Question

Three Identical spheres each of diameter $2\sqrt{3}$ m are kept on a horizontal surface such that each sphere touches the other two spheres. If one of the spheres Is removed, then the shift In the position of the centre of mass of the system Is

• A. 12 m
• B. 1 m
• C. 2 m
• D. $\frac{3}{2}$m

Answer 1

Answer: (B)

1 m

Answer 2

1. Initial Setup: Three identical spheres of mass $m$ have their centers ($C_1, C_2, C_3$) forming an equilateral triangle with side length $s = 2R = 2\sqrt{3}$ m. The center of mass ($CM_{initial}$) of this system is at the centroid ($G$) of this triangle. The distance from the centroid to each vertex is $d_{cv} = s/\sqrt{3} = (2\sqrt{3})/\sqrt{3} = 2$ m.
2. Centroid Coordinates: Let the centroid $G$ be at the origin $(0,0)$ in the horizontal plane. The position vectors of the centers satisfy $\vec{r}_1 + \vec{r}_2 + \vec{r}_3 = \vec{0}$. Thus, $CM_{initial}$ is at $(0,0)$.
3. Final Setup: When one sphere (say, sphere 3) is removed, two spheres ($m_1=m, m_2=m$) remain. Their center of mass ($CM_{final}$) is at the midpoint ($M$) of the line segment connecting $C_1$ and $C_2$.
4. Final CM Coordinates: $CM_{final} = \frac{\vec{r}_1 + \vec{r}_2}{2}$. Using $\vec{r}_1 + \vec{r}_2 = -\vec{r}_3$, we get $CM_{final} = \frac{-\vec{r}_3}{2}$.
5. Shift Calculation: The shift in position is the vector displacement from $CM_{initial}$ to $CM_{final}$: $\Delta \vec{r} = CM_{final} - CM_{initial} = \frac{-\vec{r}_3}{2} - \vec{0} = \frac{-\vec{r}_3}{2}$.
6. Magnitude of Shift: The magnitude of the shift is $|\Delta \vec{r}| = \frac{1}{2}|\vec{r}_3|$. Since $|\vec{r}_3|$ is the distance from the centroid to a vertex, $|\vec{r}_3| = d_{cv} = 2$ m. Therefore, the shift is $\frac{1}{2} \times 2$ m $= 1$ m.