A diver makes ${}^{\mathbf{2}\mathbf{.}\mathbf{5}}$revolutions on the way from a${}^{\mathbf{10}\mathbf{}\mathbf{m}}$high platform to the water. Assuming zero initial vertical velocity, find the average angular velocity during the dive.

The revolutions make by diver,${}^{\mathrm{\theta }=2.5\mathrm{rev}}$

The vertical height of platform,${}^{\mathrm{y}=10\mathrm{m}}$

The initial vertical velocity of the diver,${}^{{\mathrm{v}}_0=0\mathrm{m}/\mathrm{s}}$.

Average angular velocity is the ratio of total angular displacement to the total time interval. It is a vector quantity which has magnitude and direction as well.

The expression for angular velocity is given as:

${\mathrm{\omega }}_{\mathrm{avg}}=\frac{∆\mathrm{\theta }}{∆\mathrm{t}}$ … (i)

Here,${}^{∆\mathrm{\theta }}$is the angular displacement and${}^{∆\mathrm{t}}$is the time interval.

The expression for the second equation of motion is given as:

${}^{\mathrm{y}={\mathrm{v}}_0∆\mathrm{t}+\frac{1}{2}\mathrm{g}∆{\mathrm{t}}^2}$ … (ii)

Here,${}^{{\mathrm{v}}_0}$is the initial linear velocity and${}^g$ is the acceleration due to gravity.

Using equation (ii), the time taken to fall into the water is calculated as:

$$\begin{gathered} \mathrm{y}=\left(0\right)∆\mathrm{t}+\frac{1}{2}\mathrm{g}∆{\mathrm{t}}^2 \\ ∆\mathrm{t}=\sqrt{\frac{2\mathrm{y}}{\mathrm{g}}} \\ =\sqrt{\frac{2\times 10\mathrm{m}}{9.8\mathrm{m}/{\mathrm{s}}^2}} \\ =1.428\mathrm{s} \end{gathered}$$

Thus, the time it takes to fall on the florr is${}^{0.40\mathrm{s}}$.

Convert the angular displacement in radians.

$$\begin{gathered} \mathrm{\theta }=2.5\mathrm{rev}\times \frac{2\mathrm{\pi }}{1\mathrm{rev}} \\ =5\mathrm{\pi rad} \end{gathered}$$

Using equation (i), the average angular velocity during the dive is calculated as:

$$\begin{gathered} {\mathrm{\omega }}_{\mathrm{a}\mathrm{vg}}=\frac{5\mathrm{\pi rad}-0\mathrm{rad}}{1.428\mathrm{s}} \\ \approx 11\mathrm{rad}/\mathrm{s} \end{gathered}$$

Thus, the average angular velocity is${}^{11\mathrm{rad}/\mathrm{s}}$.