Question

problem 7: an ice skater with a moment of inertia of 0.4 kg·m² is spinning with arms stretch at 2 revolutions per second. when she tucks her arms and legs the ice skater now spins at 5 revolutions per second. find the new moment of inertia of the ice skater.

Explanation:

Step1: Convert angular velocity from revolutions per second to radians per second

Angular velocity \(\omega = 2\pi n\) (where \(n\) is revolutions per second).
Initial angular velocity \(\omega_1=2\pi\times2 = 4\pi\) rad/s.
Final angular velocity \(\omega_2 = 2\pi\times5=10\pi\) rad/s.

Step2: Apply the law of conservation of angular momentum \(L = I\omega\) (\(L_1 = L_2\))

\(I_1\omega_1=I_2\omega_2\).
Given \(I_1 = 0.4\space kg\cdot m^2\), \(\omega_1 = 4\pi\) rad/s, \(\omega_2=10\pi\) rad/s.
Substitute into the formula: \(0.4\times4\pi=I_2\times10\pi\).
Solve for \(I_2\): \(I_2=\frac{0.4\times4\pi}{10\pi}\).
Cancel out \(\pi\): \(I_2=\frac{0.4\times4}{10}\).
Calculate: \(I_2 = 0.16\space kg\cdot m^2\).

Answer:

\(0.16\space kg\cdot m^2\)