Sum

Three particles, each of mass 200 g, are kept at the corners of an equilateral triangle of side 10 cm. Find the moment of inertial of the system about an axis joining two of the particles.

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#### Solution

The distance of mass at A from the axis passing through side BC,

\[\left( AD \right) = \frac{\sqrt{3}}{2} \times 10 = 5\sqrt{3} cm\]

Therefore, we have

Moment of inertia of mass about the axis BC,

\[l = m r^2 = 200 \times \left( 5\sqrt{3} \right)^2 \]

\[ = 200 \times 25 \times 3\]

\[ = 15000 gm - {cm}^2 \]

\[ = 1 . 5 \times {10}^{- 3} kg - m^2\]

Concept: Values of Moments of Inertia for Simple Geometrical Objects (No Derivation)

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