Question

Four identical rods of mass $M=6\text{kg}$ each are welded at their ends to form a square and then welded to a massive ring having mass $m=4\text{kg}$ and radius $R=1\text{m}$. If the system is allowed to roll down the incline of inclination $\theta ={30}^{\circ }$.


The moment of inertia of the system about the centre of ring will be

• A. $20{\text{kg m}}^2$
• B. $40{\text{kg m}}^2$
• C. $10{\text{kg m}}^2$
• D. $60{\text{kg m}}^2$

Answer 1

The correct option is A $20{\text{kg m}}^2$

Moment of inertia is given by,
$I_o=I_{ring}+4I_{rod}{I}_o=m{R}^2+4I_{rod}$ ...(1)
using parallel axis theorem :
$I_{y{y}^{\prime }}=\frac{M{L}^2}{12}+M{x}^2$
$I_{\left(rod\right)}=\frac{M(\frac{2R}{\sqrt{2}}{)}^2}{12}+M(\frac{R}{\sqrt{2}}{)}^2{I}_{rod}=\frac{2}{3}M{R}^2$

For all 4 rods= $4\times \frac{2}{3}M{R}^2=\frac{8}{3}M{R}^2$

putting eq. (1)
$I_0=m{R}^2+\frac{8}{3}M{R}^2$
where, $m=4kg,M=6kg,R=1m$
$I_0=4+\frac{8}{3}6=20Kg-m^2$