Four particles each of mass m are lying symmetrically on the rim of a disc of mass M and radius R. Moment of inertia of this system about an axis passing through one of the particles and I to plane of disc is

Four particles each of mass m are lying symmetrically on the rim of a

1.	Four particles each of mass m are lying symmetrically on the rim of a disc of mass M and radius R. Moment of inertia of this system about an axis passing through one of the particles and I to plane of disc is
Answer» `16 mR^2` `(3 M + 16 m) (R^2)/(2)` `(3 m + 12 M) (R^2)/(2)` zero Solution :According to the THEOREM of parallel axes , MOMENT of inertia of DISC about an axis passing through K and `bot` to plane of disc of the given figure = `(1)/(2) MR^(2) + MR^(2) = (3)/(2) MR^(2)` Total moment of inertia of the system `= (3)/(2) MR^(2) + m (2R)^(2) + m (sqrt2 R)^(2) +m (sqrt2 R)^(2) = (3 M + 16 m ) (R^(2))/(2)`