Answer
$\alpha=-6.14\times 10^{-22}\frac{rad}{s^2}$
Work Step by Step
We know that
$\omega-\omega_{\circ}=\theta(\frac{-\Delta T}{T^2})$
We plug in the known values to obtain:
$\omega-\omega_{\circ}=(365rev\times \frac{2\pi \space rad}{rev})(\frac{-0.840}{(365d\times 24\frac{h}{d}\times 3600\frac{s}{h})^2})=-1.94\times 10^{-12}\frac{rad}{s}$
Now $\alpha_{avg}=\frac{\Delta \omega}{\Delta t}$
We plug in the known values to obtain:
$\alpha=\frac{-1.94\times 10^{-12}}{100y\times 3.16\times 10^7\frac{s}{y}}=-6.14\times 10^{-22}\frac{rad}{s^2}$
