A child is standing with hands straight down at thecentre of a platform rotating about its central axis.The kinetic energy of the system is K. The child nowstretches his arms so that the moment of inertia ofthe system doubles. The kinetic energy of thesystem now is
  1. 4k
  2. k/4
  3. k/2
  4. 2k

Let the M. I. and angular speed of the system be I and ω respectively.The kinetic energy of the system given   K=12Iω2.When the system is rotating then             Iω=2I×ω0or          ω0=ω2So new kinetic energy of the system is  K. E. = 12×2I×ω22or              K. E. = 12×2I×ω24=12×12Iω2=K2Therefore the correct answer is option 3 K2.

  • 4
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