Question 5.17: A Single-Link Robot Arm The mechanical model of a single-lin...
A Single-Link Robot Arm
The mechanical model of a single-link robot arm driven by a motor can be represented as a geartrain system, as shown in Figure 5.96, in which two rotational subsystems are coupled with a pair of gears with negligible inertia. The mass moments of inertia of the motor and the load are I_{\mathrm{m}} and I, respectively. The coefficients of torsional viscous damping of the motor and the load are B_{\mathrm{m}} and B, respectively. \tau_{\mathrm{m}} is the torque generated by the motor. Assume that the gear ratio is N=r_{1} / r_{2}. Derive the differential equation of motion in terms of the motor variable \theta_{m}.
The free-body diagrams for the motor, the load, and the gear-train are shown in Figure 5.97, where F represents the contact force between the two gears. The moments caused by the contact force on the motor and on the load are r_{1} F and r_{2} F, respectively. Applying the moment equation to the motor and the load gives
\begin{gathered} +\curvearrowleft: \sum M_{\mathrm{O}}=I_{\mathrm{O}} \alpha, \\ \tau_{\mathrm{m}}-B_{\mathrm{m}} \dot{\theta}_{\mathrm{m}}-r_{1} F=I_{\mathrm{m}} \ddot{\theta}_{\mathrm{m}}, \end{gathered}
and
\begin{gathered} +\curvearrowright: \sum M_{\mathrm{O}}=l_{\mathrm{O}} \alpha, \\ -B \dot{\theta}+r_{2} F=\ddot{\theta} . \end{gathered}
Solving for F from the equation for the load and substituting it into the equation for the motor results in
\tau_{\mathrm{m}}-B_{\mathrm{m}} \dot{\theta}_{\mathrm{m}}-\frac{r_{1}}{r_{2}}(\ddot{\theta}+B \dot{\theta})=I_{\mathrm{m}} \ddot{\theta}_{\mathrm{m}}
By the geometry of the gears,
\begin{aligned} & \dot{\theta}=\frac{r_{1}}{r_{2}} \dot{\theta}_{m}=N \dot{\theta}_{m}, \\ & \ddot{\theta}=\frac{r_{1}}{r_{2}} \ddot{\theta}_{m}=N \ddot{\theta}_{m}, \end{aligned}
where N is the gear ratio.
Substituting these into the previous differential equation gives
\tau_{\mathrm{m}}-B_{\mathrm{m}} \dot{\theta}_{\mathrm{m}}-N^{2}\left(\ddot{\theta}_{\mathrm{m}}+B \dot{\theta}_{\mathrm{m}}\right)=I_{\mathrm{m}} \ddot{\theta}_{\mathrm{m}}
which can be rearranged as
\left(I_{\mathrm{m}}+N^{2} I\right) \ddot{\theta}_{\mathrm{m}}+\left(B_{\mathrm{m}}+N^{2} B\right) \dot{\theta}_{\mathrm{m}}=\tau_{\mathrm{m}}
The equation is in terms of the angular displacement of the motor.
