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a)100 rev/min
Initial angular velocity, ω1= 40 rev/min
Final angular velocity = ω2
The moment of inertia of the boy with stretched hands = I1
The moment of inertia of the boy with folded hands = I2
The two moments of inertia are related as:
I2 = (2/5) I1
Since no external force acts on the boy, the angular momentum L is a constant.
Hence, for the two situations, we can write:
I2ω2 = I1 ω1
ω2 = (I1/I2) ω1
= [ I1 / (2/5)I1 ] × 40 = (5/2) × 40 = 100 rev/min
(b) Final K.E. = 2.5 Initial K.E.
Final kinetic rotation, EF = (1/2) I2 ω22
Initial kinetic rotation, EI = (1/2) I1 ω12
EF / EI = (1/2) I2 ω22 / (1/2) I1 ω12
= (2/5) I1 (100)2 / I1 (40)2
= 2.5
∴ EF = 2.5 E1
The increase in the rotational kinetic energy is attributed to the internal energy of the boy.
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