Rotational MotionHard
Question
Two rings of the same radius and mass are placed such that their centers are at a common point and their planes are perpendicular to each other. The moment of inertia of the system about an axis passing through the centre and perpendicular to the plane of one of the rings is (mass of the ring = m, radius = r) : -
Options
A.
mr2
mr2B.mr2
C.
mr2
mr2D.2mr2
Solution
MI of one ring about axis which is perpendicular to the plane = mr2 and this axis is the diameter of another ring
∴ MI =
∴ Total MI = mr2 + 
∴ MI =
∴ Total MI = mr2 + Create a free account to view solution
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