A man stands on a rotating platform, with his arms stretched horizontally holding a 5 \mathrm{~kg} weight in each hand. The angular speed of the platform is 30 revolutions per minute. The man then brings his arms close to his body with the distance of each weight from the axis changing from 90 \mathrm{~cm} to 20 \mathrm{~cm}. The moment of inertia of the man together with the platform may be taken to be constant and equal to 7.6 \mathrm{~kg} \mathbf{m}^{2}.
(a) What is his new angular speed? (Neglect friction.)
(b) Is kinetic energy conserved in the process? If not, from where does the change come about?
A man stands on a rotating platform, with his arms stretched horizontally holding a 5 \mathrm{~kg} weight in each hand. The angular speed of the platform is 30 revolutions per minute. The man then brings his arms close to his body with the distance of each weight from the axis changing from 90 \mathrm{~cm} to 20 \mathrm{~cm}. The moment of inertia of the man together with the platform may be taken to be constant and equal to 7.6 \mathrm{~kg} \mathbf{m}^{2}.
(a) What is his new angular speed? (Neglect friction.)
(b) Is kinetic energy conserved in the process? If not, from where does the change come about?

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.