Question 10.1: The rack-and-pinion system shown in Fig. 10.8 is to be model...

The detailed symbolic free-body diagram for this system is shown in Fig. 10.9. The elemental equations are as follows: For the pinion inertia,

T_{s} -T_{t}=J_{p}\frac{d\Omega }{dt} .          (10.10)

For the transducer,

\Omega =nv,          (10.11

T_{t} =\left({1}/{n}\right) F_{t},          (10.12)

where n={1}/{r}. For the rack mass,

F_{t}-F_{b}-F_{s}=m\frac{dv}{dt} .          (10.13)

For the rack friction,

F_{b} =bv.          (10.14)

Combine Eqs. (10.10), (10.11), and (10.12) to eliminate T_{t} and Ω:

T_{s}-\frac{1}{n} F_{t} =nJ_{p} \frac{dv}{dt} .          (10.15)

Combine Eqs. (10.13) and (10.14) to eliminate F_{b}:

F_{t} -bv-F_{s}=m\frac{dv}{dt} .          (10.16)

Combine Eqs. (10.15) and (10.16) to eliminate F_{t}:

T_{s}-\frac{1}{n} \left(m\frac{dv}{dt}+bv +F_{s}\right) =nJ_{p} \frac{dv}{dt}.          (10.17)

Rearranging yields a single state-variable equation:

\frac{dv}{dt}=\frac{1}{n^{2}J_{p} +m } \left(-bv+nT_{s}-F_{s} \right) .          (10.18)

Note that the inertia from the rotational part of the system becomes an equivalent mass equal to n^{2}J_{p}; thus, although this system has two energy-storage elements, they are coupled by the transducer so that they behave together as a single energy-storage element.
Therefore only one state-variable equation is required for describing this system.

The multiplication of a system parameter of one domain by the square of the transducer constant to produce an equivalent parameter in the other domain will be seen to occur consistently in all mixed-system analyses.