The moment of inertia of a uniform circular disc of radius `R` and mass `M` about an axis passing from the edge of the disc and normal to the disc is.

Options

A. `(1)/(2) MR^2`
B. `MR^2`
C. `(7)/(2)MR^2`
D. `(3)/(2) MR^2`

Answer

D

Explanation

(d) Moment of inertia of disc passing through its centre of gravity and perpendicular to its plane is.
`I_(AB) = (1)/(2) MR^2`
Using theorem of parallel axes, we have,
`I_(CD) = I_(AB) + MR^2`
=`(1)/(2) MR^2 + MR^2 = (3)/(2) MR^2`.
.