Question

Consider an "L"-shaped object where the vertical leg extends from the origin to $(0,5)$ and the horizontal segment extends from the origin to $(3,0)$. Assume that the object has uniform linear density throughout.
Where is the location of the center of mass $\left(CoM\right)$?

• A. $(1.5,1.5)$
  
• B. $(1.5,2.5)$
  
• C. $(2.5,2.5)$
  
• D. The center of mass for an "L" is undefined because there is no single point on the object where it can be supported, such that it will not rotate
  
• E. None of these

Answer 1

Answer: (E)

None of these

Let the mass per unit length of the object be $\rho$.

Thus mass of horizontal part $m=3\rho$

Mass of vertical part $M=5\rho$

Also center of mass of individual parts lie at their respective centres.

Horizontal direction :

$x_{cm}=\frac{ml+M\left(0\right)}{m+M}=\frac{3\rho \times \frac{3}{2}+0}{3\rho +5\rho }=\frac{9}{16}$

Vertical direction :

$y_{cm}=\frac{m\left(0\right)+M\left(L\right)}{m+M}=\frac{0+5\rho \times \frac{5}{2}}{3\rho +5\rho }=\frac{25}{16}$

Thus, COM lies at $(\frac{9}{16},\frac{25}{16})$