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

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.