Point masses 1, 2, 3 and 4 kg are lying at the points (0, 0, 0), (2, 0, 0), (0, 3, 0) and (- 2, -2, 0) respectively. The moment-of inertia of this system about X-axis will be

43 kg-m²

34 kg-m²

27 kg-m²

72 kg-m²

Solution

43 kg-m²

Moment of inertia of the whole system about the axis of rotation will be equal to the sum of the moments of inertia of all the particles.

I = I₁ + I₂ + I₃ + I₄

I = $m_{1} r_{1}^{2} + m_{2} r_{2}^{2} + m_{3} r_{3}^{2} + m_{4} r_{4}^{2}$

I = ( 1 × 0 ) + ( 2 × 0 ) + ( 3 × 3² ) + 4 (-2 )²

I = 0 + 0 + 27 + 16

I = 43 kg-m³

Some More Questions From System of Particles and Rotational Motion Chapter

For Daily Free Study Material Join wiredfaculty