Q. Point masses 1, 2, 3 and 4 kg are lying at the point (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
A) 43 kg-$m^{2}$
B) 34 kg-$m^{2}$
C) 27 kg-$m^{2}$
D) 72 kg-$m^{2}$
Explanation
(a) 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.
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$I = I_{1} + I_{2} + I_{3} + I_{4}$
$\therefore$ $I = m_{1}r_{1}^{2} + m_{2}r_{2}^{2} + m_{3}r_{3}^{2} + m_{4}r_{4}^{2}$
$I = (1 \times 0) + (2 \times 0) + \left( 3 \times 3^{2} \right) + 4( - 2)^{2}$
$I = 0 + 0 + 27 + 16 = 43\ kg - m^{2}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.