First slide

Moment of inertia

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 $I = M (\frac{R^{2}}{4} + \frac{L^{2}}{12})$ . If such a cylinder is to be made for a given mass of material, the ratio L/R for it to have minimum possible I is :

Difficult

A

$\sqrt{\frac{3}{2}}$
B

$\sqrt{\frac{2}{3}}$
C

$\frac{2}{3}$
D

$\frac{3}{2}$

Solution

$\begin{array}{l} I = \frac{M R^{2}}{4} + \frac{M L^{2}}{12} \\ V o l u m e = V = π R^{2} L \\ \Rightarrow R^{2} = \frac{V}{π L} \\ ∴ I = \frac{M V}{4 π L} + \frac{M L^{2}}{12} \\ \Rightarrow \frac{d I}{d L} = \frac{M V}{4 π} [- \frac{1}{L^{2}}] + M [\frac{2 L}{12}] \\ F o r m i n i m a, \frac{d I}{d L} = 0 \\ \Rightarrow \frac{π R^{2} L}{4 π L^{2}} = \frac{2 L}{12} \\ \Rightarrow {[\frac{R}{L}]}^{2} = \frac{2}{3} \\ \Rightarrow \frac{L}{R} = \sqrt{\frac{3}{2}} \end{array}$

Get Instant Solutions

When in doubt download our app. Now available Google Play Store- Doubts App

Recieve an sms with download link

Doubts App

OR

SCAN ME

Doubts App

Doubts App