= Solution
Under the <Boussinesq approximation>, total upper-layer <kinetic energy> is $K=\rho_lShu_U^2/2$, where $u_U$ is its volume root-mean-square speed. For <solid-body rotation> of angular speed $\omega(t)$, direct radial averaging gives $u_U^2=\omega^2R^2/2$, consistent with that energy expression. A constant $K$ implies $hu_U^2=h_0u_U(0)^2$. Define $C_K=u_U(0)/(\Omega R)$ to obtain the <fixed-energy turbulent mixed layer> relation
$$
\boxed{hu_U^2=C_K^2\Omega^2R^2h_0.}
$$
With $u_I=C_Iu_U$, the <interfacial power model for entrainment> used in (b) gives
$$
\dot h=\frac{2C_Wc_DC_I^3C_K^3\Omega^3R^3}{B}
\left(\frac{h_0}{h}\right)^{3/2}.
$$
Multiply by $h^{3/2}$ and integrate from $h_0$. This yields
$$
h^{5/2}=h_0^{5/2}
+\frac{5C_Wc_DC_I^3C_K^3\Omega^3R^3h_0^{3/2}}B\,t.
$$
The <fixed-energy two-layer entrainment law> is therefore
$$
\boxed{\frac h{h_0}=(1+A\Omega t)^{2/5},\qquad
A=\frac{5C_Wc_DC_I^3C_K^3\Omega^2R^3}{Bh_0}
=\frac{5C_Wc_DC_I^3C_K^3}{\mathrm{Ri}_{B,0}}\frac R{h_0}.}
$$
Here $\mathrm{Ri}_{B,0}=B/(\Omega^2R^2)$ uses the same lid-speed normalization as part (b). All numerical geometry factors depend on the chosen characteristic stress/speed convention. The formula ceases to apply once $h=H$.
The exponent $2/5$ requires the local power to decrease as $u_I^3\propto h^{-3/2}$. It does not follow from constant upper-layer <kinetic energy> and a fixed fraction of unspecified lid power alone. For example, an imposed constant shaft power together with a constant fraction going to <potential energy> would give constant $\dot h$, not this exponent. The appropriate distinction between local interface power and global forcing appears explicitly in the primary study https://www.whoi.edu/cms/files/entrainment_mixedlayer_201804.pdf[Entrainment and mixed layer dynamics of a surface-stress-driven stratified fluid], equations (3.6)–(3.7); the calculations above use that local interpretation.
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