From a disc of radius R, a concentric circular portion of radius r is cut out so as to leave an annular disc of mass M. The moment of inertia of this annular disc about the axis perpendicular to its plane and passing through its centre of gravity is

Options

A. `1//2M(R^(2)+r^(2))`
B. `1//2M(R^(2)-r^(2))`
C. `1//2M(R^(4)+r^(4))`
D. `1//2M(R^(4)-r^(4))`

Answer

A

Explanation

Area of the face of ring = `pi(R^(2)-r^(2))`
Mass per unit area = `(M)/(pi(R^(2)-r^(2)))`
Face area of ring of radius `x` and `x+dx=2pixdx`
Mass (dm) of this ring = `(M)/(pi(R^(2)-r^(2)))xx2pixdx`

MOI of this ring -
implies `I=intx^(2)dm=int_(r)^(R)x^(2).(2M)/((R^(2)-r^(2)))xdx`
implies `I=(2M)/((R^(2)-r^(2)))[(x^(4))/(4)]_(r)^(R)`
implies `I=(2M)/(4(R^(2)-r^(2)))[R^(4)-r^(4)]=(M)/(2)(R^(2)+r^(2))`