Question

A diver having a moment of inertia of 6⋅0 kg-m2 about an axis thorough its centre of mass rotates 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 5⋅0 kg-m2, what will be the new angular speed?

Answer 1

Explanation:

I1 = 6 kg-m2, ω1 = 2 rad/s , I2 = 5 kg-m2 Since external torque = 0 Therefore I1ω​1 = I2ω​2 ω2 = (6 × 2) / 5 = 2.4 rad/s