Question

Four cubes of side '$a$' each of mass $40gm,20gm,10gm,20gm$ are arranged in $XY$ plane as shown in figure. The coordinates of the centre of mass of the combination with respect to origin are:

• A. $19a/18,17a/18$
• B. $17a/18,11a/18$
• C. $17a/18,13a/18$
• D. $13a/18,17a/18$

Answer 1

Answer: (A)

$19a/18,17a/18$

Here, $m_4=40gm$, $m_1=20gm$, $m_2=10gm$ and $m_3=20gm$

Center of mass of each cube lies at its geometrical centre.

Also, $x_1=0.5a$, $x_2=1.5a$, $x_3=2.5a$ and $x_4=0.5a$

Also, we get $y_1=y_2=y_3=0.5a$ and $y_4=1.5a$

X-coordinate of centre of mass $X_{cm}=\frac{m_1{x}_1+m_2{x}_2+m_3{x}_3+m_4{x}_4}{m_1+m_2+m_3+m_4}$

$⟹$ $X_{cm}=\frac{\left(20\right)\left(0.5a\right)+\left(10\right)\left(1.5a\right)+\left(20\right)\left(2.5a\right)+\left(40\right)\left(0.5a\right)}{20+10+20+40}=\frac{19}{18}a$

Y-coordinate of centre of mass $Y_{cm}=\frac{m_1{y}_1+m_2{y}_2+m_3{y}_3+m_4{y}_4}{m_1+m_2+m_3+m_4}$

$⟹$ $Y_{cm}=\frac{\left(20\right)\left(0.5a\right)+\left(10\right)\left(0.5a\right)+\left(20\right)\left(0.5a\right)+\left(40\right)\left(1.5a\right)}{20+10+20+40}=\frac{17}{18}a$