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A wheel of mass 8 kg has moment of inertia equals to `0.5 "kg-m"^(2)`. Determine its radius of gyration. |
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Answer» Given, mass , `M= 8 kg` Moment of Inertia, `I=0.5 "kg-m^"(2)` `because I=MK^(2)rArrK^(2)=(I)/(M)rArrK=sqrt((I)/(M))` `rArr` Radius of gyration, `K=sqrt((0.5)/(8))rArr K=0.25 m or K=25` cm. |
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