Answer
In case 1, $~~\alpha \gt 0$
In case 2, $~~\alpha \lt 0$
In case 3, $~~\alpha \gt 0$
Work Step by Step
$\alpha = \frac{d\omega}{dt} = \frac{d^2\theta}{dt^2}$
If the $\theta$ versus $t$ curve is concave up, then $\alpha \gt 0$
If the $\theta$ versus $t$ curve is concave down, then $\alpha \lt 0$
In case 1, we can see that the $\theta$ versus $t$ curve is concave up.
Thus, in case 1, $~~\alpha \gt 0$
In case 2, we can see that the $\theta$ versus $t$ curve is concave down.
Thus, in case 2, $~~\alpha \lt 0$
In case 3, we can see that the $\theta$ versus $t$ curve is concave up.
Thus, in case 3, $~~\alpha \gt 0$
