Uncertainty Principle in the Ground State of a Rigid Rotor
As far as the uncertainty relation between $L_{z} = -i\hbar \partial/\partial\phi$ and $\phi$ is concerned, there is a simple way to understand why we cannot use the commutation relation $[\partial/\partial\phi,\, \phi] = 1$, which leads to the problematic inequality $\Delta L_z \Delta\phi \ge \hbar/2$.

Remember that the variable $\phi$ is defined on a circle, i.e., $\phi = 0$ and $\phi=2\pi$ represent the same point. A function of the form $\phi\psi(\phi)$, where $\psi(\phi)$ is $2\pi$-periodic, has a discontinuous jump by the amount $-2\pi \psi(0)$ at $\phi = 0$ (or $2\pi$). Hence, when applying $\partial/\partial\phi$ to $\phi\psi(\phi)$, one should make sure to differentiate the discontinuity as well, which leads to an additional term: $-2\pi \psi(0)$ times a Dirac delta function. Then, the correct commutation relation should be $[\partial/\partial\phi,\, \phi]\,\psi(\phi) = \psi(\phi) - 2\pi\psi(0)\delta(\phi)$, and we have the modified uncertainty principle
$$
\Delta L_z \Delta\phi \ge \frac{1}{2}\big|\langle[L_{z}, \phi]\rangle\big| = \frac{\hbar}{2}\Big|\langle\psi|[\partial/\partial\phi,\, \phi]|\psi\rangle\Big| = \frac{\hbar}{2}[ 1- 2\pi|\psi(0)|^2].
$$
For a normalized angular momentum eigenfunction $\psi_{m}(\phi) = e^{im\phi}/\sqrt{2\pi}$, the right-hand side of the above yieids zero. This is exactly what should be the case when the uncertainties in the angular momentum and the angle are respectively zero and $2\pi$.