A wooden cylinder of specific gravity 0.6 and circular in cross-section is required to float in oil of specific gravity 0.8. Calculate the ratio of length to diameter for the cylinder so that it will just float upright in water.

Stability of a Floating Wooden Cylinder – CivInnovate

Problem Statement

A wooden cylinder with a circular cross-section has the following properties:

Determine the ratio of length to diameter (\( L/D \)) required for the cylinder to float upright in oil.

The weight of the cylinder must equal the weight of the displaced oil:

\[ \gamma_{\text{cylinder}} V_{\text{cylinder}} = \gamma_{\text{oil}} V_{\text{displaced oil}} \]

\[ 0.6 \times 9810 \times \frac{\pi}{4} D^2 L = 0.8 \times 9810 \times \frac{\pi}{4} D^2 h \]

Solving for \( h \):

\[ h = \frac{0.6}{0.8} L = 0.75L \]

The centroid of the submerged volume is:

\[ OB = \frac{h}{2} = \frac{0.75L}{2} = 0.375L \]

The cylinder’s center of gravity is at its midpoint:

\[ OG = \frac{L}{2} \]

The distance between the buoyancy center and gravity center is:

\[ BG = OG – OB = \frac{L}{2} – 0.375L = 0.125L \]

The metacentric radius \( MB \) is given by:

\[ MB = \frac{I}{V} \]

For stability, \( GM > 0 \):

\[ \frac{D^2}{12L} – 0.125L > 0 \]

Rearranging:

\[ \frac{L^2}{D^2} < 0.667 \]

Required ratio for stable floating: \( L/D < 0.8167 \)

1. Stability Condition: The cylinder remains stable if the metacentric height is positive, ensuring a restoring moment.

2. Practical Impact:

1. Engineering Applications: Used in floating structures, ships, and offshore platforms.

2. Real-World Uses: