Question 3.3.6: Conveyor systems are used to produce translation of the load......
Conveyor systems are used to produce translation of the load, as shown in Figure 3.3.5. The reducer is a geared system that reduces the motor speed by a factor of 3:1. The motor inertia is I_{1} = 0.002 kg·m². Disregard the inertias of the reducer and the tachometer, which are used to measure the speed for control purposes. Also discount the inertias of the two sprockets, the chain, and all shafts. The only significant masses and inertias are the inertias of the four drive wheels (0.02 kg·m² each), the two drive chains (8 kg each), and the load mass (10 kg).
The radius of sprocket 1 is 0.05 m, and that of sprocket 2 is 0.1 m. The drive wheel has a radius of 0.1 m. The load friction torque measured at the drive shaft is 1.2 N·m.
a. Derive the equation of motion of the conveyor in terms of the motor velocity, with the motor torque T_{1} as the input.
b. Suppose the motor torque is constant at 1.6 N·m. Determine the resulting motor angular acceleration and load acceleration.
a. The speed ratio of the sprocket drive at the driving shaft is the ratio of the sprocket diameters, which 0.1/0.05 = 2. So the sprocket drive is a speed reducer. Thus, in going from the motor to the drive shaft, there is a total speed reduction of 3(2) = 6. Therefore, when referencing the significant inertias to the motor shaft, we must divide the inertias by 6² = 36. Thus, the equivalent inertia felt at the motor shaft due to the four drive wheels is 4(0.02)/36 = 0.2.
To compute the equivalent inertias of the drive chains and load mass, we first express their kinetic energies in terms of the motor speed \omega_{1}. Let v be the translational velocity of the drive chains and the load mass. Because a drive wheel has a radius of 0.1 m, the angular velocity \omega_{d} of the drive shaft is related to v as 0.1 \omega_{d} = v. We are now ready to express the kinetic energy drive chains and load mass as follows. Each 8 kg drive chain has a kinetic energy of
\mathrm{KE}_{\mathrm{chain}}=\frac{1}{2}8v^{2}=\frac{1}{2}8(0.1\omega_{d})^{2}=\frac{1}{2}(0.08)\omega_{d}^{2}But \omega_{d} is related to the motor speed \omega_{1} as \omega_{d} = \omega_{1}/6. Thus
{\mathrm{KE}}_{\mathrm{chain}}={\frac{1}{2}}\left(0.08\right)\left({\frac{\omega_{1}}{6}}\right)^{2}={\frac{1}{2}}\left({\frac{0.08}{36}}\right)\omega_{1}^{2}={\frac{1}{2}}\left({\frac{0.02}{9}}\right)\omega_{1}^{2}Because the kinetic energy of an inertia I rotating at a speed ω is I ω²/2, we see that the equivalent inertia of the four drive wheels is 4(0.02)/9 = 0.08/9.
We use a similar method to compute the equivalent inertia of the 10 kg load mass.
{\mathrm{KE}}_{\mathrm{load}}={\frac{1}{2}}10v^{2}={\frac{1}{2}}10(0.1\omega_{d})^{2}={\frac{1}{2}}(0.1)\omega_{d}^{2}={\frac{1}{2}}\left({\frac{0.1}{36}}\right)\omega_{1}^{2}Thus the equivalent inertia of the load mass 0.1/36 = 0.025/9.
The total equivalent inertia {\mathit{I}}_{e} felt at the motor shaft is the sum of the equivalent inertias of the motor, the drive wheels, the drive chain, and the load mass:
I_{e}=0.002+{\frac{0.02}{9}}+{\frac{0.08}{9}}+{\frac{0.025}{9}}=0.002+{\frac{0.125}{9}}=0.0159\,\,{\mathrm{kg}}\cdot{\mathrm{m}}^{2}The 1.2 N·m friction torque acts against the motor torque and is reduced 6 because of the net gear ratio of 3(2) = 6. Thus the equation of motion in terms of the motor speed is
0.0159\dot{\omega}_{1}=T_{1}-\frac{1.2}{6}=T_{1}-0.2b. If T_{1} = 1.6, then the motor angular acceleration is
\dot{\omega}_{1}=\frac{1.6-0.2}{0.0159}=88.05\,\mathrm{rad/s}^{2}The acceleration \dot{{v}} of the load is related to {\dot{\omega}}_{1} by the drive wheel radius (0.1) and the net gear ratio (6). Thus \dot{{v}} = 0.1 {\dot{\omega}}_{1} /6 = 1.4675 m/s².