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Grade 11Mechanics

mass of original disc is 9 m and radius r. A small disc of radius r by 3 is removed as shown in figure ..The moment of inertia of remaining disc about an Axis perpendicular to plane of the disc and passing through the centre of disc is

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Profile image of Rankush
7 Years agoGrade 11
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1 Answer

Profile image of Khimraj
7 Years ago
mass of removed part = (9m/\pi r^{2})*\pi (r/3)^{2} = m
I = Itotal – Iremoved
I = 9mr2/2 – (mr2/18 + 4mr2/9)
I = 4mr2
Hope it clears.