A man stands on a platform that is rotating with an angular speed of 1.22 rev/s;

Question

A man stands on a platform that is rotating with an angular speed of 1.22 rev/s; his arms are outstretched and he holds a weight in each hand. With his hands in this position the total rotational inertia of the man, the weights, and the platform is 6.13 kgm2 . If by moving the weights the man decreases the rotational inertia to 1.97 kgm2 , (a) what is the resulting angular speed of the platform, and (b) what is the ratio of the new kinetic energy to the original kinetic energy? Assume the platforms rotate without friction.

Explanation / Answer

According to the given problem,

Angular Momentum = I (Inertia x Angulart speed)

Rotational KE = ½ I ^2

a) 0.332 rev /sec = 1.22 x 2 x = 7.66 rad/s

Applying conservation of Momentum:
6.13 * 7.66 = 1.97 f
f  = 23.85 rad/s
23.85 / 2 = 3.796 rev/s

b) Initial KE = ½ I 2
= 0.5 * 6.13 * 7.662 = 180.098 J
Final KE = 0.5* 1.97 * 23.872 = 560.4 J
Ratio of KE's 560.4 : 180.098 = 3.1111

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