A man stands on a platform that is rotating (without friction) withan angular sp

Question

A man stands on a platform that is rotating (without friction) withan angular speed of 1.18 rev/s; his arms are outstretched and heholds a brick in each hand. The rotational inertia of the systemconsisting of the man, bricks, and platform about the central axisis 7.28 kg·m2. If by moving the bricks the mandecreases the rotational inertia of the system to 2.70kg·m2, (a) what is the resultingangular speed of the platform and (b) what is theratio of the new kinetic energy of the system to the originalkinetic energy?

Explanation / Answer

.          Initialangular speed, 1 = 1.18 rev/s .                                              = 1.18 * 2 rad/s .                                              = 7.41 rad/s .         Initial momentof inertia, I1 = 7.28 kg.m^2 .          Newmoment of inertia, I2 = 2.7 kg.m^2 .    (a) .          From the lawof conservation of angular momentum, .          I2 2 =I1 1 .          2.7 *2 = 7.28 * 7.41 .                  2 = 19.98 rad/s .                         = ( 19.98 / 2 ) rev / s .                        = 3.18 rev/s .    (b) .          New KE /Original KE = [ ( 1/2 ) I2 2 2 ] / [ ( 1/2 ) I11 2 ] .                                            = ( 2.7 / 7.28 ) ( 3.18 / 1.18 ) 2 .                                             =2.7

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