The moment of inertia of a thin rod of mass M and length L about an axis passing through its centre and normal to its length is given by ML square/12 using theorm of parallel axes find its moment of inertia about an axis passing through on end and normal to its length if L=1m and M= 0•2kg
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According to parellel axis theorem
I = Ic + Md^2
I = ML^2/12 + M(L/2)^2
= ML^2 / 3
= 0.2kg × 1^2 / 3
= 0.067 kgm^2
Moment of inertia along that axis is 0.067 kgm^2
I = Ic + Md^2
I = ML^2/12 + M(L/2)^2
= ML^2 / 3
= 0.2kg × 1^2 / 3
= 0.067 kgm^2
Moment of inertia along that axis is 0.067 kgm^2
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