A man stands on a rotating platform, with his arms stretched horizontally holding a 5 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 90cm to 20cm. The moment of inertia of the man together with the platform may be taken to be constant and equal to 7.6 kg m2.

(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?

Asked by Abhisek | 1 year ago |  83

##### Solution :-

(a) Given,

Mass of each weight = 5 kg

Moment of inertia of the man-platform system = 7.6 kg m2

Moment of inertia when his arms are fully stretched to 90 cm:
2 × m r2
= 2 × 5 × ( 0.9 )2
= 8.1 kg m2

Initial moment of inertia of the system, Ii = 7.6 + 8.1= 15.7 kg m2

Angular speed, ωi = 30 rev / min
= Angular momentum, Li = Iiωi  =  15.7 × 30

=     471    . . . . .  . . (i)
Moment of inertia when he folds his hands inward to 15 cm :
2 × mr2
= 2 × 5 x  (0.2)2 = 0.4kg m2

Final moment of inertia, If = 7.6 + 0.4 = 8.0 kg m2
let final angular speed = ωf
= Final angular momentum, Lf = Ifωf = 8.0ωf . . . . . .  (ii)
According to the principle of  conservation of angular momentum:
Iiωi  =  Ifωf
∴ ωf = $$\dfrac{471}{8}$$  =  58.88 rev/min

(b) There is a change in kinetic energy , with the decrease in the moment of inertia  kinetic energy increases. The extra kinetic energy is supplied to the system by the work done by the man in folding his arms inside.

Answered by Pragya Singh | 1 year ago

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