Discussion Forum

The moment of inertia of a uniform cylinder of length l and radius R about its perpendicular bisector is $I$. What is the ratio l/R such that the moment of inertia is minimum?

Answer:

Explanation:

MI of a solid cylinder about its perpendicular bisector of length is

                         $I= m(\frac{l^{2}}{12}+\frac{R^{2}}{4})$

$\Rightarrow I=\frac{mR^{2}}{4}+\frac{ml^{2}}{12}=\frac{m^{2}}{4\pi\rho l}+\frac{ml^{2}}{12}$

                                                                                                                  [ $\therefore \rho\pi R^{2}l=m$ ]

 For l to be maximum,

                                   $\frac{\text{d}I}{\text{dl}}=-\frac{m^{2}}{4\pi\rho}(\frac{1}{l^{2}})+\frac{ml}{6}=0$

                        $\Rightarrow \frac{m^{2}}{4\pi\rho}=\frac{ml^{3}}{6}=0$

Now, putting    $m=\rho \pi R^{2}l$

         $\therefore            l^{3}=\frac{3}{2\pi\rho}.\rho\pi R^{2}l$

    $\frac{l^{2}}{R^{2}}=\frac{3}{2}$

                         $\frac{l}{R}=\sqrt{\frac{3}{2}}$