A uniform circular disc has original mass M and radius R . A sector corresponding to an angle of 120^ is removed from it. The radius of gyration of the remaining part of the disc about an axis passing through its centre and perpendicular to its plane is:

Step-by-step solution

The mass of the remaining part of the disc is m' = 360^ - 120^ 360^ M = 2 3 M . The moment of inertia of a sector of a disc about its central axis is given by I = 1 2 ( mass of sector ) R^2 . Thus, the moment of inertia of the remaining part is I' = 1 2 m' R^2 . The radius of gyration k is defined using the mass of the actual object, so I' = m' k^2 . Equating the two expressions for I' : m' k^2 = 1 2 m' R^2 k^2 = R^2 2 k = R 2 . Note that the radius of gyration of a sector of a uniform disc is independent of the an