91Ó°ÊÓ

For symmetrical shape, the center of mass is located at the center.The cube is symmetrical in shape; so the CM of an individual cube is at its center.

The sides of the cubes are \({l_ \circ }\), \(2{l_ \circ }\), and \(3{l_ \circ }\).

Let \(\rho \)be the density of the material by which the cubes are made.

Now, the mass of the small cube is \({m_1} = \rho {\left( {{l_ \circ }} \right)^3}\).

The mass of the medium cube is:

\(\begin{array}{c}{m_2} = \rho {\left( {2{l_ \circ }} \right)^3}\\ = 8\rho {\left( {{l_ \circ }} \right)^3}\\ = 8{m_1}\end{array}\).

The mass of the large cube is:

\(\begin{array}{c}{m_3} = \rho {\left( {3{l_ \circ }} \right)^3}\\ = 27\rho {\left( {{l_ \circ }} \right)^3}\\ = 27{m_1}\end{array}\).

The cube is symmetrical in shape; therefore, the CM of an individual cube is in the middle of the cube.

So, the distance of the CM of the small cube is at \({x_1} = \frac{{{l_ \circ }}}{2}\).

The distance of the CM of the medium cube is at:

\(\begin{array}{c}{x_2} = {l_ \circ } + \frac{{2{l_ \circ }}}{2}\\ = 2{l_ \circ }\end{array}\)

The distance of the CM of the large cube is at:

\(\begin{array}{c}{x_3} = {l_ \circ } + 2{l_ \circ } + \frac{{3{l_ \circ }}}{2}\\ = \frac{9}{2}{l_ \circ }\end{array}\)

Now, the center of mass of the cube system is:

\(\begin{array}{c}{x_{{\rm{CM}}}} = \frac{{\left( {{m_1} \times {x_1}} \right) + \left( {{m_2} \times {x_2}} \right) + \left( {{m_2} \times {x_2}} \right)}}{{{m_1} + {m_2} + {m_3}}}\\ = \frac{{\left( {{m_1} \times \frac{{{l_ \circ }}}{2}} \right) + \left( {8{m_1} \times 2{l_ \circ }} \right) + \left( {27{m_1} \times \frac{9}{2}{l_ \circ }} \right)}}{{{m_1} + 8{m_1} + 27{m_1}}}\\ = \frac{{\frac{{{m_1}{l_ \circ }}}{2}\left( {1 + 32 + 243} \right)}}{{36{m_1}}}\\ = 3.83{l_ \circ }\end{array}\)

Hence, the center of mass of the cube system is at \(3.83{l_ \circ }\).