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

Question

A man stands on a platform that is rotating (without friction) with an angular speed of 0.527 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 6.13 kg.m2. If by moving the bricks the man decreases the rotational inertia of the system to 1.37 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? (a) Number Units

Explanation / Answer

using conservation of angular momentum:

6.13*0.527=1.37*w

w=2.358 rev/sec

b)

we know that rotational kinetic energy=0.5*moment of inertia*angular velocity^2

so ratio of kinetic energies=ratio of moment of inertia*(ratio of angular veolocity)^2

=(6.13/1.37)*(0.527/2.358)^2=0.2235

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