JEE Main · 2017 · 02 Apr 2017 · Offline · Rotational Mechanics

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?

(A)1
(B) $\frac{3}{2}$
(C) $\frac{3}{2}$ ✓ Correct
(D) $\frac{3}{2}$

Answer: (C)

Solution

About its perpendicular bisector the cylinder has

I = \frac{M R ^{2}}{4} + \frac{M l ^{2}}{12}

. Holding the volume

V = π R^{2} l

fixed, write

l = \frac{V}{π R ^{2}}

and minimise over

R

. Setting

\frac{d I}{d R} = 0

leads to

R^{2} = \frac{2 l ^{2}}{3}

, that is

\frac{l ^{2}}{R ^{2}} = \frac{3}{2}

, so the slimmest-inertia shape has

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

JEE Main · 2017 · 02 Apr 2017 · Offline · Single correct option

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