Question

three particles each of mass m are placed at three corner of an equilateral triangle of length l. find the position of center of mass in terms of coordinates

Answer 1

Answer: (l/2, root3 l/6)

Explanation: Step- wise explanation is attached below.. Hope it helps

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Answer 2

Answer:

The position of the center of mass in terms of coordinates is $(\frac{l}{2} , \frac{\sqrt{3} l}{6} )$.

Explanation:

Centre Of Mass: Centre of mass is that point in the body where the whole mass of the body is supposed to be concentrated.

It is given that the three particles of each mass m are placed at the three corners of an equilateral triangle of length l.

Let the three particles are $m_{1} ,m_{2}$ and $m_{3}$

So, $m_{1} =m_{2}= m_{3} =m$

And the coordinates of the equilateral triangle be $(x_{1} ,y_{1} )$ , $(x_{2} ,y_{2} )$ and $(x_{3} ,y_{3} )$

So,

$x_{1} =0 , y_{1} = 0\\ x_{2} = l, y_{2} = 0\\ x_{3} = l/2, y_{3} = \sqrt{3}l/2$

Now,

x- Coordinate for the center of mass

$x_{cm} = \frac{x_{1} m_{1} +x_{2} m_{2} +x_{3} m_{3} }{m_{1} +m_{2} +m_{3} }$

$x_{cm} = \frac{m*0 +m*l +m*\frac{l}{2} }{m +m +m} = \frac{l+\frac{l}{2} }{3} = \frac{l}{2}$

Now,

$y_{cm} = \frac{y_{1} m_{1} +y_{2} m_{2} +y_{3} m_{3} }{m_{1} +m_{2} +m_{3} }\\y_{cm} = \frac{m*0+m*0+m*\frac{\sqrt{3}l}{2} }{3m} =\frac{\sqrt{3}l }{6}$

Hence, the position of the center of mass in terms of coordinates is $(\frac{l}{2} , \frac{\sqrt{3} l}{6} )$.

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