Question

Two identical solid sphere each of mass $m=5\text{kg}$ and radius $r=1\text{m}$ are joined at the ends of a light rod of length $l=2\text{m}$ to form a system as shown in figure. Radius of gyration of the system about an axis perpendicular to the length of rod and passing through center of mass of system is

• A. $\sqrt{44}\text{m}$
• B. $\sqrt{2.4}\text{m}$
• C. $\sqrt{3.2}\text{m}$
• D. $\sqrt{4.4}\text{m}$

Answer 1

The correct option is D $\sqrt{4.4}\text{m}$


Given:
Mass of each sphare $m=5\text{kg}$
Radius of sphere $r=1\text{m}$
We know, moment of inertia of the sphere about its center of mass $I_{com\left(sphere\right)}=\frac{2}{5}m{r}^2$

Using parallel axes theorem, moment of inertia of individual sphere about the rotational axis $I_{com\left(system\right)}=I_{com\left(sphere\right)}+m{d}^2$
$\text{Here}d\text{=}(r+\frac{l}{2})=1+\frac{2}{2}=2$
$\Rightarrow I_{com\left(system\right)}=\frac{2}{5}m{r}^2+m\times 2^2$
$=\frac{2}{5}\times 5\times (1{)}^2+5\times (2{)}^2=2+20=22{\text{kg-m}}^2$

As there are two identical sphere having same axis of rotation, total moment of inertia about $I_{com\left(system\right)}$ is
$=2\times 22=44{\text{kg-m}}^2$

We know ,
Radius of gyration $K=\sqrt{\frac{I_{com\left(system\right)}}{\text{Mass of system}}}K=\sqrt{\frac{44}{2\times 5}}\text{m}=\sqrt{4.4}\text{m}$