Question

A box shaped pontoon of length 100 m, breadth 12 m and draft 10 m is floating in water of density 1 tonne/m\(^3\). The roll radius of gyration is 1% of the ship's length and the roll added mass moment of inertia is 20% of the roll mass moment of inertia. The vertical centre of buoyancy is located at half of the draft, and the vertical centre of gravity is at 6 m from the keel.
The roll natural frequency of the pontoon is ______ rad/s (rounded off to two decimal places).
Assume g = 10 m/s\(^2\).

Hint:

Find GM from the box hull's BM and KB first, then use the roll frequency formula with added inertia.

Answer 1

Correct Answer: 1.29

Step 1: Build KM from the box-shape geometry.
The waterplane inertia for a box hull is $I_{wp} = LB^3/12 = 100 \times 1728/12 = 14400$ m$^4$, and the displaced volume is $\nabla = LBT = 100 \times 12 \times 10 = 12000$ m$^3$, so the metacentric radius is $BM = I_{wp}/\nabla = 1.2$ m. Adding $KB = T/2 = 5$ m gives $KM = 6.2$ m.

Step 2: Get GM and note the mass cancels in the frequency formula.
$GM = KM - KG = 6.2 - 6 = 0.2$ m. The undamped roll frequency is $\omega_n = \sqrt{g \, GM / k_{eff}^2}$, where $k_{eff}^2$ is the effective (mass plus added) squared radius of gyration; since the added inertia is 20% extra, $k_{eff}^2 = 1.2k^2$ with the mass itself dropping out of both sides.

Step 3: Insert the radius of gyration and compute.
$k = 0.01L = 1$ m, so $k_{eff}^2 = 1.2 \times 1^2 = 1.2$ m$^2$. Then $\omega_n = \sqrt{10 \times 0.2 / 1.2} = \sqrt{5/3} = 1.291$ rad/s.

Final Answer:
Rounded to two decimals, $\omega_n = 1.29$ rad/s, within the 1.27 to 1.31 rad/s key range. \[ \boxed{\omega_n \approx 1.29 \text{ rad/s}} \]