Question: 6: 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 center of mass of the system from the 10 kg mass is
(1) 10 m
(2) 5 m
(3) \frac{10}{3} \mathrm{~m}
(4) \frac{20}{3} \mathrm{~m}
Answer: Option (4)
Explanation:
Let the 10 kg mass be placed at the origin. Then its position is x_1 = 0.
The 20 kg mass is at the other end of the rod of length 10 m, so its position is
x_2 = 10 \text{ m}.
The position of the center of mass of a system of particles in one dimension is given by
x_{cm} = \frac{m_1 x_1 + m_2 x_2}{m_1 + m_2}.
Substituting the given values,
x_{cm} = \frac{10 \times 0 + 20 \times 10}{10 + 20}.
x_{cm} = \frac{200}{30} = \frac{20}{3} \text{ m}.
Thus, the center of mass of the system is at a distance of
\frac{20}{3} \text{ m} from the 10 kg mass.