Problem — New angular velocity of a figure skater

Problem

A figure skater spins about herself with $\omega = 1{,}0\;\mathrm{rad/s}$ and moment of inertia $I = 6{,}0\;\mathrm{kg\,m^2}$ with arms outstretched. Pulling her arms to her chest, the moment of inertia becomes $I' = 2{,}0\;\mathrm{kg\,m^2}$. Calculate the new angular velocity and comment on how the rotational kinetic energy changes.

Solution

No external torque acts about the vertical axis, so angular momentum is conserved: $I\,\omega = I'\,\omega' \quad\Rightarrow\quad \omega' = \frac{I}{I'}\,\omega = \frac{6{,}0}{2{,}0}\cdot 1{,}0$ $\ev{\omega' \approx 3{,}0\;\mathrm{rad/s}}$ Rotational kinetic energy: before, $E = \tfrac{1}{2}I\omega^2 = \tfrac{1}{2}\cdot 6{,}0\cdot 1{,}0^2 = 3{,}0\;\mathrm{J}$; after, $E' = \tfrac{1}{2}I'\omega'^2 = \tfrac{1}{2}\cdot 2{,}0\cdot 3{,}0^2 = 9{,}0\;\mathrm{J}$. The energy triples: the increase comes from the muscular work done to draw the arms in against the apparent centrifugal force.

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Topics: Dinamica rotazionale Concepts: Conservazione del momento angolare · Energia cinetica rotazionale Skills: Conservazione del momento angolare