The moment of inertia of a uniform circular disc of radius $R$ and mass $M$ about an axis touching the disc…
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:
$\frac{1}{2} M R^2$
$M R^2$
$\frac{2}{5} M R^2$
$\frac{3}{2} M R^2$
Solution
M. I. of uniform circular disc about an axis and perpendicular to the plane is:
$I_C=\frac{1}{2} M R^2$
Using the theorem of parallel axis.
M. I. of uniform circular disc about an axis touching the disc at it diameter is:
$\begin{aligned}
I & =I_C+M R^2 \\
& =\frac{1}{2} M R^2+M R^2 \\
& =\frac{3}{2} M R^2 .
\end{aligned}$
.