Question

Four particles of masses \[{m_1},{m_2},{m_3}\] ​ and \[{m_4}\]are placed at the vertices A,B,C, and D as shown respectively in the square shown. Then what is the condition for the centre of mass of the system which lies at diagonal AC.


A. \[{m_1} = {m_3}\]
B. \[{m_2} = {m_4}\]
C. \[{m_1} = {m_2}\]
D. \[{m_3} = {m_4}\]

Answer 1

Hint:Here before we proceed with the problem let’s know about the centre of mass. Centre of mass of a body or a system is defined as a point where all of the system’s mass is supposed to be concentrated, and this point like a mass has the same kind of translational motion as the system as a whole, if the same net external force operates on this point like mass as on the system as a whole.

Formula Used:
The formula to find the centre of mass we have,
\[{X_{CM}} = \dfrac{{{m_1}{x_1} + {m_2}{x_2} + {m_3}{x_3} + {m_4}{x_4}}}{{{m_1} + {m_2} + {m_3} + {m_4}}}\]
Where,
\[{m_1},{m_2},{m_3}\] are the masses of three particles.
\[{x_1},{x_2},{x_3}\] are the positions of masses.

Complete step by step solution:

Image: Four particles are placed at the four vertices of a square.

Suppose, consider a square in which all the four particles are placed at all the four vertices as shown and we are going to find the condition for the centre of mass of the system which lies at diagonal AC. The centre of mass\[{X_{CM}}\] is given by,
\[{X_{CM}} = \dfrac{{{m_1}{x_1} + {m_2}{x_2} + {m_3}{x_3} + {m_4}{x_4}}}{{{m_1} + {m_2} + {m_3} + {m_4}}}\]

From diagram, we have \[\left( {{x_1},{y_1}} \right) = \left( {0,0} \right)\] , \[\left( {{x_2},{y_2}} \right) = \left( {0,a} \right)\], \[\left( {{x_3},{y_3}} \right) = \left( {a,a} \right)\] and \[\left( {{x_4},{y_4}} \right) = \left( {a,0} \right)\]
\[{X_{CM}} = \dfrac{{{m_1}\left( 0 \right) + {m_2}\left( 0 \right) + {m_3}\left( a \right) + {m_4}\left( a \right)}}{{{m_1} + {m_2} + {m_3} + {m_4}}} \\ \]
\[\Rightarrow {X_{CM}} = \dfrac{{{m_3}a + {m_4}a}}{{{m_1} + {m_2} + {m_3} + {m_4}}} \\ \]
Similarly,
\[{Y_{CM}} = \dfrac{{{m_1}{y_1} + {m_2}{y_2} + {m_3}{y_3} + {m_4}{y_4}}}{{{m_1} + {m_2} + {m_3} + {m_4}}} \\ \]
\[\Rightarrow {Y_{CM}} = \dfrac{{{m_1}\left( 0 \right) + {m_2}\left( a \right) + {m_3}\left( a \right) + {m_4}\left( 0 \right)}}{{{m_1} + {m_2} + {m_3} + {m_4}}} \\ \]
\[\Rightarrow {Y_{CM}} = \dfrac{{{m_2}\left( a \right) + {m_3}\left( a \right)}}{{{m_1} + {m_2} + {m_3} + {m_4}}} \\ \]
So, if \[{m_2} = {m_4}\] then only the centre of mass of the system will lie at diagonal AC.
Therefore, the condition for the centre of mass in which the system will lie at diagonal AC is,
\[{m_2} = {m_4}\]

Hence, option B is the correct answer.

Note: Remember that, for the square and the rectangular objects, the centre of mass lies at the point where the diagonals meet. The centre of mass is nothing but all the masses are concentrated at the centre of the object or a system. But here, it will lie where the diagonal meets.