Calculate velocity of helium baloon

[Samuelriesterer]

Problem statement:

A Helium filled balloon has a volume of 1 cubic meter and a mass of 0.200 kilograms. It is released and rises through the air because of its buoyancy. The buoyant force is equal to the weight of air displaced by the balloon.. This force is given by: F_b = [1.29e^(-1.21h)]g, where h is the altitude in km.

1) Draw a free body diagram and calculate the net force on the balloon at h=0.
2) Determine the altitude, h, at which the net force is zero..
3) Using the mass of the balloon and the expression for the buoyant force, derive an expression for the net force on the balloon as a function of altitude, h.
4) Get an expression for the velocity as a function of height by using: dv = adt = a(dt/dh) dh (Note that this is not constant acceleration, you need to integrate.)
5) At what height does the balloon stop moving upward?

Relevant equations:

f_net = ma
f_g = mg
v = integral of [a] dt

Attempt at solution:

1)
F_b at (h = 0) = 12.65N
F_g = mg = .2*9.81=1.96N
F_net = 12.65-1.96=10.69N

2)
F_net = 0 when F_b = F_g

h=1.54 km
3)
F_net = F_b-mg = 9.81[1.29e^(-1.21h)]-.2(9.81)

4)
F_net = ma
a = F_net/m = (1.29e^(-1.21h)g-mg)/m
v = 9.81[(-1.29e^(-1.21h))/(.2*-1.21) – h]

5)
v=0 when:
9.81[(-1.29e^(-1.21h))/(.2*-1.21) – h] = 0
h=1.21922 km

I am not too sure of my answer for #4. And also, what is troubling me is why the balloon will stop moving at 1.21922 km but the net force will not be zero until 1.54 km?

 

[ehild]

4)
F_net = ma
a = F_net/m = (1.29e^(-1.21h)g-mg)/m
v = 9.81[(-1.29e^(-1.21h))/(.2*-1.21) – h]

a=dv/dt, so integrating by h does not give v.
Also remember that you need to integrate between the boundaries, and the exponential is not zero at h=0.

 

[Samuelriesterer]

I reworked the integral:

dv = a dt = a dt/dh dh = a dh/v :

v delta_v = a dh
integral (v) dv = integral (a) dh
(v^2)/2 = integral (a) dh
v = sqrt[(integral (a) dh)/2]

Which will yield the same result in step 5 for finding the value of h to make v = 0.

 

[ehild]

If you use indefinite integral you need to include an integration constant and use the initial condition that v=0 at h=0.

 

[Samuelriesterer]

Thsnks for the feedback. So if I use a definite integral it would be from h = 0 to h = x? All I need is to find h when v = 0.

 

[ehild]

Thsnks for the feedback. So if I use a definite integral it would be from h = 0 to h = x? All I need is to find h when v = 0.

Answer question 4) first. What is the velocity in terms of h?
You can integrate the acceleration using a dummy variable x, from 0 to h. Or determine the indefinite integral, and fit the integration constant to the initial condition.

Also check the number of minus signs in the integral .

 

[Samuelriesterer]

OK, here is how I understand this problem:

Solve for a:

F_net = ma
a = F_net/m = (1.29e^(-1.21h)g - mg)/m

Solve for ∫ (a) dh:

∫ (a) dh = 9.8((-1.29e^(-1.21h))/(.2*-1.21) – h) + C

Solve for v:

dv = a dt = a dt/dh dh = a dh/v →
v Δv = a dh
integral (v) dv = integral (a) dh
(v^2)/2 = ∫ (a) dh
v = √[ 2*(9.81((-1.29e^(-1.21h))/(.2*-1.21) – h) + C)]

In order for the condition h = 0 and v = 0, C must equal ≈ -52.292975, which I presume would represent the force of holding the balloon down at h = 0 in order for v = 0. But when it is released, that constant is no longer there so it must equal 0.

As the balloon is released in the air, we need to find h such that v = 0. That h would be ≈ 1.21922, just as I derived previously.

 

[ehild]

OK, here is how I understand this problem:

Solve for a:

F_net = ma
a = F_net/m = (1.29e^(-1.21h)g - mg)/m

Solve for ∫ (a) dh:

∫ (a) dh = 9.8((-1.29e^(-1.21h))/(.2*-1.21) – h) + C

There are too many minuses. Why is minus in front of 1,29?

Solve for v:

dv = a dt = a dt/dh dh = a dh/v →
v Δv = a dh
integral (v) dv = integral (a) dh
(v^2)/2 = ∫ (a) dh
v = √[ 2*(9.81((-1.29e^(-1.21h))/(.2*-1.21) – h) + C)]

In order for the condition h = 0 and v = 0, C must equal ≈ -52.292975, which I presume would represent the force of holding the balloon down at h = 0 in order for v = 0. But when it is released, that constant is no longer there so it must equal 0.

No, C does not represent force when holding the balloon. That would be a different equation. C is there all time after the balloon has been released. . But its numerical value is not correct because of your sign error in the integral.

 

[Samuelriesterer]

Oh shoot that is right.

V^2/2 = g[(1.29e^(-1.21h))/(-1.21m) -h] +C

V = sqrt [2( [(1.29e^(-1.21h))/(-1.21m) -h] +C)] =

sqrt [-19.62h -104.586e^(-1.21h) +2C]

When h =0, C needs to be 52.293 for v =0.

V = sqrt [-19.62h -104.586e^(-1.21h) +104.586]

V =0 when h = 0 or 5.32207

 

[ehild]

Oh shoot that is right.

V^2/2 = g[(1.29e^(-1.21h))/(-1.21m) -h] +C

V = sqrt [2( [(1.29e^(-1.21h))/(-1.21m) -h] +C)] =

sqrt [-19.62h -104.586e^(-1.21h) +2C]

When h =0, C needs to be 52.293 for v =0.

V = sqrt [-19.62h -104.586e^(-1.21h) +104.586]

V =0 when h = 0 or 5.32207

That is much better (correct if you give the unit of h, and use only 3 significant digits. ) :)

 

[Samuelriesterer]

Thanks for your help, you rock!

 

[ehild]

You are welcome :)