---
title: "Torsional system: find J and D for 20% overshoot"
subject: "Control Systems"
chapter: 3
key: "semester_3/control_systems:torsional-j-d-overshoot"
source: https://www.bctengineer.com/q/semester_3/control_systems/torsional-j-d-overshoot
site: BCT Engineering
---

# Torsional system: find J and D for 20% overshoot

Control Systems · Chapter 3: System Transfer Function and Responses

- Asked in: 2081 Baisakh, 8 marks; 2066 Bhadra (old course), 8 marks (asked 2 times)
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/control_systems/torsional-j-d-overshoot
- Whole chapter: https://www.bctengineer.com/md/semester_3/control_systems/chapter-3

## Question

For the given mechanical system, obtain the transfer function with angular displacement θ(t) as output and torque T(t) as input. Hence find J and D to give 20% overshoot and settling time of 2 second for step input of torque T(t). [Figure: fixed wall – torsional spring K = 5 N-m/rad – inertia J (torque T(t) applied, angle θ(t)) – viscous damper D – fixed wall]

## Short answer

For 20% overshoot, ζ = 0.456; from ts = 4/(ζ ωn) = 2 s gives ζ ωn = 2, ωn = 4.386 rad/s. Then J = K/ωn^2 = 5/19.24 ≈ 0.26 kg-m^2, D = 2ζ ωn J ≈ 1.04 N-m-s/rad.

## Answer

#### 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.

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. What is θ(s)/T(s) for the torsional system?
   A. 1/(Js^2 + Ds + K)
   B. K/(Js^2 + Ds)
   C. s/(Js^2 + Ds + K)
   D. 1/(Js + D + K)
   Answer: A. Torque balance J θ'' + D θ' + K θ = T gives this second-order function. (Key point: Transfer function)

2. What ζ gives 20% overshoot?
   A. 0.456
   B. 0.2
   C. 0.707
   D. 0.8
   Answer: A. ζ = 1.6094/sqrt(π^2 + 1.6094^2) = 0.456. (Key point: Damping ratio)

3. With ts = 2 s (2% criterion) and ζ = 0.456, what is ωn?
   A. 4.386 rad/s
   B. 2 rad/s
   C. 0.456 rad/s
   D. 8.77 rad/s
   Answer: A. ζωn = 4/ts = 2, so ωn = 2/0.456 = 4.386 rad/s. (Key point: Settling time)

4. How are J and D found?
   A. J = K/ωn^2 = 0.26 and D = 2ζωn J = 1.04
   B. J = K ωn^2 and D = 2ζ J
   C. J = ωn^2/K and D = ζωn
   D. J = K/ωn and D = 2ζ
   Answer: A. ωn^2 = K/J gives J = 5/19.24 = 0.26; 2ζωn = D/J gives D = 4(0.26) = 1.04. (Key point: Finding J and D)

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