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(A) The moment of inertia $I$ of a system of particles about an axis is given by $I = \sum m_i r_i^2$,where $r_i$ is the perpendicular distance of the

Vedclass · Accepted Answer

$43$

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(A) The moment of inertia $I$ of a system of particles about an axis is given by $I = \sum m_i r_i^2$,where $r_i$ is the perpendicular distance of the $i$-th particle from the axis.For the $x$-axis,the perpendicular distance $r$ of a point $(x, y, z)$ is given by $r = \sqrt{y^2 + z^2}$.$1$. For mass $m_1 = 1 \ kg$ at $(0, 0, 0)$: $r_1 = \sqrt{0^2 + 0^2} = 0$. So,$I_1 = 1 	imes 0^2 = 0 \ kg-m^2$.$2$. For mass $m_2 = 2 \ kg$ at $(2, 0, 0)$: $r_2 = \sqrt{0^2 + 0^2} = 0$. So,$I_2 = 2 	imes 0^2 = 0 \ kg-m^2$.$3$. For mass $m_3 = 3 \ kg$ at $(0, 3, 0)$: $r_3 = \sqrt{3^2 + 0^2} = 3 \ m$. So,$I_3 = 3 	imes (3)^2 = 27 \ kg-m^2$.$4$. For mass $m_4 = 4 \ kg$ at $(-2, -2, 0)$: $r_4 = \sqrt{(-2)^2 + 0^2} = 2 \ m$. So,$I_4 = 4 	imes (-2)^2 = 4 	imes 4 = 16 \ kg-m^2$.The total moment of inertia about the $x$-axis is $I = I_1 + I_2 + I_3 + I_4 = 0 + 0 + 27 + 16 = 43 \ kg-m^2$.

Point masses of $1, 2, 3,$ and $4 \ kg$ are placed at $(0, 0, 0), (2, 0, 0), (0, 3, 0),$ and $(-2, -2, 0)$ respectively. What is the moment of inertia of this system about the $x$-axis in $kg-m^2$?

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