The Question is from neet-physics rotational-motion , and it was created by the author NAGA SAI

Question

A system consists of three particles of masses 1 kg, 2 kg, and 3 kg located at the vertices of an equilateral triangle of side $L$.  If the 1 kg mass is removed, by what distance does the center of mass of the system shift?

NAGA SAI · Accepted Answer

## CORE INFORMATION
✓ **Answer**: C
## CONCEPT FRAMEWORK
	**Primary Concept**: Center of Mass
	**Related Topics**: Vector Addition, Coordinate Systems
	**Prerequisites**: Definition of center of mass, vector components
## SOLUTION PATHWAY
**Given Information**:
	* Three masses: $m_1 = 1$ kg, $m_2 = 2$ kg, $m_3 = 3$ kg
	* Equilateral triangle of side $L$
	* $m_1$ is removed
**Method & Steps**:
	1.	Define a coordinate system and find the initial center of mass.
		Reasoning:  We need a reference frame to calculate the center of mass. Let's place the 1 kg mass at the origin (0,0), the 2 kg mass at (L,0), and the 3 kg mass at (L/2, L*sqrt(3)/2).
		Formula/Rule: Center of mass coordinates: $x_{cm} = \frac{\sum m_i x_i}{\sum m_i}$ and $y_{cm} = \frac{\sum m_i y_i}{\sum m_i}$
	2.	Calculate the initial center of mass coordinates.
		Working: $x_{cm,initial} = \frac{1(0) + 2(L) + 3(L/2)}{1+2+3} = \frac{2L + 1.5L}{6} = \frac{3.5L}{6} = \frac{7L}{12}$ and $y_{cm,initial} = \frac{1(0) + 2(0) + 3(L\sqrt{3}/2)}{1+2+3} = \frac{3L\sqrt{3}/2}{6} = \frac{L\sqrt{3}}{4}$
		Key Point: The coordinates are with respect to the chosen coordinate system.
	3.	Calculate the new center of mass after removing the 1 kg mass.
		Working: $x_{cm,final} = \frac{2L + 3(L/2)}{2+3} = \frac{2L + 1.5L}{5} = \frac{3.5L}{5} = \frac{7L}{10}$ and $y_{cm,final} = \frac{0 + 3(L\sqrt{3}/2)}{2+3} = \frac{3L\sqrt{3}/2}{5} = \frac{3L\sqrt{3}}{10}$
	4.	Find the shift in the center of mass using the distance formula.
		Working:  The shift is given by $\Delta r = \sqrt{(x_{cm,final} - x_{cm,initial})^2 + (y_{cm,final} - y_{cm,initial})^2}$. Substituting the values, we get $\Delta r = \sqrt{(\frac{7L}{10}-\frac{7L}{12})^2 + (\frac{3L\sqrt{3}}{10} - \frac{L\sqrt{3}}{4})^2} = \sqrt{(\frac{42L-35L}{60})^2 + (\frac{6L\sqrt{3}-5L\sqrt{3}}{20})^2} =  \sqrt{(\frac{7L}{60})^2 + (\frac{L\sqrt{3}}{20})^2} = \sqrt{\frac{49L^2}{3600} + \frac{3L^2}{400}} = \sqrt{\frac{49L^2 + 27L^2}{3600}} = \sqrt{\frac{76L^2}{3600}} = L\sqrt{\frac{19}{900}} = \frac{L\sqrt{19}}{30}$
		Result: The shift in the center of mass is $\frac{L\sqrt{19}}{30}$
		Verification: The calculated value matches with option C.
 ## OPTION ANALYSIS
[A] :x: Incorrect. The calculation does not simplify to L/6.
[B] :x: Incorrect. The calculated value does not simplify to $L\sqrt{3}/10$.
[C] ✓: Correct. The calculated shift matches this option.
[D] :x: Incorrect. The calculated value does not simplify to $L\sqrt{7}/15$.
 ## LEARNING AIDS
	**Common Mistakes**:
		1.[Error 1]: Forgetting to consider the vector nature of the position vectors. Use x and y components separately.
		2.[Error 2]: Incorrectly applying the center of mass formula by not taking the sum of all masses in the denominator.
	**Quick Tips**:
		1.[Helpful shortcut]: Use a well-chosen coordinate system to simplify calculations.
		2.[Memory aid]: Remember the center of mass is a weighted average of the positions of the masses.

NAGA SAI · Answer

L/6

NAGA SAI · Answer

L*sqrt(3)/10

NAGA SAI · Answer

The correct Option is "L*sqrt(19)/30"

NAGA SAI · Answer

L*sqrt(7)/15

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