Conservation Of Angular Momentum

Question

A figure skater spinning at 3 rad/s about a vertical axis pulls her outstretched arms inward, reducing her moment of inertia from 4 kg m^2 to 1.5 kg m^2. What is her resulting angular velocity?

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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, $\omega_2 = \frac{I_1 \omega_1}{I_2} = \frac{(4)(3)}{1.5} = \frac{12}{1.5} = 8	ext{ rad/s}$. 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 $8	ext{ rad/s} > 3	ext{ rad/s}$ 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.

Conservation Of Angular Momentum

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