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Consider a homogeneous right circular cylinder of length \(L,\) radius \(R,\) and specific gravity \(\mathrm{SG},\) floating in water \((\mathrm{SG}=1) .\) Show that the body will be stable with its axis vertical if \\[\frac{R}{L}>[2 \mathrm{SG}(1-\mathrm{SG})]^{1 / 2}\\]

Short Answer

Expert verified
The body is stable with its axis vertical if \(\frac{R}{L} > [2SG(1-SG)]^{1/2}\).

Step by step solution

01

Understand the Problem

We are asked to prove the stability of a floating cylinder based on given parameters: length \(L\), radius \(R\), and specific gravity \(SG\). The cylinder floats with its axis vertical if the inequality \(\frac{R}{L}>[2 \, SG \, (1-SG)]^{1/2}\) holds. The specific gravity of water is 1.
02

Analyze Buoyant and Gravitational Forces

A floating body is in equilibrium when buoyant force equals the gravitational force. For the cylinder, the buoyant force depends on the submerged volume times water's density, while the gravitational force depends on the entire volume of the cylinder times its density. Since \(SG\) is the object's density relative to water, the submerged height \(h_s\) is equal to \(SG \times L\).
03

Locate the Centers of Buoyancy and Gravity

For a floating cylinder, the center of gravity (\(G\)) and center of buoyancy (\(B\)) must be identified. \(G\) is located at the midpoint of the cylinder height. \(B\) is at the midpoint of the submerged height, at \(\frac{h_s}{2} = \frac{SG \times L}{2}\) from the base.
04

Determine Metacentric Height (GM)

The metacentric height \(GM\) is a crucial factor for stability. Calculate \(BM\), the distance from \(B\) to the metacenter \(M\), using the formula \(BM = \frac{I}{V_s}\), where \(I\) is the second moment of area about the horizontal axis through \(M\), and \(V_s\) is the submerged volume. For a cylinder, \(I = \frac{\pi R^4}{4}\) and \(V_s = \pi R^2 h_s\).
05

Compute \(GM\) and Establish Stability Criterion

\(GM = BM - BG\), where BG is \(\frac{L - h_s}{2}\). Substitute \(BM = \frac{\pi R^4}{4\pi R^2 SG L}\) into GM formula: \[ GM = \frac{R^2}{4SG L} - \frac{(1-SG) L}{2} \]. For stability, \(GM > 0\). Simplifying this inequality gives \(\frac{R}{L} > [2SG(1-SG)]^{1/2}\).
06

Verify the Solution

Verification through substitution back shows that the derived inequality matches the requirement. Therefore, the body will be stable if \(\frac{R}{L} > [2SG(1-SG)]^{1/2}\). Feel free to test with different values for assurance.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Buoyancy
Buoyancy is the upward force exerted by a fluid that opposes the weight of an object immersed in the fluid. When you place a body like a cylinder in water, it's buoyed up by a force equal to the weight of the water displaced by the submerged part of the cylinder. This principle was discovered by Archimedes and is known as Archimedes' principle.

In the given exercise, our cylinder floats on water, a perfect illustration of buoyancy in action. When a floating object is in equilibrium, the buoyant force exactly balances the gravitational force pulling it downwards. The important aspect here is the volume of water displaced, which depends on the depth to which the body sinks. For our cylinder, this depth is directly related to its specific gravity and length. Hence, the buoyant force can be calculated based on these variables, showing how the body floats with its axis vertical.
Stability of Floating Bodies
The stability of floating bodies is a key concept in fluid mechanics, deeply connected with buoyancy and balance. A body is stable if, after being slightly disturbed, it returns to its original position. For a cylindrical body floating with its axis vertical, stability is a major consideration.

In our context, stability is determined by the relationship between the center of gravity (G) and the center of buoyancy (B). When the floating body tilts slightly, the center of buoyancy shifts, creating a restoring moment that can bring the body back to its upright position. The condition \( \frac{R}{L}>[2 \mathrm{SG}(1-\mathrm{SG})]^{1 / 2} \) ensures that this restoring moment is sufficient for stability, allowing the cylinder to resettle itself straight.
Metacentric Height
The concept of metacentric height (GM) is an important factor for assessing the stability of floating objects. It is the vertical distance between the center of gravity and the metacenter (M), which is a point found above the center of buoyancy when the object is slightly tilted.

For stability, GM must be positive so that the restoring moment is present. In the exercise, GM is calculated using the formula:\[GM = BM - BG\]Where BM is the distance between the center of buoyancy and the metacenter, determined as BM = \frac{I}{V_s}, and BG is the distance from the center of buoyancy to the center of gravity. The criterion derived shows that by satisfying certain ratios of radius to length, the cylinder achieves the required GM, leading to stability.
Specific Gravity
Specific Gravity ( SG ) is a dimensionless number that denotes the ratio of the density of a substance to the density of a reference substance; usually, for liquids, this is water. An SG value of 1 indicates that the substance has the same density as water, while values less than 1 mean it's less dense.

In the context of floating bodies, SG influences how much of the object is submerged. The exercise uses specific gravity to determine the submerged height of the cylinder, which, combined with buoyancy calculations, helps establish stability criteria. The submerged volume crucially impacts both the buoyant force and the stability calculations, highlighting why understanding SG matters.

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Most popular questions from this chapter

Consider a hollow cube of side length \(22 \mathrm{cm},\) filled completely with water at \(20^{\circ} \mathrm{C}\). The top surface of the cube is horizontal. One top corner, point \(A\), is open through a small hole to a pressure of 1 atm. Diagonally opposite to point \(A\) is top corner \(B .\) Determine and discuss the various rigidbody accelerations for which the water at point \(B\) begins to cavitate, for \((a)\) horizontal motion and \((b)\) vertical motion.

Atlanta, Georgia, has an average altitude of \(1100 \mathrm{ft}\). On a standard day (Table A.6), pressure gage \(A\) in a laboratory experiment reads \(93 \mathrm{kPa}\) and gage \(B\) reads 105 kPa. Express these readings in gage pressure or vacuum pressure (Pa), whichever is appropriate.

Consider a wooden cylinder \((\mathrm{SG}=0.6) 1 \mathrm{m}\) in diameter and \(0.8 \mathrm{m}\) long. Would this cylinder be stable if placed to float with its yertical in oil \((\mathrm{SG}=0.8) ?\)

A very deep 18 -cm-diameter can contains \(12 \mathrm{cm}\) of water overlaid with \(10 \mathrm{cm}\) of \(\mathrm{SAE}\) 30 oil. If the can is rotated in rigid-body motion about its central axis at \(150 \mathrm{r} / \mathrm{min},\) what will be the shapes of the air-oil and oil-water interfaces? What will be the maximum fluid pressure in the can in \(\mathrm{Pa}\) (gage)?

A cylindrical can of weight \(W\), radius \(R\), and height \(H\) is open at one end. With its open end down, and while filled with atmospheric air \(\left(p_{\text {atm }}, T_{\text {atm }}\right),\) the can is eased down vertically into liquid, of density \(\rho,\) which enters and compresses the air isothermally. Derive a formula for the height \(h\) to which the liquid rises when the can is submerged with its top (closed) end a distance \(d\) from the surface.

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