Why are three equations needed?

There are three angles: J1, the massless shaft point and J2

The factor s in the denominator indicates:

No spring ties the system to the frame, so a free integration exists

In the T-I analogy, J maps to:

Capacitance C = J, torque to current, and angular velocity to node voltage

2075 Chaitra · 8 marks

Rotational system θ/T and T-I analogy

Short answer

Introduce angles θ₁, θ₃, θ, write three equations, eliminate θ₁ and θ₃ to find θ(s)/T(s) = k₂(b₁s+k₁)/Δ, where Δ = J₁J₂(b₁+b₂)s⁵+[J₁J₂(k₁+k₂)+J₂b₁b₂]s⁴+[(J₁+J₂)b₁k₂+J₁b₂k₂+J₂b₂k₁]s³+[(J₁+J₂)k₁k₂+b₁b₂k₂]s²+b₂k₁k₂s. T–I analogy: T→i, J→C, b→1/R, k→1/L; circuit: C₁=J₁, C₂=J₂, L₁∥R₁ between nodes v₁ and v₃, L₂ from v₃ to v₂, R₂ from v₃ to ground.

Find the transfer function θ(s)/T(s) for the mechanical rotational system shown below. Also develop T-I analogous circuit. [Figure: torque T(t) applied to inertia J1; J1 coupled through torsional spring k1 in parallel with damper b1 to a shaft point, which has viscous damper b2 to the fixed frame; that point is coupled through torsional spring k2 to inertia J2, whose angle is θ(t)]

Let $\theta_1$ = angle of $J_1$ , $\theta_3$ = angle of the massless shaft point (where $b_2$ is attached) and $\theta$ = angle of $J_2$ . Bearings of $J_1$ , $J_2$ are taken as frictionless (only the shown elements act).

Differential equations (Laplace form)

\begin{aligned} J_1s^2\theta_1 + (b_1s + k_1)(\theta_1 - \theta_3) &= T \\ (b_1s + k_1)(\theta_3 - \theta_1) + b_2s\,\theta_3 + k_2(\theta_3 - \theta) &= 0 \\ J_2s^2\theta + k_2(\theta - \theta_3) &= 0 \end{aligned}

In matrix form:

\begin{bmatrix} J_1s^2 + b_1s + k_1 & -(b_1s+k_1) & 0 \\ -(b_1s+k_1) & (b_1+b_2)s + k_1 + k_2 & -k_2 \\ 0 & -k_2 & J_2s^2 + k_2 \end{bmatrix} \begin{bmatrix}\theta_1\\ \theta_3\\ \theta\end{bmatrix} = \begin{bmatrix}T\\0\\0\end{bmatrix}

Transfer function

By Cramer's rule, the numerator for $\theta$ is the product of the off-diagonal couplings, $(b_1s+k_1)k_2$ :

\frac{\theta(s)}{T(s)} = \frac{k_2(b_1s + k_1)}{\Delta}

where

\Delta = (J_1s^2 + b_1s + k_1)\left\{\left[(b_1+b_2)s + k_1 + k_2\right](J_2s^2 + k_2) - k_2^2\right\} - (b_1s + k_1)^2(J_2s^2 + k_2)

Expanded:

\begin{aligned} \Delta =\; & J_1J_2(b_1 + b_2)s^5 + \left[J_1J_2(k_1 + k_2) + J_2b_1b_2\right]s^4 \\ & + \left[(J_1 + J_2)b_1k_2 + J_1b_2k_2 + J_2b_2k_1\right]s^3 \\ & + \left[(J_1 + J_2)k_1k_2 + b_1b_2k_2\right]s^2 + b_2k_1k_2\,s \end{aligned}

The factor $s$ in $\Delta$ shows a free integration: nothing ties the system to the frame by a spring.

T-I analogy

Torque $\to$ current, $J \to C$ , $b \to 1/R$ , $k \to 1/L$ , angular velocity $\omega \to$ node voltage.

Mechanical	Electrical
$T$	current source $i$
$J_1$ , $J_2$	$C_1 = J_1$ , $C_2 = J_2$
$b_1$ , $b_2$	$R_1 = 1/b_1$ , $R_2 = 1/b_2$
$k_1$ , $k_2$	$L_1 = 1/k_1$ , $L_2 = 1/k_2$
$\omega_1, \omega_3, \omega$	$v_1, v_3, v_2$

 v1         L1∥R1      v3  L2   v2
 +----+---+--LLLL--+---+--LLLL--+
 |    |   +-/\/\/--+   |        |
(i)   C1               R2       C2
 |    |                |        |
 +----+----------------+--------+ gnd

( $L_1$ and $R_1$ are in parallel between nodes 1 and 3.)

✓ Answer checked by Claude Opus 5.5 (high effort)

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