Question 3.5.5: A single link of a robot arm is shown in Figure 3.5.5. The a......
A single link of a robot arm is shown in Figure 3.5.5. The arm mass is m and its center of mass is located a distance L from the joint, which is driven by a motor torque T_{m} through a pair of spur gears. We model the arm as a pendulum with a concentrated mass m. Thus we take the arm’s moment of inertia I_{G} to be zero. The values of m and L depend on the payload being carried in the hand and thus can be different for each application. The gear ratio is N = 2 (the motor shaft has the greater speed). The motor and gear rotation axes are fixed by bearings.
To control the motion of the arm we need to have its equation of motion. (a) Obtain this equation in terms of the angle θ. The given values for the motor, shaft, and gear inertias are
I_{m}=0.05{\mathrm{~kg}}\cdot{\mathrm{m}}^{2}\qquad I_{G_{1}}=0.025{\mathrm{~kg}}\cdot{\mathrm{m}}^{2}\qquad I_{S_{1}}=0.01{\mathrm{~kg}}\cdot{\mathrm{m}}^{2}I_{G_{2}}=0.1\ k g\cdot{m}^{2}\qquad I_{S_{2}}=0.02\ k g\cdot{m}^{2}
b. Solve this equation for the case where T_{m}=0.5 N·m, m = 10 kg, and L = 0.3 m. Assume that the system starts from rest at θ = 0 and that the angle θ remains small.
Our approach is to model the system as a single inertia rotating about the motor shaft with a speed \omega_{1}. To find the equivalent inertia about this shaft we first obtain the expression for the kinetic energy of the total system and express it in terms of the shaft speed \omega_{1}. Note that the mass m is translating with a speed L\omega_{2}.
\mathrm{KE}={\frac{1}{2}}{\big(}I_{m}+I_{S_{1}}+I_{G_{1}}{\big)}\omega_{1}^{2}+{\frac{1}{2}}{\big(}I_{S_{2}}+I_{G_{2}}{\big)}\omega_{2}^{2}+{\frac{1}{2}}m{\big(}L\omega_{2}{\big)}^{2}But \omega_{2}=\omega_{1}/N=\omega_{1}/2. Thus,
\mathrm{KE}={\frac{1}{2}}\left[I_{m}+I_{S_{1}}+I_{G_{1}}+{\frac{1}{2^{2}}}\bigl(I_{S_{2}}+I_{G_{2}}+m L^{2}\bigr)\right]\omega_{1}^{2}Therefore, the equivalent inertia referenced to the motor shaft is
I_{e}=I_{m}+I_{S_{1}}+I_{G_{1}}+\frac{1}{2^{2}}\left(I_{S_{2}}+I_{G_{2}}+m L^{2}\right)=0.115+0.25m L^{2}The equation of motion for this equivalent inertia can be obtained in the same way as that of a pendulum, by noting that the gravity moment mg L sin θ, which acts on shaft 2, is also felt on the motor shaft, but reduced by a factor of N due to the gear pair. Thus,
{ I}_{e}\dot{\omega}_{1}=T_{m}-\frac{1}{N}m g{ L}\;\mathrm{sin}\;\thetaBut \omega_{1}=N\omega_{2}=N\dot{\theta}. Thus
{I}_{e}N\ddot{\theta}=T_{m}-\frac{1}{N}m g{ L}\sin\thetaSubstituting the given values, we have
2(0.115+0.25m L^{2})\ddot{\theta}=T_{m}-\frac{9.8}2m L\sin\thetaor
(0.23+0.5m L^{2})\ddot{\theta}=T_{m}-4.9m L\sin\theta (1)
b. If |θ| is small, then sin θ ≈ θ and the equation of motion becomes
(0.23+0.5m L^{2})\ddot{\theta}=T_{m}-4.9m L\theta (2)
For the given values, the equation is
0.68\ddot{\theta}=0.5-14.7\thetaThe characteristic equation is 0.68s² + 14.7 = 0 and the roots are s = ± 4.65j. Following the method of Example 3.1.4, we obtain the following solution:
\theta(t)={\frac{0.5}{14.7}}(1-\cos4.65t)=0.034(1-\cos4.65t) (3)
Thus, the model predicts that the arm oscillates with a frequency of 4.65 rad/s about θ = 0.034 with an amplitude of 0.034 rad. To obtain the linear model (2), we used the approximation sin θ ≈ θ. Therefore, before we accept the solution (3), we should check the validity of this assumption. The maximum value that θ will reach is twice the amplitude, or 2(0.034) = 0.068 rad. Therefore, because sin 0.068 = 0.068 to three decimal places, our assumption that sin θ ≈ θ is justified.