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·m^{2}. If by moving the bricks the man decreases the rotational inertia of the system to 2.72 kg·m^{2}, **(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?

Step 1

**a)** The law of conservation of angular momentum can be applied to the given situation. Let the initial angular momentum be *L _{1}* and the new angular momentum be

Step 2

Solve equation (2) for *ω _{2}*.

Step 3

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

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