Question

Between 1911 and 1990, the top of the leaning bell tower at Pisa, Italy, moved toward the south at an average rate of 1.2 mm/y.The tower is 55 m tall. In radians per second, what is the average angular speed of the tower’s top about its base

Answer 1

Answer:

Angular speed, $\omega=6.90\times 10^{-13}\ rad/s$

Explanation:

It is given that,    

The top of the leaning bell tower at Pisa, Italy, moved toward the south at an average rate of, v = 1.2 mm/yr

$1\ mm/yr=3.171\times 10^{-11}\ m/s$

Velocity, $v=3.80\times 10^{-11}\ m/s$

Height of the tower, h = 55 m

The height of the tower is equivalent to the radius. Let $\omega$ is the angular speed of the tower’s top about its base. The relation between the angular speed and the angular speed is given by :

$v=r\omega$

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

$\omega=\dfrac{3.80\times 10^{-11}\ m/s}{55\ m}$

$\omega=6.90\times 10^{-13}\ rad/s$

So, the average angular speed of the tower’s top about its base is $6.90\times 10^{-13}\ rad/s$. Hence, this is the required solution.