Question

In the figure given below, four identical circular rings of mass $10\text{kg}$ and radius $1\text{m}$ each, are lying in the same plane. The moment of inertia (in ${\text{kg m}}^2$) of the system about an axis through point $A$ and perpendicular to the plane of the rings is

• A. 880.00
• B. 880
• C. 880.0

Answer 1

Answer: (A), (B), (C)

Given,
Radius of each circular ring $\left(r\right)=1\text{m}$
Mass of each circular ring $\left(m\right)=10\text{kg}$

Moment of inertia of first circular ring about perpendicular axis passing through ${}^{\prime }{A}^{\prime }$
$I_{A1}=m{r}^2+m{r}^2=2m{r}^2$
(using parallel axis theorem)
$=2\times 10(1{)}^2=20{\text{kg-m}}^2$
Similarly,
Moment of inertia of 2nd circular ring about axis passing through ${}^{\prime }{A}^{\prime }$.
$I_{A2}=m{r}^2+m(3r{)}^2=10m{r}^2=100{\text{kg-m}}^2$
Moment of inertia of 3rd circular ring about axis passing through ${}^{\prime }{A}^{\prime }$
$I_{A3}=m{r}^2+m(5r{)}^2=26m{r}^2=260{\text{kg-m}}^2$
Moment of inertia of 4th circular ring about axis passing through ${}^{\prime }{A}^{\prime }$
$I_{A4}=m{r}^2+m(7r{)}^2=50m{r}^2=500{\text{kg-m}}^2$

Therefore, total moment of inertia about axis passing through ${}^{\prime }{A}^{\prime }$
$I_A=20+100+260+500=880{\text{kg-m}}^2$