Question

A wooden cylinder (specific gravity = 0.6) of length $L$ and diameter $D$ floats in water (density 1000 kg/m$^3$). Find out the minimum value of $D/L$ for which the cylinder floats with its axis vertical. (Round off to three decimal places)

Hint:

Floating stability for cylinders uses $BM = I/V_s$ and must exceed $BG$ for the cylinder to stay vertical.

Answer 1

Correct Answer: 1.38

For a floating cylinder to remain vertical, restoring moment $>$ overturning moment. Let $SG = 0.6$, so density ratio = 0.6. Fraction submerged: \[ \frac{V_s}{V} = 0.6 \] For vertical stability: \[ BM>BG \] Where $BM = \frac{I}{V_s}$ and $BG$ is distance between center of buoyancy and center of gravity. Moment of inertia for circular cross–section: \[ I = \frac{\pi D^4}{64} \] Submerged volume: \[ V_s = 0.6 \left(\frac{\pi D^2}{4} L\right) \] Thus: \[ BM = \frac{I}{V_s} = \frac{\frac{\pi D^4}{64}}{0.6\left(\frac{\pi D^2}{4}L\right)} = \frac{D^2}{9.6L} \] Distance BG for cylinder: \[ BG = \left|\frac{L}{2} - 0.6\frac{L}{2}\right| = 0.2 L \] Stability condition: \[ BM>BG \] \[ \frac{D^2}{9.6 L}>0.2 L \] \[ \Rightarrow D^2>1.92 L^2 \] \[ \Rightarrow \frac{D}{L}>\sqrt{1.92} \] \[ \boxed{\frac{D}{L} = 1.385} \]