Rotation with Constant Angular Acceleration — Question 1

Question 1

A stopping-angle specification determines the brake rate

A rotor initially turns at $12\,\mathrm{rad/s}$. It must stop after exactly $18\,\mathrm{rad}$ under constant angular acceleration.

1. Find the acceleration and stopping time.

2. If the same signed acceleration continues for two more seconds, find the additional signed displacement.

Question 1 - Solution

A stopping-angle specification determines the brake rate

(a) Derivation and calculation

Eliminate time during the stopping phase: $$\omega_f^2=\omega_0^2+2\alpha\Delta\theta, \qquad 0=144+36\alpha\Rightarrow\boxed{\alpha=-4\ \mathrm{rad/s^2}}.$$ $$t_s=\frac{0-12}{-4}=\boxed{3\ \mathrm s}.$$

(b) Result and interpretation

Start a new time interval at the stopping instant. If the same signed acceleration persists, $$\Delta\theta_{\mathrm{extra}}=0(2)+\tfrac12(-4)(2)^2=\boxed{-8\ \mathrm{rad}},\qquad \omega_f=(-4)(2)=-8\ \mathrm{rad/s}.$$ This model reverses the motion; a brake that holds the rotor at rest would instead change its torque after stopping.

Physical check

The stopping area is $\tfrac12(3)(12)=18$ rad, as required.