Question

A point mass $m_A$ is connected to a point mass $m_B$ by a massless rod of length $l$ as shown in the figure. It is observed that the ratio of the moment of inertia of the system about the two axes $BB$ and $AA$, which are parallel to each other and perpendicular to the rod is $\frac{I_{BB}}{I_{AA}}=3$
The distance of the centre of mass of the system from the $\mathrm{mass}A$ is

• A. $\frac{1}{2}l$
• B. $\frac{1}{4}l$
• C. $\frac{2}{3}l$
• D. $\frac{3}{4}l$

Answer 1

Answer: (B)

$I_{AA}=m_B{l}^2$ and $I_{BB}=m_A{l}^2$
$\frac{I_{BB}}{I_{AA}}=3($ Given $);\therefore \frac{m_A}{m_B}=3...(i)$
Let $x$ be the distance of centre of mass from mass $A$.
$\therefore m_{A}x=m_B(l-x)$ or $\frac{m_A}{m_B}x=l-x$
or $3x=l-x$ or $x=\frac{l}{4}$ (Using(i))