Question

A helium-filled balloon is tied to a 2.00-m-long,
0.050 0-kg uniform string. The balloon is spherical with a radius of 0.400 m. When released, the balloon lifts a length h of string and then remains in equilibrium. Determine the value of h. The envelope of the balloon has a mass of 0.250 kg.

Answer 1

Answer on Question #57294, Physics / Mechanics | Relativity

A helium-filled balloon is tied to a 2.00-m-long, 0.050 0-kg uniform string. The balloon is spherical with a radius of $0.400\mathrm{m}$ . When released, the balloon lifts a length $h$ of string and then remains in equilibrium. Determine the value of $h$ . The envelope of the balloon has a mass of 0.250 kg.

Solution:

Density of Helium gas $\rho_{He} = 0.179\mathrm{kg / m^3}$

Density of air $\rho_{air} = 1.29\mathrm{kg / m^3}$

1 force up: buoyant force (due to air) $F_{b}$

3 forces down: weights $W_{\text{balloon}}$ , $W_{\text{He}}$ (helium inside balloon), $W_{\text{string}}$ (part of string above the ground)

Newton's Second Law :

$$F _ {b} = W _ {\text {b a l l o o n}} + W _ {\text {H e}} + W _ {\text {s t r i n g}}$$

where

$$F _ {b} = \rho_ {a i r} V g = \rho_ {a i r} \frac {4}{3} \pi r ^ {3} g$$ $$W _ {b a l l o o n} = m _ {b a l l o o n} g$$ $$W _ {H e} = \rho_ {H e} \frac {4}{3} \pi r ^ {3} g$$ $$W _ {s t r i n g} = \frac {h}{L} M g$$

where $h$ is the unknown of this problem.

$$\rho_ {a i r} \frac {4}{3} \pi r ^ {3} = m _ {b a l l o o n} + \rho_ {H e} \frac {4}{3} \pi r ^ {3} + \frac {h}{L} M$$ $$h = \frac {L}{M} \left(\rho_ {a i r} \frac {4}{3} \pi r ^ {3} - m _ {b a l l o o n} - \rho_ {H e} \frac {4}{3} \pi r ^ {3}\right)$$ $$h = \frac {L}{M} \left(\frac {4}{3} \pi r ^ {3} \left(\rho_ {a i r} - \rho_ {H e}\right) - m _ {b a l l o o n}\right)$$ $$h = \frac {2 . 0 0}{0 . 0 5} \cdot \left(\frac {4}{3} \cdot \pi \cdot 0. 4 ^ {3} \cdot (1. 2 9 - 0. 1 7 9) - 0. 2 5 0\right) = 1. 9 1 \mathrm {m}$$

Answer: $1.91\mathrm{m}$

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