A man stands on a platform that is rotating (without friction) with an angular speed of 1.39 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 6.30 kg·m2. If by moving the bricks the man decreases the rotational inertia of the system to 2.72 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?

Question
Asked May 2, 2019
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A man stands on a platform that is rotating (without friction) with an angular speed of 1.39 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 6.30 kg·m2. If by moving the bricks the man decreases the rotational inertia of the system to 2.72 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?

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Expert Answer

Step 1

a) The law of conservation of angular momentum can be applied to the given situation. Let the initial angular momentum be L1 and the new angular momentum be L2.

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Step 2

Solve equation (2) for ω2.

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Step 3

b) The kinetic energy can be expressed in terms of angul...

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