Question

The moment of inertia of a cylinder of radius $a$, mass $M$ and height $h$ about an axis parallel to the axis of the cylinder and distance $b$ from its centre is

• A. $\frac{3}{4}M{(a-b)}^2$
• B. $\frac{2}{3}M{(a+b)}^2$
• C. $\frac{3}{2}M{a}^2$
• D. $\frac{M(a^2+2b^2)}{2}$

Answer 1

The correct option is D $\frac{M(a^2+2b^2)}{2}$

First of all the moment of inertia about the $axis$ of the cylinder will be same as that of a $disc$ with radius $a$ and mass $M$

Also, this axis will pass from the $center$ of mass so the moment of inertia can be written as $I_{cm}=\frac{M{a}^2}{2}$

$\text{Now as the given axis is *parallel* to the axis from the center of maas}$

so we may apply $\text{parallel axis theorem}$ so required I is $I=I_{cm}+M{b}^2=\frac{M(a^2+2b^2)}{2}$

Option D is correct.