Question

The M.I. of a disc about an axis passing through its centre and perpendicular to plane is $\frac{1}{2}M{R}^2$, then its M.I. about a tangent parallel to its diameter is

• A. $\frac{1}{2}M{R}^2$
• B. $\frac{4}{5}M{R}^2$
• C. $\frac{5}{4}M{R}^2$
• D. $\frac{3}{4}M{R}^2$

Answer 1

The correct option is C $\frac{5}{4}M{R}^2$

$I_{zc}=\frac{M{R}^2}{2}$

$I_{ZC}=I_{YC}{I}_{XC}+I_{YC}=I_{ZC}$

By perpendicular axis theorem,

$\Rightarrow I_{XC}=I_{YC}=\frac{I_{ZC}}{2}=\frac{M{R}^2}{4}$

$I_{target}=I_{XC}+M{R}^2$

Therefore,

$I_{target}=\frac{M{R}^2}{4}+M{R}^2=\frac{5}{4}M{R}^2$

Hence the correct answer is $\frac{5}{4}M{R}^2$