Question 5.19: A Single-Degree-of-Freedom Gear–Train System For the gear–tr...
A Single-Degree-of-Freedom Gear–Train System
For the gear–train system shown in Figure 5.96a, derive the differential equation of motion. The mass moments of inertia of the two gears about their respective fixed centers are I_{C1} and I_{C2}.
The free-body diagram is shown in Figure 5.96b. The geometric constraint is
\frac{θ_2}{θ_1} = \frac{r_1}{r_2}
Applying the moment equation to each gear gives
+ \curvearrowright : \sum M_{C1} = I_{C1}\alpha
τ_1 – r_1F = I_{C1}\ddot{θ}_1
and
+ \curvearrowleft : \sum M_{C2} = I_{C2}\alpha
r_2F = I_{C2}\ddot{θ}_2
Combining the two equations and eliminating F yields
τ_1 – \frac{r_1}{r_2} I_{C2}\ddot{θ}_2 = I_{C1}\ddot{θ}_1
Note that the angular displacements θ_1 and θ_2 are dependent through the geometric constraint. Thus, the gear–train in Figure 5.96 is a single-degree-of-freedom system, which requires only one equation of motion in terms of only one coordinate. Assume that the equation is expressed in terms of θ_1. Introducing the geometric constraint results in the equation of motion
τ_1 – \frac{r_1}{r_2} I_{C2} \left\lgroup \frac{r_1}{r_2} \ddot{θ}_1 \right\rgroup = I_{C1}\ddot{θ}_1
simplified to
\left\lgroup I_{C1} + \frac{r_1^2}{r_2^2} I_{C2} \right\rgroup \ddot{θ}_1 = τ_1
which is the dynamics seen from the input side.