Question

A light rod of length I has two masses m₁ and m₂ attached to its two ends. The moment of inertia of the system about an axis perpendicular to the rod and passing through the centre of mass is
(1) [(m₁m₂)/(m₁+m₂)]. l²
(2) [(m₁+m₂)/(m₁m₂)]. l²
(3) (m₁+m₂). l²
(4) √(m₁m₂). l²

Answer 1

(1) [(m₁m₂)/(m₁+m₂)]. l²

Answer 2

Answer:

According to formula:

I = m₁r₁² + m₂r₂²...........mₓrₓ²

I = m₁r₁² + m₂r₂²

 = m₁(m₂l/m₁ + m₂)² + m₂(m₂l/m₁+m₂)²

 = m₁m₂(m₁+m₂)l² /(m₁ +m₂)²

 = m₁m₂l²/(m₁+m₂)

Moment of inertia is the tendency to change the angular acceleration.  It is given by the sum of the product of mass of each particle with the square of distance from axis of rotation

Explanation:

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