Your team is in charge of launching a large helium weather balloon that is spher

Question

Your team is in charge of launching a large helium weather balloon that is spherical in shape, and whose radius is 2.3 m and total mass is 13.4 kg (balloon plus helium plus equipment).

(a) What is the initial upward acceleration of the balloon when it is released from sea level?

m/s^2

(b) If the drag force on the balloon is given by the equation below, where r is the balloon radius, is the density of air, and v is the balloon's ascension speed, calculate the terminal velocity of the ascending balloon.

m/s

Explanation / Answer

a)

The submerged body is replaced by an equal volume of fluid (dashed line). The detached volume is at equilibrium balancing its own weight by the buoyant force exerted on it by the rest of the fluid.

Then, the value of the buoyant force on the submerged body must be the weight of the displaced fluid. The line of action of the buoyant force passes through the center of mass of the volume.

By Newton’s second law,

Fnet = Fup – Fdown

Fnet = Fbuyont – Fggravity

Fnet = airVballoong - mballoong

Fnet = airVballoon - mballoon

Fnet = air(4r3/3) - mballoon

Fnet = 1.1644*(4*3.14*2.3^3/3) - 13.4 = 45.91 N

anet = Fnet/m = 45.91/13.4 = 3.4 m/s^2

b) at terminal velocity anet =0 m/s^2 and Fnet =0 N

By Newton’s second law ,

Fnet – Fd = 0

Fd = Fnet

1/2*r2v2 = 45.91

1/2*3.14*2.3^2*1.644*v2 = 45.91   => v = 1.83 m/s

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