Question

From a circular disc of radius R and mass 9M, a small disc of mass M and radius R/3 is removed concentrically. The moment of inertia of the remaining disc about an axis perpendicular to the plane of the disc and passing through its centre is

Answer 1

Answer:

3.55$kgm^{2}$

Explanation:

moment of inertia = mass x [ perpendicular distance ]^2

radius of the rem. disc = R - R/3

                                      = 2R/3

Mass of the disc            = 9M - M

                                       = 8M

now moment of inertia  =  8 x [2/3]^2

                                       =  8 x 4/9

                                       =  32/9

                                       = 3.55 $kg m^{2}$