Question:

Three particles of masses $1\, kg, \frac{3}{2} kg$, and $2\,kg$ are located at the vertices of an equilateral triangle of side $a$. The $x, y$ coordinates of the centre of mass are

Updated On: Jul 7, 2022
  • $\left(\frac{5a}{9}, \frac{2a}{3\sqrt{3}}\right)$
  • $\left(\frac{2a}{3\sqrt{3}}, \frac{5a}{9}\right)$
  • $\left(\frac{5a}{9}, \frac{2a}{\sqrt{3}}\right)$
  • $\left(\frac{2a}{\sqrt{3}}, \frac{5a}{9}\right)$
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The Correct Option is A

Solution and Explanation

A uniform sphere of mass $M$ and radius $R$ is placed on a rough horizontal surface (Figure). The sphere is struck horizontally at a height $h$ from the floor. Match Column $I$ with Column $II$. Let the masses $1 \,kg, \frac{3}{2}\, kg$ and $2\,kg$ are located at the vertices $A, B$ and $C$ as shown in figure. The coordinates of points $A, B$ and $C$ are$ (0,0), (a, 0),$$\frac{a}{2},$$ \frac{\sqrt{3a}}{2}$ respectively.
The coordinates of centre of mass are $X_{CM} = \frac{m_{1}x_{1} +m_{2}x_{2} +m_{3}x_{3}}{m_{1}+m_{2}+m_{3}}$ $= \frac{\left[1\times0+\frac{3}{2} \times a+2 \times\frac{a}{2}\right]}{\left[1+ \frac{3}{2}+2\right]} = \frac{5a}{9} $ $Y_{CM} = \frac{m_{1}y_{1} +m_{2}y_{2}+m_{3}y_{3}}{m_{1}+m_{2}+m_{3}} $ $ = \frac{\left[1\times 0+\frac{3}{2} \times 0+2 \times \frac{\sqrt{3}a}{2}\right]}{\left[1+ \frac{3}{2}+2\right]} $ $= \frac{2\sqrt{3}a}{9} $ $ = \frac{2a}{3\sqrt{3}}$
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Concepts Used:

Center of Mass

The center of mass of a body or system of a particle is defined as a point where the whole of the mass of the body or all the masses of a set of particles appeared to be concentrated.

The formula for the Centre of Mass:

Center of Gravity

The imaginary point through which on an object or a system, the force of Gravity is acted upon is known as the Centre of Gravity of that system. Usually, it is assumed while doing mechanical problems that the gravitational field is uniform which means that the Centre of Gravity and the Centre of Mass is at the same position.