Rotation with Constant Angular Acceleration — Question 9

Question 9

Worst-case stopping angle uses the weakest deceleration

The initial positive angular speed is bounded by $9.8\le\omega_0\le10.2\,\mathrm{rad/s}$. Constant braking magnitude is bounded independently by $1.9\le a\le2.1\,\mathrm{rad/s^2}$.

1. Find sharp bounds on stopping angle.

2. Does a $27\,\mathrm{rad}$ allowance guarantee stopping for all permitted values?

Question 9 - Solution

Worst-case stopping angle uses the weakest deceleration

(a) Derivation and calculation

The stopping angle and its monotonicities are $$\theta_s=\frac{\omega_0^2}{2a},\qquad \frac{\partial\theta_s}{\partial\omega_0}=\frac{\omega_0}{a}>0,\quad \frac{\partial\theta_s}{\partial a}=-\frac{\omega_0^2}{2a^2}<0.$$ $$\boxed{\frac{9.8^2}{2(2.1)}\le\theta_s\le\frac{10.2^2}{2(1.9)}} \quad\Rightarrow\quad\boxed{22.867\le\theta_s\le27.379\ \mathrm{rad}}.$$

(b) Result and interpretation

The largest attainable stopping angle exceeds the allowance: $$\theta_{s,\max}-27=27.3789-27=\boxed{0.3789\ \mathrm{rad}>0}.$$ Thus 27 rad does not guarantee stopping for every permitted parameter pair.

Physical check

The extreme endpoint pairs are allowed because the speed and braking bounds are independent.