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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 -
A
$\dfrac{1}{2}MR^2$
B
$\dfrac{7}{2}MR^2$
C
$\dfrac{3}{2}MR^2$
D
$MR^2$
Detailed Solution
Moment of inertia of the disc about the axis through its centre and normal to its plane: $I_{cm} = \dfrac{MR^2}{2}$
The required axis passes through the edge and is parallel to this axis, at a distance $d = R$ from it.
By the theorem of parallel axes: $I = I_{cm} + Md^2$
$I = \dfrac{MR^2}{2} + MR^2$
$I = \dfrac{3}{2}MR^2$
The required axis passes through the edge and is parallel to this axis, at a distance $d = R$ from it.
By the theorem of parallel axes: $I = I_{cm} + Md^2$
$I = \dfrac{MR^2}{2} + MR^2$
$I = \dfrac{3}{2}MR^2$