A woman stands at the center of a platform. The woman and the platform rotate with an angular speed of 5.20 rad/s. Friction is negligible. Her arms are outstretched, and she is holding a dumbbell in each hand. In this position the total moment of inertia of the rotating system (platform, woman, and dumbbells) is 5.40 kg·m2. By pulling in her arms, she reduces the moment of inertia to 4.40 kg·m2. Find her new angular speed.
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Angular momentum (=Iω) is conserved so I₁ω₁ = I₂ω₂ where 1 and 2 refer to initial and final values.
5.40 x 5.20 = 4.40ω₂
ω₂ = 6.38rad/s
5.40 x 5.20 = 4.40ω₂
ω₂ = 6.38rad/s