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Asked by meheboob1doctor | 21 Dec, 2022, 09:09: AM
Moment of inertia of rod AB,  IAB = M (L/2)2   ( whole mass is symmetric with respect to axis of rotation )

Moment of inertia of rod CD,  ICD = M ( L2 / 3 ) ( Moment of inertia about perpendicular axis at end )

Moment of inertia od rod EF , IEF = M ( L2 / 12 ) + M (L/2)2  [ parallel axis theorem ]

Total moment of inertia I = IAB + ICD + IEF = [ (1/4) + (1/3) + (1/12) + (1/4) ] M L2

Total moment of inertia I =(11/12) M L2

Answered by Thiyagarajan K | 21 Dec, 2022, 10:33: AM
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