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The moment of inertia of a uniform circular disc of radius '$R$' and mass '$M$' about an axis touching the disc at its diameter and normal to the disc is :-
A
$MR^2$
B
$\dfrac{2}{5}MR^2$
C
$\dfrac{3}{2}MR^2$
D
$\dfrac{1}{2}MR^2$
Detailed Solution
The axis is normal to the disc and passes through a point on its edge (the end of a diameter).
Moment of inertia of the disc about the axis through its centre and normal to its plane: $I_1 = \dfrac{1}{2}MR^2$
The required axis is parallel to this axis and at a distance $R$ from it.
By the theorem of parallel axes: $I_2 = I_1 + MR^2$
$I_2 = \dfrac{1}{2}MR^2 + MR^2$
$I_2 = \dfrac{3}{2}MR^2$
Moment of inertia of the disc about the axis through its centre and normal to its plane: $I_1 = \dfrac{1}{2}MR^2$
The required axis is parallel to this axis and at a distance $R$ from it.
By the theorem of parallel axes: $I_2 = I_1 + MR^2$
$I_2 = \dfrac{1}{2}MR^2 + MR^2$
$I_2 = \dfrac{3}{2}MR^2$