Answer
$\omega=6.92\times10^{-13}rad/s$
Work Step by Step
First convert the velocity of the top of the tower to meters per second using dimensional analysis:
$1.2mm/y\times\frac{1m}{1000mm}\times\frac{1y}{365days}\times\frac{1day}{24hours}\times\frac{1hour}{60min}\times\frac{1min}{60s}$ $=3.81\times10^{-11}m/s$
Determine the angular velocity of the tower by dividing the velocity of the top of the tower by the height of the tower.
$\omega=\frac{v}{r}=\frac{3.81\times10^{-11}m/s}{55m}=6.92\times10^{-13}rad/s$
