Question

Use differentials to estimate the volume of thin metal spherical shells of thickness dr. If the metal has density 20gm/cm^3, how large must the radius be so that it floats in water (1gm/cm^3)? (Recall Archimedes Principle: the buoyant force is equal to the weight of the water being displaced.)

Answer 1

r=outer radius in cm,
dr=thickness in cm
\(\rho_m = density of metal=20 g/mL\)
\(\rho_w = density of water=1 g/mL\)

Approximate volume of metal
Vm=A*dr
=\(4\pi~r^2dr\) mL

Approximate mass of metal
Mm=\(\rho_m ~Vm\)
=\(20*4\pi~r^2~dr\) g

Volume of sphere
Vs=\((4/3)\pi~r^3\) mL

If sphere is submerged, mass of water displaced
\(M_w\)=volume of sphere * mass density of water
= \((4/3)\pi~r^3 *\rho_w\) g

For sphere to float,
\(M_mg\le M_wg \)
\(20*981*4\pi~r^2~dr\times 981\le (4/3)\pi~r^3 *1\times 981\)
Solve for r in terms of dr.

Answer 2

Wow, thank you so much for your help! I really appreciate the amount of work you put into that answer!

Answer 3

I do have a question though, where did you get the number 981 from?

Answer 4

981 is the acceleration due to gravity in cm/s^2.

In this case, both densities are mass densities, so i need to multiply by g to get the weight (in force units) to satisfy Archimedes Principle. However, "g" cancels out on both sides, so the value does not really mater!

:)

Answer 5

*matter

Answer 6

you multiplied 981 twice on the left side, but only once on the right side, so how would they cancel out?

Answer 7

Oh, sorry, my bad, I wasn't paying attention, thought I missed it on both sides.
There should be one on each side! My apologies! :(

Answer 8

No problem, thanks again for being so helpful!

Answer 9

mass * g = weight

Answer 10

no problem! :)