First slide

Moment of intertia

Question

Two small spheres of masses 10kg and 30 kg are joined by a rod of length 0.5 m and of negligible mass. The M.I of the system about a normal axis through centre of mass of the system is

Easy

A

1.875 kgm²
B

2.45 kgm²
C

0.75 kgm²
D

1.75 kgm²

Solution

When two particles of masses m₁ and m₂ are at a distance d apart, M.O.I. of the system about an axis passing through the COM and perpendicular to the line joining the particles is given by
$\large I = \frac{{{m_1}{m_2}}}{{{m_1} + {m_2}}}{d^2} = \frac{{10 \times 30}}{{10 + 30}}{\left( {\frac{1}{2}} \right)^2} = 1.875kg{m^2}$

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