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(B) The moment of inertia $I$ of a system of particles about an axis is given by $I = \sum m_i r_i^2$,where $m_i$ is the mass of the $i$-th particle a

Vedclass · Accepted Answer

$1.04$

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(B) The moment of inertia $I$ of a system of particles about an axis is given by $I = \sum m_i r_i^2$,where $m_i$ is the mass of the $i$-th particle and $r_i$ is its perpendicular distance from the axis of rotation.From the figure,the axis $PQ$ passes through the center of the rod.There are two masses of $5 \ kg$ at a distance of $0.2 \ m$ from the axis $PQ$.There are two masses of $2 \ kg$ at a distance of $(0.2 + 0.2) = 0.4 \ m$ from the axis $PQ$.Calculating the total moment of inertia:$I = 2 	imes (5 \ kg) 	imes (0.2 \ m)^2 + 2 	imes (2 \ kg) 	imes (0.4 \ m)^2$$I = 2 	imes 5 	imes 0.04 + 2 	imes 2 	imes 0.16$$I = 10 	imes 0.04 + 4 	imes 0.16$$I = 0.4 + 0.64 = 1.04 \ kg-m^2$.

Four masses are fixed on a massless rod as shown in the figure. The moment of inertia about the axis $PQ$ is about ..... $kg-m^2$.

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