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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

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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.Here,the axis passes through the $100 \ cm$ mark.The distance of the $1 \ kg$ mass (at $20 \ cm$) from the $100 \ cm$ mark is $r_1 = |100 \ cm - 20 \ cm| = 80 \ cm = 0.8 \ m$.The distance of the $4 \ kg$ mass (at $70 \ cm$) from the $100 \ cm$ mark is $r_2 = |100 \ cm - 70 \ cm| = 30 \ cm = 0.3 \ m$.Therefore,the total moment of inertia is:$I = m_1 r_1^2 + m_2 r_2^2$$I = 1 \ kg 	imes (0.8 \ m)^2 + 4 \ kg 	imes (0.3 \ m)^2$$I = 1 	imes 0.64 + 4 	imes 0.09$$I = 0.64 + 0.36 = 1 \ kg \ m^2$.

Weights of $1 \ kg$ and $4 \ kg$ are placed at the $20 \ cm$ and $70 \ cm$ marks of a light meter scale,respectively. What is the moment of inertia in $kg \ m^2$ about an axis perpendicular to the scale and passing through the $100 \ cm$ mark?

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