Problem 15: Breaking Stress of Balloon Rubber
1. Analyze Pressure Conditions
The helium expansion is isothermal. We apply Boyle’s Law ($P_1V_1 = P_2V_2$) to find the internal pressure just before bursting.
- Initial State: Pressure $P_0$, Volume $V$.
- Final State: Volume $1.25V$.
The external atmospheric pressure at the bursting altitude is given as $P_{out} = 0.5 P_0$.
The excess pressure ($\Delta P$) inside the balloon is:
$$ \Delta P = P_{in} – P_{out} = 0.8 P_0 – 0.5 P_0 = 0.3 P_0 $$2. Force Balance on the Rubber Shell
Consider the equilibrium of a hemispherical section of the balloon. The force due to excess pressure pushing the two halves apart is balanced by the tension in the rubber ring (hoop stress).
$$ F_{pressure} = F_{tension} $$ $$ (\Delta P) \times (\text{Projected Area}) = (\text{Breaking Stress}) \times (\text{Cross-Sectional Area}) $$ $$ (0.3 P_0) \times (\pi r^2) = \sigma \times (2 \pi r t) $$Where $r$ is the final radius and $t$ is the thickness of the rubber.
$$ \sigma = \frac{0.3 P_0 r}{2t} \quad \text{…(Equation 1)} $$3. Relating Thickness and Mass
The volume of the rubber material remains constant. Let $m$ be the mass and $\rho$ be the density of the rubber.
$$ \text{Volume of rubber} = \text{Surface Area} \times \text{thickness} = 4 \pi r^2 t $$ $$ \frac{m}{\rho} = 4 \pi r^2 t \implies t = \frac{m}{4 \pi r^2 \rho} $$Substitute this expression for $t$ into Equation 1:
$$ \sigma = \frac{0.3 P_0 r}{2 \left( \frac{m}{4 \pi r^2 \rho} \right)} = \frac{0.3 P_0 r (4 \pi r^2 \rho)}{2m} = \frac{0.6 \pi P_0 \rho r^3}{m} $$4. Final Calculation
We know the volume of the gas at bursting is $V_{final} = \frac{4}{3} \pi r^3 = 1.25 V$.
From this, we can isolate $r^3$ or simply substitute $\pi r^3$ directly.
$$ \pi r^3 = \frac{3}{4} (1.25 V) = \frac{3}{4} \times \frac{5}{4} V = \frac{15}{16} V $$Substitute $\pi r^3$ back into the stress equation:
$$ \sigma = \frac{0.6 P_0 \rho}{m} \left( \frac{15}{16} V \right) $$ $$ \sigma = \frac{6}{10} \times \frac{15}{16} \frac{P_0 V \rho}{m} = \frac{3}{5} \times \frac{15}{16} \frac{P_0 V \rho}{m} $$ $$ \sigma = \frac{9 P_0 V \rho}{16 m} $$