A uniform circular disc of radius 8 cm rotates about an axis perpendicular to its plane. If the radius of gyration of the disc about this axis is 9 cm , the perpendicular distance of the axis from the centre of the disc is x cm . The value of x is _____.

Step-by-step solution

Let M be the mass and R be the radius of the disc. The moment of inertia of the disc about an axis passing through its centre and perpendicular to its plane is I_ cm = 1 2 MR^2 . Let the axis of rotation be at a distance d from the centre. By the parallel axis theorem, the moment of inertia about this axis is: I = I_ cm + Md^2 = 1 2 MR^2 + Md^2 We are given that the radius of gyration is k . Thus, I = Mk^2 . Equating the two expressions for I : Mk^2 = 1 2 MR^2 + Md^2 k^2 = 1 2 R^2 + d^2 Substitute R = 8 cm and k =