Answer:
[tex]\omega=7.16*10^{-13}\frac{rad}{s}[/tex]
Explanation:
The angular speed is given by:
[tex]\omega=\frac{v}{r}[/tex]
Here v is the linear speed and r is the radius of the circular motion. The height of the tower is equal to the radius of the circular motion of the top of the tower, since is rotating about its base. We need to convert the given linear speed to [tex]\frac{m}{s}[/tex]:
[tex]1.4\frac{mm}{y}*\frac{10^{-3}m}{1mm}*\frac{1y}{3.154*10^7s}=4.44*10^{-11}\frac{m}{s}[/tex]
Now, we calculate the angular speed:
[tex]\omega=\frac{4.44*10^{-11}\frac{m}{s}}{62m}\\\omega=7.16*10^{-13}\frac{rad}{s}[/tex]
