two point masses of 2kg and 10kg are connected by a weight less rod of length 1.2 m. Calculate the M.I of the system about an axis passing through the centre of mass and perpendicular to the system.

Centre of mass:

XCM=m1x1+m2x2m1+m2         =2×0+10×1.22+10         =1 m

Determination moment of inertia:

r1=XCM    = 1 mr2=L-XCM    = 1.2 -1      = 0.2 m


I=m1r12+m2r22  =212+100.22   = 2.4 kg.m2

Answer:

Moment of inertia, I=2.4 kg.m2

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