Question

A diver makes $4\text{revolutions}$ on the way from a $4.9\text{m}$ platform above the water surface. Assuming zero initial vertical velocity, calculate the average angular velocity of the diver. [Take $g=9.8m/s^2$]

• A. $8\pi \text{rad/s}$
• B. $4\pi \text{rad/s}$
• C. $2\pi \text{rad/s}$
• D. $\pi \text{rad/s}$

Answer 1

The correct option is A $8\pi \text{rad/s}$

We know that the average angular velocity is given by
$\overline{\omega }=\frac{\Delta \theta }{\Delta t}$
Since the diver makes four revolutions, the angular displacement of the diver is $\Delta \theta =4\times 2\pi =8\pi \text{rad}$

Now, as the dimension of the diver is not mentioned here, we can consider it as a small particle falling freely under gravity.
Time taken to fall is
$\Delta t=\sqrt{\frac{2h}{g}}=\sqrt{\frac{2\times 4.9}{9.8}}=1\text{sec}$
Thus, average angular velocity is
$\overline{\omega }=\frac{\Delta \theta }{\Delta t}=\frac{8\pi }{1}=8\pi \text{rad/s}$