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# A 60.0-kg woman stands at the rim of a horizontal turntable having a moment of inertia of 500 $\mathrm { kg } \cdot \mathrm { m } ^ { 2 }$ and a radius of 2.00 m. The turntable is initially at rest and is free to rotate about a frictionless, vertical axle through its center. The woman then starts walking around the rim clockwise (as viewed from above the system) at a constant speed of 1.50 m/s relative to Earth. How much work does the woman do to set herself and the turntable into motion?

Solution

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$W = \Delta K$

$W = (1/2) \ m_w \ v_w^2 + (1/2) \ I_t \ w_t^2$

$W = (1/2) \ (60.0) \ (1.50)^2 + (1/2) \ (500) \ (0.360)^2$

$W = 99.9 \ J$

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