1.

A wheel of mass 8 kg has moment of inertia equals to `0.5 "kg-m"^(2)`. Determine its radius of gyration.

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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