A circular disc A of radius r is fabricated from an iron plate of thickness t , and a second circular disc B of radius 4r is fabricated from an iron plate of thickness t/4 . The relationship between the moments of

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Correct answer: A. I_A < I_B

Let the density of iron be . The mass of disc A is M_A = π r^2 t The moment of inertia of disc A about its central axis is I_A = (1)/(2) M_A r^2 = (1)/(2) (π r^2 t ) r^2 = (1)/(2) π t r^4 The mass of disc B is M_B = π (4r)^2 ((t)/(4)) = 4 π r^2 t The moment of inertia of disc B about its central axis is I_B = (1)/(2) M_B (4r)^2 = (1)/(2) (4 π r^2 t ) (16r^2) = 32 π t r^4 Comparing I_A and I_B , we get I_B = 64 I_A Therefore, I_A < I_B

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