Question

The position vector of the centre of mass $r _{\text{cm}}$ of an asymmetric uniform bar of negligible area of cross-section as shown in figure is

• A. $r =\frac{13}{8} L \hat{ x }+\frac{5}{8} L \hat{ y }$
• B. $r =\frac{11}{8} L \hat{ x }+\frac{3}{8} L \hat{ y }$
• C. $r =\frac{3}{8} L \hat{ x }+\frac{11}{8} L \hat{ y }$
• D. $r =\frac{5}{8} L \hat{ x }+\frac{13}{8} L \hat{ y }$

Answer 1

Answer: (A)

Coordinates of centre of mass (COM) are given by
$X_{ COM }=\frac{m_{1} x_{1}+m_{2} x_{2}+m_{3} x_{3}}{m_{1}+m_{2}+m_{3}}$
$Y_{ COM }=\frac{m_{1} y_{1}+m_{2} y_{2}+m_{3} y_{3}}{m_{1}+m_{2}+m_{3}}$
For given system of rods, masses and coordinates of centre of rods are as shown.

So, $X_{ COM }=\left(\frac{2 m L+m 2 L+m \frac{5 L}{2}}{4 m}\right)=\frac{13}{8} L$ and $Y_{ COM }=\frac{2 m L+m \times \frac{L}{2}+m \times 0}{4 m}=\frac{5 L}{8}$ So, position vector of COM is
$r _{ COM } =X_{ COM } \hat{ x }+Y_{ COM } \hat{ y }$
$=\frac{13}{8} L \hat{ x }+\frac{5}{8} L \hat{ y }$