Four masses are located in the xy plane. A 4.00 kg mass is located at the origin, and a second 4.00 kg mass is located at the point (2.00 m, 2.00 m). Two 3.00 kg masses are located at (0, 2.00 m) and (2.00 m, 0). (The masses are at the corners of a square with sides 2.00 m.) (a) Calculate the moment of inertia of this arrangement for rotations about an axis perpendicular to the xy plane and passing through the origin. (b) Calculate the moment of inertia of this arrangement for rotations about an

Answer:A)    I_total = 16 m, B)   I_total = 8 m,  C)   I_total = 8 m, D)  I_total = 8 mExplanation:The moment of inertia is a scalar quantity, therefore the total moment of inertia           I_total = I₁ + I₂ + I₃ + I₄ the moment of inertia of a point mass with respect to an axis of rotation          I = m r² Let's apply this to our case A) Rotation axis at the origin      I₁ = m 0 = 0 for the second masses, we find the distance using the Pythagorean theorem      r = [tex]\sqrt{2^2 + 2^2}[/tex]      r = 2 √2      I₂ = m (2 √2) ²      I₂ = 8 m        I₃ = m 2² = 4 m      I₄ = m 2² = 4 m we substitute      I_total = 0 + 8m + 4m + 4m      I_total = 16 m B) axis of rotation in the center of the square let's find the distance to any mass      r = [tex]\sqrt{1^2+ 1^2}[/tex]      r = √2      I₁ = m 2      I₂ = m 2      i₃ = m 3      I₄ = m 4 we substitute     I_total = 4 (2m)     I_total = 8 m C) axis of rotation is the x axis        I₁ = 0        I₂ = m 2² = 4 m        I₃ = m 2² = 4 m        I₄ = 0        I_total = 8 m D) axis of rotation is the y-axis        I₁ = 0        I₂ = 4m        I₃ = 0        I₄ = 4 m        I_total = 8 m