Calculating Upward Acceleration of a Helium-Filled Sphere

[BbyBlue24]

A thin spherical shell of mass 4 kg and diameter 0.2 m is filled with helium (density 0.18 kg/m3). It is then released from rest on the bottom of a pool of water that is 4 m deep. Determine the value of the upward acceleration of the shell.

 

[Pyrrhus]

what have you done?

Apply Newton's 2nd Law to the forces acting on the shell, buoyant force and the weight.

 

[BbyBlue24]

I determined the weight to be 39.2 N. I used the formula B=ρgV = (.18)(9.8)(.004) = .007 and V=(4πr3)/3 = .004.

Acceleration = Force/Mass = (.004-39.2)/4 = -9.799

And this answer doesn't make sense to me, I think my problem is with the buoyant force. Thanks!

 

[Pyrrhus]

You should have out for the buoyant force the density of the water, remember the buoyant force is equal to the weight of fluid displaced.

 

[BbyBlue24]

Doing that I get (1)(9.8)(.004) = .0392...sorry I am not getting this at all

 

[Pyrrhus]

You should have

Applying Newton's 2nd Law

B - m_{shell}g -m_{He}g = (m_{He} + m_{shell}) a

 

[BbyBlue24]

So that makes the upward acceleration -9.79?

 

[Pyrrhus]

Check well, i get a = 0.46 m/s^2

(\rho_{water}V - \rho_{He}V - m_{shell})g = (m_{shell} + \rho_{He}V)a

 

[BbyBlue24]

These are the values that I entered into the equation, I think the V (volume) might be incorrect, I am using .004 (V=(4Πr^3)/3)


((1∙.004)-(.18∙.004)-4)9.8=(4+.18∙.004)a = -9.79

Thank you!

 

[Pyrrhus]

I still get the same answer, i will write it to the simpler form

a = g \frac{(\rho_{water} - \rho_{He}) \frac{\pi d^3}{6} - m_{shell}}{m_{shell} + \rho_{He} \frac{\pi d^3}{6}}

The density of water is 1000 kg/m^3!

 

[BbyBlue24]

Yes, thank you. Also thank you very much for helping me out with this problem and the rest of the problems I have had tonight!

 

[pack_rat2]

The upward force will be equal to the difference between the weight of the He plus the material of the shell, and that of the water the sphere displaces. But there will be a considerable drag force exerted against the sphere's upward travel by the water. The D.E. for this system will be: -m*x'' - k*x' + F = 0, where m is the mass of the sphere, m*x'' is the inertial force on the sphere (downward, hence negative), k is the coefficient of drag (the force is downward), and F is the buoyant force (upward). x' and x'' are the sphere's velocity and acceleration, respectively.

 

[pack_rat2]

Hey, check this out: http://www.ma.iup.edu/projects/CalcDEMma/drag/drag.html#drag15
It's got a complete explanation!

 

[Pyrrhus]

This problem probably was meant for a non-viscous, stable, incompresible and irrotational fluid.

 

[pack_rat2]

The problem says it's hydrogen hydroxide, aka hydrogen monoxide...HOH...H2O...