Question

What should be the $L/R$ ratio for a cylinder having radius R and length L so that Moment of Inertia about an axis passing through the middle of the rod is minimum?

Answer 1

R.E.F image

Moment of inertia of a cylinder about in perpendicular

bisector is given by :

$I=\frac{1}{4}M{R}^2+\frac{1}{12}M{L}^2$

$=\frac{M}{4}(R^2+\frac{L^2}{3})$

To minimise $I,f=R^2+\frac{L^2}{3}$ should be minimum

$\Rightarrow \frac{df}{dl}=2R\frac{dR}{dL}+\frac{2L}{3}=0$ (to obtain minima)

$\Rightarrow \frac{dr}{dL}=\frac{-L}{3R}\to \left(1\right)$

also values of cylinder,

$v=\pi R^{2}L=$ constant

$\Rightarrow dv=2\pi RLdR+\pi R^{2}dL=0$

$\Rightarrow \frac{dR}{dL}=\frac{-R}{2L}$ ___(2)

From (1) & (2)

$\frac{L}{3R}=\frac{R}{2L}\Rightarrow 2L^2=3R^2$

$\Rightarrow \frac{L}{R}=\sqrt{\frac{3}{2}}$