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(B) The moment of inertia $(I)$ of a system of point masses rotating about an axis is given by $I = \sum m_i r_i^2$,where $r_i$ is the perpendicular d

Vedclass · Accepted Answer

$92$

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(B) The moment of inertia $(I)$ of a system of point masses rotating 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.Here,the system rotates about the $x-$ axis. The perpendicular distance of a particle at position $y$ from the $x-$ axis is $|y|$.Given masses and their positions:$1$. $m_1 = 4 \, kg$ at $y_1 = 3 \, m$,so $r_1 = 3 \, m$.$2$. $m_2 = 2 \, kg$ at $y_2 = -2 \, m$,so $r_2 = 2 \, m$.$3$. $m_3 = 3 \, kg$ at $y_3 = -4 \, m$,so $r_3 = 4 \, m$.Calculating the total moment of inertia:$I = m_1 r_1^2 + m_2 r_2^2 + m_3 r_3^2$$I = 4(3)^2 + 2(2)^2 + 3(4)^2$$I = 4(9) + 2(4) + 3(16)$$I = 36 + 8 + 48$$I = 92 \, kg \cdot m^2$.

Three particles are connected by massless rods lying along the $y-$ axis. If the system rotates about the $x-$ axis with an angular speed of $2 \, rad/s$,the moment of inertia $(M.I.)$ of the system is ......... $kg \cdot m^2$.

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