A solid cylinder has a moment of inertia 2 kg m 2 and is rotating freely at a fi

Question

A solid cylinder has a moment of inertia 2 kg m2and is rotating freely at a fixed angular velocity of 12 rad/s. Asecond cylinder has a moment of inertia 5 kg m2 andinitially at rest is dropped onto the rotating cylinder. 1. What was the initial angular momentum of the initialcylinder before the dropping? 2. What was the angular velocity of the cylinder pair afterthe droping? 3. What was the magnitude of the change in the kineticenergy of the system? A solid cylinder has a moment of inertia 2 kg m2and is rotating freely at a fixed angular velocity of 12 rad/s. Asecond cylinder has a moment of inertia 5 kg m2 andinitially at rest is dropped onto the rotating cylinder. 1. What was the initial angular momentum of the initialcylinder before the dropping? 2. What was the angular velocity of the cylinder pair afterthe droping? 3. What was the magnitude of the change in the kineticenergy of the system?

Explanation / Answer

   Li = Ii x i
   = (2)(12 rad /s )
   = 24 kgm^2 /s
Final angualr momentum
   Lf = If x f
   = ( 2 + 5)f
  basing on conservation of angularmomentum
     24 kgm^2 /s = ( 2 + 5)f
   f = 3.4285 rad /s
Change in kinetic energy is
   K = (1/2)Iff -(1/2)Iii
   = (1/2)(7)(3.4285) -(1/2)(2)(24)
   = 12.5 J

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