Conservation Of Angular Momentum

Because no external torque acts about the vertical spin axis, the skater's angular momentum $L=I\omega$ is conserved throughout the manoeuvre, even though her kinetic energy changes. Conservation gives $I_1{\omega }_1=I_2{\omega }_2$. Solving for the final angular velocity, ${\omega }_2=\frac{I_1{\omega }_1}{I_2}=\frac{(4)(3)}{1.5}=\frac{12}{1.5}=8\text{ rad/s}$. The value 1.125 rad/s inverts the ratio of the moments of inertia and would correspond to slowing down, which contradicts the physics. The value 6 rad/s wrongly assumes the spin rate merely doubles regardless of the inertia values. The value 12 rad/s mistakenly keeps only the numerator $I_1{\omega }_1$ without dividing by $I_2$. This is the classic NCERT illustration of angular momentum conservation, the rotational counterpart of linear momentum conservation. As a plausibility check, pulling the arms in lowers the moment of inertia, so the spin rate must increase, and $8\text{ rad/s}>3\text{ rad/s}$ confirms the expected speed-up. Note that although angular momentum stays fixed, the rotational kinetic energy $\frac{1}{2}I{\omega }^2$ actually rises, because the skater performs internal muscular work pulling her arms inward against their outward tendency, and that work is precisely what feeds the increase in spin energy.