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Rotational Motion question

2003 · Q159
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Rotational Motion question

2003 · Q159

NEETPhysicsRotational MotionMCQ+4 / −1
A ball rolls without slipping. The radius of gyration of the ball about an axis passing through its centre of mass is K. If radius of the ball be R. then the fraction of total energy associated with its rotational energy will be
  1. A
    K2+R2R2{{{K^2} + {R^2}} \over {{R^2}}}R2K2+R2​
  2. B
    K2R2{{{K^2}} \over {{R^2}}}R2K2​
  3. C
    K2K2+R2{{{K^2}} \over {{K^2} + {R^2}}}K2+R2K2​
  4. D
    R2K2+R2{{{R^2}} \over {{K^2} + {R^2}}}K2+R2R2​
View written solutionFree

Correct answer: C

Rotational energy =12I(ω)2=12(mK2)ω2{1 \over 2}I{\left( \omega \right)^2} = {1 \over 2}\left( {m{K^2}} \right){\omega ^2}21​I(ω)2=21​(mK2)ω2

Linear kinetic energy = 12mω2R2{1 \over 2}m{\omega ^2}{R^2}21​mω2R2

∴\therefore∴ Required fraction

=12(mK2)ω212(mK2)ω2+12mω2R2=K2K2+R2 = {{{1 \over 2}\left( {m{K^2}} \right){\omega ^2}} \over {{1 \over 2}\left( {m{K^2}} \right){\omega ^2} + {1 \over 2}m{\omega ^2}{R^2}}} = {{{K^2}} \over {{K^2} + {R^2}}}=21​(mK2)ω2+21​mω2R221​(mK2)ω2​=K2+R2K2​

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