First slide

Location of centre on mass for multiple point masses

Question

The co-ordinates of centre of mass of particles of mass 10, 20 and 30 gm are (1, 1, 1) cm.The position co-ordinates of mass 40 gm which when added to the system, the position of combined centre of mass be at (0, 0, 0) are,

Easy

A

(3/2, 3/2, 3/2)
B

(-3/2, -3/2, -3/2)
C

(3/4, 3/4, 3/4)
D

(-3/4, -3/4, -3/4)

Solution

$\large {m_1} = 10,\,{m_2} = 20,\,\,{m_3} = 30\,,\,$
$\large \left( {{x_{cm}},{y_{cm}},{z_{cm}}} \right) = \left( {1,1,1} \right)$
$\large {x_{cm}} = \frac{{{m_1}{x_1} + {m_2}{x_2} + {m_3}{x_3}}}{{{m_1} + {m_2} + {m_3}}}$
$\large 1 = \frac{{10{x_1} + 20{x_2} + 30{x_3}}}{{60}}$
$\large \Rightarrow 10{x_1} + 20{x_2} + 30{x_3} = 60$
for 4 particle system m₄= 40
$\large \left( {{x^1}cm,{y^1}cm,{z^1}cm} \right) = \left( {0,0,0} \right)$
$\large {x^1}cm = \frac{{{m_1}{x_1} + {m_2}{x_2} + {m_3}{x_3} + {m_4}{x_4}}}{{{m_1} + {m_2} + {m_3} + {m_4}}}$
$\large 0 = \frac{{10{x_1} + 20{x_2} + 30{x_3} + 40{x_4}}}{{100}}$
$\large \Rightarrow 10{x_1} + 20{x_2} + 30{x_3} + 40{x_4} = 0$
$\large \Rightarrow 60 + 40{x_4} = 0$
$\large 40{x_4} = - 60\,\,\, \Rightarrow {x_4} = \frac{{ - 60}}{{40}} = \frac{{ - 3}}{2}$
$\large {y_4} = \frac{{ - 3}}{2},\,\,{z_4} = \frac{{ - 3}}{2}\$
$\large \left( {\frac{{ - 3}}{2},\frac{{ - 3}}{2},\frac{{ - 3}}{2}} \right)$

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