Question 5.16: A Single-Degree-of-Freedom Gear–Train System For the gear–tr...

The free-body diagram is shown in Figure 5.95b. The geometric constraint is

\frac{\theta_{2}}{\theta_{1}}=\frac{r_{1}}{r_{2}} .

Applying the moment equation to each gear gives

\begin{gathered} +\curvearrowright: \sum M_{\mathrm{C} 1}=I_{\mathrm{C} 1} \alpha, \\ \tau_{1}-r_{1} F=I_{\mathrm{C} 1} \ddot{\theta}_{1}, \end{gathered}

and

\begin{gathered} +\curvearrowleft: \sum M_{\mathrm{C} 2}=I_{\mathrm{C} 2} \alpha, \\ r_{2} F=I_{\mathrm{C} 2} \ddot{\theta}_{2} .\end{gathered}

Combining the two equations and eliminating F yields

\tau_{1}-\frac{r_{1}}{r_{2}} I_{\mathrm{C} 2} \ddot{\theta}_{2}=I_{\mathrm{C} 1} \ddot{\theta}_{1}

Note that the angular displacements \theta_{1} and \theta_{2} are dependent through the geometric constraint. Thus, the gear-train in Figure 5.95 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 \theta_{1}. Introducing the geometric constraint results in the equation of motion

\tau_{1}-\frac{r_{1}}{r_{2}} I_{\mathrm{C} 2}\left(\frac{r_{1}}{r_{2}} \ddot{\theta}_{1}\right)=I_{\mathrm{C} 1} \ddot{\theta}_{1^{\prime}}

simplified to

\left(I_{\mathrm{C} 1}+\frac{r_{1}^{2}}{r_{2}^{2}} I_{\mathrm{C} 2}\right) \ddot{\theta}_{1}=\tau_{1^{\prime}}

which is the dynamics seen from the input side.