A solid cylinder of diameter 3m has a height of 2m. Find the meta-centric height of cylinder when it is floating in water with its axis vertical. The specific gravity of cylinder is 0.7.

A solid cylinder of diameter 3m has a height of 2m. Find the meta-centric height of cylinder when it is floating in water with its axis vertical. The specific gravity of cylinder is 0.7.

A solid cylinder of diameter 3m has a height of 2m. Find the meta-centric height of cylinder when it is floating in water with its axis vertical. The specific gravity of cylinder is 0.7.
Metacentric Height of a Floating Cylinder

Problem Statement

A solid cylinder is floating vertically in water with the following properties:

  • Diameter: 3m
  • Height: 2m
  • Specific gravity: 0.7

Determine the metacentric height (\(GM\)) of the cylinder.

Solution

1. Calculate the Depth of Immersion (\(h\))

The weight of the body is equal to the weight of the water displaced: \[ \gamma_{\text{cylinder}} V_{\text{cylinder}} = \gamma_{\text{water}} V_{\text{displaced water}} \] \[ 0.7 \times 9810 \times \pi \times (1.5)^2 \times 2 = 9810 \times \pi \times (1.5)^2 \times h \] Cancelling \( \pi \times (1.5)^2 \times 9810 \): \[ 0.7 \times 2 = h \] \[ h = 1.4 \text{ m} \]

2. Calculate the Center of Buoyancy (\(OB\))

The center of buoyancy is at the centroid of the submerged volume: \[ OB = \frac{h}{2} = \frac{1.4}{2} \] \[ OB = 0.7 \text{ m} \]

3. Calculate the Center of Gravity (\(OG\))

The center of gravity for a uniform cylinder is at its midpoint: \[ OG = \frac{\text{Total Height}}{2} = \frac{2}{2} \] \[ OG = 1 \text{ m} \] \[ BG = OG – OB = 1 – 0.7 = 0.3 \text{ m} \]

4. Calculate the Metacentric Height (\(GM\))

The metacentric radius (\(MB\)) is given by: \[ MB = \frac{I}{V} \] The moment of inertia about the vertical axis is: \[ I = \frac{1}{4} \pi r^4 \] \[ = \frac{1}{4} \pi (1.5)^4 \] \[ = \frac{1}{4} \pi \times 5.0625 = 3.976 \text{ m}^4 \] The displaced volume: \[ V = \pi \times (1.5)^2 \times 1.4 \] \[ V = \pi \times 2.25 \times 1.4 \] \[ V = 9.9 \text{ m}^3 \] \[ MB = \frac{3.976}{9.9} \] \[ MB = 0.4017 \text{ m} \] \[ GM = MB – BG \] \[ GM = 0.4017 – 0.3 \] \[ GM = 0.1017 \text{ m} \]
Final Results:
  • Depth of immersion: 1.4 m
  • Metacentric height (\(GM\)): 0.1017 m

Explanation

1. Floating Stability:
A floating body is stable if the metacentric height (\(GM\)) is positive. A small \(GM\) value means the object is close to tipping over.

2. Calculation of Buoyancy and Stability:
– The center of buoyancy (\(OB\)) is at the midpoint of the submerged volume.
– The center of gravity (\(OG\)) is at the midpoint of the full height.
– The metacentric height (\(GM\)) is derived using the moment of inertia and volume of displaced water.

3. Importance of Metacentric Height:
– If \(GM\) is large, the floating body is highly stable but may have poor maneuverability.
– If \(GM\) is too small or negative, the object will be unstable and may tip over.

Physical Meaning

1. Ship and Boat Stability:
A similar method is used in naval architecture to ensure that ships and boats remain upright while floating.

2. Offshore Platform Design:
Floating offshore structures, such as oil rigs, rely on metacentric height calculations to maintain stability in waves and strong winds.

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