Question (14h0002)
A helium balloon in air.
You hold a string that is tied to a helium balloon that is approximately spherical with a radius of 10 cm.
The balloon is in equilibrium. At atmospheric pressure and a temperature of
0◦ C,
helium has a density of
4
3
3 πR .) Assume that
◦
the densities of helium and air at room temperature are close to their densities at 0 C. The mass of the
0.1786 kg/m
3
and air has a density of
1.28 kg/m
3
. (Note: the volume of a sphere is
latex balloon before being inated is 2 g.
(a) What is the buoyant force of air on the balloon?
(b) What is the tension in the string?
(c) Suppose that you cut the string from the balloon and watch the balloon rise vertically. As it speeds
up, it eventually reaches a terminal velocity as it is rising. Assume constant density for helium and
air as the balloon rises at its terminal velocity. What is the force of air resistance on the balloon as it
rises at terminal velocity?
Solution
(a)
The buoyant force by air on the balloon can be calculated from the fact that the upward buoyant force
by air on the balloon is equal in magnitude to the weight of an equivalent volume of air, i.e. the weight of
the volume of air displaced" by the balloon.
Fbuoyant
by air on balloon
= Fgrav
by Earth on air displaced by balloon
= mair g
Write the mass of air in terms of its density and the volume of air displaced by the balloon, which is in this
case just the volume of the balloon since the balloon is completely immersed in the air.
Density is mass
divided by volume, so mass is density times volume.
Fbuoyant
by air on balloon
=
=
=
=
(ρair Vballoon )g
4
(ρair )( πR3 )g
3
4
(1.28 kg/m3 ) π(0.10 m)3 (9.8 N/kg)
3
0.0525 N
Express the buoyant force as a vector. Buoyant force is due to the pressure dierence as a function of height
between the top and bottom of the balloon. Therefore, the buoyant force is upward.
F~buoyant
(b)
by air on balloon
= < 0, 0.0525, 0 > N
Dene the system as the balloon (including both the latex and helium). Sketch a picture of the situation
and draw a free-body diagram. There are three objects in the surroundings that interact with the system:
Figure 1: A free-body diagram for the balloon.
(1) air; (2) string; and (3) Earth. Note that the balloon is in equilibrium; therefore, the sum of the forces
on the balloon must be zero.
Apply the momentum principle. Since the balloon remains at rest, its momentum is constant and therefore
it is in equilibrium.
F~net
=
=
d~
p
dt
0
Apply the denition of net force by adding all the forces from the free-body diagram and solve for the
unknown force.
F~net
0
= F~buoyant
= F~buoyant
+ F~tension
~
balloon + Ftension
+ F~grav
~
balloon + Fgrav
by air on balloon
by string on balloon
by Earth on balloon
by air on
by string on
by Earth on balloon
Substitute expressions for buoyant force and gravitational force. Note that the mass of the balloon includes
the latex and the helium. Use the density and volume of the helium to get its mass. First, calculate the
gravitational force by Earth on the balloon.
F~grav
by Earth on balloon
= < 0, −mg, 0 >
= < 0, −(0.002 kg + ρVballoon )(9.8 N/kg), 0 >
4
= < 0, −(0.002 kg + (0.1786 kg/m3 )( π(0.10 m)3 ))(9.8 N/kg), 0 >
3
= < 0, −0.0269 N, 0 >
Now, calculate the tension in the string.
0
F~tension
by string on balloon
= F~buoyant by air on balloon + F~tension by string on balloon + F~grav
= −(F~buoyant by air on balloon + F~grav by Earth on balloon )
by Earth on balloon
= −(< 0, 0.0525, 0 > N+ < 0, −0.0269 N, 0 >)
= −(< 0, 0.0256, 0 > N)
= < 0, −0.0256, 0 > N)
Note that it comes out negative, in agreement with the free-body diagram and what we know from experience
about the force of tension in the string when holding a helium-inated balloon.
(c)
If you release the balloon from rest, then it will speed up. However, eventually, it will reach a terminal
velocity because of the downward force of air resistance on the balloon. A picture showing the balloon moving
with a constant momentum and the forces on the balloon is shown below. Though only air and Earth exert
forces on the balloon, it is convenient to separate the force by air on the balloon into two components: (1)
the buoyant force that is due to a pressure dierence between the top and bottom of the balloon and (2) the
drag force, or air resistance, that is due to the motion of the balloon through the air.
Figure 2: A free-body diagram for the balloon as it travels upward at terminal velocity.
Apply the momentum principle to the balloon. Because it is at terminal velocity, its momentum is constant.
Thus, even though it is moving, it is still in equilibrium, just not static equilibrium.
F~net
=
=
d~
p
dt
0
Apply the denition of net force by adding all the forces from the free-body diagram and solve for the force
by air resistance.
F~net
0
= F~buoyant
= F~buoyant
+ F~drag
~
balloon + Fdrag
+ F~grav by
~
balloon + Fgrav
by air on balloon
by air on balloon
Earth on balloon
by air on
by string on
by Earth on balloon
This equation looks exactly like what we had before when holding the string, except this time, it's the
downward force of air resistance (drag) that balance out the buoyant and gravitational forces on the balloon.
Though the balloon is moving, it is still in equilibrium, just as before when it remained at rest. Thus, the
force of air resistance is
F~drag
by air on balloon
= < 0, −0.0256, 0 > N)
just as before for tension. The math comes out the same because the balloon is in equilibrium in both cases
and the buoyant and gravitational forces on the balloon are the same in both cases.
Note that as a real balloon rises, the density of the air around it will decrease. This will cause the balloon
to expand. The mass of the helium will remain constant, but the buoyant force on the balloon will change
as the balloon reaches more signicant elevations.