Question

Two rods $\text{AB}$ and $\text{CD}$ of the same mass and lengths $1\text{m}$ and $\frac{3}{2}\text{m}$ are joined at ${}^{\prime }{\text{O}}^{\prime }$ to form an $L$-shaped frame $\text{AOD}$ as shown in figure. Assuming $COM$ of a rod is at the mid point of the rod, the location of centre of mass of the $L$-shaped frame from the point of intersection ${}^{\prime }{\text{O}}^{\prime }$ is

• A. $(\frac{3}{8},\frac{3}{8})\text{m}$
• B. $(\frac{1}{4},\frac{1}{4})\text{m}$
• C. $(\frac{1}{4},\frac{3}{8})\text{m}$
• D. $(\frac{3}{8},\frac{1}{4})\text{m}$

Answer 1

The correct option is D $(\frac{3}{8},\frac{1}{4})\text{m}$


From the figure ${}^{\prime }{\text{O}}^{\prime }$ is considered as the origin of $x-y$ co-ordinate system.
Given that rods have same mass $\therefore m_1=m_2=m\left(\text{say}\right)$
COM of rod $AB(x_1,y_1)=(0,\frac{1}{2})$
COM of rod $CD(x_2,y_2)=(\frac{3}{4},0)$

Let $C$ be the centre of mass of the L - shaped frame
$x-$ coordinate of COM: $x_{cm}=\frac{m_{1}x+m_2{x}_2}{m_1+m_2}=\frac{m\left(0\right)+m\left(\frac{3}{4}\right)}{2m}=\frac{3}{8}\text{m}$
$y-$ coordinate of COM: $y_{cm}=\frac{m_1{y}_1+m_2{y}_2}{m_1+m_2}=\frac{m\left(\frac{1}{2}\right)+m\left(0\right)}{2m}=\frac{1}{4}\text{m}$
$\therefore (x_{cm},y_{cm})=(\frac{3}{8},\frac{1}{4})$$\text{m}$