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Units And Measurement

Question
CBSEENPH11020680

A circular disk of moment of inertia It is rotating in a horizontal plane, about its symmetry axis, with a constant angular speed.straight omega subscript straight i Another disk of moment of inertia Ib is dropped coaxially onto the rotating disk. Initially, the second disk has zero angular speed. Eventually, both the disks rotate with constant angular speed.straight omega subscript straight fThe energy lost by the initially rotating disc due to friction is 

  • 1 half fraction numerator straight I subscript straight b superscript 2 over denominator left parenthesis straight I subscript straight t plus straight I subscript straight b right parenthesis end fraction space straight omega subscript straight i superscript 2
  • 1 half fraction numerator straight I subscript straight t superscript 2 over denominator left parenthesis straight I subscript straight t plus straight I subscript straight b right parenthesis end fraction straight omega subscript straight i superscript 2
  • 1 half fraction numerator straight I subscript straight b minus straight I subscript straight t over denominator left parenthesis straight I subscript straight t plus straight I subscript straight b right parenthesis end fraction straight omega subscript straight i superscript 2
  • 1 half fraction numerator straight I subscript straight b straight I subscript straight t over denominator left parenthesis straight I subscript straight t plus straight I subscript straight b right parenthesis end fraction straight omega subscript straight i superscript 2

Solution

D.

1 half fraction numerator straight I subscript straight b straight I subscript straight t over denominator left parenthesis straight I subscript straight t plus straight I subscript straight b right parenthesis end fraction straight omega subscript straight i superscript 2

Loss of energy, 
increment straight E space equals space 1 half space straight I subscript straight t space straight omega subscript straight i superscript 2 minus 1 half fraction numerator straight I subscript straight t superscript 2 straight omega subscript straight i superscript 2 over denominator left parenthesis straight I subscript straight t plus straight I subscript straight b right parenthesis end fraction

equals space 1 half fraction numerator straight I subscript straight b straight I subscript straight t straight omega subscript straight i superscript 2 over denominator 2 left parenthesis straight I subscript straight t plus straight I subscript straight b right parenthesis end fraction