Calculate the moment of inertia of the system | vedclass

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(B) The axis of rotation $XX'$ is at $100 \ cm$ $(1 \ m)$.The position of the $4 \ kg$ mass is at $70 \ cm$ $(0.7 \ m)$. The distance from the axis is

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(B) The axis of rotation $XX'$ is at $100 \ cm$ $(1 \ m)$.The position of the $4 \ kg$ mass is at $70 \ cm$ $(0.7 \ m)$. The distance from the axis is $r_1 = 100 \ cm - 70 \ cm = 30 \ cm = 0.3 \ m$.The position of the $1 \ kg$ mass is at $20 \ cm$ $(0.2 \ m)$. The distance from the axis is $r_2 = 100 \ cm - 20 \ cm = 80 \ cm = 0.8 \ m$.The moment of inertia $I$ is given by $I = \sum m_i r_i^2$.$I = (4 \ kg) 	imes (0.3 \ m)^2 + (1 \ kg) 	imes (0.8 \ m)^2$.$I = 4 	imes 0.09 + 1 	imes 0.64$.$I = 0.36 + 0.64 = 1 \ kg \cdot m^2$.

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

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