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 inertia I_A and I_B is

Step-by-step solution

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