Question 340: Centre of Mass and Rotational Motion

Question

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

Accepted Answer

The volume of the cylinder V = $$\pi {R^2}l$$
$$	herefore$$ $${R^2} = {V \over {\pi l}}$$
We know, moment of inertia of a uniform cylinder of length $$l$$
 and radius R about its perpendicular bisector is,
$$I = {{M{l^2}} \over {12}} + {{M{R^2}} \over 4}$$
[ Putting $${R^2} = {V \over {\pi l}}$$ in this equation]
$$ \Rightarrow $$ $$I = {{M{l^2}} \over {12}} + {{MV} \over {4\pi l}}$$
Here $$I$$ is a function of $$l$$ as M and V are constant.
$$I$$ will be maximum or minimum when $${{{dI} \over {dl}}}$$ = 0.
$$ \Rightarrow {{Ml} \over 6} - {{MV} \over {4\pi {l^2}}} = 0$$
$$ \Rightarrow {{Ml} \over 6} = {{MV} \over {4\pi {l^2}}}$$
$$ \Rightarrow {l \over 6} = {{\pi {R^2}l} \over {4\pi {l^2}}}$$  [ as $${V = \pi {R^2}l}$$ ]
$$ \Rightarrow {{{R^2}} \over {{l^2}}} = {4 \over 6}$$
$$ \Rightarrow {l \over R} = \sqrt {{3 \over 2}} $$

Answer

$${3 \over {\sqrt 2 }}$$

Answer

$$\sqrt {{3 \over 2}} $$

Answer

$${{\sqrt 3 } \over 2}$$

Answer

1

Centre of Mass and Rotational Motion

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

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