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Grade: 11
        
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
6 months ago

Answers : (1)

Khimraj
2873 Points
							
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.
6 months ago
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