Question

Four masses, $m_1=1\text{Kg},m_2=3\text{Kg},$
$m_3=4\text{Kg}\text{and}{m}_4=2\text{Kg},$ are fixed on the circumference of a light circular frame of radius $1\text{m}$. The radius of gyration of this system about an axis passing through the centre of the circular frame and perpendicular to its plane would be,

• A. $1\text{m}$
• B. $10\text{m}$
• C. $0.5\text{m}$
• D. $1.5\text{m}$

Answer 1

The correct option is A $1\text{m}$

Given, $r=1\text{m}$
The masses, $m_1=1\text{Kg}$
$m_2=3\text{Kg}$
$m_3=4\text{Kg}$
$m_4=2\text{Kg}$
Moment of inertia of mass $m_1$ $I_1=m_1{r}^2=1\times (1{)}^2=1{\text{kg m}}^2$

Moment of inertia of mass $m_1$ $I_2=m_2{r}^2=3\times (1{)}^2=3{\text{kg m}}^2$

Moment of inertia of mass $m_1$ $I_3=m_3{r}^2=4\times (1{)}^2=4{\text{kg m}}^2$

Moment of inertia of mass $m_1$ $I_4=m_4{r}^2=2\times (1{)}^2=2{\text{kg m}}^2$


MOI of the system is,
$I=I_1+I_2+I_3+I_4=10{\text{kg m}}^2$

Mass of the system is,
$m=m_1+m_2+m_3+m_4=1+3+4+2=10\text{kg}$

And radius of gyration is given by,
$K=\sqrt{\frac{I}{m}}=\sqrt{\frac{10}{10}}=1\text{m}$

Hence, $\left(A\right)$ is the correct answer.