如图，睿睿同学用圆规$BOA$画一个半径为$8cm$的圆，测得此时$\angle O=90^{\circ}$，为了画一个半径更大的同心圆，固定$A$端不动，将$B$端向左移至${B'}$处，此时测得$\angle {O'}=120^{\circ}$，则$BB'$的长为______.

作${O'}C\bot BA$于点$C$.如图：

根据已知可得：$BA=8cm.OB=OA$.

$\because \angle O=90^{\circ}$.

$\therefore OB=\frac{AB}{\sqrt{2}}=4\sqrt{2}cm$.

$\because \angle {O'}=120^{\circ}$，${O'}B={O'}A$.

$\therefore \angle {O'}{B'}C=30^{\circ}$.

$\because O′B′=OB=4\sqrt{2}cm$.

$\therefore {B'}C=\cos \angle {O'}{B'}C\cdot {O'}{B'}=2\sqrt{6}cm$.

$\therefore AB′=4\sqrt{6}cm$.

$\therefore B′B=AB′-BA=(4\sqrt{6}-8)cm$.

故答案为：（$4\sqrt{6}-8)cm$.