A diver having a moment of inertia of 6.0 kg-m^2 about an axis through its centre of mass rotates at an angular speed
A diver having a moment of inertia of 6.0 kg-m2 about an axis through 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 ?
5 views
1 Answers
I1 = 6 kg-m2, ω1 = 2 rad/s , I2 = 5 kg-m2
Since external torque = 0
Therefore I1ω1 = I2ω2
ω2 = (6 × 2) / 5 = 2.4 rad/s
5 views
Answered