Rotation with Constant Angular Acceleration — Question 2

Question 2

Two arrival times correspond to opposite crossing directions

A rotor has $\theta_0=0$, $\omega_0=6\,\mathrm{rad/s}$ and constant $\alpha=-2\,\mathrm{rad/s^2}$.

1. Find every nonnegative time when its unwrapped angle equals $5\,\mathrm{rad}$.

2. Give angular velocity at each crossing and find the largest angle reached.

Question 2 - Solution

Two arrival times correspond to opposite crossing directions

(a) Derivation and calculation

For constant angular acceleration, $$\theta(t)=\theta_0+\omega_0t+\tfrac12\alpha t^2=6t-t^2.$$ $$6t-t^2=5\Rightarrow t^2-6t+5=(t-1)(t-5)=0, \qquad\boxed{t=1,5\ \mathrm s}.$$

(b) Result and interpretation

Evaluate the signed velocity at each root: $$\omega(t)=6-2t,\qquad\boxed{\omega(1)=4,\quad\omega(5)=-4\ \mathrm{rad/s}}.$$ The turning point satisfies $$\omega=0\Rightarrow t=3\ \mathrm s,\qquad\boxed{\theta_{\max}=6(3)-3^2=9\ \mathrm{rad}}.$$

Physical check

The equal-angle crossings occur symmetrically about the maximum at 3 s.