91Ó°ÊÓ

The revolution of the skater is\(\omega = 1\;{\rm{rev}}\).

The final rate is\(\omega ' = 2.5\;{\rm{rev/s}}\).

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

The moment of inertia is \(I = 4.6\;{\rm{kg}} \cdot {{\rm{m}}^2}\).

In this problem, the skater’s angular momentum will be constant because the torque applied on the skater is equivalent to zero.

The relation of angular momentum can be written as:

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

Here, \(L\) and \(L'\) are the initial and final angular momentum, and \(I'\) is the final moment of inertia of the diver.

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

\(\begin{aligned}{l}I' &= \left( {4.6\;{\rm{kg}} \cdot {{\rm{m}}^2}} \right)\left( {\frac{{\left( {\frac{{1\,\;{\rm{rev}}}}{{1.5\,{\rm{s}}}}} \right)}}{{\,2.5\;{\rm{rev/s}}}}} \right)\\I' &= 1.22\;{\rm{kg}} \cdot {{\rm{m}}^2}\end{aligned}\)

Thus, \(I' = 1.22\;{\rm{kg}} \cdot {{\rm{m}}^2}\) is the final moment of inertia. Therefore, she will accomplish this by starting with her arms extended and then pulling them toward the center of her body.