Question

Ice skaters often end their performances with spin turns, where they spin very fast about their center of mass with their arms folded in and legs together. Upon ending, their arms extend outward, proclaiming their finish. Not quite as noticeably, one leg goes out as well.
Suppose that the moment of inertia of a skater with arms out and one leg extended is 3.5 kg*m^2 and for arms and legs in is 0.70 kg*m^2. If she starts out spinning at 6.0 rev/s, what is her angular speed (in rev/s) when her arms and one leg open outward?

Answer 1

Answer:

$\large \boxed{\text{30 rev/s}}$

Explanation:

This question is based on the Law of Conservation of Angular Momentum.

Angular momentum (L) equals the moment of inertia (I) times the angular speed (ω).

L = Iω

If momentum is conserved,

I₁ω₁ = I₂ω₂

Data:

 I₁ = 3.5    kg·m²s⁻¹

ω₁ = 6.0    rev·s⁻¹

 I₂ = 0.70 kg·m²s⁻¹

Calculation:

$\begin{array}{rcl}I_{1}\omega_{1} &= &I_{2}\omega_{2}\\\text{3.5 kg \cdot m ^{2} }\times \text{6.0 rev/s} &= &\text{0.70 kg \cdot m ^{2} }\times\omega_{2}\\\text{21 rev/s} &= &0.70\omega_{2}\\\omega_{2} & = & \dfrac{\text{21 rev/s}}{0.70}\\\\&=&\textbf{30 rev/s}\\\end{array}\\\text{The skater's final rotational speed is \large \boxed{\textbf{30 rev/s}} }$