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प्रश्न
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 ______.
विकल्प
`"K"^2/"R"^2`
`"K"^2/("K"^2+"R"^2)`
`"R"^2/("K"^2+"R"^2)`
`("K"^2+"R"^2)/"R"^2`
MCQ
रिक्त स्थान भरें
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उत्तर
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 `underlinebb("K"^2/("K"^2+"R"^2))`.
Explanation:
`"Rotational KE"/"Total KE" =(1/2"mv"^2("K"^2/"R"^2))/(1/2"mv"^2(1+"K"^2/"R"^2)`
= `"K"^2/("K"^2+"R"^2)`
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