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Two objects of mass 10 kg and 20 kg respectively are connected to the two ends of a rigid rod of length 10 m with negligible mass. The distance of the centre of mass of the system from the 10 kg mass is :
A
5 m
B
$\frac{10}{3}\ m$
C
$\frac{20}{3}\ m$
D
10 m
Detailed Solution
Taking the 10 kg mass at the origin and the 20 kg mass at 10 m:
$X_{CM} = \frac{10 \times 0 + 20 \times 10}{10 + 20} = \frac{200}{30} = \frac{20}{3}$ m