Question

A 3.0 g latex balloon is filled with 2.8 g of helium. When filled, the balloon is a 32-cm-diameter sphere. When released, the balloon accelerates upward until it reaches a terminal speed. What is this speed? Assume an air density of 1.2 kg/m3 and a drag coefficient for sphere of 0.47.

Answer 1

Given,

Mass of the balloon, m = 3.0 g

mass of helium filled, m'=2.8 g

total mass of balloon when helium is filled in it, M = 3.0 + 2.8 = 5.8 gm = 0.0058 kg

Diameter of filled balloon, D = 32 cm = 0.32 m

Volume of filled balloon, V = 4/3**(D/2)³ = 1.33*3.14*(0.16)³

                                           = 0.0172 m³

Density of air, _air = 1.2 kg/m³

drag coefficient of sphere, C_D = 0.47

Now,

terminal velocity is attained when

buoyant force = weight of balloon + drag force

=> _air*V*g = Mg + 1/2*_air*v²*C_D**(D/2)²

=> 1.2*0.0172*9.8 = 0.0058*9.8 + 0.5*1.2*v²*0.47*3.14*(0.16)²

=> 0.2023 = 0.0568 + 0.0227*v²

=> 0.0227*v² = 0.2023 - 0.0568 = 0.1455

=> v² = 0.1455 / 0.0227

=> v² = 6.4097

=> v = = 2.532 m/s

Thus, terminal velocity is 2.532 m/s