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\(^2\).
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\(^2\).
The roll natural frequency of the pontoon is ______ rad/s (rounded off to two decimal places).
Assume g = 10 m/s\(^2\).
Show Hint
Find GM from the box hull's BM and KB first, then use the roll frequency formula with added inertia.
Updated On: Jul 28, 2026
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Verified By Collegedunia
Correct Answer: 1.29
Solution and Explanation
Step 1: Find the metacentric height GM.
For a box shape, \(BM = \dfrac{I_{wp}}{\nabla}\), where \(I_{wp} = \dfrac{L B^3}{12}\) is the waterplane moment of inertia and \(\nabla = LBT\) is the displaced volume. Here \(I_{wp} = \dfrac{100 \times 12^3}{12} = 14400\) m\(^4\) and \(\nabla = 100 \times 12 \times 10 = 12000\) m\(^3\), so \(BM = 14400/12000 = 1.2\) m.
Step 2: Combine KB, BM and KG.
\(KB = T/2 = 5\) m, so \(KM = KB + BM = 5 + 1.2 = 6.2\) m. With \(KG = 6\) m, \(GM = KM - KG = 6.2 - 6 = 0.2\) m.
Step 3: Apply the roll natural frequency formula.
\(\omega_n = \sqrt{\dfrac{\Delta g \, GM}{I_{roll} + I_{added}}}\). Since \(I_{roll} = m k^2\) and \(I_{added} = 0.2 \, I_{roll}\), the denominator becomes \(1.2 \, m k^2\), and the mass \(m\) cancels against the \(\Delta = mg\) term, giving \(\omega_n = \sqrt{\dfrac{g \, GM}{1.2 \, k^2}}\), with \(k = 0.01 \times 100 = 1\) m.
Final Answer:
\(\omega_n = \sqrt{\dfrac{10 \times 0.2}{1.2 \times 1^2}} = \sqrt{1.667} \approx 1.29\) rad/s, inside the official 1.27 to 1.31 rad/s band. \[ \boxed{\omega_n \approx 1.29 \text{ rad/s}} \]
For a box shape, \(BM = \dfrac{I_{wp}}{\nabla}\), where \(I_{wp} = \dfrac{L B^3}{12}\) is the waterplane moment of inertia and \(\nabla = LBT\) is the displaced volume. Here \(I_{wp} = \dfrac{100 \times 12^3}{12} = 14400\) m\(^4\) and \(\nabla = 100 \times 12 \times 10 = 12000\) m\(^3\), so \(BM = 14400/12000 = 1.2\) m.
Step 2: Combine KB, BM and KG.
\(KB = T/2 = 5\) m, so \(KM = KB + BM = 5 + 1.2 = 6.2\) m. With \(KG = 6\) m, \(GM = KM - KG = 6.2 - 6 = 0.2\) m.
Step 3: Apply the roll natural frequency formula.
\(\omega_n = \sqrt{\dfrac{\Delta g \, GM}{I_{roll} + I_{added}}}\). Since \(I_{roll} = m k^2\) and \(I_{added} = 0.2 \, I_{roll}\), the denominator becomes \(1.2 \, m k^2\), and the mass \(m\) cancels against the \(\Delta = mg\) term, giving \(\omega_n = \sqrt{\dfrac{g \, GM}{1.2 \, k^2}}\), with \(k = 0.01 \times 100 = 1\) m.
Final Answer:
\(\omega_n = \sqrt{\dfrac{10 \times 0.2}{1.2 \times 1^2}} = \sqrt{1.667} \approx 1.29\) rad/s, inside the official 1.27 to 1.31 rad/s band. \[ \boxed{\omega_n \approx 1.29 \text{ rad/s}} \]
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