91Ó°ÊÓ

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

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,

$\mathbf{Ó\neg }\mathbf{=}\mathbf{r}$

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

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

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

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