Mass of each weight is given as 5 kg
The moment of inertia of the man-platform system is given as 7.6 kg m2
So, the moment of inertia when his arms are fully stretched to 90 cm can be calculated as,
2 × m r2
= 2 × 5 × (0.9)2
= 8.1 kg m2
The initial moment of inertia of the system will be
Ii = 7.6 + 8.1= 15.7 kg m2
Angular speed is given as ωi = 30 rev/min
Angular momentum can be calculated as,
Li = Iiωi = 15.7 × 30
= 471……(i)
The moment of inertia when he folds his hands inward to 15 cm can be calculated as,
2 × mr2
= 2 × 5 x (0.2)2 = 0.4kg m2
The final moment of inertia will be,
If = 7.6 + 0.4 = 8.0 kg m2
let ωf be the final angular speed
Final angular momentum will be
Lf = Ifωf = 8.0ωf . . . . . . (ii)
According to the principle of conservation of angular momentum, we can write,
Iiωi = Ifωf
∴ ωf = 471/ 8 = 58.88 rev/min
(b) There is a change in kinetic energy; when the moment of inertia decreases, kinetic energy rises. The man’s work in folding his arms inside provides more kinetic energy to the system.