Figure shows a single degree of freedom system consisting of a gear of radius r and moment inertia, I, a rack of mass m a linear spring of stiffness K, and a torsional spring with stiffness Kt, obtain the equivalent inertia and stiffness:

1] With $\theta$ as the chosen co ordinate and

2] With X as the chosen co ordinate.

KE = (KE) rod kt (KE) pinion

$= \frac{1}{2} mx^2 + \frac{1}{2} I_o \theta^2$

$= \frac{1}{2} m.r^2 \theta^2 + \frac{1}{2} I_o \theta^2$

$= \frac{1}{2} [ m.r^2 + I_o ] \theta^2$

$Iq = m.r^2 + I_o$

$PE = (PE)_{sp} + (PE)_{tor.sp}$

$= \frac{1}{2} k. (def)^2 + \frac{1}{2} kt (def^n)^2$

$= \frac{1}{2} k. x^2 + \frac{1}{2} kt \theta^2$

$= \frac{1}{2} k. r^2 \theta^2 + \frac{1}{2} kt \theta^2$

$= \frac{1}{2} [ kr^2 + kt] \theta^2$

X co-ordinate:

$KE = (KE)_{rod} + (KE)_{pinion}$

$= \frac{1}{2} \times m \times x^2 + \frac{1}{2} \times I_o \theta^2$

$= \frac{1}{2} \times m \times x^2 + \frac{1}{2} \times I_o \times \frac{x^2}{r^2}$

$= \frac{1}{2} [ m + \frac{Io}{r^2} x^2$

$PE = (PE)_sp + (PE) Tor.sp$

$= \frac{1}{2} k (def^n)^2 + \frac{1}{2} kt. (def)^2$

$= \frac{1}{2} \times k \times x^2 + \frac{1}{2} \times kt \theta^2$

$= \frac{1}{2} \times k \times x^2 + \frac{1}{2} \times kt \times \frac{x^2}{r^2}$

$= \frac{1}{2} [ k + \frac{kt}{r^2}] x^2$

$k_{eg} = k + \frac{kt}{r^2}$