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A turntable turns about a fixed vertical...

A turntable turns about a fixed vertical axis, making one revolution in `10 s`. The moment of inertia of the turntable about the axis is `1200 kgm^(2)`. A man of mass `80 kg`, initially standing at centre of the turnable, runs out along the radius. What is the angular velocity of the turtable when the man is `2 m` from the centre?

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`I_0`=initial moment of inertia of the system `I_0=I_(mm+)I_(tabl e)` `I_0=0+1200=1200kgm^2` (`I_(man)`=0 final moment of inertia of the system )`I=I_(man+I_(tabl e)``I=mr^2+1200` `I=80(2)^2+1200=1520kgm^2` By conservation of angular momentum:`I_0omega_0=Iomega` Now `omega_0=(2pi)/T_0=(2pi)/10=pi/(rad5)/s`.
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