A man stands on a frictionless platform that is rotating with an angular speed o

Question

A man stands on a frictionless platform that is rotating with an angular speed of 3.3 rad/s; his arms are outstretched and he holds a weight in each hand. With his hands in this position the rotational inertial of the system of man, weights, and platform is 4.3 kg*m2. If by moving the weights the man decreases the rotational inertial of the system to 1.8 kg*m2, what is the resulting angular speed of the platform? which is = 7.8833 rad/s

1)What is the change of the kinetic energy when the man moves his arms?

Explanation / Answer

Initial angular momentum = I11 = 4.3*3.3

Final angular momentun = I22

Since momentum wll remain conserved in absence of external forces

I22 = 4.3*3.3

=> 2 = 4.3*3.3/1.8 = 7.8833 rad/s

Initial K.E. = (1/2)*I112 = 0.5*4.3*3.32 = 23.4135 J

Final K.E. = (1/2)* I222 = 0.5*1.8*7.88332 = 55.9318 J

K.E. = 55.9318-23.4135 = 32.5183 32.52 J

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