A reel of massless thread unrolls itself falling down under gravity. The acceleration of its fall is 

$\frac{2}{3}g$

Use energy conservation

$\mathrm{mgh}=\frac{1}{2} \mathrm{mv}^2+\frac{1}{2} \mathrm{I} \omega^2=\frac{1}{2} \mathrm{mv}^2+\frac{1}{2} \times \frac{\mathrm{mr}^2}{2} \omega^2$

$\mathrm{mgh}=\frac{1}{2} \mathrm{mv}^2+\frac{1}{4} \mathrm{mv}^2=\frac{3}{4} \mathrm{mv}^2$

$\mathrm{v}^2=\frac{4}{3} \mathrm{gh}=2 \mathrm{ah} ; \mathrm{a}=\frac{2}{3} \mathrm{~g}$