Question

Find the moment of inertia of ring having mass $2\text{kg}$ and radius $1\text{m}$ about an axis passing along the diameter of the ring.

• A. $2{\text{kg m}}^2$
• B. $1{\text{kg m}}^2$
• C. $0.5{\text{kg m}}^2$
• D. $4{\text{kg m}}^2$

Answer 1

The correct option is B $1{\text{kg m}}^2$

Given, mass of the ring $M=2\text{kg}$
Radius of the ring $R=1\text{m}$

Here, $I_x$ and $I_y$ is the moment of inertia of the ring about an axis passing along the diameter in $x$ and $y$ direction respectively.
$I_z$ is the moment of inertia about an axis passing through center of mass of ring along $z-$ axis, perpendicular to the plane of ring.

Since, $x-$ axis, $y-$ axis and $z-$ axis are mutually perpendicular to each other.
So, from perpendicular axis theorem,
$I_z=I_x+I_y$
$I_z=2I_x\because I_x=I_y$
or $I_z=I_{COM}=2I_x$

But, for uniform ring $I_z=I_{COM}=M{R}^2$

$\Rightarrow I_x=\frac{I_z}{2}=\frac{I_{COM}}{2}=\frac{M{R}^2}{2}$
$\Rightarrow I_x=\frac{M{R}^2}{2}$
Put $M=2\text{kg}$, and $R=1\text{m}$
$\Rightarrow I_x=\frac{\left(2\right)(1{)}^2}{2}=1$
$\Rightarrow I_x=1{\text{kg m}}^2$
So, moment of inertia about an axis passing along the diameter of the ring is $1{\text{kg m}}^2$