The moment of inertia of a meter scale of mas | vedclass

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(B) Given: Total mass $M = 0.6 \ kg$, Length $L = 1 \ m = 100 \ cm$.Mass per unit length $\lambda = \frac{0.6 \ kg}{100 \ cm} = 0.006 \ kg/cm$.The axi

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$0.104$

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(B) Given: Total mass $M = 0.6 \ kg$, Length $L = 1 \ m = 100 \ cm$.Mass per unit length $\lambda = \frac{0.6 \ kg}{100 \ cm} = 0.006 \ kg/cm$.The axis is at $20 \ cm$. This divides the scale into two parts: $AB$ $(20 \ cm)$ and $BC$ $(80 \ cm)$.Mass of part $AB$, $m_1 = \lambda 	imes 20 = 0.006 	imes 20 = 0.12 \ kg$.Mass of part $BC$, $m_2 = \lambda 	imes 80 = 0.006 	imes 80 = 0.48 \ kg$.The moment of inertia of a rod about an axis passing through one end is $I = \frac{1}{3} m l^2$.For part $AB$ (axis at end $B$): $I_1 = \frac{1}{3} m_1 (l_1)^2 = \frac{1}{3} 	imes 0.12 	imes (0.2)^2 = 0.04 	imes 0.04 = 0.0016 \ kg-m^2$.For part $BC$ (axis at end $B$): $I_2 = \frac{1}{3} m_2 (l_2)^2 = \frac{1}{3} 	imes 0.48 	imes (0.8)^2 = 0.16 	imes 0.64 = 0.1024 \ kg-m^2$.Total moment of inertia $I = I_1 + I_2 = 0.0016 + 0.1024 = 0.104 \ kg-m^2$.

The moment of inertia of a meter scale of mass $0.6 \ kg$ about an axis perpendicular to the scale and located at the $20 \ cm$ position on the scale in $kg-m^2$ is: (Breadth of the scale is negligible)

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