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Dmitry [639]
3 years ago
13

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.T

he tower is 55 m tall. In radians per second, what is the average angular speed of the tower’s top about its base
Physics
1 answer:
Anit [1.1K]3 years ago
6 0

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.

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