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A diver having a moment of inertia of 6....

A diver having a moment of inertia of `6.0 kg-m^2` about an axis through its centre of mass at an angular speed of 2 rad/s about this axis. If he folds his hands and feet to decrease the moment of inertia to 50.0 kg-m^2` what will be the new angular speed?

Text Solution

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The correct Answer is:
A, B, D
`I=6kg-m^2,omega=2rad/s, I_2=5kgk-m^2`
since exterN/Al torque=0
Therefore `I_1omega_1=I_2omega_2`
rarr omega_2=(6xx2)/5`
`=12/5=2.4rad/s`
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