Question

Between${}^{\mathbf{1911}}$and${}^{\mathbf{1990}}$, the top of the leaning bell tower at Pisa, Italy, moved toward the south at an average rate of${}^{\mathbf{1}\mathbf{.}\mathbf{2}\mathbf{}\mathbf{mm}\mathbf{/}\mathbf{y}}$. The tower is${}^{\mathbf{55}\mathbf{m}}$ tall. In radians per second, what is the average angular speed of the tower’s top about its base?

Short answer

The average angular speed of the tower’s top to its base is${}^{6.9\times {10}^{-13\mathrm{rad}/\mathrm{s}}}$..

Step-by-step solution

Understanding the given information

1. The height of the tower${}^{\mathbf{r}}$is${}^{\mathbf{55}\mathbf{m}}$.
2. The rate of leaning of the tower${}^{\mathbf{v}\mathbf{}\mathbf{is}\mathbf{}\mathbf{1}\mathbf{.}\mathbf{2}\mathbf{}\mathbf{mm}\mathbf{/}\mathbf{yr}}\mathbf{}$.

Concept and Formula used for the given question

The top of the leaning bell tower is moving towards southfrom its base. It is like the top of the tower is rotatingon its base and the height of the tower is equivalent to its radius. Hence, we can use the relation between the linear variables and the angular variables to determine the angular speed.

The relation between the linear velocity and the angular velocity is given as,

$\mathbf{\omega }\mathbf{=}\mathbf{r}$

Calculation for the average angular speed of the tower’s top about its base

Convert the rate of leaning of the tower to the units of m/s.

$$\begin{gathered} \mathrm{v}=1.2\mathrm{mm}/\mathrm{yr} \\ =1.2\frac{\mathrm{mm}}{\mathrm{yr}}\times \frac{{10}^{-3}\mathrm{m}}{1\mathrm{mm}}\times \frac{1\mathrm{yr}}{365\times 24\times 3600\mathrm{s}} \\ =3.8\times {10}^{-11}\mathrm{m}/\mathrm{s} \end{gathered}$$

Rearrange equation (i) to write the formula for angular velocity and substitute the values. Therefore,

localid="1654582822603" $$\begin{gathered} \mathrm{\omega }=\frac{\mathrm{v}}{\mathrm{r}} \\ =\frac{3.8\times {10}^{-11}\mathrm{m}/\mathrm{s}}{55\mathrm{m}} \\ =6.9\times {10}^{-13}\mathrm{rad}/\mathrm{s} \\ \mathrm{Thus},\mathrm{the}\mathrm{average}\mathrm{angular}\mathrm{speed}\mathrm{of}\mathrm{the}\mathrm{tower}\text{'}\mathrm{s}\mathrm{and}\mathrm{top}\mathrm{about}\mathrm{it}\mathrm{base}\mathrm{is} \\ 6.9\times {10}^{-13}\mathrm{rad}/\mathrm{s} \end{gathered}$$