First slide

Location of centre on mass for multiple point masses

Question

Two particles of mass 5kg and 10kg respectively are attached to the two ends of a rigid rod of length 1m with negligible mass.
The centre of mass of the system from the 5kg particle is nearly at a distance of

Easy

A

50 cm
B

67 cm
C

80 cm
D

33 cm

Solution

$\begin{array}{l} r_{1} = \frac{m_{2} d}{m_{1} + m_{2}} = \frac{10 \times 1}{5 + 10} = \frac{10}{15} = \frac{2}{3} m \\ r_{1} = 0.67 m = 67 cm \end{array}$

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