A Single-Link Robot Arm The mechanical model of a single-link robot arm driven by a motor can be represented as a gear–train 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

Question

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 [latex]I_{\mathrm{m}}[/latex] and [latex]I[/latex], respectively. The coefficients of torsional viscous damping of the motor and the load are [latex]B_{\mathrm{m}}[/latex] and [latex]B[/latex], respectively. [latex] au_{\mathrm{m}}[/latex] is the torque generated by the motor. Assume that the gear ratio is [latex]N=r_{1} / r_{2}[/latex]. Derive the differential equation of motion in terms of the motor variable [latex] heta_{m}[/latex].

Accepted Answer

The free-body diagrams for the motor, the load, and the gear-train are shown in Figure 5.97, where [latex]F[/latex] represents the contact force between the two gears. The moments caused by the contact force on the motor and on the load are [latex]r_{1} F[/latex] and [latex]r_{2} F[/latex], respectively. Applying the moment equation to the motor and the load gives

[latex] \begin{gathered} +\curvearrowleft: \sum M_{\mathrm{O}}=I_{\mathrm{O}} \alpha, \ au_{\mathrm{m}}-B_{\mathrm{m}} \dot{ heta}_{\mathrm{m}}-r_{1} F=I_{\mathrm{m}} \ddot{ heta}_{\mathrm{m}}, \end{gathered} [/latex]

and

[latex] \begin{gathered} +\curvearrowright: \sum M_{\mathrm{O}}=l_{\mathrm{O}} \alpha, \ -B \dot{ heta}+r_{2} F=\ddot{ heta} . \end{gathered} [/latex]

Solving for [latex]F[/latex] from the equation for the load and substituting it into the equation for the motor results in

[latex] au_{\mathrm{m}}-B_{\mathrm{m}} \dot{ heta}_{\mathrm{m}}-\frac{r_{1}}{r_{2}}(\ddot{ heta}+B \dot{ heta})=I_{\mathrm{m}} \ddot{ heta}_{\mathrm{m}} [/latex]

By the geometry of the gears,

[latex] \begin{aligned} & \dot{ heta}=\frac{r_{1}}{r_{2}} \dot{ heta}_{m}=N \dot{ heta}_{m}, \ & \ddot{ heta}=\frac{r_{1}}{r_{2}} \ddot{ heta}_{m}=N \ddot{ heta}_{m}, \end{aligned} [/latex]

where [latex]N[/latex] is the gear ratio.

Substituting these into the previous differential equation gives

[latex] au_{\mathrm{m}}-B_{\mathrm{m}} \dot{ heta}_{\mathrm{m}}-N^{2}\left(\ddot{ heta}_{\mathrm{m}}+B \dot{ heta}_{\mathrm{m}} ight)=I_{\mathrm{m}} \ddot{ heta}_{\mathrm{m}} [/latex]

which can be rearranged as

[latex] \left(I_{\mathrm{m}}+N^{2} I ight) \ddot{ heta}_{\mathrm{m}}+\left(B_{\mathrm{m}}+N^{2} B ight) \dot{ heta}_{\mathrm{m}}= au_{\mathrm{m}} [/latex]

The equation is in terms of the angular displacement of the motor.

Question 5.17: A Single-Link Robot Arm The mechanical model of a single-lin...

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