velocity vector makes with the positive $x$-axis at time $t$ is $\alpha(t)=\frac{\pi}{2} - 5t^2$.
$\alpha(t)=\frac{\pi}{2} - 5t^2 \implies \frac{d\alpha}{dt} = -10t$, which is always negative because $t \ge 0$. That would normally imply that the angle is decreasing without bound, but intuitively I don't see how it can go below $-\pi$, so I'm not sure how to interpret that angle.
Note: This is the exercise $14.15.11$ in Tom Apostol's Calculus vol. $1$.
Clarification: We know the curvature is $\kappa(t) = |\frac{d \alpha}{ds}|$, where $s$ is arc length. To compute $\alpha$, I would like to know the sign of the $\frac{d \alpha}{ds}$. If we take $\alpha$ is always decreasing, we know $\kappa(t) = -\frac{d \alpha}{ds}$, and I can integrate to find $\alpha$. However, if the angle is constrained to be in the range $[-\pi, \pi]$, then I'm not sure how to assign the sign to the absolute value (and even the derivative is not continuous).
You almost said it yourself: "if the angle is constrained to be in the range...".
Therefore, by contrapositive (or contradiction), it must be that the angle is not constrained to any bounded range.
This is acceptable because an angle like $-5\pi$ is coterminal/equivalent to $\pi$. And it's actually extremely helpful for calculus so that an angle can decrease past $-\pi$ or increase past $\pi$ in a continuous or differentiable way, without having to suddenly jump by $2\pi$.