Consider a uniform horizontal solid cylinder | vedclass

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(C) Given, mass of solid cylinder, $M = 10 \,kg$.Radius, $R = 40 \,cm = 0.4 \,m$.Length, $L = 9R = 9 	imes 0.4 = 3.6 \,m$.The moment of inertia of a

Vedclass · Accepted Answer

$43.6 \,kg-m^2$

Vedclass · Answer

(C) Given, mass of solid cylinder, $M = 10 \,kg$.Radius, $R = 40 \,cm = 0.4 \,m$.Length, $L = 9R = 9 	imes 0.4 = 3.6 \,m$.The moment of inertia of a solid cylinder about an axis passing through its centre of mass and perpendicular to its axis is given by:$I_{COM} = M \left( \frac{L^2}{12} + \frac{R^2}{4} ight)$Substituting the values:$I_{COM} = 10 \left( \frac{(3.6)^2}{12} + \frac{(0.4)^2}{4} ight)$$I_{COM} = 10 \left( \frac{12.96}{12} + \frac{0.16}{4} ight)$$I_{COM} = 10 (1.08 + 0.04) = 10 	imes 1.12 = 11.2 \,kg-m^2$.According to the parallel axis theorem, the moment of inertia $I'$ about an axis passing through its edge and perpendicular to its axis is:$I' = I_{COM} + M \left( \frac{L}{2} ight)^2$$I' = 11.2 + 10 \left( \frac{3.6}{2} ight)^2$$I' = 11.2 + 10 (1.8)^2$$I' = 11.2 + 10 (3.24) = 11.2 + 32.4 = 43.6 \,kg-m^2$.

Consider a uniform horizontal solid cylinder of mass $10 \,kg$ such that its length is $9$ times its radius. Let the radius be $40 \,cm$. Calculate the moment of inertia of the cylinder about a line passing through its edge and perpendicular to its axis.

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