Question

Two particles of masses 2 kg and 4 kg are separated by a distance of 3 m. The moment of inertia of the system of the two particles about an axis passing through the centre of mass of the system and perpendicular to the line joining the two particles is

• A. $24 \times 10^3$ kg m$^2$
• B. $12 \times 10^3$ kg m$^2$
• C. $6 \times 10^3$ kg m$^2$
• D. $30 \times 10^2$ kg m$^2$

Hint:

For two-particle systems, use $I = m_1 x_1^2 + m_2 x_2^2$ after determining distances from the center of mass.

Answer 1

The Correct Option is B

Let $m_1 = 2$ kg, $m_2 = 4$ kg, and the distance between them $d = 3$ m.
The center of mass (COM) lies closer to the heavier mass. Let the distance of $m_1$ from COM be $x_1$ and $m_2$ from COM be $x_2$, then:\ $m_1 x_1 = m_2 x_2$, and $x_1 + x_2 = 3$
So, $2x_1 = 4x_2$ ⇒ $x_1 = 2x_2$ ⇒ $2x_2 + x_2 = 3$ ⇒ $x_2 = 1$ m, $x_1 = 2$ m
Now, $I = m_1 x_1^2 + m_2 x_2^2 = 2 \times (2)^2 + 4 \times (1)^2 = 8 + 4 = 12$ kg m$^2$
Converting to standard form: $12 \times 10^3$ g cm$^2$ = $12 \times 10^3$ kg m$^2$