A uniform thin circular ring of mass M and radius R is cut and a sector of angle is removed. If the moment of inertia of the remaining arc about an axis passing through its centre and perpendicular to its plane is 3 5 MR^2 , then the value of the removed angle is

Step-by-step solution

The moment of inertia of a circular arc of mass m' and radius R about its central axis is given by I = m'R^2 . Given that the moment of inertia of the remaining arc is 3 5 MR^2 , we can write: m'R^2 = 3 5 MR^2 m' = 3 5 M This means the mass of the remaining arc is 3 5 of the original mass. The mass of the removed sector must therefore be: m_ removed = M - 3 5 M = 2 5 M For a uniform ring, the mass of a sector is directly proportional to the angle it subtends at the centre. Thus, the angle of the removed sector is: