Question

A diver makes 2.5 revolutions on the way from a 9.3-m-high platform to the water. Assuming zero initial vertical velocity, find the diver's average angular velocity during a dive.

Answer 1

Answer:

11.4 rad/s

Explanation:

The motion of the diver is a free-fall motion, so its center of mass falls down with constant acceleration of

$g=9.8 m/s^2$ towards the water

Therefore, we can use the following suvat equation:

$s=ut-\frac{1}{2}gt^2$

where:

s = 9.3 m is the vertical displacement of the diver

u = 0 is the initial vertical velocity

$g=9.8 m/s^2$

And t is the total time of flight. Solving for t,

$t=\sqrt{\frac{2s}{g}}=\sqrt{\frac{2(9.3)}{9.8}}=1.38 s$

So, the diver takes 1.38 s to reach the water.

During this time, the diver makes 2.5 revolutions; since 1 revolution is equal to an angle of $2\pi$ radians, then the total angular displacement is

$\theta=2.5\cdot 2\pi =15.7 rad$

Therefore, the average angular velocity of the diver is the ratio between the total angular displacement and the time taken:

$\omega=\frac{\theta}{t}=\frac{15.7}{1.38}=11.4 rad/s$