Question

Four spheres of radii $R$, $2R$, $2R$ and $3R$ respectively, are placed such that their respective centres lie at $x=2\text{m}$, $x=6\text{m}$, $x=10\text{m}$ and $x=20\text{m}$ respectively. If density of each sphere is the same, then find the $x$ coordinate of $\text{COM}$ of the system of spheres.

• A. $x=10\text{m}$
• B. $x=16\text{m}$
• C. $x=16.44\text{m}$
• D. $x=15.22\text{m}$

Answer 1

The correct option is D $x=15.22\text{m}$

Let mass of $1^{\text{st}}$ sphere be $m$, density of material $={\rho }_0$
Then, $m_1=m={\rho }_0\times \frac{4}{3}\pi R^3$
Mass of second sphere is $m_2={\rho }_0\times \frac{4}{3}\pi (2R{)}^3=\frac{8{\rho }_0\pi R^3}{3}$
$\Rightarrow m_2=8m$
Similarly,
$m_3=8m,m_4=27\times \frac{4\pi R^3}{3}=27m$

Replacing the spheres by equivalent point masses placed at their respective $\text{COM}$ and applying formula for system's $\text{COM}$:
$x_{\text{CM}}=\frac{m_1{x}_1+m_2{x}_2+m_3{x}_3+m_4{x}_4}{m_1+m_2+m_3+m_4}$
$\Rightarrow x_{\text{CM}}=\frac{(m\times 2)+(8m\times 6)+(8m\times 10)+(27m\times 20)}{m+8m+8m+27m}$
$\therefore x_{\text{CM}}=\frac{670m}{44m}=15.22\text{m}$