How Do You Calculate the Density of a Partially Submerged Sphere?

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SUMMARY

The discussion focuses on calculating the density of a partially submerged sphere using Archimedes' Principle. Given a sphere of uniform density and radius R, floating with the top submerged at a distance of R/2 from the water surface, the density of the sphere is determined to be 27/32 times the density of water. The calculation involves finding the volume of the immersed part of the sphere, which is derived through integration, yielding a volume of 9πR³/8. The method is validated as legitimate and correct for solving the problem.

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Homework Statement


a sphere of uniform density and radius R is floating on water , partially immersed such that the distance between the top of the sphere and the water surface is R/2
find the density of the sphere

Homework Equations


Archimedes Principle

The Attempt at a Solution


One can deduce from the Archimedes Principle ,that the weight of the displaced water = the weight of the object

ρWaterVDisplaced waterg=ρObject VObjectg

which basically turns the problem into a mathematical problem involving finding the volume of the immersed part of the sphere.

Consider a circle of radius R centered at the origin ,

the required volume is ∫π(R2-x2)dx from -R to R/2 = 9πR3/8

Thus , ρObject=(9πR3/8)/(4πR3/3) * ρWater

=27/32 ρWaterI don't know if it's a legitimate method . It is suggested that I utilize the concept of hydrostatic pressure instead , but i have no idea how to do that.[/SUB]
 
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It is legitimate and correct. The buoyant force is equal to the weight of the displaced fluid.
 

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