Torsional system: find J and D for 20% overshoot

Transfer function

Torque balance on $J$ (spring $K$ and damper $D$ both to the frame):

$$J\ddot\theta + D\dot\theta + K\theta = T(t)$$ $$\frac{\theta(s)}{T(s)} = \frac{1}{Js^2 + Ds + K} = \frac{1/J}{s^2 + \frac{D}{J}s + \frac{K}{J}}, \qquad K = 5\ \text{N-m/rad}$$

Comparing with the standard form: $\omega_n^2 = \dfrac{K}{J}$, $2\zeta\omega_n = \dfrac{D}{J}$.

Damping ratio for 20% overshoot

$$\zeta = \frac{-\ln(0.20)}{\sqrt{\pi^2 + \ln^2(0.20)}} = \frac{1.6094}{\sqrt{9.8696 + 2.5903}} = \frac{1.6094}{3.5299} = 0.456$$

Natural frequency from settling time (2% criterion)

$$t_s = \frac{4}{\zeta\omega_n} = 2 \;\Rightarrow\; \zeta\omega_n = 2$$ $$\omega_n = \frac{2}{0.456} = 4.386\ \text{rad/s}$$

Solve for $J$ and $D$

$$\begin{aligned} J &= \frac{K}{\omega_n^2} = \frac{5}{(4.386)^2} = \frac{5}{19.24} = 0.260\ \text{kg-m}^2 \\ D &= 2\zeta\omega_nJ = 2\times 2\times 0.260 = 1.04\ \text{N-m-s/rad} \end{aligned}$$

Answer: $J \approx 0.26\ \text{kg-m}^2$ and $D \approx 1.04\ \text{N-m-s/rad}$.

With these values, $\dfrac{\theta(s)}{T(s)} = \dfrac{3.85}{s^2 + 4s + 19.24}$ (poles at $s = -2 \pm j3.90$), giving 20% overshoot and $t_s = 2$ s. The steady-state angle for a unit step torque is $1/K = 0.2$ rad.