Question

a skater is initially spinning at a rate of 18.0 rad/s with a rotational inertia of 3.60 kg·m² when her arms are extended. what is her angular velocity after she pulls her arms in and reduces her rotational inertia to 1.60 kg·m²?
□ rad/s

Explanation:

Step1: Recall the law of conservation of angular momentum

The law of conservation of angular momentum states that \( L_i = L_f \), where \( L = I\omega \) (angular momentum \( L \), rotational inertia \( I \), angular velocity \( \omega \)). So \( I_i\omega_i = I_f\omega_f \).

Step2: Rearrange the formula to solve for \( \omega_f \)

From \( I_i\omega_i = I_f\omega_f \), we get \( \omega_f=\frac{I_i\omega_i}{I_f} \).

Step3: Substitute the given values

We know \( I_i = 3.60\space kg\cdot m^2 \), \( \omega_i = 18.0\space rad/s \), and \( I_f = 1.60\space kg\cdot m^2 \). Substituting these values into the formula: \( \omega_f=\frac{3.60\times18.0}{1.60} \).

Step4: Calculate the result

First, calculate the numerator: \( 3.60\times18.0 = 64.8 \). Then divide by the denominator: \( \frac{64.8}{1.60}=40.5 \space rad/s \).

Answer:

40.5