FLUIDS BYU 14

Problem 14: Compressible Balloons on a Rod

Physical Analysis:

Let the length of the rod be $L$ meters. Since balloons are attached “at every metre except at the top end,” a rod of integer length $n$ meters will have $n$ balloons attached at depths $1\,\text{m}, 2\,\text{m}, \dots, n\,\text{m}$ (assuming the top of the rod is at the water surface).

Data Given:

• Linear mass density of rod (including balloons) $\lambda = 2.7\,\text{kg/m}$.
• Initial volume of balloon $V_0 = 0.003\,\text{m}^3$.
• Volume reduction rate: $\Delta V = 100\,\text{cm}^3 = 10^{-4}\,\text{m}^3$ per $10\,\text{kPa}$.
• Pressure increase in water: $10\,\text{kPa}$ corresponds to a depth of $h = \frac{P}{\rho g} = \frac{10^4}{1000 \times 10} = 1\,\text{m}$.

Therefore, the volume of a balloon at depth $y$ meters is:

$$ V(y) = V_0 – (10^{-4})y $$

Equilibrium Condition:

For the rod to float, the total buoyant force must be greater than or equal to the total weight.

$$ F_B \ge Mg $$ $$ \rho_{w} g \sum_{k=1}^{n} V(y_k) \ge (\lambda n) g $$

Canceling $g$ and substituting values ($\rho_w = 1000$):

$$ 1000 \sum_{k=1}^{n} (0.003 – 0.0001 k) \ge 2.7 n $$

Dividing by $n$ (where $\sum V_k$ is the average volume $\times n$):

$$ \sum_{k=1}^{n} (3 – 0.1 k) \ge 2.7 n $$ $$ 3n – 0.1 \sum_{k=1}^{n} k \ge 2.7 n $$ $$ 3n – 0.1 \frac{n(n+1)}{2} \ge 2.7 n $$

Simplifying:

$$ 3 – 0.05(n+1) \ge 2.7 $$ $$ 0.3 \ge 0.05(n+1) $$ $$ 6 \ge n + 1 $$ $$ n \le 5 $$