Question

Two steel spheres of radius $r$ and $2r$ are made to touch each other. The distance of their center of mass from their point of contact is:
A) at a distance $\dfrac{{8r}}{3}$ in the bigger space
B) at a distance $\dfrac{r}{3}$ in the smaller sphere
C) at a distance $\dfrac{{5r}}{3}$ in the bigger sphere
D) at a distance $\dfrac{r}{3}$ in the bigger sphere

Answer 1

Hint: Use the formula of the centre of mass and substitute the centre of mass of the two spheres in it. Use the formula to calculate the value of the mass and get substituted in the above formed equation. The simplification of it provides the result.

Formula used:
(1) The density is given by
$\rho = \dfrac{m}{V}$
Where $\rho $ is the density of the spheres, $m$ is the mass of the spheres and $V$ is the volume of the spheres.
(2) The centre of mass is given by
$CM = \dfrac{{{m_1}{x_1} + {m_2}{x_2}}}{{{m_1} + {m_2}}}$
Where ${m_1}\,and\,{m_2}$ are the masses of the two spheres and the ${x_1}\,and\,{x_2}$ are the distance of the centre of mass from the point of contact.

Complete step by step solution:
It is given that the
Radius of the first sphere is $r$
Radius of the second sphere is $2r$
It is known that the centre of mass of the first sphere is $0$ and the centre of mas of the second sphere is $3r$ .
Using the formula (2) of the centre of mass,
$\Rightarrow CM = \dfrac{{{m_1}{x_1} + {m_2}{x_2}}}{{{m_1} + {m_2}}}$
Substituting the known values in the above step,
$\Rightarrow CM = \dfrac{{{m_1}\left( 0 \right) + {m_2}\left( {3r} \right)}}{{{m_1} + {m_2}}}$
Substituting the formula (1) by rearranging it in the above step.
$\Rightarrow CM = \dfrac{{\rho \dfrac{4}{3}\pi {{\left( {2r} \right)}^3}\left( {3r} \right)}}{{\rho \dfrac{4}{3}\pi {{\left( {2r} \right)}^3} + \rho \dfrac{4}{3}\pi {{\left( r \right)}^3}}}$
By simplifying the above equation, we get
$\Rightarrow CM = \dfrac{3}{9}r$
By further simplification,
$\Rightarrow CM = \dfrac{r}{3}$
Hence the centre of mass is located at a distance of $\dfrac{r}{3}$ from the centre.

Hence the option (D) is correct.

Note: The mass is the product of the density and the volume (from formula (1)) . The given figure is the sphere, hence the volume is $\dfrac{4}{3}\pi {r^3}$. The radius is substituted in it and simplified to obtain the value of the coordinates of the centre of mass.