Question

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

\( y_{A} \) \( \begin{array}{|l|l|l|} \hline 40 & \multicolumn{2}{|c|}{} \\ \hline 20 & 10 & 20 \\ \hline \end{array} \) \( \vec{x} \)

• A. $\frac{13}{18}a,\frac{17}{18}a$
• B. $\frac{17}{18}a,\frac{13}{18}a$
• C. $\frac{19}{18}a,\frac{17}{18}a$
• D. $\frac{17}{18}a,\frac{11}{18}a$

Answer 1

The correct option is C

$\frac{19}{18}a,\frac{17}{18}a$

Step 1: Given data

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

[where $m_1=massforcube1,m_2=massforcube2,m_3=massforcube3,m_4=massforcube4$ ]

Step 2: Calculating coordinates of the center of mass with respect to the origin

According to the given figure, we can get the value of the x-axis point and y-axis points for the center of the cubes,

$$\begin{gathered} x_1=0.5a,x_2=1.5a,x_3=2.5a,x_4=0.5a \\ {y}_1=y_2=y_3=0.5a,y_4=1.5a \end{gathered}$$

X-coordinate of the center of mass

$$\begin{gathered} X_{cm}=\frac{m_1{x}_1+m_2{x}_2+m_3{x}_3+m_4{x}_4}{m_1+m_2+m_3+m_4} \\ {X}_{cm}=\frac{\left(20\right)(0.5a)+\left(10\right)(1.5a)+\left(20\right)(2.5a)+\left(40\right)(0.5a)}{20+10+20+40} \\ {X}_{cm}=\frac{19}{18}a \end{gathered}$$

Y-coordinate of the center of mass

$$\begin{gathered} Y_{cm}=\frac{m_1{y}_1+m_2{y}_2+m_3{y}_3+m_4{y}_4}{m_1+m_2+m_3+m_4} \\ {Y}_{cm}=\frac{\left(20\right)(0.5a)+\left(10\right)(0.5a)+\left(20\right)(0.5a)+\left(40\right)(1.5a)}{20+10+20+40} \\ {Y}_{cm}=\frac{17}{18}a \end{gathered}$$

Hence, option C is the correct answer.