Question

The top of a tower much like the leaning bell tower at Pisa, Italy, moves toward the south at an average rate of 1.4 mm/y. The tower is 62 m tall. In radians per second, what is the average angular speed of the tower's top about its base

Answer 1

Answer:

$\omega=7.16*10^{-13}\frac{rad}{s}$

Explanation:

The angular speed is given by:

$\omega=\frac{v}{r}$

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 $\frac{m}{s}$:

$1.4\frac{mm}{y}*\frac{10^{-3}m}{1mm}*\frac{1y}{3.154*10^7s}=4.44*10^{-11}\frac{m}{s}$

Now, we calculate the angular speed:

$\omega=\frac{4.44*10^{-11}\frac{m}{s}}{62m}\\\omega=7.16*10^{-13}\frac{rad}{s}$