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

Options

A. `16 mR^2`
B. `(3 M + 16 m) (R^2)/(2)`
C. `(3 m + 12 M) (R^2)/(2)`
D. zero

Answer

B

Explanation

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)`