Two hoops or rings (I = MR²)are centered,lying on a turntable.The smaller ring has radius = 0.05 m;the larger has radius = 0.10 m.Both have a mass of 5.0 kg.What is the total moment of inertia as the turntable spins? Ignore the mass of the turntable. A)7.5E-1 kg·m² B)1.1E-1 kg·m² C)1.3E-2 kg·m² D)6.3E-2 kg·m² E)5.0E-2 kg·m²

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The moment of inertia of each hoop can be found using the formula I = mr^2, where m is the mass of the hoop and r is the radius. For the smaller ring, I = (5.0 kg)(0.05 m)^2 = 0.0125 kg·m^2. For the larger ring, I = (5.0 kg)(0.10 m)^2 = 0.0500 kg·m^2. Since the rings are on the same axis of rotation, the total moment of inertia is the sum of the moment of inertia of each hoop: 0.0125 kg·m^2 + 0.0500 kg·m^2 = 0.0625 kg·m^2. Rounded to the nearest significant figure, the answer is 6.3E-2 kg·m^2, which is choice D.

Two Hoops or Rings (I = MR2)are Centered,lying on a Turntable.The

Two Hoops or Rings (I = MR²)are Centered,lying on a Turntable.The