A wheel is rotating freely at angular speed 800 rev/min on ashaft whose rotational inertia is negligible. Suddenly linked to the same shaft is a second wheel that is initially at rest and has double the rotational inertia of the first.A) What is the combined angular speed of the shaft and the two wheels that results?B)What fraction of the orginial rotational kinetic energy is lost?

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Clelioo · Accepted Answer

(A) Initial MI = I  Angular speed ωi=800r±=800×60×2π rad. / s  MI of second wheel = 2I  So final MI after coupling = 3I  Final angular speed = ωf  Angular momentum is conserved in the process. So,   |ωi=3|ωf  ωf=ωi3=800×60××2π3  =1.005×105 rad. /s   (or) 266.7 rpm  (B) Initial Rotational KE = 0.5 ∣(ωi)2  Final Rotational K.E. = 0.5(3|) (ωf)2  Fraction of the orginial rotational kinetic energy lostis  =(R.K.E.)i−(R.K.E.)i(R.K.E.)i=0.667

A wheel is rotating freely at angular speed 800 rev/min on ashaft whose rotational inertia is negligible. A secod wheel,initially rest and with twice

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