The moment of inertia of a uniform circular disc of mass M and of radius R about one of its diameter is

Options

A. `(1)/(4) MR^(2)`
B. `(1)/(2) MR^(2)`
C. `(2)/(3) MR^(2)`
D. `(2)/(5) MR^(2)`

Answer

A

Explanation

Moment of inertia of the disc about an axis perpendicular to it and through its centre is `(MR^(2))/(2)`

x and y axes lie in the plane of the disc and z is perpendicular to it .
By the theorem of perpendicular axes , `I_(z) = I_(x) +I_(y)`
As x and y axes are along two diameters of the disc . By symmetry , the moment of inertia of the disc is the same about any diameter . Hence
`I_(x) = I_(y) " " therefore = I_(z) = 2 I_(x)`
`because I_(z) = (1)/(2) MR^(2) " " therefore I_(x) = (I_(z))/(2) = (1)/(4) MR^2`
Thus , the moment of inertia of the disc about any of its diameter is `(1)/(4) MR^(2)`.