Question
The moment of inertia of a uniform cylinder of length $$\ell $$ and radius $$R$$ about its perpendicular bisector is $$I.$$ What is the ratio $$\frac{\ell }{R}$$ such that the moment of inertia is minimum?
A.
$$1$$
B.
$$\frac{3}{{\sqrt 2 }}$$
C.
$$\sqrt {\frac{3}{2}} $$
D.
$$\frac{{\sqrt 3 }}{2}$$
Answer :
$$\sqrt {\frac{3}{2}} $$
Solution :
As we know, moment of inertia of a solid cylinder about an axis which is perpendicular bisector
$$\eqalign{ & I = \frac{{m{R^2}}}{4} + \frac{{m{l^2}}}{{12}} \cr & I = \frac{m}{4}\left[ {{R^2} + \frac{{{l^2}}}{3}} \right] \cr & {\text{Let }}V = {\text{ volume of cylinder }} = \pi {R^2}l \cr & = \frac{m}{4}\left[ {\frac{V}{{\pi l}} + \frac{{{l^2}}}{3}} \right] \cr & \Rightarrow \frac{{dl}}{{dl}} = \frac{m}{4}\left[ {\frac{{ - V}}{{\pi {l^2}}} + \frac{{2l}}{3}} \right] = 0 \cr & \frac{V}{{\pi {l^2}}} + \frac{{2l}}{3}\,\,\,\,\,\,\, \Rightarrow V = \frac{{2\pi {l^3}}}{3} \cr & \pi {R^2}l = \frac{{2\pi {l^3}}}{3}\,\,\,\, \Rightarrow \frac{{{l^2}}}{{{R^2}}} = \frac{3}{2}\;{\text{ or, }}\frac{l}{R} = \sqrt {\frac{3}{2}} \cr} $$
As we know, moment of inertia of a solid cylinder about an axis which is perpendicular bisector
$$\eqalign{ & I = \frac{{m{R^2}}}{4} + \frac{{m{l^2}}}{{12}} \cr & I = \frac{m}{4}\left[ {{R^2} + \frac{{{l^2}}}{3}} \right] \cr & {\text{Let }}V = {\text{ volume of cylinder }} = \pi {R^2}l \cr & = \frac{m}{4}\left[ {\frac{V}{{\pi l}} + \frac{{{l^2}}}{3}} \right] \cr & \Rightarrow \frac{{dl}}{{dl}} = \frac{m}{4}\left[ {\frac{{ - V}}{{\pi {l^2}}} + \frac{{2l}}{3}} \right] = 0 \cr & \frac{V}{{\pi {l^2}}} + \frac{{2l}}{3}\,\,\,\,\,\,\, \Rightarrow V = \frac{{2\pi {l^3}}}{3} \cr & \pi {R^2}l = \frac{{2\pi {l^3}}}{3}\,\,\,\, \Rightarrow \frac{{{l^2}}}{{{R^2}}} = \frac{3}{2}\;{\text{ or, }}\frac{l}{R} = \sqrt {\frac{3}{2}} \cr} $$
