Question

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. MR2
• B. $\bigg(\frac{2}{5}\bigg)$MR2
• C. $\bigg(\frac{3}{2}\bigg)$MR2
• D. $\bigg(\frac{5}{6}\bigg)$MR2

Answer 1

Answer: (C)

$\bigg(\frac{3}{2}\bigg)$MR2

Answer 2

The moment of inertia about an axis passing through centre of mass of disc and perpendicular to its plane is $\text{I}_{CM}$ = \frac{1}{2}$$\text{MR}^2
where M is the mass of disc and R its radius. According to theorem of parallel axis, moment of inertia of circular disc about an axis touching the disc at its diameter and normal to the disc is
$\text {I}$ = $\text{I}_{CM}$ + $\text{MR}^2$

= $\frac{1}{2}$ $\text{MR}^2$ + $\text{MR}^2$

= $\frac{3}{2}$ $\text{MR}^2$

Therefore, the correct option is (C): $(\frac{3}{2})\text{MR}^2$