A Cylinder Has A Mass ‘M’, Length ‘l’ And Radius ‘a’. Find The Ratio Of ‘l’ To ‘a’ If The Moment Of Inertia About An Axis Through The Centre And Perpendicular To The Length Is Minimum

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Ans.

Here, the mass of the cylinder is M, radius is a and length is l. If \( \rho \) be the density of material of the cylinder, then the mass of the cylinder is \( M=\pi{a^2}l\rho \). or, \( a^2l=constant=k \)(say)

The moment of inertia of this cylinder about an axis passing through the centre and perpendicular to the length of the cylinder is

\( \displaystyle{I=M\left(\frac{a^2}{4}+\frac{l^2}{12}\right)} \)
[To know the derivation of the moment of inertia of the cylinder about an axis passing through the centre and perpendicular to its length (CLICK HERE) ]

or, \( \displaystyle{I=M\left(\frac{k}{4l}+\frac{l^2}{12}\right)=f(t)} \)

For, \( I \) to be a minimum,

\( \frac{dI}{dl}=0 \).

or, \( M\left[-\frac{k}{4l^2}+\frac{l}{6}\right]=0 \)

or, \( \displaystyle{\frac{k}{4l^2}=\frac{l}{6}} \)

or, \( \displaystyle{\frac{l}{a}=\frac{\sqrt{3}}{\sqrt{2}}} \)

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