Two hoops or rings (I = MR²)are centered,lying on a turntable.The smaller ring has radius = 0.050 m;the larger has radius = 0.10 m.Both have a mass of 3.0 kg.What is the total moment of inertia as the turntable spins? Ignore the mass of the turntable. A)0.030 kg*m² B)0.007 5 kg*m² C)0.038 kg*m² D)0.075 kg*m²

Question

Quizplus · Accepted Answer

The moment of inertia of a ring is given by I = MR^2, where M is the mass of the ring and R is the radius of the ring. For the smaller ring, I = 3.0 kg x (0.050 m)^2 = 0.0075 kg·m^2. For the larger ring, I = 3.0 kg x (0.10 m)^2 = 0.030 kg·m^2.

To find the total moment of inertia, we can add the individual moments of inertia since they are centered on the same axis of rotation. So, I_total = I_small + I_large = 0.0075 kg·m^2 + 0.030 kg·m^2 = 0.038 kg·m^2. Therefore, the best choice is C.

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