Question

Two concentric rings, one with radius of $0.5\text{m}$ and mass $1\text{kg}$ and other having a radius of $1\text{m}$ and mass $2\text{kg}$ are placed together on a horizontal table. Find out net moment of inertia of the system, about an axis perpendicular to their plane and passing through their common centre.

• A. $1.75{\text{kg-m}}^2$
• B. $1.2{\text{kg-m}}^2$
• C. $2.25{\text{kg-m}}^2$
• D. $1.5{\text{kg-m}}^2$

Answer 1

The correct option is C $2.25{\text{kg-m}}^2$

considering mass and radius of both rings
$m_1=\text{1 kg}$
$r_1=\text{0.5 m}$
$m_2=\text{2 kg}$
$r_2=\text{1 m}$

$\text{MOI}$ of first ring about the axis $X{X}^{\prime }$ passing through it's centre:
$I_1=m_1{r}_1^2=\left(1\right)(0.5{)}^2$
$\therefore I_1=0.25{\text{kg-m}}^2$
$\text{MOI}$ of second ring about the axis $X{X}^{\prime }$ passing through it's centre:
$I_2=m_2{r}_2^2=\left(2\right)(1{)}^2$
$I_2=2{\text{kg-m}}^2$
$\Rightarrow$ Net moment of inertia of system is given by:
$I_{\text{net}}=I_1+I_2=0.25+2$
$\therefore I_{\text{net}}=2.25{\text{kg-m}}^2$