The moment of inertia of a semicircular ring about an axis through its centre and perpendicular to its plane is (1/x)MR², where R is radius and M is mass. Find x.

Question

G1 Physics · Accepted Answer

For a semicircular ring, every mass element is at the same distance R from the axis passing through the centre and perpendicular to the plane. Therefore, I = ∫R² dm = R²∫dm = MR² The question writes I = (1/x)MR² Hence, 1/x = 1 so x = 1. Final value: 1