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?
Given:
(a) Moment of Inertia of the man-platform system = 7.6 Kgm2
Moment of inertia when the man stretches his hands to a distance of 90 cm.
We know, I = 2 × mr2
Where,
m = mass
r = distance of mass from axis of rotation
Using the above formula, we get,
⇒ I = 2 × 5Kg × (0.9m)2
⇒ I = 8.1 Kgm2
Now, the Initial moment of inertia of the system,
Ii = 7.6 + 8.1 = 15.7 Kgm2
Angular speed, ωi = 30 rev/min
Angular momentum, Li = Iiωi
Li = 15.7Kgm2 × 30rev/min
⇒ Li = 471 Kgm2/sec …(i)
Moment of inertia when the man folds his hands to a distance of 20 cm,
I = 2 × mr2
Where,
m = mass
r = distance of mass from axis of rotation
⇒ I = 2 × 5Kg × (0.2m)2
⇒ I = 0.4 Kgm2
Final moment of inertia, If = 7.6 + 0.4 = 8 Kgm2
Final angular speed = ωf
Final angular momentum = Lf = Ifωf
Lf = 0.79 ωf …(ii)
From the conservation of angular momentum, We have
Iiωi = Ifωf
From equation (i) and (ii),
ωf = 15.7 Kg m2 × ((30rev/sec)/8 Kgm2)
ωf = 58.88 rev/min
(b) Kinetic energy is not conserved in the given process. In fact, with the decrease in moment of inertia, kinetic energy increases. The additional kinetic energy comes from the work done by the man to fold his hands towards himself.