Rotation with Constant Angular Acceleration — Question 10

Question 10

A shortest-time turn with acceleration and speed limits

A rotor must advance by $20\,\mathrm{rad}$ from rest to rest without reversal. Its controls allow $|\alpha|\le2\,\mathrm{rad/s^2}$ and $0\le\omega\le4\,\mathrm{rad/s}$.

1. Find the minimum possible time and a motion that attains it.

2. Find the angle below which the optimal motion has no constant-speed phase.

Question 10 - Solution

A shortest-time turn with acceleration and speed limits

(a) Derivation and calculation

The fastest admissible ramps reach the speed cap in $$t_a=t_b=\frac{\omega_{\max}}{\alpha_{\max}}=\frac42=2\ \mathrm s,\qquad \theta_a=\theta_b=\frac{4^2}{2(2)}=4\ \mathrm{rad}.$$ $$\theta_c=20-4-4=12\ \mathrm{rad},\quad t_c=\frac{12}{4}=3\ \mathrm s, \qquad\boxed{T_{\min}=2+3+2=7\ \mathrm s}.$$ For any duration $T$, the speed cannot exceed $\min(2t,4,2(T-t))$; the constructed profile attains this upper envelope and hence maximizes angular travel.

(b) Result and interpretation

The two full ramps alone require $$\boxed{\theta_* =2\frac{\omega_{\max}^2}{2\alpha_{\max}} =\frac{16}{2}=8\ \mathrm{rad}}.$$ For $\theta<8$ rad, a triangular profile has peak speed $\sqrt{\alpha_{\max}\theta}<4$ and duration $2\sqrt{\theta/\alpha_{\max}}$.

Physical check

The complete profile has area $4+12+4=20$ rad and respects both limits throughout.