Perpendicular Axis Theorem

Question

A thin circular ring of mass M and radius R has a moment of inertia MR² about its central axis perpendicular to its plane. What is its moment of inertia about a diameter lying in its plane?

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From NCERT Class 11, Chapter 7 (System of Particles and Rotational Motion), the perpendicular axis theorem states that for a planar lamina the moment of inertia about an axis perpendicular to the plane equals the sum of the moments of inertia about two mutually perpendicular axes that lie in the plane and intersect the perpendicular axis at the same point: I_z = I_x + I_y. This theorem applies only to flat, two-dimensional bodies such as rings, discs, and laminae, which is why it suits a thin circular ring. For the ring, the moment of inertia about the central axis perpendicular to its plane is given as I_z = MR². Because the ring is perfectly symmetric, the two in-plane diametric axes are physically equivalent, so I_x = I_y. Substituting into the theorem gives MR² = I_x + I_x = 2I_x, which yields I_x = MR²/2. The option MR² ignores the symmetry split and simply reuses the perpendicular-axis value. The option 2MR² doubles the value instead of halving it. The option MR²/4 incorrectly divides by four rather than two. A check confirms that the two equal in-plane contributions must add up to the perpendicular-axis value MR², so each one is exactly half, matching the answer MR²/2 and respecting the geometry of the ring.

Perpendicular Axis Theorem

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