A wheel is rotating freely at angular speed 760 rev/min on ashaft whose rotation

Question

A wheel is rotating freely at angular speed 760 rev/min on ashaft whose rotational inertia is negligible. A second wheel,initially at rest and with 3 times the rotational inertia of thefirst, is suddenly coupled to the same shaft. (a)What is the angular speed of the resultant combination of the shaftand two wheels? (b) What fraction of the originalrotational kinetic energy is lost?

Explanation / Answer

According to conservation of angular momentum we have            Li = Lf The initial angular momentum of the wheel is            Li =I11 = I1 *760rev/min       ..............1 The final angular momentum when second wheel is suddenlycoupled to the same shaft is           Lf = (I1 + I2 )2                      Given that I2 = 3I1          Lf = 4I1 * 2                          .................2 Therefore from eq 1 and 2 we get           I1 * 760rev/min = 4I1 *2            ==>2 = 760 / 4                        = 190 rev/min The fraction of rotational kinetic energy lost =K / K1                                                                    = ( 0.5I121 - 0.5(I1 + I2)22 ) /I112                                                                    = ( 0.5I121 - 2*I1 22 ) /I112                                                                    = ( 0.521 - 222 )/ 12                                                                    = (0.5*760*760 - 2*190*190 ) / (760*760)                                                                    = 0.375

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