Question

Figure shows a coordinate system in which 8 uniform cubes of side a are placed as shown. If masses of cubes 1, 2, 3 & 4 is m of each & 2m for remaining cubes, find location of centre of mass of this system.

Answer 1

The centre of mass of the system is at

$$\left(\frac{3a}{2},\frac{4a}{3}\right).$$

Answer 2

Solution:

1. Assign centers:

   • Let the bottom row have cubes 1, 2, 3, 4 with centers at

     Cube 1: $\left(\frac{a}{2},\frac{a}{2}\right)$

     Cube 2: $\left(\frac{3a}{2},\frac{a}{2}\right)$

     Cube 3: $\left(\frac{5a}{2},\frac{a}{2}\right)$

     Cube 4: $\left(\frac{7a}{2},\frac{a}{2}\right)$

   • The cubes on top:

     Cube 5 on cube 1: $\left(\frac{a}{2},\frac{3a}{2}\right)$

     Cube 6 on cube 2: $\left(\frac{3a}{2},\frac{3a}{2}\right)$

     Cube 7 on cube 3: $\left(\frac{5a}{2},\frac{3a}{2}\right)$

     Cube 8 on cube 5: $\left(\frac{a}{2},\frac{5a}{2}\right)$

2. Masses:

   • Cubes 1, 2, 3, 4 each have mass $m$.
   • Cubes 5, 6, 7, 8 each have mass $2m$.

3. Total mass:

   $$M = 4m + 4(2m) = 12m.$$

4. Calculate $x$-coordinate of centre of mass:

   • Bottom row:

     $$\sum x_{\text{bottom}} = \frac{a}{2} + \frac{3a}{2} + \frac{5a}{2} + \frac{7a}{2} = \frac{16a}{2} = 8a.$$

   • Top cubes:

     $$\sum x_{\text{top}} = \frac{a}{2} + \frac{3a}{2} + \frac{5a}{2} + \frac{a}{2} = \frac{10a}{2} = 5a.$$

   • Weighted sum:

     $$x_{cm} = \frac{m(8a) + 2m(5a)}{12m} = \frac{8ma + 10ma}{12m} = \frac{18a}{12} = \frac{3a}{2}.$$

5. Calculate $y$-coordinate of centre of mass:

   • Bottom row contribution (each at $\frac{a}{2}$)

     $$\text{Bottom moment } = 4m\left(\frac{a}{2}\right) = 2ma.$$

   • Top cubes:

     Cube 5,6,7 each at $y=\frac{3a}{2}$ and Cube 8 at $y=\frac{5a}{2}$

     $$\text{Top moment } = 2m\left(\frac{3a}{2}+\frac{3a}{2}+\frac{3a}{2}+\frac{5a}{2}\right) = 2m\left(\frac{14a}{2}\right) = 14ma.$$

   • Total moment:

     $$y_{cm} = \frac{2ma+14ma}{12m} = \frac{16a}{12} = \frac{4a}{3}.$$

Summary:

Assign cube centers using geometry. Compute weighted average:

$x_{cm} = \frac{\sum m_i x_i}{M}=\frac{18a m}{12m}=\frac{3a}{2}$.

$y_{cm} = \frac{\sum m_i y_i}{M}=\frac{16a m}{12m}=\frac{4a}{3}$.