Question 226: Centre of Mass and Rotational Motion

Question

Moment of inertia of a cylinder of mass M,
length L and radius R about an axis passing
through its centre and perpendicular to the
axis of the cylinder is I = $$M\left( {{{{R^2}} \over 4} + {{{L^2}} \over {12}}} \right)$$. If such a
cylinder is to be made for a given mass of a
material, the ratio $${L \over R}$$ for it to have minimum
possible I is

Accepted Answer

Given I = $$M\left( {{{{R^2}} \over 4} + {{{L^2}} \over {12}}} ight)$$
M = $$ho $$.V = $$ho $$$$\pi $$R
2
L
$$ \Rightarrow $$ R
2
= $${M \over {ho \pi L}}$$
$$ 	herefore $$ I = $$M\left( {{M \over {4ho \pi L}} + {{{L^2}} \over {12}}} ight)$$
$$ \Rightarrow $$ $${{dI} \over {dL}}$$ = $$M\left( {{M \over {4ho \pi }}\left( { - {1 \over {{L^2}}}} ight) - {{2L} \over {12}}} ight)$$
For minimum I, $${{dI} \over {dL}} = 0$$
$$ 	herefore $$ $$M\left( {{M \over {4ho \pi }}\left( { - {1 \over {{L^2}}}} ight) - {{2L} \over {12}}} ight)$$ = 0
$$ \Rightarrow $$ $${{{M^2}} \over {4ho \pi {L^2}}} = {{2LM} \over {12}}$$
$$ \Rightarrow $$ $${M \over {4ho \pi {L^2}}} = {L \over 6}$$
$$ \Rightarrow $$ $${{ho \pi {R^2}L} \over {4ho \pi {L^2}}} = {L \over 6}$$
$$ \Rightarrow $$ $${{{R^2}} \over {{L^2}}} = {2 \over 3}$$
$$ \Rightarrow $$ $${R \over L} = \sqrt {{2 \over 3}} $$
$$ \Rightarrow $$ $${L \over R} = \sqrt {{3 \over 2}} $$

Answer

$${3 \over 2}$$

Answer

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

Answer

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

Answer

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

Centre of Mass and Rotational Motion

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