Question

Moment of inertia of a uniform semicircular disc about an axis passing through center of mass and parallel to diameter as shown in figure is: [take${\pi }^2=10$]

• A. $\frac{3M{R}^2}{172}$
• B. $\frac{M{R}^2}{12}$
• C. $\frac{13M{R}^2}{180}$
• D. $\frac{7}{90}M{R}^2$

Answer 1

The correct option is C $\frac{13M{R}^2}{180}$


Since $I=I_C+M{y}^2$
$I_C=I-M{y}^2$
$I_C=\frac{M{R}^2}{4}-M{R}^2\times \frac{16}{90}=\frac{M{R}^2}{1}[\frac{1}{4}-\frac{16}{90}]$
$I_C=\frac{13}{180}M{R}^2$