Time of reversal of a drive

Initial steady state. Motor torque equals load torque:

$$100 - 0.1N = 0.05N \;\Rightarrow\; N_0 = \frac{100}{0.15} = 666.67\ \text{rpm}$$

After the change. Motor characteristic $T = -100 - 0.1N$ (valid for positive and negative speed). The passive load torque $0.05N$ always opposes motion, so the same expression holds for negative speed. New steady state:

$$-100 - 0.1N = 0.05N \;\Rightarrow\; N_f = -666.67\ \text{rpm}$$

Dynamic equation ($\omega = \frac{2\pi}{60}N$):

$$J\frac{2\pi}{60}\frac{dN}{dt} = T - T_l = -100 - 0.15N$$ $$\frac{dN}{dt} = -\frac{0.15}{J(2\pi/60)}\left(N + 666.67\right)$$

Time constant:

$$\tau = \frac{J(2\pi/60)}{0.15} = \frac{10\times0.10472}{0.15} = 6.981\ \text{s}$$

Solution:

$$N(t) = N_f + (N_0 - N_f)e^{-t/\tau} = -666.67 + 1333.33\,e^{-t/6.981}$$

Since the speed approaches $-666.67$ rpm only asymptotically, reversal is taken as complete when the speed reaches 95 % of the final value, i.e. $N = -633.33$ rpm (usual textbook assumption, Dubey).

$$\begin{aligned} t_r &= \tau\ln\frac{N_0 - N_f}{N - N_f} = 6.981\ln\frac{666.67+666.67}{-633.33+666.67} \\ &= 6.981\ln\frac{1333.33}{33.33} = 6.981\ln 40 = 25.75\ \text{s} \end{aligned}$$

(Braking to zero speed alone takes $6.981\ln 2 = 4.84$ s.)

Answer: reversal time $\approx 25.75$ s (to 95 % of the reverse steady-state speed of $-666.67$ rpm).