Four particles of masses 5 kg, 2 kg, 3 kg, | vedclass

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(B) The moment of inertia $I$ of a system of particles is given by $I = \sum m_i r_i^2$,where $r_i$ is the perpendicular distance of the $i$-th partic

Vedclass · Accepted Answer

$43, 24, 67$ units

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(B) The moment of inertia $I$ of a system of particles 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.For the $x$-axis,the distance $r_x = \sqrt{y^2 + z^2}$.$I_x = 5(0^2+0^2) + 2(0^2+0^2) + 3(3^2+0^2) + 4((-2)^2+0^2) = 0 + 0 + 27 + 16 = 43$ units.For the $y$-axis,the distance $r_y = \sqrt{x^2 + z^2}$.$I_y = 5(0^2+0^2) + 2(2^2+0^2) + 3(0^2+0^2) + 4((-2)^2+0^2) = 0 + 8 + 0 + 16 = 24$ units.For the $z$-axis,the distance $r_z = \sqrt{x^2 + y^2}$.$I_z = 5(0^2+0^2) + 2(2^2+0^2) + 3(0^2+3^2) + 4((-2)^2+(-2)^2) = 0 + 8 + 27 + 4(4+4) = 35 + 32 = 67$ units.Thus,the moments of inertia are $43, 24, 67$ units.

Four particles of masses $5 \ kg, 2 \ kg, 3 \ kg$,and $4 \ kg$ are placed at $(0, 0, 0), (2, 0, 0), (0, 3, 0)$,and $(-2, -2, 0)$ respectively. The moments of inertia about the $x$-axis,$y$-axis,and $z$-axis respectively are:

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