Question

The moment of inertia of a uniform circular disc of mass M and radius R about any of its diameter is (1/4) MR², what is the moment of inertia of the disc about an axis passing through its centre and normal to the disc? 

(A) MR² 

(B) (1/2) MR² 

(C) (3/2) MR² 

(D) 2MR²

Answer 1

Correct option: (B) (1/2) MR²

Explanation:

Consider two perpendicular diameter. One along x axis and other along y axis.

I_x = I_y = (1/4)MR² 

by perpendicular axes theorem. The MI of disc about an axis passing through centre is

I_c = I_x + I_y 

= [(MR²) / 4] + [(MR²) / 4]

= [(MR²) / 2]