Calculate the moment of inertia of the system | vedclass

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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

$92 \ kg \cdot m^2$

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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 of rotation.In this case,the axis of rotation is the $XX'$ axis (the $x$-axis).The perpendicular distance of a point $(x, y)$ from the $x$-axis is $|y|$.$1$. For the $4 \ kg$ mass at $(0, 3)$,the distance $r_1 = |3| = 3 \ m$.$I_1 = 4 	imes (3)^2 = 4 	imes 9 = 36 \ kg \cdot m^2$.$2$. For the $2 \ kg$ mass at $(0, -2)$,the distance $r_2 = |-2| = 2 \ m$.$I_2 = 2 	imes (2)^2 = 2 	imes 4 = 8 \ kg \cdot m^2$.$3$. For the $3 \ kg$ mass at $(0, -4)$,the distance $r_3 = |-4| = 4 \ m$.$I_3 = 3 	imes (4)^2 = 3 	imes 16 = 48 \ kg \cdot m^2$.The total moment of inertia is $I = I_1 + I_2 + I_3 = 36 + 8 + 48 = 92 \ kg \cdot m^2$.

Calculate the moment of inertia of the system of particles shown in the figure about the axis of rotation $XX'$.

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