A man is standing on the center of a rotating table with arms stretched outwards the table is rotating freely with angular speed of 24 revolution per minute. Now the man withdraws his hand towards his chest thereby reducing moment of inertia 3/5 times his original. Calculate angular speed of man when he withdraws his hand ?
Let I be the initial moment of inertia, ω be the initial angular velocity.
ω = 24 rev/min
So, angular momentum is, L = Iω
When he withdraws his hand, let the moment of inertia be I/, angular velocity be ω/.
I/ = (3/5)I
Now, by law of conservation of angular momentum,
Iω = I/ω/
=> I(24)= (3/5)Iω/
=> ω/ = 40 rev/min