Question

If the moment of inertia of a circular disc about an axis tangentially and parallel to its surface be $I$, then the moment of inertia about the axis tangential but perpendicular to the surface will be :

• A. $\frac{6}{5}I$
• B. $\frac{3}{4}I$
• C. $\frac{3}{2}I$
• D. $\frac{5}{4}I$

Answer 1

The correct option is A $\frac{6}{5}I$

$I_{A{A}^1}=I$ also $I=\frac{M{R}^2}{4}+M{R}^2=\frac{5}{4}M{R}^2$

$I_{B{B}^1}=I-M{R}^2$

$I_{C{C}^1}=2I_{B{B}^1}$ ( parallel axis thereon )

$=2(I-M{R}^2)$

$I_{00}=I_{C{C}^1}+M{R}^2$

$=2I-2M{R}^2+M{R}^2$

$=2I-M{R}^2$

$=2\times \frac{5}{4}M{R}^2-M{R}^2$

$=\frac{3}{2}M{R}^2$

$=\frac{3}{2}\times \frac{4}{5}\times \left(\frac{5}{4}M{R}^2\right)$

$=\frac{6}{5}I.$