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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

User Tybalt
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1 Answer

3 votes

Answer:

Angular speed,
\omega=6.90* 10^(-13)\ rad/s

Step-by-step 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* 10^(-11)\ m/s

Velocity,
v=3.80* 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=(v)/(r)


\omega=(3.80* 10^(-11)\ m/s)/(55\ m)


\omega=6.90* 10^(-13)\ rad/s

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

User Tom Prats
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