Point masses $1,2,3$ and $4 \mathrm{~kg}$ are lying at the points $(0,0,0),(2,0,0),(0,3,0)$ and $(-2,-2,0)$…

Point masses $1,2,3$ and $4 \mathrm{~kg}$ 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
  1. $43 \mathrm{~kg} \mathrm{~m}^{2}$
  2. $34 \mathrm{~kg} \mathrm{~m}^{2}$
  3. $27 \mathrm{~kg} \mathrm{~m}^{2}$
  4. $72 \mathrm{~kg} \mathrm{~m}^{2}$

Solution

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.
$=0+0+27+16=43 \mathrm{~kg} \mathrm{~m}^{2}$ *

Asked in: JEE Mains - Rotational Motion - Test 1

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