A man stands on a platform that is rotating (without friction) with the angular

Question

A man stands on a platform that is rotating (without friction) with the angular speed of 4.5 rev/s, his arms are outstretched and he holds a brick in each hand. The rotational interia of the system consisting of the man, bricks the man decreases the rotational inertia of the system by 20.0 kg/m^2, a)what is the physical basis for solving this problem? b) what is the angular speed of the platform? c)what is the original kinetic energy of the system in Joules d)What is the new kinetic energy of the system in Joules? e)which kinetic energy is larger and explain why.

Explanation / Answer

A man stands on a platform that is rotating (without friction) with an angular speed of 2.27 rev/s; his arms are outstretched and he holds a brick in each hand. The rotational inertia of the system consisting of the man, bricks, and platform about the central axis is 6.53 kg·m2. If by moving the bricks the man decreases the rotational inertia of the system to 1.22 kg·m2, (a) what is the resulting angular speed of the platform and (b) what is the ratio of the new kinetic energy of the system to the original kinetic energy? For this problem I followed the equation. (Initial Interia/Final Interia)*Initial Angular Speed and got 12.1501. But the answer has to be in rad/s. I'm not sure how to do now 12.15 rev/s * (2 pi) rad/rev = 76.34 rad/s btw: 2.27 rev/s is really spinning fast for energy use KE = ½ I ?^2 (use rad/s)

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