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

Answer

A, B, D

Explanation

`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`