= Solution
Let $a_c\ll R$ be the contact radius of the adhering <neutrophil>. Indentation is of order $a_c^2/R$, so the elastic <strain> is of order $a_c/R$ within a region of volume $a_c^3$. The elastic <energy> is consequently $G a_c^5/R^2$, while adhesion lowers the <energy> by order $Ja_c^2$. Balancing these terms gives
$$
a_c\sim\left(\frac{JR^2}{G}\right)^{1/3},\qquad \frac{a_c}{R}\sim\left(\frac{J}{GR}\right)^{1/3}.
$$
This estimate requires weak adhesion, $J\ll GR$.
For an <adhesive-cell rolling threshold>, assume peeling the rear bonds dissipates work of order $J$ per unit contact area. Rolling through angle $d\theta$ sweeps contact area of order $a_cR\,d\theta$, giving resisting moment $M_r\sim Ja_cR$. A surrounding fluid of <dynamic viscosity> $\eta$ supplies a moment of order $\eta\dot\gamma R^3$. Equating moments yields
$$
\boxed{\dot\gamma_c\sim\frac{Ja_c}{\eta R^2}\sim\frac{J^{4/3}}{\eta G^{1/3}R^{4/3}}.}
$$
A dimensionless numerical coefficient depends on the contact and near-wall <Stokes flow>. Fluid viscosity is indispensable to a rate estimate, even though the contact model alone specifies only $R,G,J$.
A perfectly reversible elastic adhesive contact does not lose net <energy> when an unchanged contact footprint translates. It therefore does not, by itself, justify a finite rolling threshold. The above estimate assumes adhesive hysteresis or irreversible bond peeling. If equilibrium adhesion is $J$ but the work lost during peeling and rebinding is $\Delta J$, retain $a_c\sim(JR^2/G)^{1/3}$ and replace $J$ by $\Delta J$ only in the resisting moment: $\dot\gamma_c\sim\Delta J\,a_c/(\eta R^2)$.
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