Question 9.1.1: Speed Response of a Rotational System A certain rotational s...
Speed Response of a Rotational System
A certain rotational system has an inertia I = 50 kg · m² and a viscous damping constant c = 10 N · m · s/rad. The torque T (t) is applied by an electric motor (Figure 9.1.6a). From the free body diagram shown in part (b) of the figure, the equation of motion is
50 \frac{d ω}{d t} + 10ω = T (t) = K_{f} i_{f} (1)
The model of the motor’s field current i_{f} in amperes is
0.001 \frac{d i_{f}}{d t} + 5 i_{f} = v(t) (2)
where v(t) is the voltage applied to the motor. The motor torque constant is K_{T} =25 N·m/A.
Suppose the applied voltage is 10 V. Determine the steady-state speed of the inertia and estimate the time required to reach that speed.
From equation (2) we see that the time constant of the motor circuit is 0.001/5 = 2 × 10^{−4} s. Thus the current will reach a steady-state value of 10/5 = 2 A in approximately 4(2 × 10^{−4}) = 8 × 10^{−4} s. The resulting steady-state torque is K_{T} (2) = 25(2) = 50 N · m.
From equation (1) we find the time constant of the rotational system to be 50/10 = 5 s. Since this is much larger than the time constant of the circuit (2 × 10^{−4} s), we conclude that the motor torque may be modeled as a step function. The magnitude of the step function is 50 N · m. The steady-state speed is ω = 50/10 = 5 rad/s, and therefore it will take approximately 4(5) = 20 s
to reach this speed.