A solid wood cylinder has a diameter of 0.6m and a height of 1.2m. The sp.gr. of the wood is 0.6. If the cylinder is placed vertically in oil of sp.gr. 0.85, would it be stable?

The weight of the cylinder is equal to the weight of the displaced oil: \[ \gamma_{\text{cylinder}} V_{\text{cylinder}} = \gamma_{\text{oil}} V_{\text{displaced oil}} \] \[ 0.6 \times 9810 \times \pi \times (0.3)^2 \times 1.2 = 0.85 \times 9810 \times \pi \times (0.3)^2 \times h \] Cancelling \( \pi \times (0.3)^2 \times 9810 \): \[ 0.6 \times 1.2 = 0.85 \times h \] \[ h = \frac{0.6 \times 1.2}{0.85} \] \[ h = 0.847 \text{ m} \]

The center of buoyancy is at the centroid of the submerged volume: \[ OB = \frac{h}{2} = \frac{0.847}{2} \] \[ OB = 0.4235 \text{ m} \]

Since the cylinder is uniform, its center of gravity is at the midpoint of its height: \[ OG = \frac{1.2}{2} \] \[ OG = 0.6 \text{ m} \] \[ BG = OG – OB = 0.6 – 0.4235 \] \[ BG = 0.1765 \text{ m} \]

The metacentric radius (\(MB\)) is given by: \[ MB = \frac{I}{V} \] The moment of inertia about the vertical axis is: \[ I = \frac{1}{4} \pi r^4 \] \[ = \frac{1}{4} \pi (0.3)^4 \] \[ = \frac{1}{4} \pi \times 0.0081 = 0.00636 \text{ m}^4 \] The displaced volume: \[ V = \pi \times (0.3)^2 \times 0.847 \] \[ V = \pi \times 0.09 \times 0.847 \] \[ V = 0.2395 \text{ m}^3 \] \[ MB = \frac{0.00636}{0.2395} \] \[ MB = 0.0265 \text{ m} \] \[ GM = MB – BG \] \[ GM = 0.0265 – 0.1765 \] \[ GM = -0.15 \text{ m} \]