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(R²/4 + L²/12). If such a cylinder is to be made for a given mass of a material, the ratio L/R for it to have minimum possible I is (a) 3/2 (b) √(3/2) (c) √(2/3) (d) 2/3

Answer: (b). Fixed mass means fixed volume V = π R² L, so R² = (V)/(π L). I = M((V)/(4π L) + L²/12) dI/dL = 0 ⇒ -(V)/(4π L²) + L/6 = 0 ⇒ L³ = (3V)/(2π) = 3/2R² L L² = 3/2R² So the answer is (b) √(3/2).

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Moment of Inertia & Radius of Gyration

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

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Answer: (b)

Fixed mass means fixed volume $V = \pi R^2 L$ , so $R^2 = \frac{V}{\pi L}$ .
$I = M\left(\frac{V}{4\pi L} + \frac{L^2}{12}\right)$
$\frac{dI}{dL} = 0 \Rightarrow -\frac{V}{4\pi L^2} + \frac{L}{6} = 0 \Rightarrow L^3 = \frac{3V}{2\pi} = \frac{3}{2}R^2 L$
$L^2 = \frac{3}{2}R^2$
So the answer is (b) $\sqrt{\frac{3}{2}}$ .

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