A 60.0-kg woman stands at the western rim of a horizontal turntable.Moment of inertia of the turntable system is 500 kg. $m^{2}$ and a radius of 2.0 m. The turntable is initially at rest and is free to rotate about a frictionless, vertical axle through its centre. 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 the Earth. The final angular velocity of the woman and the turntable system is

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$0.36 rad / s (counterclockwise)$

$1.8 rad / s (counterclockwise)$

$3.6 rad / s (clockwsie)$

$0.36 rad / s (clockwise)$

answer is .

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Detailed Solution

From conservation of angular momentum for the system of the woman and the turntable, we have $L_{f} = L_{i} = 0$

So $L_{f} = I_{woman} ω_{woman} + I_{table} ω_{table} = 0$

and $ω_{table} = (- \frac{I_{woman}}{I_{table}}) ω_{woman}$

= $(\frac{m_{woman} r^{2}}{I_{table}}) (\frac{v_{woman}}{r})$

= $- \frac{m_{woman} {rv}_{woman}}{I_{table}}$

$ω_{t a b l e} = - \frac{60.0 k g (2.00 m) (1.50 m / s)}{500 k g . m^{2}}$

= -0.360 rad/s

$or ω_{table} = 0.360 rad / s (counterclockwise)$

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