Rotation with Constant Angular Acceleration — Question 4

Question 4

Accelerate then brake to a specified orientation

A rotor starts and finishes at rest while advancing by $24\,\mathrm{rad}$. It accelerates at $+3\,\mathrm{rad/s^2}$, then switches once to $-6\,\mathrm{rad/s^2}$, without coasting.

1. Find the peak angular speed and both phase durations.

2. Find the angle at which braking begins.

Question 4 - Solution

Accelerate then brake to a specified orientation

(a) Derivation and calculation

Let $w$ be the positive switching speed. The acceleration and braking distances satisfy $$\theta_a=\frac{w^2}{2(3)},\qquad\theta_b=\frac{w^2}{2(6)},\qquad 24=\frac{w^2}{6}+\frac{w^2}{12}=\frac{w^2}{4}.$$ $$\boxed{w=4\sqrt6=9.798\ \mathrm{rad/s}},\quad \boxed{t_a=\frac{4\sqrt6}{3}=3.266\ \mathrm s},\quad \boxed{t_b=\frac{2\sqrt6}{3}=1.633\ \mathrm s}.$$

(b) Result and interpretation

The switching angle measured from the initial orientation is $$\boxed{\theta_a=\frac{96}{6}=16\ \mathrm{rad}},\qquad \theta_b=\frac{96}{12}=8\ \mathrm{rad}.$$ The braking phase has half the duration and half the angle because its acceleration magnitude is twice as large.

Physical check

The two triangular areas add to 24 rad and end with zero speed.