= Solution
For local cold <magnetocentrifugal acceleration>, the field must guide the gas and enforce approximate corotation with its disk footpoint in the sub-Alfvénic launching region. Let that footpoint be $R=R_f$, and keep its <field-line angular velocity> $\Omega_f=\Omega(R_f)$ constant along the field. In this rotating frame the relevant <effective potential> is
$$
V(R,z)=\Phi(R,z)-\frac12\Omega_f^2R^2.
$$
The footpoint is a critical point because $\partial_R\Phi(R_f,0)=R_f\Omega_f^2$ and $\partial_z\Phi(R_f,0)=0$. The meridional <Hessian matrix> there is
$$
V_{RR}=-(\beta+2)\Omega_f^2,\qquad V_{zz}=\lambda^2\Omega_f^2,\qquad V_{Rz}=0.
$$
Take distance $\ell$ along a smooth outward <magnetic field line> whose initial tangent is $(\sin\alpha,\cos\alpha)$ in the $(R,z)$ plane. Since the gradient of $V$ vanishes at the footpoint, field-line curvature does not contribute to its second derivative. Consequently
$$
V(\ell)-V(0)=\frac12\Omega_f^2\left[\lambda^2\cos^2\alpha-(\beta+2)\sin^2\alpha\right]\ell^2+O(\ell^3).
$$
A negative quadratic coefficient gives a downhill displacement and acceleration without thermal assistance. The <magnetocentrifugal launching criterion in a flattened power-law potential> is therefore
$$
\boxed{\tan^2\alpha>\frac{\lambda^2}{\beta+2},\qquad\alpha>\alpha_{\rm crit}=\arctan\left(\frac{|\lambda|}{\sqrt{\beta+2}}\right).}
$$
Angles are measured from the vertical, with $0\leq\alpha<\pi/2$. Below the critical angle the footpoint is a local minimum along the field, so a cold particle cannot cross the initial potential barrier. At equality the quadratic test is marginal; higher derivatives and the unspecified field curvature can decide the outcome. The threshold is a local condition for launch, not proof that the wind escapes along an arbitrary global field geometry. The spherical point-mass limit $\lambda=1$, $\beta\to1$ recovers $\alpha_{\rm crit}=30^\circ$, consistent with https://arxiv.org/pdf/1604.03835[Ogilvie's discussion of cold disk winds, section 9.9].
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