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Kelvi

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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 center of mass of the system from the $10$ kg mass is:

easy

System of Particles and Rotational Motion

2022

physics

$\frac{10}{3}$ m

$\frac{20}{3}$

Explanation

To find the center of mass of the system, we use the formula: $\\ x_{cm} = \frac{m_1 x_1 + m_2 x_2}{m_1 + m_2} \\$

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m

$10$ m

$5$ m

Given:

\\

•

m_1 = 10

kg (mass of the first object)

\\

•

m_2 = 20

kg (mass of the second object)

\\

• Length of the rod

= 10

m

\\

Assume the 10 kg mass is at position

x_1 = 0

m (left end of the rod).

\\

Then the 20 kg mass is at position

x_2 = 10

m (right end of the rod).

\\

Substitute the values into the center of mass formula:

\\ x_{cm} = \frac{10 \times 0 + 20 \times 10}{10 + 20} \\ x_{cm} = \frac{0 + 200}{30} \\ x_{cm} = \frac{200}{30} \\ x_{cm} = \frac{20}{3}

m

\\

Therefore, the distance of the center of mass from the 10 kg mass is

\frac{20}{3}

m.

\\

This corresponds to Option 2.