Question

Short answer

The angular speed of the diver in the straight position is \(0.38\;{\rm{rev/s}}\).

Step-by-step solution

Given data

The factor by which the moment of inertia reduces is \(\frac{{I'}}{I} = 3.5\).

The number of rotations is \(\omega = 2\;{\rm{rev}}\).

The time is \(t = 1.5\;{\rm{s}}\).

Understanding conservation of angular momentum

In this problem, the external force acting on the diver is the force due to gravity passing through the center of the mass. No other type of force takes place. So, the angular momentum is conserved.

Determine the angular speed of the diver

he relation of angular momentum can be written as:

\(\begin{aligned}{c}L &= L'\\I\omega &= I'\omega '\\\omega ' &= \left( {\frac{I}{{I'}}} \right)\omega \end{aligned}\)

Here, \(L\) and \(L'\) are the initial and final angular momentum, and \(\omega '\) is the final angular velocity of the diver.

On plugging the values in the above relation, you get:

\(\begin{aligned}{l}\omega ' &= \left( {\frac{1}{{3.5}}} \right)\left( {\frac{{2\;{\rm{rev}}}}{{1.5\;{\rm{s}}}}} \right)\\\omega ' &= 0.38\;{\rm{rev/s}}\end{aligned}\)

Thus, \(\omega ' = 0.38\;{\rm{rev/s}}\) is the final angular speed of the diver.