First slide

Location of centre on mass for multiple point masses

Question

The centre of mass of a system of two particles of masses $m_{1} and m_{2}$ is at distance 8 cm from $m_{1}$ and at a distance 12 cm from $m_{2}$ . Then

Easy

A

$\frac{m_{1}}{m_{2}} = \frac{2}{3}$
B

$\frac{m_{1}}{m_{2}} = \frac{1}{4}$
C

$\frac{m_{1}}{m_{2}} = \frac{1}{2}$
D

$\frac{m_{1}}{m_{2}} = \frac{3}{2}$

Solution

Distance of centre of mass from $C, r_{c} = \frac{m_{2} \times 12 + m_{1} \times (- 8)}{m_{1} + m_{2}}$
Since C is the centre of mass, $r_{c} = 0$
$∴ 12 m_{2} - 8 m_{1} = 0 \Rightarrow \frac{m_{1}}{m_{2}} = \frac{12}{8} = \frac{3}{2}$

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