Review problem. The Abbott of Aberbrothock paid to have a...

Review problem. The Abbott of Aberbrothock paid to have a bell moored to the Inchcape Rock to warn scamen of the hazard. Assume the bell was $3.00 \mathrm{m}$ in diameter, cast from brass with a bulk modulus of $14.0 \times 10^{10} \mathrm{N} / \mathrm{m}^{2} .$ The pirate Ralph the Rover cut loose the warning bell and threw it into the ocean. By how much did the diameter of the bell decrease as it sank to a depth of $10.0 \mathrm{km}^{2}$ Years later, Ralph drowned when his ship collided with the rock. Note: The brass is compresed uniformly, so you may model the bell as a sphere of diameter $3.00 \mathrm{m}$

Key Concepts

Uniform Compression

Uniform compression assumes that pressure is applied evenly in all directions on an object, leading to a uniform reduction in its volume. This simplification allows for the use of bulk modulus and related formulas to model how the object's dimensions change under pressure. It is often used to analyze material behavior without the complexities of directional stresses.

Volume-Diameter Relationship in Spheres

For objects with spherical symmetry, the volume is mathematically related to the cube of the diameter or radius. This relationship is used to translate a calculated change in volume (due to pressure) into a corresponding change in linear dimensions such as the diameter. Understanding this relationship is essential in problems where a volumetric compression needs to be interpreted in terms of changes in size.

Hydrostatic Pressure

Hydrostatic pressure refers to the pressure exerted by a fluid at equilibrium due to the force of gravity. It increases linearly with depth in a fluid, which is crucial when considering the compression of objects submerged at great depths. This concept is fundamental in fluid mechanics, affecting the design and analysis of submerged structures and vessels.

Bulk Modulus

Bulk modulus is a property of materials that measures their resistance to uniform compression. It is defined as the ratio of the applied pressure change to the relative volume change, indicating how incompressible a material is. In many physics and engineering applications, knowing the bulk modulus helps predict how much a material will compress when subjected to high pressures.

Transcript

00:01 So the question here is that we have a bell that is dropped into the ocean depth of 10 kilometers, and we can approximate that bell as a sphere with a diameter of 3 meters.

00:11 Now, the question is, what is the diameter of our bell going to be at the bottom of the ocean? so how much would it shrink by due to the pressure of the water? and another thing to note is that the bulk modulus of the material of our bell is given, which is 14 times the 10, the 10 newtons per meter squared.

00:35 Now with this information, we know that the change in volume will be equal to the negative initial volume times the change in pressure divided by the bulk modulus.

00:49 Now what is the change in pressure? this bellow experience.

00:55 Well, we have an expression for that, which is just the density of water, row times gravitational acceleration, times the depth.

01:04 In this case, h equals 10 kilometers, divided by the bulk modulus.

01:11 Now we want to find out the final diameter, so we should figure out what the final volume is before that to relate that, find out the final diameter.

01:25 Well, what is our final volume going to be? well, that's just going to be our initial volume plus the change in our volume.

01:38 Now we can rewrite this as v0 minus v -not row g -h over b.

01:49 Pull out the v -0, we get 1 minus row gh over b.

01:59 Well, what's our initial volume? we know that the expression for a sphere, a volume of a sphere is just 4 over 3 pi are cubed.

02:13 So we have 4 over 3 pi r not cubed, and then 1 minus row gh.

02:23 Over b.

02:26 Then what will be our final diameter? well, again, we know that our volume is going to be equal to 4 .3 pi r cubed.

02:38 So this whole thing should be equal to 4 over 3 pi are final cubed, right? equals 4 over 3 pi r not cubed 1 minus row g h over b.

02:57 I can see that 4 .3 pies cancel, and let's rewrite this in terms of our diameter.

03:07 So it should be diameter over 2 cubed.

03:11 So it should be d cubed over 8, 2 cubed.

03:17 And then same thing here.

03:19 We have initial diameter cubed over 8.

03:24 1 minus row gh root b...

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