Problem — Initial ω of a skater

Problem

A skater has $I_1 = 10\;\mathrm{kg\,m^2}$. By pulling in their arms they reach $I_2 = 6\;\mathrm{kg\,m^2}$ and $\omega_2 = 5\;\mathrm{rad/s}$. Calculate $\omega_1$ (use conservation of $L$) and comment on the change in kinetic energy.

Solution

Conservation of angular momentum $I_1\omega_1 = I_2\omega_2$, solved for $\omega_1$: $\omega_1 = \frac{I_2\,\omega_2}{I_1} = \frac{6\cdot 5}{10} = \ev{3\;\mathrm{rad/s}}$ Rotational kinetic energy: before $E_1 = \tfrac{1}{2}I_1\omega_1^2 = \tfrac{1}{2}\cdot 10\cdot 3^2 = 45\;\mathrm{J}$; after $E_2 = \tfrac{1}{2}I_2\omega_2^2 = \tfrac{1}{2}\cdot 6\cdot 5^2 = 75\;\mathrm{J}$. $\ev{E_\text{kin}: 45\;\mathrm{J} \rightarrow 75\;\mathrm{J}}$ The energy increases by $30\;\mathrm{J}$: this is the muscular work done to bring the arms closer to the axis.

Links

Topics: Rotational dynamics Concepts: Conservation of angular momentum · Rotational kinetic energy Skills: Conservation of angular momentum