power available for motility is going into driving the cluster over the substrate, then we can write: $P = -\int d^2 r \, { {\mathbf v}}\cdot{ {\mathbf f}}_\textrm{drag} = \xi \int d^2 r |{ {\mathbf v}}|^2$. If the cluster is traveling as a rigid, circular cluster with its maximum possible velocity, ${ {\mathbf v}} = v_{\textrm{max}} \hat{{ {\mathbf x}}}$ then $P = \xi \pi R^2 v_\textrm{max}^2 \equiv \gamma_t v_\textrm{max}^2$ where $\gamma_t = \xi \pi R^2$ is the translational drag coefficient of the cluster. If, instead, the cluster puts its entire power into rigid-body rotation with ${ {\mathbf v}}({\ensuremath{{ {\mathbf r}}}\xspace}) = \Omega_{\textrm{max}} r (-\sin \theta,\cos \theta)$ (in polar coordinates), then $P = \xi \Omega_\textrm{max}^2 \int d^2 r r^2 = \xi\frac{\pi}{2} R^4 \Omega_{\textrm{max}}^2 \equiv \gamma_r \Omega_\textrm{max}^2$ where $\gamma_r = \frac{\xi \pi}{2} R^4$ is the rotational drag coefficient of the cluster. In general, the power dissipated if the cluster is moving rigidly with velocity ${ {\mathbf v}}$ and angular speed $\Omega$ is $P = \gamma_t v^2 + \gamma_r \Omega^2$ and hence, we find that the speed $v(\Omega$) that a cluster rotating with angular velocity $\Omega$ is able to travel to obtain is $$v(\Omega) = \sqrt{v_\textrm{max}^2 - \frac{\gamma_r}{\gamma_t} \Omega^2} \label{eq:tradeoff}$$ This quantifies one 