Question

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:

• A. \(\frac{10}{3} m\)
• B. \(\frac{20}{3} m\)
• C. \(10 m\)
• D. \(5 m\)

Answer 1

The Correct Option is B

To find the center of mass of the system consisting of two objects attached to a rigid rod, we use the formula for the center of mass \(x_{cm}\) of a system of particles:

 \(x_{cm} = \frac{\sum m_ix_i}{\sum m_i}\)

where \(m_i\) is the mass and \(x_i\) is the position of each particle.  
Given:
Mass of first object \(m_1 = 10 \text{ kg}\)

Mass of second object \(m_2 = 20 \text{ kg}\)

Length of the rod \(L = 10 \text{ m}\)

Placing the 10 kg mass at the origin (0 m) and the 20 kg mass at 10 m, the positions are:

\(x_1 = 0 \text{ m}\) for 10 kg mass,

\(x_2 = 10 \text{ m}\) for 20 kg mass.

Thus, the center of mass is calculated as:

\(x_{cm} = \frac{m_1x_1 + m_2x_2}{m_1 + m_2} = \frac{10 \times 0 + 20 \times 10}{10 + 20}\)

\(= \frac{0 + 200}{30} = \frac{200}{30} = \frac{20}{3} \text{ m}\)

Thus, the distance from the 10 kg mass to the center of mass is \(\frac{20}{3} \text{ m}\).

Answer 2

\(x = [\frac{m_{2r}}{(m_1+m_2)}]\)

\(= [\frac{20(20)}{(20+10)}]\)

\(= \frac{200}{30}\)

\(= \frac{20}{3}\)

The correct option is (B) : \(\frac{20}{3} m\).