Actuator 2 has a motor 1/2 the diameter and the same length as the motor in actuator 1. At the same power dissipation, the motor produces $\frac{1}{2\sqrt{2}}$ times the torque, so it needs a gear ratio of $2\sqrt{2}$ to produce the same torque at the output of the actuator. The rotor inertia is 1/8th that of the motor in actuator 1, since $j\propto r^{3}$. The reflected inertia after the gearing is $g^{2}j = \frac{(2\sqrt{2})^{2}}{8}= 1$.
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