A uniform circular disc has mass M and radius R . Let K₁ be its radius of gyration about an axis passing through its centre and perpendicular to its plane. Let K₂ be its radius of gyration about an axis tangential to its edge and lying in the plane of the disc. The ratio K₁ : K₂ is:

Step-by-step solution

The moment of inertia of the disc about an axis passing through its centre and perpendicular to its plane is: I₁ = 1 2 M R^2 Since I₁ = M K₁^2 , the radius of gyration K₁ is: K₁ = R 2 The moment of inertia of the disc about its diameter is I_d = 1 4 M R^2 . Using the parallel axis theorem, the moment of inertia about a tangent in the plane of the disc is: I₂ = I_d + M R^2 = 1 4 M R^2 + M R^2 = 5 4 M R^2 Since I₂ = M K₂^2 , the radius of gyration K₂ is: K₂ = 5 2 R The ratio K₁ : K₂ is: K₁ K₂ = R 2 5 R 2 = 2 10 = 2 5