Question

The moment of inertia of a solid cylinder of mass $M$, length $L$ and radius $R$ about the diameter of one of its faces will be :

• A. $M(\frac{L^2}{12}+\frac{R^2}{4})$
• B. $M(\frac{L^2}{4}+\frac{R^2}{4})$
• C. $Zero$
• D. $\frac{M{R}^2}{2}$

Answer 1

The correct option is B $M(\frac{L^2}{4}+\frac{R^2}{4})$

The moment of inertia of the cylinder about its rotational axis is $\frac{M{R}^2}{2}$. Now let the moment of inertia of any two perpendicular axes on the face with the centroidal axis be $I$. Thus using perpendicular axes theorem we have $2I=\frac{M{R}^2}{2}$
or
$I=\frac{M{R}^2}{4}$
Now using parallel axis theorem we get moment of inertia about the diameter of one of its faces will be $\frac{M{R}^2}{4}+M(\frac{L}{2}{)}^2=\frac{M{R}^2}{4}+\frac{M{L}^2}{4}=M(\frac{R^2}{4}+\frac{L^2}{4})$