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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 `80 kg`, initially standing at centre of the turnable, runs out along the radius. What is the angular velocity of the turnable when the man is `2 m` from the centre?

Text Solution

Verified by Experts

Let `I_(0)` be the initial moment of inertia of the system (man`+`table)
`I_(0)=I_("man")+I_("table")`
`0+1200=1200kgm^(2)`
`(I_("man")=0` as the man is at the axis)
`I=`final moment of inertia of the system
`=I_("man")+I_("table")`
`=mr^(2)+1200`
`==80(2)^(2)+1200=15200kgm^(2)`

As no external torque about axis we can conserve angular momentum.
By conservation of angular momentum `I_(0)omega_(0)=Iomega`
Now `omega_(0)=2pi//T_(0)=2pi/10=pi/5rad//s`
`implies omega=(I_(0)omega_(0))/I=(1200xxpi)/(1520xx5)=0.51rad//s`
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