Rotation with Constant Angular Acceleration — Question 6

Question 6

A speed-only formula loses the reversal sign

A rotor starts at $\omega_0=-5\,\mathrm{rad/s}$ with constant $\alpha=2\,\mathrm{rad/s^2}$. It later has net angular displacement $-4\,\mathrm{rad}$ from its starting orientation.

1. Find the two possible angular velocities and their corresponding times.

2. Explain why $\omega^2=\omega_0^2+2\alpha\Delta\theta$ cannot choose between them.

Question 6 - Solution

A speed-only formula loses the reversal sign

(a) Derivation and calculation

The time-eliminated relation gives both signs: $$\omega^2=(-5)^2+2(2)(-4)=9\Rightarrow\boxed{\omega=\pm3\ \mathrm{rad/s}}.$$ $$t=\frac{\omega-\omega_0}{\alpha}=\frac{\omega+5}{2} \Rightarrow\boxed{(t,\omega)=(1,-3),\ (4,+3)}\quad\text{in SI units}.$$

(b) Result and interpretation

Check the two branches against the signed angle history: $$\theta(t)-\theta_0=-5t+t^2,\qquad\theta(1)-\theta_0=\theta(4)-\theta_0=-4.$$ $$\omega=0\Rightarrow t_r=\frac52\ \mathrm s.$$ Squaring removes direction; the time equation identifies the crossing before and after reversal.

Physical check

The minimum displacement is $-6.25$ rad, below $-4$ rad, allowing two crossings.