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 `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?

Explanation

`I_(0)` = intial moment of inertia of the system

`I_(0)=I_("man")+I_("table") `
`I_(0) = 0 +1200 = 1200 kgm^(2) (I_(man) = 0 "as the man is at the axis ")`
I = Final moment of inertia of the system
`I=I_(man) +I_(table) " " I = mr^(2) +1200`
`I = 80 (2)^(2) +1200 = 1520 "kgm"^(2)`
By conservation of angular momentum
`I_(0) omega_(0) = Iomega`
Now `omega_(0) = 2pi //T_(0)=2pi//10 = pi//5 ` rad/s
` omega = (I_(0)omega_(0))/I = (1200 xx pi)/(1520 xx5) = 0.51 ` rad/s