A rigid body has a radius of gyration K about an axis passing through its centre of mass. The radius of gyration of the body about a parallel axis located at a perpendicular distance K from the centre of mass axis is

Step-by-step solution

Let the mass of the rigid body be M . The moment of inertia of the body about the axis passing through its centre of mass is I_ cm = MK^2 . According to the theorem of parallel axes, the moment of inertia I' about a parallel axis at a perpendicular distance d is given by: I' = I_ cm + Md^2 Given that the distance d = K , we have: I' = MK^2 + MK^2 = 2MK^2 If K' is the new radius of gyration about this parallel axis, then I' = M(K')^2 . Equating the two expressions for I' : M(K')^2 = 2MK^2 (K')^2 = 2K^2 K' = 2 K . An