If an object has a moment of inertia \(19\,\,{\rm{kg}} \cdot {{\rm{m}}^{\rm{2}}}\) and the magnitude of its rotational angular momentum is \(36\,\,\,{\rm{kg}} \cdot {{\rm{m}}^{\rm{2}}}{\rm{/s}},\) what is its rotational kinetic energy?

Moment of inertia:\(19\,\,{\rm{kg}} \cdot {{\rm{m}}^{\rm{2}}}\)

Rotational angular momentum: \(36\,\,\,{\rm{kg}} \cdot {{\rm{m}}^{\rm{2}}}{\rm{/s}}\)

The rotating analogue of linear momentum is angular momentum (rarely referred to as moment of momentum or rotational momentum). It's a crucial quantity in physics since it's a conserved quantity, meaning the total angular momentum of a closed system remains constant.

In terms of moment of inertia, the rotational angular momentum of an object rotating with angular speed is as follows:

\({L_{rot}} = I\omega .......(1)\)

Here,\(I\)Is the moment of inertia of the object.

The rotational kinetic energy of an object rotating with angular speed \(\left( \omega \right)\) is,

\({K_{rot}} = \frac{1}{2}I{\omega ^2}.......(2)\)

The rotational kinetic energy of the object in terms of moment of inertia \(\left( I \right)\) and rotational angular momentum \(\left( {{L_{rot}}} \right)\) can be written as follows,

\(\begin{aligned}{c}{K_{rot}} = \frac{1}{2}I{\omega ^2}\\ = \frac{1}{2}\frac{{{{\left( {I\omega } \right)}^2}}}{I}\\ = \frac{1}{2}\frac{{L_{rot}^2}}{I}\end{aligned}\)

Therefore,

\({K_{rot}} = \frac{{L_{rot}^2}}{{2I}}......(3)\)

Substitute \(19\,\,{\rm{kg}} \cdot {{\rm{m}}^{\rm{2}}}\) for \(I\) and \(36\,\,{\rm{kg}} \cdot {\rm{m}}{}^{\rm{2}}{\rm{/s}}\) for in the equation (3),

\(\begin{aligned}{}{K_{rot}} &= \frac{{L_{rot}^2}}{{2I}}\\ &= \frac{{{\rm{36}}\,\,{\rm{kg}} \cdot {{\rm{m}}^{\rm{2}}}{\rm{/s}}}}{{{\rm{2}}\left( {{\rm{19kg}} \cdot {\rm{m}}{}^{\rm{2}}} \right)}}\\ &= 0.947{\rm{J}}\end{aligned}\)

Therefore, the rotational energy is \({\rm{0}}{\rm{.947J}}\).